Advances in Inequalities of the Schwarz, Grüss and Bessel Type in

Inner Product Spaces

Sever Silvestru Dragomir

School of Computer Science & Mathematics, Victoria University, Melbourne, Victoria, Australia

E-mail address: sever.dragomir@vu.edu.au
URL: http://rgmia.vu.edu.au/SSDragomirWeb.html

2000 Mathematics Subject Classification. Primary 46C05, 46E30; Secondary 25D15, 26D10

Abstract. The main aim of this monograph is to survey some recent results obtained by the author related to reverses of the Schwarz, triangle and Bessel inequalities. Some Grüss’ type inequalities for orthonormal families of vectors in real or complex inner product spaces are presented as well. Generalizations of the Boas-Bellman, Bombieri, Selberg, Heilbronn and Pečarić inequalities for finite sequences of vectors that are not necessarily orthogonal are also provided. Two extensions of the celebrated Ostrowski’s inequalities for sequences or real numbers and the generalization of Wagner’s inequality in inner product spaces are pointed out. Finally, some Grüss type inequalities for n-tuples of vectors in inner product spaces and their natural applications for the approximation of the discrete Fourier and Mellin transforms are given as well.

Contents

Preface v

Part 1. Reverse Inequalities 1

Chapter 1. Reverses for the Schwarz Inequality 1 1. Introduction 1 2. An Additive Reverse of the Schwarz Inequality 2 3. A Generalisation of the Cassels Inequality 9 4. Quadratic Reverses of Schwarz’s Inequality 18 5. More Reverses of Schwarz’s Inequality 27

Bibliography 35

Chapter 2. Inequalities of the Grüss Type 37 1. Introduction 37 2. Grüss’ Inequality in Inner Product Spaces 41 3. Companions of Grüss’ Inequality 52 4. Other Grüss Type Inequalities 61

Bibliography 69

Chapter 3. Reverses of Bessel’s Inequality 71 1. Introduction 71 2. Reverses of Bessel’s Inequality 72 3. Another Reverse for Bessel’s Inequality 84 4. More Reverses of Bessel’s Inequality 97 5. General Reverses of Bessel’s Inequality 109

Bibliography 121

Part 2. Other Inequalities in Inner Product Spaces 123

Chapter 4. Generalisations of Bessel’s Inequality 125 1. Boas-Bellman Type Inequalities 125 2. Bombieri Type Inequalities 134 3. Pečarić Type Inequalities 148 Bibliography 159

Chapter 5. Some Grüss’ Type Inequalities for n-Tuples of Vectors161 1. Introduction 161 2. The Version for Norms 163 3. The Version for Inner-Products 177 4. More Grüss’ Type Inequalities 184 5. Some Inequalities for Forward Difference 193 6. Bounds for a Pair of n-Tuples of Vectors 201

Bibliography 215

Chapter 6. Other Inequalities in Inner Product Spaces 217 1. The Ostrowski Inequality 217 2. Another Ostrowski Type Inequality 221 3. The Wagner Inequality in Inner Product Spaces 224 4. A Monotoniciy Property of Bessel’s Inequality 227 5. Other Bombieri Type Inequalities 230 6. Some Pre-Grüss Inequalities 235

Bibliography 245

Index 247

Preface

The theory of Hilbert spaces plays a central role in contemporary mathematics with numerous applications for Linear Operators, Partial Differential Equations, in Nonlinear Analysis, Approximation Theory, Optimization Theory, Numerical Analysis, Probability Theory, Statistics and other fields.

The Schwarz, triangle, Bessel, Gram and most recently, Grüss type inequalities have been frequently used as powerful tools in obtaining bounds or estimating the errors for various approximation formulae occurring in the domains mentioned above. Therefore, any new advancement related to these fundamental facts will have a flow of important consequences in the mathematical fields where these inequalities have been used before.

The main aim of this monograph is to survey some recent results obtained by the author related to reverses of the Schwarz, triangle and Bessel inequalities. Some Grüss type inequalities for orthonormal families of vectors in real or complex inner product spaces are presented as well. Generalizations of the Boas-Bellman, Bombieri, Selberg, Heilbronn and Pečarić inequalities for finite sequences of vectors that are not necessarily orthogonal are also provided. Two extensions of the celebrated Ostrowski inequalities for sequences of real numbers and the generalization of Wagner’s inequality in inner product spaces are pointed out. Finally, some Grüss type inequalities for n-tuples of vectors in inner product spaces and their natural applications for the approximation of the discrete Fourier and Mellin transforms are given as well.

The monograph may be used by researchers in different branches of Mathematical and Functional Analysis where the theory of Hilbert spaces is of relevance. Since it is self-contained and all the results are completely proved, the work may be also used by graduate students interested in Theory of Inequalities and its Applications.

The Author,

February, 2004.

Part 1

Reverse Inequalities

CHAPTER 1 Reverses for the Schwarz Inequality

1. Introduction

Let H be a linear space over the real or complex number field 𝕂. The functional ⟨·, ·⟩ : H × H → 𝕂 is called an inner product on H if it satisfies the conditions

(i) ⟨x, x⟩ ≥ 0 for any x ∈ H and ⟨x, x⟩ = 0 iff x = 0; (ii) ⟨αx + βy, z⟩ = α ⟨x, z⟩+β ⟨y, z⟩ for any α, β ∈ 𝕂 and x, y, z ∈

H;

(iii) ⟨y, x⟩ = \overline{⟨x, y⟩} for any x, y ∈ H.

A first fundamental consequence of the properties (i)-(iii) above, is the Schwarz inequality:

(1.1)  |⟨x, y⟩|^{2} ≤ ⟨x, x⟩ ⟨y, y⟩ ,

for any x, y ∈ H. The equality holds in (1.1) if and only if the vectors x and y are linearly dependent, i.e., there exists a nonzero constant α ∈ 𝕂 so that x = αy.

If we denote ∥x∥ := \sqrt{⟨x, x⟩}, x ∈ H, then one may state the following properties

(n) ∥x∥ ≥ 0 for any x ∈ H and ∥x∥ = 0 iff x = 0;

(nn) ∥αx∥ = |α| ∥x∥ for any α ∈ 𝕂 and x ∈ H; (nnn) ∥x + y∥ ≤ ∥x∥+∥y∥ for any x, y ∈ H (the triangle inequality);

i.e., ∥·∥ is a norm on H.

In this chapter we present some recent reverse inequalities for the Schwarz and the triangle inequalities. More precisely, we point out upper bounds for the nonnegative quantities

∥x∥ ∥y∥ − |⟨x, y⟩| , ∥x∥^{2} ∥y∥^{2} − |⟨x, y⟩|^{2}

and

∥x∥ + ∥y∥ − ∥x + y∥

under various assumptions for the vectors x, y ∈ H.

If the vectors x, y ∈ H are not orthogonal, i.e., ⟨x, y⟩ ≠ 0, then some upper bounds for the supra-unitary quantities

\begin{matrix}∥x∥ & ∥y∥ \\ | & ⟨x, y⟩ & | \end{matrix}, \begin{matrix}∥x∥ & 2 & ∥y∥ & 2 \\ |⟨x, y⟩| & 2 \end{matrix}

are provided as well.

2. An Additive Reverse of the Schwarz Inequality

2.1. Introduction. Let \overline{a} = (a_{1}, …, a_{n}) and \overline{b} = (b_{1}, …, b_{n}) be two positive n−tuples with

(2.1)  0 < m_{1} ≤ a_{i} ≤ M_{1} < ∞ and 0 < m_{2} ≤ b_{i} ≤ M_{2} < ∞;

for each i ∈ {1, …, n} , and some constants m_{1}, m_{2}, M_{1}, M_{2}.

The following reverses of the Cauchy-Bunyakovsky-Schwarz inequality for positive sequences of real numbers are well known:

(1) Pólya-Szegö’s inequality [20]

\frac{∑_{k=1}^{n} a_{k}^{2} ∑_{k=1}^{n} b_{k}^{2}}{(∑_{k=1}^{n} a_{k}b_{k})^{2}} ≤ \frac{1}{4} (\sqrt{\frac{M_{1}M_{2}}{m_{1}m_{2}}} + \sqrt{\frac{m_{1}m_{2}}{M_{1}M_{2}}} )^{2} .

(2) Shisha-Mond’s inequality [22]

\frac{∑_{k=1}^{n} a_{k}^{2}}{∑_{k=1}^{n} a_{k}b_{k}} − \frac{∑_{k=1}^{n} a_{k}b_{k}}{∑_{k=1}^{n} b_{k}^{2}} ≤ [(\frac{M_{1}}{m_{2}} )^{\frac{1}{2}} − (\frac{m_{1}}{M_{2}} )^{\frac{1}{2}} ]^{2} .

(3) Ozeki’s inequality [19]

∑_{k=1}^{n} a_{k}^{2} ∑_{k=1}^{n} b_{k}^{2} − (∑_{k=1}^{n} a_{k}b_{k})^{2} ≤ \frac{n^{2}}{4} (M_{1}M_{2} − m_{1}m_{2})^{2} .

(4) Diaz-Metcalf’s inequality [2]

∑_{k=1}^{n} b_{k}^{2} + \frac{m_{2}M_{2}}{m_{1}M_{1}} ∑_{k=1}^{n} a_{k}^{2} ≤ (\frac{M_{2}}{m_{1}} + \frac{m_{2}}{M_{1}} ) ∑_{k=1}^{n} a_{k}b_{k}.

If \overline{w} = (w_{1}, …, w_{n}) is a positive sequence, then the following weighted inequalities also hold:

(1) Cassel’s inequality [23]. If the positive real sequences \overline{a} = (a_{1}, …, a_{n}) and \overline{b} = (b_{1}, …, b_{n}) satisfy the condition

(2.2)  0 < m ≤ \frac{a_{k}}{b_{k}} ≤ M < ∞ for each k ∈ \{1, …, n\} ,

then

\begin{matrix}(∑_{k=1}^{n} w_{k}a_{k}^{2}) & (∑_{k=1}^{n} w_{k}b_{k}^{2}) \\ (∑_{k=1}^{n} w_{k}a_{k}b_{k}) & 2 \end{matrix} ≤ \frac{(M + m)^{2}}{4mM} .

(2) Greub-Reinboldt’s inequality [12]. We have

(∑_{k=1}^{n} w_{k}a_{k}^{2}) (∑_{k=1}^{n} w_{k}b_{k}^{2}) ≤ \frac{(M_{1}M_{2} + m_{1}m_{2})^{2}}{4m_{1}m_{2}M_{1}M_{2}} (∑_{k=1}^{n} w_{k}a_{k}b_{k})^{2} ,

provided \overline{a} = (a_{1}, …, a_{n}) and \overline{b} = (b_{1}, …, b_{n}) satisfy the condition (2.1) .

(3) Generalised Diaz-Metcalf’s inequality [2], see also [17, p. 123]. If u, v ∈ [0, 1] and v ≤ u, u + v = 1 and (2.2) holds, then one has the inequality

u ∑_{k=1}^{n} w_{k}b_{k}^{2} + vMm ∑_{k=1}^{n} w_{k}a_{k}^{2} ≤ (vm + uM) ∑_{k=1}^{n} w_{k}a_{k}b_{k}.

(4) Klamkin-McLenaghan’s inequality [15]. If \overline{a}, \overline{b} satisfy (2.2), then

(2.3) (∑_{i=1}^{n} w_{i}a_{i}^{2}) (∑_{i=1}^{n} w_{i}b_{i}^{2}) − (∑_{i=1}^{n} w_{i}a_{i}b_{i})^{2}

≤ (M^{\frac{1}{2}} − m^{\frac{1}{2}} )^{2} ∑_{i=1}^{n} w_{i}a_{i}b_{i} ∑_{i=1}^{n} w_{i}a_{i}^{2}.

For other recent results providing discrete reverse inequalities, see the recent monograph online [5].

In this section, by following [3], we point out a new reverse of Schwarz’s inequality in real or complex inner product spaces. Particular cases for isotonic linear functionals, integrals and sequences are also given.

2.2. An Additive Reverse Inequality. The following reverse of Schwarz’s inequality in inner product spaces holds [3].

Theorem 1. Let A, a ∈ 𝕂 (𝕂 = ℂ, ℝ) and x, y ∈ H. If

(2.4)  Re ⟨Ay − x, x − ay⟩ ≥ 0,

or, equivalently,

(2.5)  ∥x − \frac{a + A}{2} · y∥ ≤ \frac{1}{2} |A − a| ∥y∥ ,

holds, then one has the inequality

(2.6)  0 ≤ ∥x∥^{2} ∥y∥^{2} − |⟨x, y⟩|^{2} ≤ \frac{1}{4} |A − a|^{2} ∥y∥^{4} .

The constant \frac{1}{4} is sharp in (2.6).

Proof. The equivalence between (2.4) and (2.5) can be easily proved, see for example [10].

Let us define

I_{1} := Re [(A ∥y∥^{2} − ⟨x, y⟩) (\overline{⟨x, y⟩} − \overline{a} ∥y∥^{2})]

and

I_{2} := ∥y∥^{2} Re ⟨Ay − x, x − ay⟩ .

Then

I_{1} = ∥y∥^{2} Re [A\overline{⟨x, y⟩} + \overline{a} ⟨x, y⟩] − |⟨x, y⟩|^{2} − ∥y∥^{4} Re (A\overline{a})

and

I_{2} = ∥y∥^{2} Re [A\overline{⟨x, y⟩} + \overline{a} ⟨x, y⟩] − ∥x∥^{2} ∥y∥^{2} − ∥y∥^{4} Re (A\overline{a}) ,

which gives

I_{1} − I_{2} = ∥x∥^{2} ∥y∥^{2} − |⟨x, y⟩|^{2} ,

for any x, y ∈ H and a, A ∈ 𝕂. This is an interesting identity in itself as well.

If (2.4) holds, then I_{2} ≥ 0 and thus

(2.7) ∥x∥^{2} ∥y∥^{2} −|⟨x, y⟩|^{2} ≤ Re [(A ∥y∥^{2} − ⟨x, y⟩) (\overline{⟨x, y⟩} − \overline{a} ∥y∥^{2})] .

Further, if we use the elementary inequality for u, v ∈ 𝕂 (𝕂 = ℂ, ℝ)

Re (u\overline{v}) ≤ \frac{1}{4} |u + v|^{2} ,

then we have, for

u := A ∥y∥^{2} − ⟨x, y⟩ , v := ⟨x, y⟩ − a ∥y∥^{2} ,

that

(2.8) Re [(A ∥y∥^{2} − ⟨x, y⟩) (\overline{⟨x, y⟩} − \overline{a} ∥y∥^{2})] ≤ \frac{1}{4} |A − a|^{2} ∥y∥^{4} .

Making use of the inequalities (2.7) and (2.8) , we deduce (2.6). Now, assume that (2.6) holds with a constant C > 0, i.e.,

(2.9)  ∥x∥^{2} ∥y∥^{2} − |⟨x, y⟩|^{2} ≤ C |A − a|^{2} ∥y∥^{4} ,

where x, y, a, A satisfy (2.4).

Consider y ∈ H, ∥y∥ = 1, a ≠ A and m ∈ H, ∥m∥ = 1 with m ⊥ y. Define

x := \frac{A + a}{2}y + \frac{A − a}{2}m.

Then

⟨Ay − x, x − ay⟩ = |\frac{A − a}{2}|^{2} ⟨y − m, y + m⟩ = 0,

and thus the condition (2.4) is fulfilled. From (2.9) we deduce

(2.10) ∥\frac{A + a}{2}y + \frac{A − a}{2}m∥^{2} − |⟨\frac{A + a}{2}y + \frac{A − a}{2}m, y⟩|^{2}

≤ C |A − a|^{2} ,

and since

∥\frac{A + a}{2}y + \frac{A − a}{2}m∥^{2} = |\frac{A + a}{2}|^{2} + |\frac{A − a}{2}|^{2}

and

|⟨\frac{A + a}{2}y + \frac{A − a}{2}m, y⟩|^{2} = |\frac{A + a}{2}|^{2}

then, by (2.10), we obtain

\frac{|A − a|^{2}}{4} ≤ C |A − a|^{2} ,

which gives C ≥ \frac{1}{4}, and the theorem is completely proved.

2.3. Applications for Isotonic Linear Functionals. Let F (T) be an algebra of real functions defined on T and L a subclass of F (T) satisfying the conditions:

(i) f, g ∈ L implies f + g ∈ L; (ii) f ∈ L, α ∈ ℝ implies αf ∈ L. A functional A defined on L is an isotonic linear functional on L

provided that (a) A (αf + βg) = αA (f) + βA (g) for all α, β ∈ ℝ and f, g ∈ L; (aa) f ≥ g, that is, f (t) ≥ g (t) for all t ∈ T, implies A (f) ≥ A (g) . The functional A is normalised on L, provided that 1 ∈ L, i.e.,

1 (t) = 1 for all t ∈ T, implies A (1) = 1.

Usual examples of isotonic linear functionals are integrals, sums, etc.

Now, suppose that h ∈ F (T) , h ≥ 0 is given and satisfies the properties that fgh ∈ L, fh ∈ L, gh ∈ L for all f, g ∈ L. For a given isotonic linear functional A : L → ℝ with A (h) > 0, define the mapping (·, ·)_{A,h} : L × L → ℝ by

(f, g)_{A,h} := \frac{A (fgh)}{A (h)}.

This functional satisfies the following properties:

(s) (f, f)_{A,h} ≥ 0 for all f ∈ L; (ss) (αf + βg, k)_{A,h} = α (f, k)_{A,h} +β (g, k)_{A,h} for all f, g, k ∈ L and

α, β ∈ ℝ;

(sss) (f, g)_{A,h} = (g, f)_{A,h} for all f, g ∈ L.

The following reverse of Schwarz’s inequality for positive linear functionals holds [3].

Proposition 1. Let f, g, h ∈ F (T) be such that fgh ∈ L, f^{2}h ∈ L, g^{2}h ∈ L. If m, M are real numbers such that

(2.11)  mg ≤ f ≤ Mg on F (T) ,

then for any isotonic linear functional A : L → ℝ, with A (h) > 0, we have the inequality

(2.12) 0 ≤ A (hf^{2}) A (hg^{2}) − [A (hfg)]^{2} ≤ \frac{1}{4} (M − m)^{2} A^{2} (hg^{2}) .

The constant \frac{1}{4} in (2.12) is sharp.

Proof. We observe that

(Mg − f, f − mg)_{A,h} = A [h (Mg − f) (f − mg)] ≥ 0.

Applying Theorem 1 for (·, ·)_{A,h} , we get

0 ≤ (f, f)_{A,h} (g, g)_{A,h} − (f, g)_{A,h}^{2} ≤ \frac{1}{4} (M − m)^{2} (g, g)_{A,h}^{2} ,

which is clearly equivalent to (2.12).

The following corollary holds.

Corollary 1. Let f, g ∈ F (T) such that fg, f^{2}, g^{2} ∈ F (T) . If m, M are real numbers such that (2.11) holds, then

(2.13)  0 ≤ A (f^{2}) A (g^{2}) − A^{2} (fg) ≤ \frac{1}{4} (M − m)^{2} A^{2} (g^{2}) .

The constant \frac{1}{4} is sharp in (2.13).

Remark 1. The condition (2.11) may be replaced with the weaker assumption

(Mg − f, f − mg)_{A,h} ≥ 0.

2.4. Applications for Integrals. Let (Ω, Σ, μ) be a measure space consisting of a set Ω, Σ a σ−algebra of subsets of Ω and μ a countably additive and positive measure on Σ with values in ℝ ∪ {∞} .

Denote by L_{ρ}^{2} (Ω, 𝕂) the Hilbert space of all 𝕂-valued functions f defined on Ω that are 2−ρ−integrable on Ω, i.e., ∫_{Ω} ρ (t) |f (s)|^{2} dμ (s) < ∞, where ρ : Ω → [0, ∞) is a measurable function on Ω.

The following proposition contains a reverse of the weighted Cauchy-Bunyakovsky-Schwarz’s integral inequality [3].

Proposition 2. Let A, a ∈ 𝕂 (𝕂 = ℂ, ℝ) and f, g ∈ L_{ρ}^{2} (Ω, 𝕂) . If

(2.14) ∫_{Ω} Re [(Ag (s) − f (s)) (\overline{f (s)} − \overline{a} \overline{g} (s))] ρ (s) dμ (s) ≥ 0

or, equivalently,

∫_{Ω} ρ (s) |f (s) − \frac{a + A}{2}g (s)|^{2} dμ (s) ≤ \frac{1}{4} |A − a|^{2} ∫_{Ω} ρ (s) |g (s)|^{2} dμ (s) ,

holds, then one has the inequality

0 ≤ ∫_{Ω} ρ (s) |f (s)|^{2} dμ (s) ∫_{Ω} ρ (s) |g (s)|^{2} dμ (s)
− |∫_{Ω} ρ (s) f (s) \overline{g (s)}dμ (s)|^{2}

≤ \frac{1}{4} |A − a|^{2} (∫_{Ω} ρ (s) |g (s)|^{2} dμ (s))^{2} .

The constant \frac{1}{4} is best possible.

Proof. Follows by Theorem 1 applied for the inner product ⟨·, ·⟩_{ρ} : L_{ρ}^{2} (Ω, 𝕂) × L_{ρ}^{2} (Ω, 𝕂) → 𝕂,

⟨f, g⟩_{ρ} := ∫_{Ω} ρ (s) f (s) \overline{g (s)}dμ (s) .

Remark 2. A sufficient condition for (2.14) to hold is

Re [(Ag (s) − f (s)) (\overline{f (s)} − \overline{a} \overline{g} (s))] ≥ 0, for μ − a.e. s ∈ Ω.

In the particular case ρ = 1, we have the following reverse of the Cauchy-Bunyakovsky-Schwarz inequality.

Corollary 2. Let a, A ∈ 𝕂 (𝕂 = ℂ, ℝ) and f, g ∈ L^{2} (Ω, 𝕂) . If

(2.15)  ∫_{Ω} Re [(Ag (s) − f (s)) (\overline{f (s)} − \overline{a} \overline{g} (s))] dμ (s) ≥ 0,

8 1. REVERSES FOR THE SCHWARZ INEQUALITY

or, equivalently,

∫_{Ω} |f (s) − \frac{a + A}{2}g (s)|^{2} dμ (s) ≤ \frac{1}{4} |A − a|^{2} ∫_{Ω} |g (s)|^{2} dμ (s) ,

holds, then one has the inequality

0 ≤ ∫_{Ω} |f (s)|^{2} dμ (s) ∫_{Ω} |g (s)|^{2} dμ (s) − |∫_{Ω} f (s) \overline{g (s)}dμ (s)|^{2}

≤ \frac{1}{4} |A − a|^{2} (∫_{Ω} |g (s)|^{2} dμ (s))^{2} .

The constant \frac{1}{4} is best possible

Remark 3. If 𝕂 = ℝ, then a sufficient condition for either (2.14) or (2.15) to hold is

ag (s) ≤ f (s) ≤ Ag (s) , for μ − a.e. s ∈ Ω,

where, in this case, a, A ∈ ℝ with A > a > 0.

2.5. Applications for Sequences. For a given sequence (w_{i})_{i∈ℕ} of nonnegative real numbers, consider the Hilbert space ℓ_{w}^{2} (𝕂) , (𝕂 = ℂ, ℝ) , where

ℓ_{w}^{2} (𝕂) := \{\overline{x} = (x_{i})_{i∈ℕ} ⊂ 𝕂 |∑_{i=0}^{∞} w_{i} |x_{i}|^{2} < ∞\} .

The following proposition that provides a reverse of the weighted Cauchy-Bunyakovsky-Schwarz inequality for complex numbers holds.

Proposition 3. Let a, A ∈ 𝕂 and \overline{x}, \overline{y} ∈ ℓ_{w}^{2} (𝕂) . If

(2.16)  ∑_{i=0}^{∞} w_{i} Re [(Ay_{i} − x_{i}) (\overline{x_{i}} − \overline{a} \overline{y_{i}})] ≥ 0,

then one has the inequality

0 ≤ ∑_{i=0}^{∞} w_{i} |x_{i}|^{2} ∑_{i=0}^{∞} w_{i} |y_{i}|^{2} − |∑_{i=0}^{∞} w_{i}x_{i}\overline{y_{i}}|^{2} ≤ \frac{1}{4} |A − a|^{2} (∑_{i=0}^{∞} w_{i} |y_{i}|^{2})^{2} .

The constant \frac{1}{4} is sharp.

Proof. Follows by Theorem 1 applied for the inner product ⟨·, ·⟩_{w} : ℓ_{w}^{2} (𝕂) × ℓ_{w}^{2} (𝕂) → 𝕂,

⟨\overline{x}, \overline{y}⟩_{w} := ∑_{i=0}^{∞} w_{i}x_{i}\overline{y_{i}}.

Remark 4. A sufficient condition for (2.16) to hold is

Re [(Ay_{i} − x_{i}) (\overline{x_{i}} − \overline{a}\overline{y_{i}})] ≥ 0, for all i ∈ ℕ.

In the particular case w_{i} = 1, i ∈ ℕ, we have the following reverse of the Cauchy-Bunyakovsky-Schwarz inequality.

Corollary 3. Let a, A ∈ 𝕂 (𝕂 = ℂ, ℝ) and \overline{x}, \overline{y} ∈ ℓ^{2} (𝕂) . If

(2.17)  ∑_{i=0}^{∞} Re [(Ay_{i} − x_{i}) (\overline{x_{i}} − \overline{a}\overline{y_{i}})] ≥ 0,

then one has the inequality

0 ≤ ∑_{i=0}^{∞} |x_{i}|^{2} ∑_{i=0}^{∞} |y_{i}|^{2} − |∑_{i=0}^{∞} x_{i}\overline{y_{i}}|^{2} ≤ \frac{1}{4} |A − a|^{2} (∑_{i=0}^{∞} |y_{i}|^{2})^{2} .

Remark 5. If 𝕂 = ℝ, then a sufficient condition for either (2.16) or (2.17) to hold is

ay_{i} ≤ x_{i} ≤ Ay_{i} for each i ∈ ℕ,

with A > a > 0.

3. A Generalisation of the Cassels Inequality

3.1. Introduction. The following result was proved by J.W.S. Cassels in 1951 (see Appendix 1 of [23]).

Theorem 2. Let \overline{a} = (a_{1}, …, a_{n}) , \overline{b} = (b_{1}, …, b_{n}) be sequences of positive real numbers and \overline{w} = (w_{1}, …, w_{n}) a sequence of nonnegative real numbers. Suppose that

(3.1)  m = min_{i=\overline{1,n}} \{\frac{a_{i}}{b_{i}} \} and M = max_{i=\overline{1,n}} \{\frac{a_{i}}{b_{i}} \} .

Then one has the inequality

(3.2)  \frac{∑_{i=1}^{n} w_{i}a_{i}^{2} ∑_{i=1}^{n} w_{i}b_{i}^{2}}{(∑_{i=1}^{n} w_{i}a_{i}b_{i})^{2}} ≤ \frac{(m + M)^{2}}{4mM} .

The equality holds in (3.2) when w_{1} = \frac{1}{a_{1}b_{1}} , w_{n} = \frac{1}{a_{n}b_{n}} , w_{2} = ⋯ = w_{n−1} = 0, m = \frac{a_{n}}{b_{1}} and M = \frac{a_{1}}{b_{n}} .

If one assumes that 0 < a ≤ a_{i} ≤ A < ∞ and 0 < b ≤ b_{i} ≤ B < ∞ for each i ∈ {1, …, n} , then by (3.2) we may obtain Greub-Reinboldt’s inequality [12]

\frac{∑_{i=1}^{n} w_{i}a_{i}^{2} ∑_{i=1}^{n} w_{i}b_{i}^{2}}{(∑_{i=1}^{n} w_{i}a_{i}b_{i})^{2}} ≤ \frac{(ab + AB)^{2}}{4abAB} .

The following “unweighted” Cassels’ inequality also holds

\frac{∑_{i=1}^{n} a_{i}^{2} ∑_{i=1}^{n} b_{i}^{2}}{(∑_{i=1}^{n} a_{i}b_{i})^{2}} ≤ \frac{(m + M)^{2}}{4mM} ,

provided \overline{a} and \overline{b} satisfy (3.1). This inequality will produce the well known Pólya-Szegö inequality [20, pp. 57, 213-114], [17, pp. 71-72, 253-255]:

\frac{∑_{i=1}^{n} a_{i}^{2} ∑_{i=1}^{n} b_{i}^{2}}{(∑_{i=1}^{n} a_{i}b_{i})^{2}} ≤ \frac{(ab + AB)^{2}}{4abAB} ,

provided 0 < a ≤ a_{i} ≤ A < ∞ and 0 < b ≤ b_{i} ≤ B < ∞ for each i ∈ {1, …, n} .

In [18], C.P. Niculescu proved, amongst others, the following generalisation of Cassels’ inequality:

Theorem 3. Let E be a vector space endowed with a Hermitian product ⟨·, ·⟩ . Then

(3.3)  \begin{matrix}Re & ⟨x, y⟩ \\ ⟨x, x⟩ & \frac{1}{2} & ⟨y, y⟩ & \frac{1}{2} \end{matrix} ≥ \frac{2}{\sqrt{\frac{ω}{Ω}} + \sqrt{\frac{Ω}{ω}}} ,

for every x, y ∈ E and every ω, Ω > 0 for which Re ⟨x − ωy, x − Ωy⟩ ≤ 0.

In this section, by following [4], we establish a generalisation of (3.3) for the complex numbers ω and Ω for which Re (\overline{ω}Ω) > 0. Applications for isotonic linear functionals, integrals and sequences are also given.

3.2. An Inequality in Real or Complex Inner Product Spaces. The following reverse of Schwarz’s inequality in inner product spaces holds [4].

Theorem 4. Let a, A ∈ 𝕂 (𝕂 = ℂ, ℝ) so that Re (\overline{a}A) > 0. If x, y ∈ H are such that

(3.4)  Re ⟨Ay − x, x − ay⟩ ≥ 0,

then one has the inequality

(3.5) ∥x∥ ∥y∥ ≤ \frac{1}{2} · \begin{matrix}Re & [A\overline{⟨x, y⟩} + \overline{a} ⟨x, y⟩] \\ [Re (\overline{a}A)] & \frac{1}{2} \end{matrix} ≤ \frac{1}{2} · \frac{|A| + |a|}{[Re (\overline{a}A)]^{\frac{1}{2}}} |⟨x, y⟩| .

The constant \frac{1}{2} is sharp in both inequalities.

Proof. We have, obviously, that

I := Re ⟨Ay − x, x − ay⟩
= Re [A\overline{⟨x, y⟩} + \overline{a} ⟨x, y⟩] − ∥x∥^{2} − [Re (\overline{a}A)] ∥y∥^{2}

and, thus, by (3.4), one has

∥x∥^{2} + [Re (\overline{a}A)] · ∥y∥^{2} ≤ Re [A\overline{⟨x, y⟩} + \overline{a} ⟨x, y⟩] ,

which gives

(3.6) \frac{1}{[Re (\overline{a}A)]^{\frac{1}{2}}} ∥x∥^{2} + [Re (\overline{a}A)]^{\frac{1}{2}} ∥y∥^{2} ≤ \begin{matrix}Re & [A\overline{⟨x, y⟩} + \overline{a} ⟨x, y⟩] \\ [Re (\overline{a}A)] & \frac{1}{2} \end{matrix}.

On the other hand, by the elementary inequality

αp^{2} + \frac{1}{α}q^{2} ≥ 2pq,

valid for p, q ≥ 0 and α > 0, we deduce

(3.7)  2 ∥x∥ ∥y∥ ≤ \frac{1}{[Re (\overline{a}A)]^{\frac{1}{2}}} ∥x∥^{2} + [Re (\overline{a}A)]^{\frac{1}{2}} ∥y∥^{2} .

Utilizing (3.6) and (3.7) we deduce the first part of (3.5).

The second part is obvious by the fact that for z ∈ ℂ, |Re (z)| ≤ |z| . Now, assume that the first inequality in (3.5) holds with a constant

c > 0, i.e.,

(3.8)  ∥x∥ ∥y∥ ≤ c · \begin{matrix}Re & [A\overline{⟨x, y⟩} + \overline{a} ⟨x, y⟩] \\ [Re (\overline{a}A)] & \frac{1}{2} \end{matrix},

where a, A, x and y satisfy (3.5).

If we choose a = A = 1, y = x ≠ 0, then obviously (3.4) holds and from (3.8) we obtain

∥x∥^{2} ≤ 2c ∥x∥^{2} ,

giving c ≥ \frac{1}{2}.

The theorem is completely proved.

The following corollary is a natural consequence of the above theorem [4].

Corollary 4. Let m, M > 0. If x, y ∈ H are such that then one has the inequality

(3.9)  ∥x∥ ∥y∥ ≤ \frac{1}{2} · \frac{M + m}{\sqrt{mM}} Re ⟨x, y⟩ ≤ \frac{1}{2} · \frac{M + m}{\sqrt{mM}} |⟨x, y⟩| .

The constant \frac{1}{2} is sharp in (3.9).

Remark 6. The inequality (3.9) is equivalent to Niculescu’s inequality (3.3).

The following corollary providing an additive reverse for the Schwarz inequality is also obvious [4].

Corollary 5. With the assumptions of Corollary 4, we have

(3.10)  0 ≤ ∥x∥ ∥y∥ − |⟨x, y⟩| ≤ ∥x∥ ∥y∥ − Re ⟨x, y⟩

≤ \frac{(\sqrt{M} − \sqrt{m})^{2}}{2\sqrt{mM}} Re ⟨x, y⟩ ≤ \frac{(\sqrt{M} − \sqrt{m})^{2}}{2\sqrt{mM}} |⟨x, y⟩|

and

(3.11)  0 ≤ ∥x∥^{2} ∥y∥^{2} − |⟨x, y⟩|^{2} ≤ ∥x∥^{2} ∥y∥^{2} − [Re ⟨x, y⟩]^{2}

≤ \frac{(M − m)^{2}}{4mM} [Re ⟨x, y⟩]^{2} ≤ \frac{(M − m)^{2}}{4mM} |⟨x, y⟩|^{2} .

The constants \frac{1}{2} and \frac{1}{4} are sharp.

Proof. If we subtract Re ⟨x, y⟩ ≥ 0 from the first inequality in (3.9), we get

∥x∥ ∥y∥ − Re ⟨x, y⟩ ≤ (\frac{1}{2} · \frac{M + m}{\sqrt{mM}} − 1) Re ⟨x, y⟩

= \frac{(\sqrt{M} − \sqrt{m})^{2}}{2\sqrt{mM}} Re ⟨x, y⟩ ,

which proves the third inequality in (3.10). The other ones are obvious.

Now, if we square the first inequality in (3.9) and then subtract [Re ⟨x, y⟩]^{2} , we get

∥x∥^{2} ∥y∥^{2} − [Re ⟨x, y⟩]^{2} ≤ [\frac{(M + m)^{2}}{4mM} − 1] [Re ⟨x, y⟩]^{2}

= \frac{(M − m)^{2}}{4mM} [Re ⟨x, y⟩]^{2} ,

which proves the third inequality in (3.11). The other ones are obvious.

3.3. Applications for Isotonic Linear Functionals. The following proposition holds [4].

Proposition 4. Let f, g, h ∈ F (T) be such that fgh ∈ L, f^{2}h ∈ L, g^{2}h ∈ L. If m, M > 0 are such that

(3.12)  mg ≤ f ≤ Mg on F (T) ,

then for any isotonic linear functional A : L → ℝ with A (h) > 0, we have the inequality

(3.13)  1 ≤ \frac{A (f^{2}h) A (g^{2}h)}{A^{2} (fgh)} ≤ \frac{(M + m)^{2}}{4mM} .

The constant \frac{1}{4} in (3.13) is sharp.

Proof. We observe that

(Mg − f, f − mg)_{A,h} = A [h (Mg − f) (f − mg)] ≥ 0.

Applying Corollary 4 for (·, ·)_{A,h} we get

1 ≤ \begin{matrix}(f, f) & A,h & (g, g) & A,h \\ (f, g) & {2 \atop A,h} \end{matrix} ≤ \frac{(M + m)^{2}}{4mM} ,

which is clearly equivalent to (3.13).

The following additive versions of (3.13) also hold [4].

Corollary 6. With the assumption in Proposition 4, one has

0 ≤ [A (f^{2}h) A (g^{2}h)]^{\frac{1}{2}} − A (hfg)

≤ \frac{(\sqrt{M} − \sqrt{m})^{2}}{2\sqrt{mM}} A (hfg)

and

0 ≤ A (f^{2}h) A (g^{2}h) − A^{2} (fgh)

≤ \frac{(M − m)^{2}}{4mM} A^{2} (fgh) .

The constants \frac{1}{2} and \frac{1}{4} are sharp.

Remark 7. The condition (3.12) may be replaced with the weaker assumption

(3.14)  (Mg − f, f − mg)_{A,h} ≥ 0.

Remark 8. With the assumption (3.12) or (3.14) and if f, g ∈ F (T) with fg, f^{2}, g^{2} ∈ L, then one has the inequalities

1 ≤ \frac{A (f^{2}) A (g^{2})}{A^{2} (fg)} ≤ \frac{(M + m)^{2}}{4mM} ,

0 ≤ [A (f^{2}) A (g^{2})]^{\frac{1}{2}} − A (fg)

≤ \frac{(\sqrt{M} − \sqrt{m})^{2}}{2\sqrt{mM}} A (fg)

and

0 ≤ A (f^{2}) A (g^{2}) − A^{2} (fg) ≤ \frac{(M − m)^{2}}{4mM} A^{2} (fg) .

3.4. Applications for Integrals. The following proposition contains a reverse of the weighted Cauchy-Bunyakovsky-Schwarz integral inequality.

Proposition 5. Let A, a ∈ 𝕂 (𝕂 = ℂ, ℝ) with Re (\overline{a}A) > 0 and f, g ∈ L_{ρ}^{2} (Ω, 𝕂) . If

(3.15) ∫_{Ω} Re [(Ag (s) − f (s)) (\overline{f (s)} − \overline{a} \overline{g} (s))] ρ (s) dμ (s) ≥ 0,

then one has the inequality

[∫_{Ω} ρ (s) |f (s)|^{2} dμ (s) ∫_{Ω} ρ (s) |g (s)|^{2} dμ (s)]^{\frac{1}{2}} (3.16)

≤ \frac{1}{2} · \frac{∫_{Ω} ρ (s) Re [A\overline{f (s)}g (s) + \overline{a}f (s) \overline{g (s)}] dμ (s)}{[Re (\overline{a}A)]^{\frac{1}{2}}}

≤ \frac{1}{2} · \frac{|A| + |a|}{[Re (\overline{a}A)]^{\frac{1}{2}}} |∫_{Ω} ρ (s) f (s) \overline{g (s)}dμ (s)| .

The constant \frac{1}{2} is sharp in (3.16).

Proof. Follows by Theorem 4 applied for the inner product ⟨·, ·⟩_{ρ} : L_{ρ}^{2} (Ω, 𝕂) × L_{ρ}^{2} (Ω, 𝕂) → 𝕂,

⟨f, g⟩ := ∫_{Ω} ρ (s) f (s) \overline{g (s)}dμ (s) .

Remark 9. A sufficient condition for (3.15) to hold is

Re [(Ag (s) − f (s)) (\overline{f (s)} − \overline{a}\overline{g (s)})] ≥ 0, for μ-a.e. s ∈ Ω.

3. A GENERALISATION OF THE CASSELS INEQUALITY 15

In the particular case ρ = 1, we have the following result. Corollary 7. Let a, A ∈ 𝕂 (𝕂 = ℂ, ℝ) with Re (\overline{a}A) > 0 and

f, g ∈ L^{2} (Ω, 𝕂) . If

(3.17)  ∫_{Ω} Re [(Ag (s) − f (s)) (\overline{f (s)} − \overline{a}\overline{g (s)})] dμ (s) ≥ 0,

then one has the inequality

[∫_{Ω} |f (s)|^{2} dμ (s) ∫_{Ω} |g (s)|^{2} dμ (s)]^{\frac{1}{2}}

≤ \frac{1}{2} · \frac{∫_{Ω} Re [A\overline{f (s)}g (s) + \overline{a}f (s) \overline{g (s)}] dμ (s)}{[Re (\overline{a}A)]^{\frac{1}{2}}}

≤ \frac{1}{2} · \frac{|A| + |a|}{[Re (\overline{a}A)]^{\frac{1}{2}}} |∫_{Ω} f (s) \overline{g (s)}dμ (s)| .

Remark 10. If 𝕂 = ℝ, then a sufficient condition for either (3.15) or (3.17) to hold is

ag (s) ≤ f (s) ≤ Ag (s) , for μ-a.e. s ∈ Ω,

where, in this case, a, A ∈ ℝ with A > a > 0.

If a, A are real positive constants, then the following proposition holds.

Proposition 6. Let m, M > 0. If f, g ∈ L_{ρ}^{2} (Ω, 𝕂) such that

∫_{Ω} ρ (s) Re [(Mg (s) − f (s)) (\overline{f (s)} − m\overline{g} (s))] dμ (s) ≥ 0, then one has the inequality

[∫_{Ω} ρ (s) |f (s)|^{2} dμ (s) ∫_{Ω} ρ (s) |g (s)|^{2} dμ (s)]^{\frac{1}{2}}

≤ \frac{1}{2} · \frac{M + m}{\sqrt{mM}} ∫_{Ω} ρ (s) Re [f (s) \overline{g (s)}] dμ (s) .

The proof follows by Corollary 4 applied for the inner product

⟨f, g⟩_{ρ} := ∫_{Ω} ρ (s) f (s) \overline{g (s)}dμ (s) .

The following additive versions also hold [4].

Corollary 8. With the assumptions in Proposition 6, one has the inequalities

0 ≤ [∫_{Ω} ρ (s) |f (s)|^{2} dμ (s) ∫_{Ω} ρ (s) |g (s)|^{2} dμ (s)]^{\frac{1}{2}}
− ∫_{Ω} ρ (s) Re [f (s) \overline{g (s)}] dμ (s)

≤ \frac{(\sqrt{M} − \sqrt{m})^{2}}{2\sqrt{mM}} ∫_{Ω} ρ (s) Re [f (s) \overline{g (s)}] dμ (s)

and

0 ≤ ∫_{Ω} ρ (s) |f (s)|^{2} dμ (s) ∫_{Ω} ρ (s) |g (s)|^{2} dμ (s)
− (∫_{Ω} ρ (s) Re [f (s) \overline{g (s)}] dμ (s))^{2}

≤ \frac{(M − m)^{2}}{4mM} (∫_{Ω} ρ (s) Re [f (s) \overline{g (s)}] dμ (s))^{2} .

Remark 11. If 𝕂 = ℝ, a sufficient condition for (3.15) to hold is

mg (s) ≤ f (s) ≤ Mg (s) , for μ-a.e. s ∈ Ω,

where M > m > 0.

3.5. Applications for Sequences. For a given sequence (w_{i})_{i∈ℕ} of nonnegative real numbers, consider the Hilbert space ℓ_{w}^{2} (𝕂) , (𝕂 = ℂ, ℝ) , where

ℓ_{w}^{2} (𝕂) := \{\overline{x} = (x_{i})_{i∈ℕ} ⊂ 𝕂 |∑_{i=0}^{∞} w_{i} |x_{i}|^{2} < ∞\} .

The following proposition that provides a reverse of the weighted Cauchy-Bunyakovsky-Schwarz inequality for complex numbers holds [4].

Proposition 7. Let a, A ∈ 𝕂 with Re (\overline{a}A) > 0 and \overline{x}, \overline{y} ∈ ℓ_{w}^{2} (𝕂) . If

(3.18)  ∑_{i=0}^{∞} w_{i} Re [(Ay_{i} − x_{i}) (\overline{x_{i}} − \overline{a}\overline{y_{i}})] ≥ 0,

then one has the inequality

(3.19) [∑_{i=0}^{∞} w_{i} |x_{i}|^{2} ∑_{i=0}^{∞} w_{i} |y_{i}|^{2}]^{\frac{1}{2}} ≤ \frac{1}{2} · \frac{∑_{i=0}^{∞} w_{i} Re [A\overline{x_{i}}y_{i} + \overline{a}x_{i}\overline{y_{i}}]}{[Re (\overline{a}A)]^{\frac{1}{2}}}

≤ \frac{1}{2} · \frac{|A| + |a|}{[Re (\overline{a}A)]^{\frac{1}{2}}} |∑_{i=0}^{∞} w_{i}x_{i}\overline{y_{i}}| .

The constant \frac{1}{2} is sharp in (3.19).

Proof. Follows by Theorem 4 applied for the inner product ⟨·, ·⟩_{w} : ℓ_{w}^{2} (𝕂) × ℓ_{w}^{2} (𝕂) → 𝕂,

⟨\overline{x}, \overline{y}⟩_{w} := ∑_{i=0}^{∞} w_{i}x_{i}\overline{y_{i}}.

Remark 12. A sufficient condition for (3.18) to hold is

(3.20)  Re [(Ay_{i} − x_{i}) (\overline{x_{i}} − \overline{a}\overline{y_{i}})] ≥ 0 for all i ∈ ℕ.

In the particular case ρ = 1, we have the following result.

Corollary 9. Let a, A ∈ 𝕂 with Re (\overline{a}A) > 0 and \overline{x}, \overline{y} ∈ ℓ^{2} (𝕂) . If

∑_{i=0}^{∞} Re [(Ay_{i} − x_{i}) (\overline{x_{i}} − \overline{a}\overline{y_{i}})] ≥ 0,

then one has the inequality

[∑_{i=0}^{∞} |x_{i}|^{2} ∑_{i=0}^{∞} |y_{i}|^{2}]^{\frac{1}{2}} ≤ \frac{1}{2} · \begin{matrix}∑ & {∞ \atop i=0} & Re & [A\overline{x_{i}}y_{i} + \overline{a}x_{i}\overline{y_{i}}] \\ [Re (\overline{a}A)] & \frac{1}{2} \end{matrix}

≤ \frac{1}{2} · \frac{|A| + |a|}{[Re (\overline{a}A)]^{\frac{1}{2}}} |∑_{i=0}^{∞} x_{i}\overline{y_{i}}| .

Remark 13. If 𝕂 = ℝ, then a sufficient condition for either (3.18) or (3.20) to hold is

ay_{i} ≤ x_{i} ≤ Ay_{i} for each i ∈ {1, …, n} ,

where, in this case, a, A ∈ ℝ with A > a > 0.

For a = m, A = M, then the following proposition also holds.

Proposition 8. Let m, M > 0. If \overline{x}, \overline{y} ∈ ℓ_{w}^{2} (𝕂) such that

(3.21)  ∑_{i=0}^{∞} w_{i} Re [(My_{i} − x_{i}) (\overline{x_{i}} − m\overline{y_{i}})] ≥ 0,

then one has the inequality

[∑_{i=0}^{∞} w_{i} |x_{i}|^{2} ∑_{i=0}^{∞} w_{i} |y_{i}|^{2}]^{\frac{1}{2}} ≤ \frac{1}{2} · \frac{M + m}{\sqrt{mM}} ∑_{i=0}^{∞} w_{i} Re (x_{i}\overline{y_{i}}) .

The proof follows by Corollary 4 applied for the inner product

⟨\overline{x}, \overline{y}⟩_{w} := ∑_{i=0}^{∞} w_{i}x_{i}, \overline{y_{i}}.

The following additive version also holds [4].

Corollary 10. With the assumptions in Proposition 8, one has the inequalities

0 ≤ [∑_{i=0}^{∞} w_{i} |x_{i}|^{2} ∑_{i=0}^{∞} w_{i} |y_{i}|^{2}]^{\frac{1}{2}} − ∑_{i=0}^{∞} w_{i} Re (x_{i}\overline{y_{i}})

≤ \frac{(\sqrt{M} − \sqrt{m})^{2}}{2\sqrt{mM}} ∑_{i=0}^{∞} w_{i} Re (x_{i}\overline{y_{i}})

and

0 ≤ ∑_{i=0}^{∞} w_{i} |x_{i}|^{2} ∑_{i=0}^{∞} w_{i} |y_{i}|^{2} − [∑_{i=0}^{∞} w_{i} Re (x_{i}\overline{y_{i}})]^{2}

≤ \frac{(M − m)^{2}}{4mM} [∑_{i=0}^{∞} w_{i} Re (x_{i}\overline{y_{i}})]^{2} .

Remark 14. If 𝕂 = ℝ, a sufficient condition for (3.21) to hold is

my_{i} ≤ x_{i} ≤ My_{i} for each i ∈ ℕ,

where M > m > 0.

4. Quadratic Reverses of Schwarz’s Inequality

4.1. Two Better Reverse Inequalities. It has been proven in [7], that

(4.1)  0 ≤ ∥x∥^{2} − |⟨x, e⟩|^{2} ≤ \frac{1}{4} |ϕ − φ|^{2} − |\frac{ϕ + φ}{2} − ⟨x, e⟩|^{2} ,

provided, either

(4.2)  Re ⟨ϕe − x, x − φe⟩ ≥ 0,

or, equivalently,

(4.3)  ∥x − \frac{ϕ + φ}{2}e∥ ≤ \frac{1}{2} |ϕ − φ| ,

holds, where e ∈ H, ∥e∥ = 1. The constant \frac{1}{4} in (4.1) is best possible.

If we choose e = \frac{y}{∥y∥}, ϕ = Γ ∥y∥ , φ = γ ∥y∥ (y ≠ 0) , Γ, γ ∈ 𝕂, then by (4.2) and (4.3) we have,

(4.4)  Re ⟨Γy − x, x − γy⟩ ≥ 0,

or, equivalently,

(4.5)  ∥x − \frac{Γ + γ}{2} y∥ ≤ \frac{1}{2} |Γ − γ| ∥y∥ ,

implying the following reverse of Schwarz’s inequality:

(4.6)  0 ≤ ∥x∥^{2} ∥y∥^{2} − |⟨x, y⟩|^{2}

≤ \frac{1}{4} |Γ − γ|^{2} ∥y∥^{4} − |\frac{Γ + γ}{2} ∥y∥^{2} − ⟨x, y⟩|^{2} .

The constant \frac{1}{4} in (4.6) is sharp.

Note that, this inequality is an improvement of (2.6), but it may not be very convenient for applications.

In [10], it has also been proven that

(4.7)  0 ≤ ∥x∥^{2} − |⟨x, e⟩|^{2} ≤ \frac{1}{4} |ϕ − φ|^{2} − Re ⟨Φe − x, x − φe⟩

provided either (4.2) or (4.3) holds true.

If we make the same choice for e, Φ and φ as above, then we deduce the inequality

(4.8)  0 ≤ ∥x∥^{2} ∥y∥^{2} − |⟨x, y⟩|^{2}

≤ \frac{1}{4} |Γ − γ|^{2} ∥y∥^{4} − ∥y∥^{2} Re ⟨Γy − x, x − γy⟩

provided either (4.4) or (4.5) holds true.

The constant \frac{1}{4} is best possible in (4.8).

One may easily realise that the bounds provided by (4.6) and (4.8) cannot be compared in general, meaning that for different choices of variables one may be better than the other.

4.2. A Reverse of Schwarz’s Inequality Under More General Assumptions. The following result holds [8].

Theorem 5. Let (H; ⟨·, ·⟩) be an inner product space over the real or complex number field 𝕂 (𝕂 = ℝ, 𝕂 = ℂ) and x, a ∈ H, r > 0 are such that

x ∈ \overline{B} (a, r) := \{z ∈ H| ∥z − a∥ ≤ r\} .

(i) If ∥a∥ > r, then we have the inequalities

(4.9) 0 ≤ ∥x∥^{2} ∥a∥^{2} − |⟨x, a⟩|^{2} ≤ ∥x∥^{2} ∥a∥^{2} − [Re ⟨x, a⟩]^{2} ≤ r^{2} ∥x∥^{2} .

The constant C = 1 in front of r^{2} is best possible in the sense that it cannot be replaced by a smaller one.

(ii) If ∥a∥ = r, then

(4.10)  ∥x∥^{2} ≤ 2 Re ⟨x, a⟩ ≤ 2 |⟨x, a⟩| .

The constant 2 is best possible in both inequalities. (iii) If ∥a∥ < r, then

(4.11)  ∥x∥^{2} ≤ r^{2} − ∥a∥^{2} + 2 Re ⟨x, a⟩ ≤ r^{2} − ∥a∥^{2} + 2 |⟨x, a⟩| .

Here the constant 2 is also best possible.

Proof. Since x ∈ \overline{B} (a, r) , then obviously ∥x − a∥^{2} ≤ r^{2}, which is equivalent to

(4.12)  ∥x∥^{2} + ∥a∥^{2} − r^{2} ≤ 2 Re ⟨x, a⟩ .

(i) If ∥a∥ > r, then we may divide (4.12) by \sqrt{∥a∥^{2} − r^{2}} > 0 getting

(4.13)  \begin{matrix}∥x∥ & 2 \\ √ & \overline{∥a∥^{2} − r^{2}} \end{matrix} + \sqrt{∥a∥^{2} − r^{2}} ≤ \frac{2 Re ⟨x, a⟩}{\sqrt{∥a∥^{2} − r^{2}}} .

Using the elementary inequality

αp + \frac{1}{α}q ≥ 2\sqrt{pq}, α > 0, p, q ≥ 0,

we may state that

(4.14)  2 ∥x∥ ≤ \begin{matrix}∥x∥ & 2 \\ √ & \overline{∥a∥^{2} − r^{2}} \end{matrix} + \sqrt{∥a∥^{2} − r^{2}}.

Making use of (4.13) and (4.14), we deduce

(4.15)  ∥x∥ \sqrt{∥a∥^{2} − r^{2}} ≤ Re ⟨x, a⟩ ,

which is an interesting inequality in itself as well.

Taking the square in (4.15) and re-arranging the terms, we deduce the third inequality in (4.9). The others are obvious.

To prove the sharpness of the constant, assume, under the hypothesis of the theorem, that, there exists a constant c > 0 such that

(4.16)  ∥x∥^{2} ∥a∥^{2} − [Re ⟨x, a⟩]^{2} ≤ cr^{2} ∥x∥^{2} ,

provided x ∈ \overline{B} (a, r) and ∥a∥ > r.

Let r = \sqrt{ε} > 0, ε ∈ (0, 1) , a, e ∈ H with ∥a∥ = ∥e∥ = 1 and a ⊥ e. Put x = a + \sqrt{ε}e. Then obviously x ∈ \overline{B} (a, r) , ∥a∥ > r and ∥x∥^{2} = ∥a∥^{2}+ε ∥e∥^{2} = 1+ε, Re ⟨x, a⟩ = ∥a∥^{2} = 1, and thus ∥x∥^{2} ∥a∥^{2} − [Re ⟨x, a⟩]^{2} = ε. Using (4.16), we may write that

ε ≤ cε (1 + ε) , ε > 0

giving

(4.17)  c + cε ≥ 1 for any ε > 0.

Letting ε → 0+, we get from (4.17) that c ≥ 1, and the sharpness of the constant is proved.

(ii) The inequality (4.10) is obvious by (4.12) since ∥a∥ = r. The

best constant follows in a similar way to the above. (iii) The inequality (4.11) is obvious. The best constant may be

proved in a similar way to the above. We omit the details.

The following reverse of Schwarz’s inequality holds [8].

Theorem 6. Let (H; ⟨·, ·⟩) be an inner product space over 𝕂 and x, y ∈ H, γ, Γ ∈ 𝕂 such that either

(4.18)  Re ⟨Γy − x, x − γy⟩ ≥ 0,

or, equivalently,

(4.19)  ∥x − \frac{Γ + γ}{2} y∥ ≤ \frac{1}{2} |Γ − γ| ∥y∥ ,

holds.

(i) If Re (Γ\overline{γ}) > 0, then we have the inequalities

(4.20)  ∥x∥^{2} ∥y∥^{2} ≤ \frac{1}{4} · \begin{matrix}\{Re [(\overline{Γ} + \overline{γ}) ⟨x, y⟩]\} & 2 \\ Re & (Γ\overline{γ}) \end{matrix}

≤ \frac{1}{4} · \begin{matrix}|Γ + γ| & 2 \\ Re & (Γ\overline{γ}) \end{matrix} |⟨x, y⟩|^{2} .

The constant \frac{1}{4} is best possible in both inequalities.

(ii) If Re (Γ\overline{γ}) = 0, then

∥x∥^{2} ≤ Re [(\overline{Γ} + \overline{γ}) ⟨x, y⟩] ≤ |Γ + γ| |⟨x, y⟩| .

(iii) If Re (Γ\overline{γ}) < 0, then

∥x∥^{2} ≤ − Re (Γ\overline{γ}) ∥y∥^{2} + Re [(\overline{Γ} + \overline{γ}) ⟨x, y⟩]
≤ − Re (Γ\overline{γ}) ∥y∥^{2} + |Γ + γ| |⟨x, y⟩| .

Proof. The proof of the equivalence between the inequalities (4.18) and (4.19) follows by the fact that in an inner product space

Re ⟨Z − x, x − z⟩ ≥ 0

for x, z, Z ∈ H is equivalent with

∥x − \frac{z + Z}{2} ∥ ≤ \frac{1}{2} ∥Z − z∥

(see for example [9]).

Consider, for y ≠ 0, a = \frac{γ+Γ}{2}y and r = \frac{1}{2} |Γ − γ| ∥y∥ . Then

∥a∥^{2} − r^{2} = \frac{|Γ + γ|^{2} − |Γ − γ|^{2}}{4} ∥y∥^{2} = Re (Γ\overline{γ}) ∥y∥^{2} .

(i) If Re (Γ\overline{γ}) > 0, then the hypothesis of (i) in Theorem 5 is satisfied, and by the second inequality in (4.9) we have

∥x∥^{2} \frac{|Γ + γ|^{2}}{4} ∥y∥^{2} − \frac{1}{4} \{Re [(\overline{Γ} + \overline{γ}) ⟨x, y⟩]\}^{2}

≤ \frac{1}{4} |Γ − γ|^{2} ∥x∥^{2} ∥y∥^{2}

from where we derive

\frac{|Γ + γ|^{2} − |Γ − γ|^{2}}{4} ∥x∥^{2} ∥y∥^{2} ≤ \frac{1}{4} \{Re [(\overline{Γ} + \overline{γ}) ⟨x, y⟩]\}^{2} ,

giving the first inequality in (4.20).

The second inequality is obvious.

To prove the sharpness of the constant \frac{1}{4}, assume that the first inequality in (4.20) holds with a constant c > 0, i.e.,

(4.21)  ∥x∥^{2} ∥y∥^{2} ≤ c · \begin{matrix}\{Re [(\overline{Γ} + \overline{γ}) ⟨x, y⟩]\} & 2 \\ Re & (Γ\overline{γ}) \end{matrix} ,

provided Re (Γ\overline{γ}) > 0 and either (4.18) or (4.19) holds.

Assume that Γ, γ > 0, and let x = γy. Then (4.18) holds and by (4.21) we deduce

γ^{2} ∥y∥^{4} ≤ c · \frac{(Γ + γ)^{2} γ^{2} ∥y∥^{4}}{Γγ}

giving

(4.22)  Γγ ≤ c (Γ + γ)^{2} for any Γ, γ > 0.

Let ε ∈ (0, 1) and choose in (4.22), Γ = 1 + ε, γ = 1 − ε > 0 to get 1 − ε^{2} ≤ 4c for any ε ∈ (0, 1) . Letting ε → 0+, we deduce c ≥ \frac{1}{4}, and the sharpness of the constant is proved. (ii) and (iii) are obvious and we omit the details.

Remark 15. We observe that the second bound in (4.20) for ∥x∥^{2} ∥y∥^{2} is better than the second bound provided by (3.5).

The following corollary provides a reverse inequality for the additive version of Schwarz’s inequality [8].

Corollary 11. With the assumptions of Theorem 6 and if Re (Γ\overline{γ}) > 0, then we have the inequality:

(4.23)  0 ≤ ∥x∥^{2} ∥y∥^{2} − |⟨x, y⟩|^{2} ≤ \frac{1}{4} · \begin{matrix}|Γ − γ| & 2 \\ Re & (Γ\overline{γ}) \end{matrix} |⟨x, y⟩|^{2} .

The constant \frac{1}{4} is best possible in (4.23).

The proof is obvious from (4.20) on subtracting in both sides the same quantity |⟨x, y⟩|^{2} . The sharpness of the constant may be proven in a similar manner to the one incorporated in the proof of (i), Theorem 6. We omit the details.

For other recent results in connection to Schwarz’s inequality, see [1], [11] and [13].

4.3. Reverses of the Triangle Inequality. The following reverse of the triangle inequality holds [8].

Proposition 9. Let (H; ⟨·, ·⟩) be an inner product space over the real or complex number field 𝕂 (𝕂 = ℝ, ℂ) and x, a ∈ H, r > 0 are such that

∥x − a∥ ≤ r < ∥a∥ .

Then we have the inequality

(4.24)  0 ≤ ∥x∥ + ∥a∥ − ∥x + a∥

≤ \sqrt{2}r · \sqrt{\begin{matrix}Re & ⟨x, a⟩ \\ \sqrt{∥a∥^{2} − r^{2}} & (\sqrt{∥a∥^{2} − r^{2}} + ∥a∥) \end{matrix}}.

Proof. Using the inequality (4.15), we may write that

∥x∥ ∥a∥ ≤ \begin{matrix}∥a∥ & Re & ⟨x, a⟩ \\ √ & \overline{∥a∥^{2} − r^{2}} \end{matrix},

which gives

(4.25)  0 ≤ ∥x∥ ∥a∥ − Re ⟨x, a⟩

≤ \frac{∥a∥ − \sqrt{∥a∥^{2} − r^{2}}}{\sqrt{∥a∥^{2} − r^{2}}} Re ⟨x, a⟩

= \frac{r^{2} Re ⟨x, a⟩}{\sqrt{∥a∥^{2} − r^{2}} (\sqrt{∥a∥^{2} − r^{2}} + ∥a∥)}.

Since

(∥x∥ + ∥a∥)^{2} − ∥x + a∥^{2} = 2 (∥x∥ ∥a∥ − Re ⟨x, a⟩) ,

then by (4.25), we have

∥x∥ + ∥a∥ ≤ \sqrt{∥x + a∥^{2} + \frac{2r^{2} Re ⟨x, a⟩}{\sqrt{∥a∥^{2} − r^{2}} (\sqrt{∥a∥^{2} − r^{2}} + ∥a∥)}}

≤ ∥x + a∥ + \sqrt{2}r · \sqrt{\begin{matrix}Re & ⟨x, a⟩ \\ \sqrt{∥a∥^{2} − r^{2}} & (\sqrt{∥a∥^{2} − r^{2}} + ∥a∥) \end{matrix}},

giving the desired inequality (4.24).

The following proposition providing a simpler reverse for the triangle inequality also holds [8].

Proposition 10. Let (H; ⟨·, ·⟩) be an inner product space over 𝕂 and x, y ∈ H, M > m > 0 such that either or, equivalently,

∥x − \frac{M + m}{2} · y∥ ≤ \frac{1}{2} (M − m) ∥y∥ ,

holds. Then we have the inequality

(4.26)  0 ≤ ∥x∥ + ∥y∥ − ∥x + y∥ ≤ \frac{\sqrt{M} − \sqrt{m}}{\sqrt[4]{mM}}\sqrt{Re ⟨x, y⟩}.

Proof. Choosing in (4.15), a = \frac{M+m}{2}y, r = \frac{1}{2} (M − m) ∥y∥ we get

∥x∥ ∥y∥ \sqrt{Mm} ≤ \frac{M + m}{2} Re ⟨x, y⟩ ,

giving

0 ≤ ∥x∥ ∥y∥ − Re ⟨x, y⟩ ≤ \frac{(\sqrt{M} − \sqrt{m})^{2}}{2\sqrt{mM}} Re ⟨x, y⟩ .

Following the same argument as in the proof of Proposition 9, we deduce the desired inequality (4.26).

For some results related to triangle inequality in inner product spaces, see [2], [14], [16] and [21].

4.4. Integral Inequalities. Denote by L_{ρ}^{2} (Ω, 𝕂) the Hilbert space of all real or complex valued functions defined on Ω and 2−ρ−integrable on Ω, i.e.,

∫_{Ω} ρ (s) |f (s)|^{2} dμ (s) < ∞.

It is obvious that the following inner product

⟨f, g⟩_{ρ} := ∫_{Ω} ρ (s) f (s) \overline{g (s)}dμ (s) ,

generates the norm ∥f∥_{ρ} := (∫_{Ω} ρ (s) |f (s)|^{2} dμ (s))^{\frac{1}{2}} of L_{ρ}^{2} (Ω, 𝕂) , and all the above results may be stated for integrals.

It is important to observe that, if

Re [f (s) \overline{g (s)}] ≥ 0, for μ − a.e. s ∈ Ω,

then, obviously,

(4.27)  Re ⟨f, g⟩_{ρ} = Re [∫_{Ω} ρ (s) f (s) \overline{g (s)}dμ (s)]
= ∫_{Ω} ρ (s) Re [f (s) \overline{g (s)}] dμ (s) ≥ 0.

The reverse is evidently not true in general.

26 1. REVERSES FOR THE SCHWARZ INEQUALITY

Moreover, if the space is real, i.e., 𝕂 = ℝ, then a sufficient condition for (4.27) to hold is:

f (s) ≥ 0, g (s) ≥ 0, for μ − a.e. s ∈ Ω.

We now provide, by the use of certain results obtained above, some integral inequalities that may be used in practical applications.

Proposition 11. Let f, g ∈ L_{ρ}^{2} (Ω, 𝕂) and r > 0 with the properties that

(4.28)  |f (s) − g (s)| ≤ r ≤ |g (s)| , for μ − a.e. s ∈ Ω.

Then we have the inequalities

(4.29) 0 ≤ ∫_{Ω} ρ (s) |f (s)|^{2} dμ (s) ∫_{Ω} ρ (s) |g (s)|^{2} dμ (s)
− |∫_{Ω} ρ (s) f (s) \overline{g (s)}dμ (s)|^{2}
≤ ∫_{Ω} ρ (s) |f (s)|^{2} dμ (s) ∫_{Ω} ρ (s) |g (s)|^{2} dμ (s)

− [∫_{Ω} ρ (s) Re (f (s) \overline{g (s)}) dμ (s)]^{2}

≤ r^{2} ∫_{Ω} ρ (s) |g (s)|^{2} dμ (s) .

The constant c = 1 in front of r^{2} is best possible.

The proof follows by Theorem 5 and we omit the details [8].

Proposition 12. Let f, g ∈ L_{ρ}^{2} (Ω, 𝕂) and γ, Γ ∈ 𝕂 such that Re (Γ\overline{γ}) > 0 and

Re [(Γg (s) − f (s)) (\overline{f (s)} − \overline{γ}\overline{g (s)})] ≥ 0, for μ − a.e. s ∈ Ω. Then we have the inequalities ∫_{Ω} ρ (s) |f (s)|^{2} dμ (s) ∫_{Ω} ρ (s) |g (s)|^{2} dμ (s) (4.30)

≤ \frac{1}{4} · \begin{matrix}\{Re [(\overline{Γ} + \overline{γ}) ∫_{Ω} ρ (s) f (s) \overline{g (s)}dμ (s)]\} & 2 \\ Re & (Γ\overline{γ}) \end{matrix}

≤ \frac{1}{4} · \begin{matrix}|Γ + γ| & 2 \\ Re & (Γ\overline{γ}) \end{matrix} |∫_{Ω} ρ (s) f (s) \overline{g (s)}dμ (s)|^{2} .

The constant \frac{1}{4} is best possible in both inequalities.

The proof follows by Theorem 6 and we omit the details.

Corollary 12. With the assumptions of Proposition 12, we have the inequality

(4.31)  0 ≤ ∫_{Ω} ρ (s) |f (s)|^{2} dμ (s) ∫_{Ω} ρ (s) |g (s)|^{2} dμ (s)
− |∫_{Ω} ρ (s) f (s) \overline{g (s)}dμ (s)|^{2}

≤ \frac{1}{4} · \begin{matrix}|Γ − γ| & 2 \\ Re & (Γ\overline{γ}) \end{matrix} |∫_{Ω} ρ (s) f (s) \overline{g (s)}dμ (s)|^{2} .

The constant \frac{1}{4} is best possible.

Remark 16. If the space is real and we assume, for M > m > 0, that

mg (s) ≤ f (s) ≤ Mg (s) , for μ − a.e. s ∈ Ω,

then, by (4.30) and (4.31), we deduce the inequalities

(4.32) ∫_{Ω} ρ (s) [f (s)]^{2} dμ (s) ∫_{Ω} ρ (s) [g (s)]^{2} dμ (s)

≤ \frac{1}{4} · \frac{(M + m)^{2}}{mM} [∫_{Ω} ρ (s) f (s) g (s) dμ (s)]^{2}

and

(4.33)  0 ≤ ∫_{Ω} ρ (s) [f (s)]^{2} dμ (s) ∫_{Ω} ρ (s) [g (s)]^{2} dμ (s)
− [∫_{Ω} ρ (s) f (s) g (s) dμ (s)]^{2}

≤ \frac{1}{4} · \frac{(M − m)^{2}}{mM} [∫_{Ω} ρ (s) f (s) g (s) dμ (s)]^{2} .

The inequality (4.32) is known in the literature as Cassel’s inequality.

5. More Reverses of Schwarz’s Inequality

5.1. General Results. The following result holds [6].

Theorem 7. Let (H; ⟨·, ·⟩) be an inner product space over the real or complex number field 𝕂, x, a ∈ H and r > 0. If

(5.1)  x ∈ B̄ (a, r) := {z ∈ H| ∥z − a∥ ≤ r} ,

then we have the inequalities:

(5.2)  0 ≤ ∥x∥ ∥a∥ − |⟨x, a⟩| ≤ ∥x∥ ∥a∥ − |Re ⟨x, a⟩|

≤ ∥x∥ ∥a∥ − Re ⟨x, a⟩ ≤ \frac{1}{2}r^{2}.

The constant \frac{1}{2} is best possible in (5.2) in the sense that it cannot be replaced by a smaller constant.

Proof. The condition (5.1) is clearly equivalent to

(5.3)  ∥x∥^{2} + ∥a∥^{2} ≤ 2 Re ⟨x, a⟩ + r^{2}.

Using the elementary inequality

2 ∥x∥ ∥a∥ ≤ ∥x∥^{2} + ∥a∥^{2} , a, x ∈ H

and (5.3), we deduce

2 ∥x∥ ∥a∥ ≤ 2 Re ⟨x, a⟩ + r^{2},

giving the last inequality in (5.2). The other inequalities are obvious. To prove the sharpness of the constant \frac{1}{2}, assume that

(5.4)  0 ≤ ∥x∥ ∥a∥ − Re ⟨x, a⟩ ≤ cr^{2}

for any x, a ∈ H and r > 0 satisfying (5.1).

Assume that a, e ∈ H, ∥a∥ = ∥e∥ = 1 and e ⊥ a. If r = \sqrt{ε}, ε > 0 and if we define x = a + \sqrt{ε}e, then ∥x − a∥ = \sqrt{ε} = r showing that the condition (5.1) is fulfilled.

On the other hand,

∥x∥ ∥a∥ − Re ⟨x, a⟩ = \sqrt{∥a + \sqrt{ε}e∥^{2}} − Re ⟨a + \sqrt{ε}e, a⟩
= \sqrt{∥a∥^{2} + ε ∥e∥^{2}} − ∥a∥^{2}
= \sqrt{1 + ε} − 1.

Utilising (5.4), we conclude that

(5.5)  \sqrt{1 + ε} − 1 ≤ cε for any ε > 0.

Multiplying (5.5) by \sqrt{1 + ε} + 1 > 0 and then dividing by ε > 0, we get

(5.6)  (\sqrt{1 + ε} + 1) c ≥ 1 for any ε > 0.

Letting ε → 0+ in (5.6), we deduce c ≥ \frac{1}{2}, and the theorem is proved.

The following result also holds [6].

Theorem 8. Let (H; ⟨·, ·⟩) be an inner product space over 𝕂 and x, y ∈ H, γ, Γ ∈ 𝕂 (Γ ≠ −γ) so that either

(5.7)  Re ⟨Γy − x, x − γy⟩ ≥ 0,

or, equivalently,

(5.8)  ∥x − \frac{γ + Γ}{2}y∥ ≤ \frac{1}{2} |Γ − γ| ∥y∥ ,

holds. Then we have the inequalities

(5.9)  0 ≤ ∥x∥ ∥y∥ − |⟨x, y⟩|

≤ ∥x∥ ∥y∥ − |Re [\frac{Γ̄ + γ̄}{|Γ + γ|} ⟨x, y⟩]|

≤ ∥x∥ ∥y∥ − Re [\frac{Γ̄ + γ̄}{|Γ + γ|} ⟨x, y⟩]

≤ \frac{1}{4} · \frac{|Γ − γ|^{2}}{|Γ + γ|} ∥y∥^{2} .

The constant \frac{1}{4} in the last inequality is best possible.

Proof. Consider for a, y ≠ 0, a = \frac{Γ+γ}{2} · y and r = \frac{1}{2} |Γ − γ| ∥y∥ . Thus from (5.2), we get

0 ≤ ∥x∥ |\frac{Γ + γ}{2} | ∥y∥ − |\frac{Γ + γ}{2} | |⟨x, y⟩|

≤ ∥x∥ |\frac{Γ + γ}{2} | ∥y∥ − |Re [\frac{Γ̄ + γ̄}{2} ⟨x, y⟩]|

≤ ∥x∥ |\frac{Γ + γ}{2} | ∥y∥ − Re [\frac{Γ̄ + γ̄}{2} ⟨x, y⟩]

≤ \frac{1}{8} · |Γ − γ|^{2} ∥y∥^{2} .

Dividing by \frac{1}{2} |Γ + γ| > 0, we deduce the desired inequality (5.9).

To prove the sharpness of the constant \frac{1}{4}, assume that there exists a c > 0 such that:

(5.10)  ∥x∥ ∥y∥ − Re [\frac{Γ̄ + γ̄}{|Γ + γ|} ⟨x, y⟩] ≤ c · \frac{|Γ − γ|^{2}}{|Γ + γ|} ∥y∥^{2} ,

provided either (5.7) or (5.8) holds.

Consider the real inner product space (ℝ^{2}, ⟨·, ·⟩) with ⟨x̄, ȳ⟩ = x_{1}y_{1} +x_{2}y_{2}, x̄ = (x_{1}, x_{2}) , ȳ = (y_{1}, y_{2}) ∈ ℝ^{2}. Let ȳ = (1, 1) and Γ, γ > 0 with Γ > γ. Then, by (5.10), we deduce

(5.11)  \sqrt{2}\sqrt{x_{1}^{2} + x_{2}^{2}} − (x_{1} + x_{2}) ≤ 2c · \frac{(Γ − γ)^{2}}{Γ + γ} .

If x_{1} = Γ, x_{2} = γ, then

⟨Γȳ − x̄, x̄ − γȳ⟩ = (Γ − x_{1}) (x_{1} − γ) + (Γ − x_{2}) (x_{2} − γ) = 0,

showing that the condition (5.7) is valid. Replacing x_{1} and x_{2} in (5.11), we deduce

(5.12)  \sqrt{2}\sqrt{Γ^{2} + γ^{2}} − (Γ + γ) ≤ 2c\frac{(Γ − γ)^{2}}{Γ + γ} .

If in (5.12) we choose Γ = 1 + ε, γ = 1 − ε with ε ∈ (0, 1) , then we have

2\sqrt{1 + ε^{2}} − 2 ≤ 2c\frac{4ε^{2}}{2} ,

giving

(5.13)  \sqrt{1 + ε^{2}} − 1 ≤ 2cε^{2}.

Finally, multiplying (5.13) with \sqrt{1 + ε^{2}} + 1 > 0 and thus dividing by ε^{2}, we deduce

(5.14)  1 ≤ 2c (\sqrt{1 + ε^{2}} + 1) for any ε ∈ (0, 1) .

Letting ε → 0+ in (5.14) we get c ≥ \frac{1}{4}, and the sharpness of the constant is proved.

5.2. Reverses of the Triangle Inequality. The following reverse of the triangle inequality in inner product spaces holds [6].

Proposition 13. Let (H; ⟨·, ·⟩) be an inner product space over the real or complex number field 𝕂, x, a ∈ H and r > 0. If ∥x − a∥ ≤ r, then we have the inequality

(5.15)  0 ≤ ∥x∥ + ∥a∥ − ∥x + a∥ ≤ r.

Proof. Since

(∥x∥ + ∥a∥)^{2} − ∥x + a∥^{2} ≤ 2 (∥x∥ ∥a∥ − Re ⟨x, a⟩) ,

then by Theorem 7 we deduce

(∥x∥ + ∥a∥)^{2} − ∥x + a∥^{2} ≤ r^{2},

from where we obtain

∥x∥ + ∥a∥ ≤ \sqrt{r^{2} + ∥x + a∥^{2}} ≤ r + ∥x + a∥ ,

giving the desired result (5.15).

We may state the following result [6].

Proposition 14. Let (H; ⟨·, ·⟩) be an inner product space over 𝕂 and x, y ∈ H, M > m > 0 such that either

Re ⟨My − x, x − my⟩ ≥ 0,

or, equivalently,

∥x − \frac{M + m}{2}y∥ ≤ \frac{1}{2} (M − m) ∥y∥ ,

holds. Then we have the inequality

(5.16)  0 ≤ ∥x∥ + ∥y∥ − ∥x + y∥ ≤ \frac{\sqrt{2}}{2} · \frac{M − m}{\sqrt{M + m}} ∥y∥ .

Proof. By Theorem 8 for Γ = M, γ = m, we have the inequality

∥x∥ ∥y∥ − Re ⟨x, y⟩ ≤ \frac{1}{4} · \frac{(M − m)^{2}}{M + m} ∥y∥^{2} .

Then we may state that

(∥x∥ + ∥y∥)^{2} − ∥x + y∥^{2} = 2 (∥x∥ ∥y∥ − Re ⟨x, y⟩)

≤ \frac{1}{2} · \frac{(M − m)^{2}}{M + m} ∥y∥^{2} ,

from where we get

∥x∥ + ∥y∥ ≤ \sqrt{\frac{1}{2} · \frac{(M − m)^{2}}{M + m} ∥y∥^{2} + ∥x + y∥^{2}}

≤ ∥x + y∥ + \frac{M − m}{\sqrt{2 (M + m)}} ∥y∥ ,

giving the desired inequality (5.16).

5.3. Integral Inequalities. We provide now, by the use of certain results obtained above, some integral inequalities that may be used in practical applications.

Proposition 15. Let f, g ∈ L_{ρ}^{2} (Ω, 𝕂) and r > 0 with the property that

|f (s) − g (s)| ≤ r for μ − a.e. s ∈ Ω.

32 1. REVERSES FOR THE SCHWARZ INEQUALITY

Then we have the inequalities

(5.17)  0 ≤ [∫_{Ω} ρ (s) |f (s)|^{2} dμ (s) ∫_{Ω} ρ (s) |g (s)|^{2} dμ (s)]^{\frac{1}{2}}
− |∫_{Ω} ρ (s) f (s) \overline{g (s)}dμ (s)|

≤ [∫_{Ω} ρ (s) |f (s)|^{2} dμ (s) ∫_{Ω} ρ (s) |g (s)|^{2} dμ (s)]^{\frac{1}{2}}

− |∫_{Ω} ρ (s) Re [f (s) \overline{g (s)}] dμ (s)|

≤ [∫_{Ω} ρ (s) |f (s)|^{2} dμ (s) ∫_{Ω} ρ (s) |g (s)|^{2} dμ (s)]^{\frac{1}{2}}

− ∫_{Ω} ρ (s) Re [f (s) \overline{g (s)}] dμ (s)

≤ \frac{1}{2}r^{2}.

The constant \frac{1}{2} is best possible in (5.17).

The proof follows by Theorem 7, and we omit the details.

Proposition 16. Let f, g ∈ L_{ρ}^{2} (Ω, 𝕂) and γ, Γ ∈ 𝕂 so that Γ ≠ −γ, and

Re [(Γg (s) − f (s)) (\overline{f (s)} − \overline{γ}\overline{g (s)})] ≥ 0, for μ − a.e. s ∈ Ω. Then we have the inequalities

(5.18)  0 ≤ [∫_{Ω} ρ (s) |f (s)|^{2} dμ (s) ∫_{Ω} ρ (s) |g (s)|^{2} dμ (s)]^{\frac{1}{2}}

− |∫_{Ω} ρ (s) f (s) \overline{g (s)}dμ (s)|

≤ [∫_{Ω} ρ (s) |f (s)|^{2} dμ (s) ∫_{Ω} ρ (s) |g (s)|^{2} dμ (s)]^{\frac{1}{2}}

− |Re [\frac{Γ̄ + γ̄}{|Γ + γ|} ∫_{Ω} ρ (s) f (s) \overline{g (s)}dμ (s)]|

≤ [∫_{Ω} ρ (s) |f (s)|^{2} dμ (s) ∫_{Ω} ρ (s) |g (s)|^{2} dμ (s)]^{\frac{1}{2}}

− Re [\frac{Γ̄ + γ̄}{|Γ + γ|} ∫_{Ω} ρ (s) f (s) \overline{g (s)}dμ (s)]

≤ \frac{1}{4} · \frac{|Γ − γ|^{2}}{|Γ + γ|} ∫_{Ω} ρ (s) |g (s)|^{2} dμ (s) .

The constant \frac{1}{4} is best possible.

Remark 17. If the space is real and we assume, for M > m > 0, that

(5.19)  mg (s) ≤ f (s) ≤ Mg (s) , for μ − a.e. s ∈ Ω,

then, by (5.18), we deduce the inequality:

0 ≤ [∫_{Ω} ρ (s) |f (s)|^{2} dμ (s) ∫_{Ω} ρ (s) |g (s)|^{2} dμ (s)]^{\frac{1}{2}}
− |∫_{Ω} ρ (s) f (s) \overline{g (s)}dμ (s)|
≤ \frac{1}{4} · \frac{(M − m)^{2}}{M + m} ∫_{Ω} ρ (s) |g (s)|^{2} dμ (s) .

The constant \frac{1}{4} is best possible.

The following reverse of the triangle inequality for integrals holds. Proposition 17. Assume that the functions f, g ∈ L_{ρ}^{2} (Ω, 𝕂) sat-

isfy (5.19). Then we have the inequality

0 ≤ (∫_{Ω} ρ (s) |f (s)|^{2} dμ (s))^{\frac{1}{2}} + (∫_{Ω} ρ (s) |g (s)|^{2} dμ (s))^{\frac{1}{2}}
− (∫_{Ω} ρ (s) |f (s) + g (s)|^{2} dμ (s))^{\frac{1}{2}}

≤ \frac{\sqrt{2}}{2} · \frac{M − m}{\sqrt{M + m}} (∫_{Ω} ρ (s) |g (s)|^{2} dμ (s))^{\frac{1}{2}} .

The proof follows by Proposition 14.

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CHAPTER 2 Inequalities of the Grüss Type

1. Introduction

Over the last five years, the development of Grüss type inequalities has experienced a surge, having been stimulated by their applications in different branches of Applied Mathematics including: in perturbed quadrature rules (see for example [5], [2]) and in the approximation of integral transforms (see [18], [21]) and the references therein.

For two Lebesgue integrable functions f, g : [a, b] → ℝ, consider the Čebyšev functional:

T (f, g) := \frac{1}{b − a} ∫_{a}^{b} f (t) g (t) dt − \frac{1}{b − a} ∫_{a}^{b} f (t) dt · \frac{1}{b − a} ∫_{a}^{b} g (t) dt.

In 1934, G. Grüss [20] showed that

(1.1)  |T (f, g)| ≤ \frac{1}{4} (M − m) (N − n) ,

provided m, M, n, N are real numbers with the property

(1.2)  −∞ < m ≤ f ≤ M < ∞,
−∞ < n ≤ g ≤ N < ∞ a.e. on [a, b] .

The quantity \frac{1}{4} is best possible in (1.1) in the sense that it cannot be replaced by a smaller one.

Another less well known inequality for T (f, g) was derived in 1882 by Čebyšev [4] under the assumption that f′, g′ exist and are continuous in [a, b] and is given by

(1.3)  |T (f, g)| ≤ \frac{1}{12} ∥f′∥_{∞} ∥g′∥_{∞} (b − a)^{2} ,

where ∥f′∥_{∞} := sup_{t∈[a,b]} |f′ (t)| .

The constant \frac{1}{12} cannot be improved in the general case.

Čebyšev’s inequality (1.3) also holds if f, g : [a, b] → ℝ are assumed to be absolutely continuous and f′, g′ ∈ L_{∞} [a, b] .

In 1970, A.M. Ostrowski [23] proved, amongst others, the following result that is in a sense a combination of the Čebyšev and Grüss results

|T (f, g)| ≤ \frac{1}{8} (b − a) (M − m) ∥g′∥_{∞} ,

provided f is Lebesgue integrable on [a, b] and satisfying (1.2) with g : [a, b] → ℝ being absolutely continuous and g′ ∈ L_{∞} [a, b] . Here the constant \frac{1}{8} is also sharp.

Finally, let us recall a result by Lupaş (see for example [24, p. 210]), which states that:

|T (f, g)| ≤ \frac{1}{π^{2}} ∥f′∥_{2} ∥g′∥_{2} (b − a) ,

provided f, g are absolutely continuous and f′, g′ ∈ L_{2} [a, b]. The constant \frac{1}{π^{2}} is the best possible here also.

For other Grüss type integral inequalities, see the books [22], [24], and the papers [6] – [15] and [19], where further references are given.

In [1], P. Cerone has obtained the following identity that involves a Stieltjes integral (Lemma 2.1, p. 3):

Lemma 1. Let f, g : [a, b] → ℝ, where f is of bounded variation and g is continuous on [a, b] , then the T (f, g) satisfies the identity,

(1.4)  T (f, g) = \frac{1}{(b − a)^{2}} ∫_{a}^{b} Ψ (t) df (t) ,

where

Ψ (t) := (t − a) A (t, b) − (b − t) A (a, t) ,

with

A (c, d) := ∫_{c}^{d} g (x) dx.

Using this representation and the properties of Stieltjes integrals he obtained the following result in bounding the functional T (·, ·) (Theorem 2.5, p. 4):

Theorem 9. With the assumptions in Lemma 1, we have:

|T (f, g)|

≤ \frac{1}{(b − a)^{2}} ·\{ \begin{matrix}sup_{t∈[a,b]} |Ψ (t)| ⋁_{a}^{b} (f) , \\ L ∫_{a}^{b} |Ψ (t)| dt, & for L − Lipschitzian; \\ ∫_{a}^{b} |Ψ (t)| df (t) , & for f monotonic nondecreasing, \end{matrix}

where ⋁_{a}^{b} (f) denotes the total variation of f on [a, b] .

Now, if we use the function φ : (a, b) → ℝ,

(1.5)  φ (t) := D (g; a, t, b) = \frac{∫_{t}^{b} g (x) dx}{b − t} − \frac{∫_{a}^{t} g (x) dx}{t − a},

then by (1.4) we may obtain the identity:

T (f, g) = \frac{1}{(b − a)^{2}} ∫_{a}^{b} (t − a) (b − t) φ (t) df (t) .

In [3] various upper bounds for |T (f, g)| have been given, from which we would like to mention only the following ones:

Theorem 10. Let f : [a, b] → ℝ be a function of bounded variation and g : [a, b] → ℝ an absolutely continuous function so that φ is bounded on (a, b) . Then one has the inequality:

|T (f, g)| ≤ \frac{1}{4} ∥φ∥_{∞} ⋁_{a}^{b} (f) ,

where φ is as given by (1.5) and

∥φ∥_{∞} := sup_{t∈(a,b)} |φ (t)| .

The case of Lipschitzian functions f : [a, b] → ℝ is embodied in the following theorem [3].

Theorem 11. Let f : [a, b] → ℝ be an L−Lipschitzian function on [a, b] and g : [a, b] → ℝ an absolutely continuous function on [a, b] . Then

|T (f, g)|

≤ \{ \begin{matrix}L\frac{(b−a)^{3}}{6} ∥φ∥_{∞} & if φ ∈ L_{∞} [a, b] ; \\ L (b − a)^{\frac{1}{q}} [B (q + 1, q + 1)]^{\frac{1}{q}} ∥φ∥_{p} , & p > 1, \frac{1}{p} + \frac{1}{q} = 1 \\ & if φ ∈ L_{p} [a, b] ; \\ \frac{L}{4} ∥φ∥_{1} , & if φ ∈ L_{1} [a, b] , \end{matrix}

where ∥·∥_{p} are the usual Lebesgue p−norms on [a, b] and B (·, ·) is Euler’s Beta function.

Finally, the following result containing Stieltjes integral holds [3]:

Theorem 12. Let f : [a, b] → ℝ be a monotonic nondecreasing function on [a, b] . If g is continuous, then one has the inequality:

|T (f, g)|

≤ \{ \begin{matrix}\frac{1}{4} ∫_{a}^{b} |φ (t)| df (t) \\ \frac{1}{(b−a)^{2}} (∫_{a}^{b} [(b − t) (t − a)]^{q} df (t))^{\frac{1}{q}} (∫_{a}^{b} |φ (t)|^{p} df (t))^{\frac{1}{p}} , \\ p > 1, \frac{1}{p} + \frac{1}{q} = 1; \\ \frac{1}{(b−a)^{2}} sup_{t∈[a,b]} |φ (t)| ∫_{a}^{b} (t − a) (b − t) df (t) . \end{matrix}

In [16], the authors have considered the following functional

D (f; u) := ∫_{a}^{b} f (x) du (x) − [u (b) − u (a)] · \frac{1}{b − a} ∫_{a}^{b} f (t) dt,

provided that the involved integrals exist.

In the same paper, the following result in estimating the above functional has been obtained.

Theorem 13. Let f, u : [a, b] → ℝ be such that u is Lipschitzian on [a, b] , i.e.,

|u (x) − u (y)| ≤ L |x − y| for any x, y ∈ [a, b] (L > 0)

and f is Riemann integrable on [a, b] . If m, M ∈ ℝ are such that

m ≤ f (x) ≤ M for any x, y ∈ [a, b] ,

then we have the inequality

|D (f; u)| ≤ \frac{1}{2}L (M − m) (b − a) .

The constant \frac{1}{2} is sharp in the sense that it cannot be replaced by a smaller constant.

In [15], the following result complementing the above one was obtained.

Theorem 14. Let f, u : [a, b] → ℝ be such that u : [a, b] → ℝ is of bounded variation in [a, b] and f : [a, b] → ℝ is K−Lipschitzian (K > 0) . Then we have the inequality

|D (f; u)| ≤ \frac{1}{2}K (b − a) ⋁_{a}^{b} (u) .

The constant \frac{1}{2} is sharp in the above sense.

The main aim of this chapter is to survey some recent inequalities of the Grüss type holding in the general setting of inner product spaces. Natural applications for Lebesgue integrals in measure spaces are presented as well.

2. Grüss’ Inequality in Inner Product Spaces

2.1. Introduction. In [6], the author has proved the following Grüss’ type inequality in real or complex inner product spaces.

Theorem 15. Let (H, ⟨·, ·⟩) be an inner product space over 𝕂 and e ∈ H, ∥e∥ = 1. If φ, γ, Φ, Γ are real or complex numbers and x, y are vectors in H such that the conditions

(2.1)  Re ⟨Φe − x, x − φe⟩ ≥ 0 and Re ⟨Γe − y, y − γe⟩ ≥ 0

hold, then we have the inequality

(2.2)  |⟨x, y⟩ − ⟨x, e⟩ ⟨e, y⟩| ≤ \frac{1}{4} |Φ − φ| · |Γ − γ| .

The constant \frac{1}{4} is best possible in the sense that it cannot be replaced by a smaller quantity.

Some particular cases of interest for integrable functions with real or complex values and the corresponding discrete versions are listed below.

Corollary 13. Let f, g : [a, b] → 𝕂 (𝕂 = ℝ, ℂ) be Lebesgue integrable and such that

Re [(Φ − f (x)) (\overline{f (x)} − \overline{φ})] ≥ 0, Re [(Γ − g (x)) (\overline{g (x)} − \overline{γ})] ≥ 0

for a.e. x ∈ [a, b] , where φ, γ, Φ, Γ are real or complex numbers and z̄ denotes the complex conjugate of z. Then we have the inequality

|\frac{1}{b − a} ∫_{a}^{b} f (x) \overline{g (x)}dx − \frac{1}{b − a} ∫_{a}^{b} f (x) dx · \frac{1}{b − a} ∫_{a}^{b} \overline{g (x)}dx|

≤ \frac{1}{4} |Φ − φ| · |Γ − γ| .

The constant \frac{1}{4} is best possible.

The discrete case is embodied in

Corollary 14. Let x, y ∈𝕂^{n} and φ, γ, Φ, Γ are real or complex numbers such that

Re [(Φ − x_{i}) (\overline{x_{i}} − \overline{φ})] ≥ 0, Re [(Γ − y_{i}) (\overline{y_{i}} − \overline{γ})] ≥ 0

for each i ∈ {1, …, n} . Then we have the inequality

|\frac{1}{n} ∑_{i=1}^{n} x_{i}\overline{y_{i}} − \frac{1}{n} ∑_{i=1}^{n} x_{i} · \frac{1}{n} ∑_{i=1}^{n} \overline{y_{i}}| ≤ \frac{1}{4} |Φ − φ| · |Γ − γ| .

The constant \frac{1}{4} is best possible.

For other applications of Theorem 15, see the recent paper [17]. In the present section, by following [13], we show that the condition

(2.1) may be replaced by an equivalent but simpler assumption and a new proof of Theorem 15 is produced. A refinement of the Grüss type inequality (2.2) , some companions and applications for integrals are pointed out as well.

2.2. An Equivalent Assumption. The following lemma holds [13].

Lemma 2. Let a, x, A be vectors in the inner product space (H, ⟨·, ·⟩) over 𝕂 (𝕂 = ℝ,ℂ) with a ≠ A. Then

Re ⟨A − x, x − a⟩ ≥ 0

if and only if

∥x − \frac{a + A}{2}∥ ≤ \frac{1}{2} ∥A − a∥ .

Proof. Define

I_{1} := Re ⟨A − x, x − a⟩ , I_{2} := \frac{1}{4} ∥A − a∥^{2} − ∥x − \frac{a + A}{2}∥^{2} .

A simple calculation shows that

I_{1} = I_{2} = Re [⟨x, a⟩ + ⟨A, x⟩] − Re ⟨A, a⟩ − ∥x∥^{2}

and thus, obviously, I_{1} ≥ 0 iff I_{2} ≥ 0 showing the required equivalence.

The following corollary is obvious

Corollary 15. Let x, e ∈ H with ∥e∥ = 1 and δ, Δ ∈ 𝕂 with δ ≠ Δ. Then

Re ⟨Δe − x, x − δe⟩ ≥ 0

iff

∥x − \frac{δ + Δ}{2} · e∥ ≤ \frac{1}{2} |Δ − δ| .

Remark 18. If H = ℂ, then

Re [(A − x) (x̄ − ā)] ≥ 0

if and only if

|x − \frac{a + A}{2}| ≤ \frac{1}{2} |A − a| ,

where a, x, A ∈ ℂ. If H = ℝ, and A > a then a ≤ x ≤ A if and only if |x − \frac{a+A}{2} | ≤ \frac{1}{2} |A − a| .

The following lemma is of interest [13].

Lemma 3. Let x, e ∈ H with ∥e∥ = 1. Then one has the following representation

(2.3)  ∥x∥^{2} − |⟨x, e⟩|^{2} = inf_{λ∈𝕂}∥x − λe∥^{2} ≥ 0.

Proof. Observe, for any λ ∈ 𝕂, that

⟨x − λe, x − ⟨x, e⟩ e⟩ = ∥x∥^{2} − |⟨x, e⟩|^{2} − λ [⟨e, x⟩ − ⟨e, x⟩ ∥e∥^{2}]
= ∥x∥^{2} − |⟨x, e⟩|^{2} .

Using Schwarz’s inequality, we have

[∥x∥^{2} − |⟨x, e⟩|^{2}]^{2} = |⟨x − λe, x − ⟨x, e⟩ e⟩|^{2}
≤ ∥x − λe∥^{2} ∥x − ⟨x, e⟩ e∥^{2}
= ∥x − λe∥^{2} [∥x∥^{2} − |⟨x, e⟩|^{2}] ,

giving the bound

(2.4)  ∥x∥^{2} − |⟨x, e⟩|^{2} ≤ ∥x − λe∥^{2} , λ ∈ 𝕂.

Taking the infimum in (2.4) over λ ∈ 𝕂, we deduce

∥x∥^{2} − |⟨x, e⟩|^{2} ≤ inf_{λ∈𝕂}∥x − λe∥^{2} .

Since, for λ_{0} = ⟨x, e⟩ , we get ∥x − λ_{0}e∥^{2} = ∥x∥^{2} − |⟨x, e⟩|^{2} , then the representation (2.3) is proved.

We are now able to provide a different proof for the Grüss type inequality in inner product spaces (mentioned in the Introduction), than the one from the paper [6].

Theorem 16. Let (H, ⟨·, ·⟩) be an inner product space over 𝕂 and e ∈ H, ∥e∥ = 1. If φ, γ, Φ, Γ are real or complex numbers and x, y are vectors in H such that the conditions (2.1) hold, or, equivalently, the following assumptions

(2.5) ∥x − \frac{φ + Φ}{2} · e∥ ≤ \frac{1}{2} |Φ − φ| , ∥y − \frac{γ + Γ}{2} · e∥ ≤ \frac{1}{2} |Γ − γ|

are valid, then one has the inequality

(2.6)  |⟨x, y⟩ − ⟨x, e⟩ ⟨e, y⟩| ≤ \frac{1}{4} |Φ − φ| · |Γ − γ| .

The constant \frac{1}{4} is best possible.

Proof. It can be easily shown (see for example the proof of Theorem 1 from [6]) that

(2.7) |⟨x, y⟩ − ⟨x, e⟩ ⟨e, y⟩| ≤ [∥x∥^{2} − |⟨x, e⟩|^{2}]^{\frac{1}{2}} [∥y∥^{2} − |⟨y, e⟩|^{2}]^{\frac{1}{2}} ,

for any x, y ∈ H and e ∈ H, ∥e∥ = 1. Using Lemma 3 and the conditions (2.5) we obviously have that

[∥x∥^{2} − |⟨x, e⟩|^{2}]^{\frac{1}{2}} = inf_{λ∈𝕂}∥x − λe∥ ≤ ∥x − \frac{φ + Φ}{2} · e∥ ≤ \frac{1}{2} |Φ − φ|

and

[∥y∥^{2} − |⟨y, e⟩|^{2}]^{\frac{1}{2}} = inf_{λ∈𝕂}∥y − λe∥ ≤ ∥y − \frac{γ + Γ}{2} · e∥ ≤ \frac{1}{2} |Γ − γ|

and by (2.7) the desired inequality (2.6) is obtained.

The fact that \frac{1}{4} is the best possible constant, has been shown in [6] and we omit the details.

2.3. A Refinement of the Grüss Inequality. The following result improving (2.1) holds [13].

Theorem 17. Let (H, ⟨·, ·⟩) be an inner product space over 𝕂 and e ∈ H, ∥e∥ = 1. If φ, γ, Φ, Γ are real or complex numbers and x, y are vectors in H such that the conditions (2.1), or, equivalently, (2.5) hold, then we have the inequality

|⟨x, y⟩ − ⟨x, e⟩ ⟨e, y⟩| (2.8)

≤ \frac{1}{4} |Φ − φ| · |Γ − γ|

− [Re ⟨Φe − x, x − φe⟩]^{\frac{1}{2}} [Re ⟨Γe − y, y − γe⟩]^{\frac{1}{2}}

≤ (\frac{1}{4} |Φ − φ| · |Γ − γ|) .

The constant \frac{1}{4} is best possible.

Proof. As in [6], we have

(2.9) |⟨x, y⟩ − ⟨x, e⟩ ⟨e, y⟩|^{2} ≤ [∥x∥^{2} − |⟨x, e⟩|^{2}] [∥y∥^{2} − |⟨y, e⟩|^{2}] ,

(2.10) ∥x∥^{2} − |⟨x, e⟩|^{2}

= Re [(Φ − ⟨x, e⟩) (\overline{⟨x, e⟩} − \overline{φ})] − Re ⟨Φe − x, x − φe⟩

and

(2.11) ∥y∥^{2} − |⟨y, e⟩|^{2}

= Re [(Γ − ⟨y, e⟩) (\overline{⟨y, e⟩} − \overline{γ})] − Re ⟨Γe − x, x − γe⟩ .

Using the elementary inequality

4 Re (a\overline{b}) ≤ |a + b|^{2} ; a, b ∈ 𝕂 (𝕂 = ℝ,ℂ) ,

we may state that

(2.12)  Re [(Φ − ⟨x, e⟩) (\overline{⟨x, e⟩} − \overline{φ})] ≤ \frac{1}{4} |Φ − φ|^{2}

and

(2.13)  Re [(Γ − ⟨y, e⟩) (\overline{⟨y, e⟩} − \overline{γ})] ≤ \frac{1}{4} |Γ − γ|^{2} .

Consequently, by (2.9) − (2.13) we may state that

(2.14) |⟨x, y⟩ − ⟨x, e⟩ ⟨e, y⟩|^{2}

≤ [\frac{1}{4} |Φ − φ|^{2} − ([Re ⟨Φe − x, x − φe⟩]^{\frac{1}{2}} )^{2}]

× [\frac{1}{4} |Γ − γ|^{2} − ([Re ⟨Γe − y, y − γe⟩]^{\frac{1}{2}} )^{2}] .

Finally, using the elementary inequality for positive real numbers

(m^{2} − n^{2}) (p^{2} − q^{2}) ≤ (mp − nq)^{2} ,

we have

[\frac{1}{4} |Φ − φ|^{2} − ([Re ⟨Φe − x, x − φe⟩]^{\frac{1}{2}} )^{2}]

× [\frac{1}{4} |Γ − γ|^{2} − ([Re ⟨Γe − y, y − γe⟩]^{\frac{1}{2}} )^{2}]

≤ (\frac{1}{4} |Φ − φ| · |Γ − γ| − [Re ⟨Φe − x, x − φe⟩]^{\frac{1}{2}} [Re ⟨Γe − y, y − γe⟩]^{\frac{1}{2}} )^{2} ,

giving the desired inequality (2.8) .

The fact that \frac{1}{4} is the best constant can be proven in a similar manner to the one in the Grüss inequality (2.2) (see for instance [6]) and we omit the details.

2.4. Some Companion Inequalities. The following companion of the Grüss inequality in inner product spaces holds [13].

Theorem 18. Let (H, ⟨·, ·⟩) be an inner product space over 𝕂 and e ∈ H, ∥e∥ = 1. If γ, Γ ∈ 𝕂 and x, y ∈ H are such that

(2.15)  Re ⟨Γe − \frac{x + y}{2} , \frac{x + y}{2} − γe⟩ ≥ 0,

or, equivalently,

∥\frac{x + y}{2} − \frac{γ + Γ}{2} · e∥ ≤ \frac{1}{2} |Γ − γ| ,

then we have the inequality

(2.16)  Re [⟨x, y⟩ − ⟨x, e⟩ ⟨e, y⟩] ≤ \frac{1}{4} |Γ − γ|^{2} .

The constant \frac{1}{4} is best possible in the sense that it cannot be replaced by a smaller constant.

Proof. Start with the obvious inequality

(2.17)  Re ⟨z, u⟩ ≤ \frac{1}{4} ∥z + u∥^{2} ; z, u ∈ H.

Since

⟨x, y⟩ − ⟨x, e⟩ ⟨e, y⟩ = ⟨x − ⟨x, e⟩ e, y − ⟨y, e⟩ e⟩ ,

then using (2.17) we may write

(2.18) Re [⟨x, y⟩ − ⟨x, e⟩ ⟨e, y⟩] = Re [⟨x − ⟨x, e⟩ e, y − ⟨y, e⟩ e⟩]

≤ \frac{1}{4} ∥x − ⟨x, e⟩ e + y − ⟨y, e⟩ e∥^{2}

= ∥\frac{x + y}{2} − ⟨\frac{x + y}{2} , e⟩ · e∥^{2}

= ∥\frac{x + y}{2} ∥^{2} − |⟨\frac{x + y}{2} , e⟩|^{2} .

If we apply Grüss’ inequality in inner product spaces for, say, a = b = \frac{x+y}{2}, we get

(2.19)  ∥\frac{x + y}{2} ∥^{2} − |⟨\frac{x + y}{2} , e⟩|^{2} ≤ \frac{1}{4} |Γ − γ|^{2} .

Making use of (2.18) and (2.19) we deduce (2.16) .

The fact that \frac{1}{4} is the best possible constant in (2.16) follows by the fact that if in (2.15) we choose x = y, then it becomes

Re ⟨Γe − x, x − γe⟩ ≥ 0,

implying that 0 ≤ ∥x∥^{2} − |⟨x, e⟩|^{2} ≤ \frac{1}{4} |Γ − γ|^{2} , for which, by Grüss’ inequality in inner product spaces, we know that the constant \frac{1}{4} is best possible.

The following corollary might be of interest if one wanted to evaluate the absolute value of

Re [⟨x, y⟩ − ⟨x, e⟩ ⟨e, y⟩] .

Corollary 16. Let (H, ⟨·, ·⟩) be an inner product space over 𝕂 and e ∈ H, ∥e∥ = 1. If γ, Γ ∈ 𝕂 and x, y ∈ H are such that

Re ⟨Γe − \frac{x ± y}{2} , \frac{x ± y}{2} − γe⟩ ≥ 0,

or, equivalently,

∥\frac{x ± y}{2} − \frac{γ + Γ}{2} · e∥ ≤ \frac{1}{2} |Γ − γ| ,

holds, then we have the inequality

(2.20)  |Re [⟨x, y⟩ − ⟨x, e⟩ ⟨e, y⟩]| ≤ \frac{1}{4} |Γ − γ|^{2} .

If the inner product space H is real, then (for m, M ∈ ℝ, M > m)

⟨Me − \frac{x ± y}{2} , \frac{x ± y}{2} − me⟩ ≥ 0,

or, equivalently,

∥\frac{x ± y}{2} − \frac{m + M}{2} · e∥ ≤ \frac{1}{2} (M − m) ,

implies

(2.21)  |⟨x, y⟩ − ⟨x, e⟩ ⟨e, y⟩| ≤ \frac{1}{4} (M − m)^{2} .

In both inequalities (2.20) and (2.21) , the constant \frac{1}{4} is best possible.

Proof. We only remark that, if

Re ⟨Γe − \frac{x − y}{2} , \frac{x − y}{2} − γe⟩ ≥ 0

holds, then by Theorem 18, we get

Re [− ⟨x, y⟩ + ⟨x, e⟩ ⟨e, y⟩] ≤ \frac{1}{4} |Γ − γ|^{2} ,

showing that

(2.22)  Re [⟨x, y⟩ − ⟨x, e⟩ ⟨e, y⟩] ≥ −\frac{1}{4} |Γ − γ|^{2} .

Making use of (2.16) and (2.22) we deduce the desired result (2.20) .

Finally, we may state and prove the following dual result as well [13].

Proposition 18. Let (H, ⟨·, ·⟩) be an inner product space over 𝕂 and e ∈ H, ∥e∥ = 1. If φ, Φ ∈ 𝕂 and x, y ∈ H are such that

(2.23)  Re [(Φ − ⟨x, e⟩) (\overline{⟨x, e⟩} − \overline{φ})] ≤ 0,

then we have the inequalities

(2.24)  ∥x − ⟨x, e⟩ e∥ ≤ [Re ⟨x − Φe, x − φe⟩]^{\frac{1}{2}}

≤ \frac{\sqrt{2}}{2} [∥x − Φe∥^{2} + ∥x − φe∥^{2}]^{\frac{1}{2}} .

Proof. We know that the following identity holds true (see (2.10))

(2.25) ∥x∥^{2} − |⟨x, e⟩|^{2}

= Re [(Φ − ⟨x, e⟩) (\overline{⟨x, e⟩} − \overline{φ})] + Re ⟨x − Φe, x − φe⟩ .

Using the assumption (2.23) and the fact that

∥x∥^{2} − |⟨x, e⟩|^{2} = ∥x − ⟨x, e⟩ e∥^{2} ,

by (2.25) , we deduce the first inequality in (2.24) .

The second inequality in (2.24) follows by the fact that for any v, w ∈ H one has

Re ⟨w, v⟩ ≤ \frac{1}{2} (∥w∥^{2} + ∥v∥^{2}) .

The proposition is thus proved.

2.5. Integral Inequalities. The following proposition holds [13].

Proposition 19. If f, g, h ∈ L^{2} (Ω, 𝕂) and φ, Φ, γ, Γ ∈ 𝕂, are such that ∫_{Ω} |h (s)|^{2} dμ (s) = 1 and

(2.26)  ∫_{Ω} Re [(Φh (s) − f (s)) (\overline{f (s)} − \overline{φ}\overline{h (s)})] dμ (s) ≥ 0,
∫_{Ω} Re [(Γh (s) − g (s)) (\overline{g (s)} − \overline{γ}\overline{h (s)})] dμ (s) ≥ 0

or, equivalently,

(∫_{Ω} |f (s) − \frac{Φ + φ}{2}h (s)|^{2} dμ (s))^{\frac{1}{2}} ≤ \frac{1}{2} |Φ − φ| ,

(∫_{Ω} |g (s) − \frac{Γ + γ}{2} h (s)|^{2} dμ (s))^{\frac{1}{2}} ≤ \frac{1}{2} |Γ − γ| ,

hold, then we have the following refinement of the Grüss integral inequality

|∫_{Ω} f (s) \overline{g (s)}dμ (s) − ∫_{Ω} f (s) \overline{h (s)}dμ (s) ∫_{Ω} h (s) \overline{g (s)}dμ (s)|

≤ \frac{1}{4} |Φ − φ|·|Γ − γ|−[∫_{Ω} Re [(Φh (s) − f (s)) (\overline{f (s)} − \overline{φ}\overline{h (s)})] dμ (s)

× ∫_{Ω} Re [(Γh (s) − g (s)) (\overline{g (s)} − \overline{γ}\overline{h (s)})] dμ (s)]^{\frac{1}{2}} .

The constant \frac{1}{4} is best possible.

The proof follows by Theorem 17 on choosing H = L^{2} (Ω, 𝕂) with the inner product

⟨f, g⟩ := ∫_{Ω} f (s) \overline{g (s)}dμ (s) .

We omit the details.

Remark 19. It is obvious that a sufficient condition for (2.26) to hold is

Re [(Φh (s) − f (s)) (\overline{f (s)} − \overline{φ}\overline{h (s)})] ≥ 0,

and

Re [(Γh (s) − g (s)) (\overline{g (s)} − \overline{γ}\overline{h (s)})] ≥ 0,

for μ−a.e. s ∈ Ω, or, equivalently,

|f (s) − \frac{Φ + φ}{2}h (s)| ≤ \frac{1}{2} |Φ − φ| |h (s)| and
|g (s) − \frac{Γ + γ}{2} h (s)| ≤ \frac{1}{2} |Γ − γ| |h (s)| ,

for μ−a.e. s ∈ Ω.

The following result may be stated as well [13].

50 2. INEQUALITIES OF THE GRÜSS TYPE

Corollary 17. If z, Z, t, T ∈ 𝕂, μ (Ω) < ∞ and f, g ∈ L^{2} (Ω, 𝕂) are such that:

Re [(Z − f (s)) (\overline{f (s)} − z̄)] ≥ 0,
Re [(T − g (s)) (\overline{g (s)} − t̄)] ≥ 0,

for a.e. s ∈ Ω, or, equivalently,

|f (s) − \frac{z + Z}{2} | ≤ \frac{1}{2} |Z − z| ,
|g (s) − \frac{t + T}{2} | ≤ \frac{1}{2} |T − t| ,

for a.e. s ∈ Ω, then we have the inequality

(2.27) |\frac{1}{μ (Ω)} ∫_{Ω} f (s) \overline{g (s)}dμ (s)

− \frac{1}{μ (Ω)} ∫_{Ω} f (s) dμ (s) · \frac{1}{μ (Ω)} ∫_{Ω} \overline{g (s)}dμ (s)|

≤ \frac{1}{4} |Z − z| |T − t| − \frac{1}{μ (Ω)} [∫_{Ω} Re [(Z − f (s)) (\overline{f (s)} − z̄)] dμ (s)

× ∫_{Ω} Re [(T − g (s)) (\overline{g (s)} − t̄)] dμ (s)]^{\frac{1}{2}} .

Using Theorem 18 we may state the following result as well [13]. Proposition 20. If f, g, h ∈ L^{2} (Ω, 𝕂) and γ, Γ ∈ 𝕂 are such that

∫_{Ω} |h (s)|^{2} dμ (s) = 1 and

(2.28) ∫_{Ω} Re \{[Γh (s) − \frac{f (s) + g (s)}{2}]

× [\frac{\overline{f (s)} + \overline{g (s)}}{2} − γ̄h̄ (s)]\} dμ (s) ≥ 0

or, equivalently,

(2.29) (∫_{Ω} |\frac{f (s) + g (s)}{2} − \frac{γ + Γ}{2}h (s)|^{2} dμ (s))^{\frac{1}{2}} ≤ \frac{1}{2} |Γ − γ| ,

holds, then we have the inequality

I := ∫_{Ω} Re [f (s) \overline{g (s)}] dμ (s)
− Re [∫_{Ω} f (s) \overline{h (s)}dμ (s) · ∫_{Ω} h (s) \overline{g (s)}dμ (s)]

≤ \frac{1}{4} |Γ − γ|^{2} .

If (2.28) and (2.29) hold with “ ± ” instead of “ + ”, then

|I| ≤ \frac{1}{4} |Γ − γ|^{2} .

Remark 20. It is obvious that a sufficient condition for (2.28) to hold is

Re \{[Γh (s) − \frac{f (s) + g (s)}{2}] · [\frac{\overline{f (s)} + \overline{g (s)}}{2} − γ̄h̄ (s)]\} ≥ 0,

for a.e. s ∈ Ω, or, equivalently,

|\frac{f (s) + g (s)}{2} − \frac{γ + Γ}{2}h (s)| ≤ \frac{1}{2} |Γ − γ| |h (s)| ,

for a.e. s ∈ Ω.

Finally, the following corollary holds.

Corollary 18. If Z, z ∈ 𝕂, μ (Ω) < ∞ and f, g ∈ L^{2} (Ω, 𝕂) are such that

(2.30)  Re [(Z − \frac{f (s) + g (s)}{2}) (\frac{\overline{f (s)} + \overline{g (s)}}{2} − \overline{z})] ≥ 0,

for a.e. s ∈ Ω, or, equivalently,

(2.31)  |\frac{f (s) + g (s)}{2} − \frac{z + Z}{2} | ≤ \frac{1}{2} |Z − z| ,

for a.e. s ∈ Ω, then we have the inequality

J := \frac{1}{μ (Ω)} ∫_{Ω} Re [f (s) \overline{g (s)}] dμ (s)

− Re [\frac{1}{μ (Ω)} ∫_{Ω} f (s) dμ (s) · \frac{1}{μ (Ω)} ∫_{Ω} \overline{g (s)}dμ (s)]

≤ \frac{1}{4} |Z − z|^{2} .

If (2.30) and (2.31) hold with “ ± ” instead of “ + ”, then

|J| ≤ \frac{1}{4} |Z − z|^{2} .

Remark 21. It is obvious that if one chooses the discrete measure above, then all the inequalities in this section may be written for sequences of real or complex numbers. We omit the details.

3. Companions of Grüss’ Inequality

3.1. A General Result. The following Grüss type inequality in inner product spaces holds [12].

Theorem 19. Let x, y, e ∈ H with ∥e∥ = 1, and the scalars a, A, b, B ∈ 𝕂 such that Re (āA) > 0 and Re (b̄B) > 0. If

Re ⟨Ae − x, x − ae⟩ ≥ 0 and Re ⟨Be − y, y − be⟩ ≥ 0

or, equivalently,

∥x − \frac{a + A}{2}e∥ ≤ \frac{1}{2} |A − a| and ∥y − \frac{b + B}{2} e∥ ≤ \frac{1}{2} |B − b| ,

holds, then we have the inequality

(3.1)  |⟨x, y⟩ − ⟨x, e⟩ ⟨e, y⟩| ≤ \frac{1}{4}M (a, A) M (b, B) |⟨x, e⟩ ⟨e, y⟩| ,

where M (·, ·) is defined by

M (a, A) := [\frac{(|A| − |a|)^{2} + 4 [|Aā| − Re (Aā)]}{Re (āA)}]^{\frac{1}{2}} .

The constant \frac{1}{4} is best possible in the sense that it cannot be replaced by a smaller constant.

Proof. Start with the inequality

(3.2) |⟨x, y⟩ − ⟨x, e⟩ ⟨e, y⟩|^{2} ≤ (∥x∥^{2} − |⟨x, e⟩|^{2}) (∥y∥^{2} − |⟨y, e⟩|^{2}) .

Now, assume that u, v ∈ H, and c, C ∈ 𝕂 with Re (c̄C) > 0 and Re ⟨Cv − u, u − cv⟩ ≥ 0. This last inequality is equivalent to

∥u∥^{2} + Re (c̄C) ∥v∥^{2} ≤ Re [C\overline{⟨u, v⟩} + c̄⟨u, v⟩] .

Dividing this inequality by [Re (Cc̄)]^{\frac{1}{2}} > 0, we deduce

(3.3) \frac{1}{[Re (c̄C)]^{\frac{1}{2}}} ∥u∥^{2} + [Re (c̄C)]^{\frac{1}{2}} ∥v∥^{2} ≤ \begin{matrix}Re & [C\overline{⟨u, v⟩} + c̄⟨u, v⟩] \\ [Re (c̄C)] & \frac{1}{2} \end{matrix}.

On the other hand, by the elementary inequality

αp^{2} + \frac{1}{α}q^{2} ≥ 2pq, α > 0, p, q ≥ 0,

we deduce

(3.4)  2 ∥u∥ ∥v∥ ≤ \frac{1}{[Re (c̄C)]^{\frac{1}{2}}} ∥u∥^{2} + [Re (c̄C)]^{\frac{1}{2}} ∥v∥^{2} .

Making use of (3.3) and (3.4) and the fact that for any z ∈ ℂ, Re (z) ≤ |z| , we get

∥u∥ ∥v∥ ≤ \frac{Re [C\overline{⟨u, v⟩} + c̄⟨u, v⟩]}{2 [Re (c̄C)]^{\frac{1}{2}}} ≤ \frac{|c| + |C|}{2 [Re (c̄C)]^{\frac{1}{2}}} |⟨u, v⟩| .

Consequently

∥u∥^{2} ∥v∥^{2} − |⟨u, v⟩|^{2} (3.5)

≤ [\frac{(|c| + |C|)^{2}}{4 [Re (c̄C)]} − 1] |⟨u, v⟩|^{2}

= \frac{1}{4} \frac{(|c| − |C|)^{2} + 4 [|c̄C| − Re (c̄C)]}{Re (c̄C)} |⟨u, v⟩|^{2}

= \frac{1}{4}M^{2} (c, C) |⟨u, v⟩|^{2} .

Now, if we write (3.5) for the choices u = x, v = e and u = y, v = e respectively and use (3.2), we deduce the desired result (3.1). The sharpness of the constant will be proved in the case where H is a real inner product space.

The following corollary which provides a simpler Grüss type inequality for real constants (and in particular, for real inner product spaces) holds [12].

Corollary 19. With the assumptions of Theorem 19 and if a, b, A, B ∈ ℝ are such that A > a > 0, B > b > 0 and

(3.6) ∥x − \frac{a + A}{2}e∥ ≤ \frac{1}{2} (A − a) and ∥y − \frac{b + B}{2} e∥ ≤ \frac{1}{2} (B − b) ,

then we have the inequality

(3.7)  |⟨x, y⟩ − ⟨x, e⟩ ⟨e, y⟩| ≤ \frac{1}{4} · \begin{matrix}(A − a) & (B − b) \\ √ & \overline{abAB} \end{matrix} |⟨x, e⟩ ⟨e, y⟩| .

Proof. To prove the sharpness of the constant \frac{1}{4} assume that the inequality (3.7) holds in real inner product spaces with x = y and for a constant k > 0, i.e.,

(3.8)  ∥x∥^{2} − |⟨x, e⟩|^{2} ≤ k · \frac{(A − a)^{2}}{aA} |⟨x, e⟩|^{2} (A > a > 0) ,

provided that ∥x − \frac{a+A}{2}e∥ ≤ \frac{1}{2} (A − a) , or, equivalently,

⟨Ae − x, x − ae⟩ ≥ 0.

We choose H = ℝ^{2}, x = (x_{1}, x_{2}) ∈ ℝ^{2}, e = (\frac{1}{\sqrt{2}}, \frac{1}{\sqrt{2}}) . Then we have

∥x∥^{2} − |⟨x, e⟩|^{2} = x_{1}^{2} + x_{2}^{2} − \frac{(x_{1} + x_{2})^{2}}{2} = \frac{(x_{1} − x_{2})^{2}}{2} ,
|⟨x, e⟩|^{2} = \frac{(x_{1} + x_{2})^{2}}{2} ,

and by (3.8) we get

(3.9)  \frac{(x_{1} − x_{2})^{2}}{2} ≤ k · \frac{(A − a)^{2}}{aA} · \frac{(x_{1} + x_{2})^{2}}{2} .

Now, if we let x_{1} = \frac{a}{\sqrt{2}}, x_{2} = \frac{A}{\sqrt{2}} (A > a > 0) , then obviously

⟨Ae − x, x − ae⟩ = ∑_{i=1}^{2} (\frac{A}{\sqrt{2}} − x_{i}) (x_{i} − \frac{a}{\sqrt{2}}) = 0,

which shows that the condition (3.6) is fulfilled, and by (3.9) we get

\frac{(A − a)^{2}}{4} ≤ k · \frac{(A − a)^{2}}{aA} · \frac{(a + A)^{2}}{4}

for any A > a > 0. This implies

(3.10)  (A + a)^{2} k ≥ aA

for any A > a > 0.

Finally, let a = 1 − q, A = 1 + q, q ∈ (0, 1) . Then from (3.10) we get 4k ≥ 1 − q^{2} for any q ∈ (0, 1) which produces k ≥ \frac{1}{4}.

Remark 22. If ⟨x, e⟩ , ⟨y, e⟩ are assumed not to be zero, then the inequality (3.1) is equivalent to

|\frac{⟨x, y⟩}{⟨x, e⟩ ⟨e, y⟩} − 1| ≤ \frac{1}{4}M (a, A) M (b, B) ,

while the inequality (3.7) is equivalent to

|\frac{⟨x, y⟩}{⟨x, e⟩ ⟨e, y⟩} − 1| ≤ \frac{1}{4} · \begin{matrix}(A − a) & (B − b) \\ √ & \overline{abAB} \end{matrix}.

3. COMPANIONS OF GRÜSS’ INEQUALITY 55

The constant \frac{1}{4} is best possible in both inequalities.

3.2. Some Related Results. The following result holds [12]. Theorem 20. Let (H; ⟨·, ·⟩) be an inner product space over 𝕂. If

γ, Γ ∈ 𝕂, e, x, y ∈ H with ∥e∥ = 1 and λ ∈ (0, 1) are such that

(3.11)  Re ⟨Γe − (λx + (1 − λ) y) , (λx + (1 − λ) y) − γe⟩ ≥ 0,

or, equivalently,

∥λx + (1 − λ) y − \frac{γ + Γ}{2}e∥ ≤ \frac{1}{2} |Γ − γ| ,

then we have the inequality

(3.12)  Re [⟨x, y⟩ − ⟨x, e⟩ ⟨e, y⟩] ≤ \frac{1}{16} · \frac{1}{λ (1 − λ)} |Γ − γ|^{2} .

The constant \frac{1}{16} is the best possible constant in (3.12) in the sense that it cannot be replaced by a smaller one.

Proof. We know that for any z, u ∈ H one has

Re ⟨z, u⟩ ≤ \frac{1}{4} ∥z + u∥^{2} .

Then for any a, b ∈ H and λ ∈ (0, 1) one has

(3.13)  Re ⟨a, b⟩ ≤ \frac{1}{4λ (1 − λ)} ∥λa + (1 − λ) b∥^{2} .

Since

⟨x, y⟩ − ⟨x, e⟩ ⟨e, y⟩ = ⟨x − ⟨x, e⟩ e, y − ⟨y, e⟩ e⟩ (as ∥e∥ = 1), using (3.13), we have Re [⟨x, y⟩ − ⟨x, e⟩ ⟨e, y⟩] (3.14)

= Re [⟨x − ⟨x, e⟩ e, y − ⟨y, e⟩ e⟩]

≤ \frac{1}{4λ (1 − λ)} ∥λ (x − ⟨x, e⟩ e) + (1 − λ) (y − ⟨y, e⟩ e)∥^{2}

= \frac{1}{4λ (1 − λ)} ∥λx + (1 − λ) y − ⟨λx + (1 − λ) y, e⟩ e∥^{2} .

Since, for m, e ∈ H with ∥e∥ = 1, one has the equality ∥m − ⟨m, e⟩ e∥^{2} = ∥m∥^{2} − |⟨m, e⟩|^{2} , then, by (3.14), we deduce the inequality

(3.15) Re [⟨x, y⟩ − ⟨x, e⟩ ⟨e, y⟩]

≤ \frac{1}{4λ (1 − λ)} [∥λx + (1 − λ) y∥^{2} − |⟨λx + (1 − λ) y, e⟩|^{2}] .

Now, if we apply Grüss’ inequality

0 ≤ ∥a∥^{2} − |⟨a, e⟩|^{2} ≤ \frac{1}{4} |Γ − γ|^{2} ,

provided

Re ⟨Γe − a, a − γe⟩ ≥ 0,

for a = λx + (1 − λ) y, we have

(3.16)  ∥λx + (1 − λ) y∥^{2} − |⟨λx + (1 − λ) y, e⟩|^{2} ≤ \frac{1}{4} |Γ − γ|^{2} .

Utilising (3.15) and (3.16) we deduce the desired inequality (3.12).

To prove the sharpness of the constant \frac{1}{16}, assume that (3.12) holds with a constant C > 0, provided that (3.11) is valid, i.e.,

(3.17)  Re [⟨x, y⟩ − ⟨x, e⟩ ⟨e, y⟩] ≤ C · \frac{1}{λ (1 − λ)} |Γ − γ|^{2} .

If in (3.17) we choose x = y, given that (3.11) holds with x = y and λ ∈ (0, 1) , then

(3.18)  ∥x∥^{2} − |⟨x, e⟩|^{2} ≤ C · \frac{1}{λ (1 − λ)} |Γ − γ|^{2} ,

provided

Re ⟨Γe − x, x − γe⟩ ≥ 0.

Since we know, in Grüss’ inequality, that the constant \frac{1}{4} is best possible, then by (3.18), one has

\frac{1}{4} ≤ \frac{C}{λ (1 − λ)} for λ ∈ (0, 1) ,

giving, for λ = \frac{1}{2}, C ≥ \frac{1}{16}.

The theorem is completely proved.

The following corollary is a natural consequence of the above result [12].

Corollary 20. Assume that γ, Γ, e, x, y and λ are as in Theorem 20. If

Re ⟨Γe − (λx ± (1 − λ) y) , (λx ± (1 − λ) y) − γe⟩ ≥ 0,

3. COMPANIONS OF GRÜSS’ INEQUALITY 57

or, equivalently,

∥λx ± (1 − λ) y − \frac{γ + Γ}{2}e∥ ≤ \frac{1}{2} |Γ − γ|^{2} ,

then we have the inequality

(3.19)  |Re [⟨x, y⟩ − ⟨x, e⟩ ⟨e, y⟩]| ≤ \frac{1}{16} · \frac{1}{λ (1 − λ)} |Γ − γ|^{2} .

The constant \frac{1}{16} is best possible in (3.19).

Proof. Using Theorem 20 for (−y) instead of y, we have that

Re ⟨Γe − (λx − (1 − λ) y) , (λx − (1 − λ) y) − γe⟩ ≥ 0, which implies that

Re [− ⟨x, y⟩ + ⟨x, e⟩ ⟨e, y⟩] ≤ \frac{1}{16} · \frac{1}{λ (1 − λ)} |Γ − γ|^{2}

giving

(3.20)  Re [⟨x, y⟩ − ⟨x, e⟩ ⟨e, y⟩] ≥ −\frac{1}{16} · \frac{1}{λ (1 − λ)} |Γ − γ|^{2} .

Consequently, by (3.12) and (3.20) we deduce the desired inequality (3.19).

Remark 23. If M, m ∈ ℝ with M > m and, for λ ∈ (0, 1) ,

(3.21) ∥λx + (1 − λ) y − \frac{M + m}{2}e∥ ≤ \frac{1}{2} (M − m) ,

then

⟨x, y⟩ − ⟨x, e⟩ ⟨e, y⟩ ≤ \frac{1}{16} · \frac{1}{λ (1 − λ)} (M − m)^{2} .

If (3.21) holds with “±” instead of “+” , then

|⟨x, y⟩ − ⟨x, e⟩ ⟨e, y⟩| ≤ \frac{1}{16} · \frac{1}{λ (1 − λ)} (M − m)^{2} .

Remark 24. If λ = \frac{1}{2} in (3.11), then we obtain the result from [13], i.e.,

Re ⟨Γe − \frac{x + y}{2} , \frac{x + y}{2} − γe⟩ ≥ 0,

or, equivalently,

∥\frac{x + y}{2} − \frac{γ + Γ}{2}e∥ ≤ \frac{1}{2} |Γ − γ| ,

58 2. INEQUALITIES OF THE GRÜSS TYPE

implies

(3.22)  Re [⟨x, y⟩ − ⟨x, e⟩ ⟨e, y⟩] ≤ \frac{1}{4} |Γ − γ|^{2} .

The constant \frac{1}{4} is best possible in (3.22).

For λ = \frac{1}{2}, Corollary 20 and Remark 23 will produce the corresponding results obtained in [13]. We omit the details.

3.3. Integral Inequalities. The following proposition holds [12]. Proposition 21. If f, g, h ∈ L^{2} (Ω, 𝕂) and φ, Φ, γ, Γ ∈ 𝕂, are

such that Re (Φ\overline{φ}) > 0, Re (Γ\overline{γ}) > 0, ∫_{Ω} |h (s)|^{2} dμ (s) = 1 and

(3.23)  ∫_{Ω} Re [(Φh (s) − f (s)) (\overline{f (s)} − \overline{φ}\overline{h (s)})] dμ (s) ≥ 0,
∫_{Ω} Re [(Γh (s) − g (s)) (\overline{g (s)} − \overline{γ}\overline{h (s)})] dμ (s) ≥ 0,

or, equivalently,

(∫_{Ω} |f (s) − \frac{Φ + φ}{2}h (s)|^{2} dμ (s))^{\frac{1}{2}} ≤ \frac{1}{2} |Φ − φ| ,
(∫_{Ω} |g (s) − \frac{Γ + γ}{2} h (s)|^{2} dμ (s))^{\frac{1}{2}} ≤ \frac{1}{2} |Γ − γ| ,

then we have the following Grüss type integral inequality

(3.24) |∫_{Ω} f (s) \overline{g (s)}dμ (s) − ∫_{Ω} f (s) \overline{h (s)}dμ (s) ∫_{Ω} h (s) \overline{g (s)}dμ (s)|

≤ \frac{1}{4}M (φ, Φ) M (γ, Γ) |∫_{Ω} f (s) \overline{h (s)}dμ (s) ∫_{Ω} h (s) \overline{g (s)}dμ (s)| ,

where

M (φ, Φ) := [\frac{(|Φ| − |φ|)^{2} + 4 [|Φ\overline{φ}| − Re (Φ\overline{φ})]}{Re (Φ\overline{φ})}]^{\frac{1}{2}} .

The constant \frac{1}{4} is best possible.

The proof follows by Theorem 20 on choosing H = L^{2} (Ω, 𝕂) with the inner product

⟨f, g⟩ := ∫_{Ω} f (s) \overline{g (s)}dμ (s) .

We omit the details.

Remark 25. It is obvious that a sufficient condition for (3.23) to hold is

Re [(Φh (s) − f (s)) (\overline{f (s)} − \overline{φ}\overline{h (s)})] ≥ 0,

and

Re [(Γh (s) − g (s)) (\overline{g (s)} − \overline{γ}\overline{h (s)})] ≥ 0,

for μ−a.e. s ∈ Ω, or, equivalently,

|f (s) − \frac{Φ + φ}{2}h (s)| ≤ \frac{1}{2} |Φ − φ| |h (s)|

and

|g (s) − \frac{Γ + γ}{2} h (s)| ≤ \frac{1}{2} |Γ − γ| |h (s)| ,

for μ−a.e. s ∈ Ω.

The following result may be stated as well.

Corollary 21. If z, Z, t, T ∈ 𝕂, μ (Ω) < ∞ and f, g ∈ L^{2} (Ω, 𝕂) are such that:

Re [(Z − f (s)) (\overline{f (s)} − z̄)] ≥ 0,
Re [(T − g (s)) (\overline{g (s)} − t̄)] ≥ 0,

for a.e. s ∈ Ω, or, equivalently,

|f (s) − \frac{z + Z}{2} | ≤ \frac{1}{2} |Z − z| ,
|g (s) − \frac{t + T}{2} | ≤ \frac{1}{2} |T − t| ,

for a.e. s ∈ Ω, then we have the inequality

(3.25) |\frac{1}{μ (Ω)} ∫_{Ω} f (s) \overline{g (s)}dμ (s)

− \frac{1}{μ (Ω)} ∫_{Ω} f (s) dμ (s) · \frac{1}{μ (Ω)} ∫_{Ω} \overline{g (s)}dμ (s)|

≤ \frac{1}{4}M (z, Z) M (t, T) |\frac{1}{μ (Ω)} ∫_{Ω} f (s) dμ (s) · \frac{1}{μ (Ω)} ∫_{Ω} \overline{g (s)}dμ (s)| .

Remark 26. The case of real functions incorporates the following interesting inequality

|\frac{μ (Ω) ∫_{Ω} f (s) g (s) dμ (s)}{∫_{Ω} f (s) dμ (s) ∫_{Ω} g (s) dμ (s)} − 1| ≤ \frac{1}{4} · \begin{matrix}(Z − z) & (T − t) \\ √ & \overline{ztZT} \end{matrix},

provided μ (Ω) < ∞,

z ≤ f (s) ≤ Z, t ≤ g (s) ≤ T

for μ−a.e. s ∈ Ω, where z, t, Z, T are real numbers and the integrals at the denominator are not zero. Here the constant \frac{1}{4} is best possible in the sense mentioned above.

Using Theorem 20 we may state the following result as well [12].

Proposition 22. If f, g, h ∈ L^{2} (Ω, 𝕂) and γ, Γ ∈ 𝕂 are such that ∫_{Ω} |h (s)|^{2} dμ (s) = 1 and

(3.26) ∫_{Ω} {Re [Γh (s) − (λf (s) + (1 − λ) g (s))]
× [λ\overline{f (s)} + (1 − λ) \overline{g (s)} − γ̄h̄ (s)]\} dμ (s) ≥ 0,

or, equivalently, (3.27)

(∫_{Ω} |λf (s) + (1 − λ) g (s) − \frac{γ + Γ}{2}h (s)|^{2} dμ (s))^{\frac{1}{2}} ≤ \frac{1}{2} |Γ − γ| ,

then we have the inequality

I := ∫_{Ω} Re [f (s) \overline{g (s)}] dμ (s)
− Re [∫_{Ω} f (s) \overline{h (s)}dμ (s) · ∫_{Ω} h (s) \overline{g (s)}dμ (s)]

≤ \frac{1}{16} · \frac{1}{λ (1 − λ)} |Γ − γ|^{2} .

The constant \frac{1}{16} is best possible.

If (3.26) and (3.27) hold with “ ± ” instead of “ + ” (see Corollary 20), then

|I| ≤ \frac{1}{16} · \frac{1}{λ (1 − λ)} |Γ − γ|^{2} .

Remark 27. It is obvious that a sufficient condition for (3.26) to hold is

Re{ [Γh (s) − (λf (s) + (1 − λ) g (s))]
× [λ\overline{f (s)} + (1 − λ) \overline{g (s)} − γ̄h̄ (s)]\} ≥ 0

for a.e. s ∈ Ω, or, equivalently,

|λf (s) + (1 − λ) g (s) − \frac{γ + Γ}{2}h (s)| ≤ \frac{1}{2} |Γ − γ| |h (s)| ,

for a.e. s ∈ Ω.

Finally, the following corollary holds.

Corollary 22. If Z, z ∈ 𝕂, μ (Ω) < ∞ and f, g ∈ L^{2} (Ω, 𝕂) are such that

(3.28) Re [ (Z − (λf (s) + (1 − λ) g (s)))
× (λ\overline{f (s)} + (1 − λ) \overline{g (s)} − \overline{z})] ≥ 0

for a.e. s ∈ Ω, or, equivalently

(3.29)  |λf (s) + (1 − λ) g (s) − \frac{z + Z}{2} | ≤ \frac{1}{2} |Z − z|,

for a.e. s ∈ Ω, then we have the inequality

J := \frac{1}{μ (Ω)} ∫_{Ω} Re [f (s) \overline{g (s)}] dμ (s)

− Re [\frac{1}{μ (Ω)} ∫_{Ω} f (s) dμ (s) · \frac{1}{μ (Ω)} ∫_{Ω} \overline{g (s)}dμ (s)]

≤ \frac{1}{16} · \frac{1}{λ (1 − λ)} |Z − z|^{2} .

If (3.28) and (3.29) hold with “ ± ” instead of “ + ”, then

|J| ≤ \frac{1}{16} · \frac{1}{λ (1 − λ)} |Z − z|^{2} .

Remark 28. It is obvious that if one chooses the discrete measure above, then all the inequalities in this section may be written for sequences of real or complex numbers. We omit the details.

4. Other Grüss Type Inequalities

4.1. General Results. We may state the following result [10].

Theorem 21. Let (H; ⟨·, ·⟩) be an inner product space over the real or complex number field 𝕂 and x, y, e ∈ H with ∥e∥ = 1. If r_{1}, r_{2} ∈ (0, 1) and

∥x − e∥ ≤ r_{1}, ∥y − e∥ ≤ r_{2},

then we have the inequality

(4.1)  |⟨x, y⟩ − ⟨x, e⟩ ⟨e, y⟩| ≤ r_{1}r_{2} ∥x∥ ∥y∥ .

The inequality (4.1) is sharp in the sense that the constant c = 1 in front of r_{1}r_{2} cannot be replaced by a smaller quantity.

Proof. Start with the inequality

(4.2) |⟨x, y⟩ − ⟨x, e⟩ ⟨e, y⟩|^{2} ≤ (∥x∥^{2} − |⟨x, e⟩|^{2}) (∥y∥^{2} − |⟨y, e⟩|^{2}) .

Using Theorem 5 for a = e, we may state that

(4.3) ∥x∥^{2} − |⟨x, e⟩|^{2} ≤ r_{1}^{2} ∥x∥^{2} ,  ∥y∥^{2} − |⟨y, e⟩|^{2} ≤ r_{2}^{2} ∥y∥^{2} .

Utilizing (4.2) and (4.3), we deduce

|⟨x, y⟩ − ⟨x, e⟩ ⟨e, y⟩|^{2} ≤ r_{1}^{2}r_{2}^{2} ∥x∥^{2} ∥y∥^{2} ,

which is clearly equivalent to the desired inequality (4.1).

The sharpness of the constant follows by the fact that for x = y, r_{1} = r_{2} = r, we get from (4.1) that

(4.4)  ∥x∥^{2} − |⟨x, e⟩|^{2} ≤ r^{2} ∥x∥^{2} ,

provided ∥e∥ = 1 and ∥x − e∥ ≤ r < 1. The inequality (4.4) is sharp, as shown in Theorem 5, and the proof is completed.

Another companion of the Grüss inequality may be stated as well [10].

Theorem 22. Let (H; ⟨·, ·⟩) be an inner product space over 𝕂 and x, y, e ∈ H with ∥e∥ = 1. Suppose also that a, A, b, B ∈ 𝕂 such that Re (A\overline{a}) , Re (B\overline{b}) > 0. If either

Re ⟨Ae − x, x − ae⟩ ≥ 0, Re ⟨Be − y, y − be⟩ ≥ 0,

or, equivalently,

∥x − \frac{a + A}{2}e∥ ≤ \frac{1}{2} |A − a| , ∥y − \frac{b + B}{2} e∥ ≤ \frac{1}{2} |B − b| ,

holds, then we have the inequality

(4.5) |⟨x, y⟩ − ⟨x, e⟩ ⟨e, y⟩| ≤ \frac{1}{4} · \begin{matrix}|A − a| & |B − b| \\ √ & \overline{Re (A\overline{a}) Re (B\overline{b})} \end{matrix} |⟨x, e⟩ ⟨e, y⟩| .

The constant \frac{1}{4} is best possible.

Proof. We know, that

(4.6) |⟨x, y⟩ − ⟨x, e⟩ ⟨e, y⟩|^{2} ≤ (∥x∥^{2} − |⟨x, e⟩|^{2}) (∥y∥^{2} − |⟨y, e⟩|^{2}) .

If we use Corollary 11, then we may state that

(4.7)  ∥x∥^{2} − |⟨x, e⟩|^{2} ≤ \frac{1}{4} · \begin{matrix}|A − a| & 2 \\ Re & (A\overline{a}) \end{matrix} |⟨x, e⟩|^{2}

and

(4.8)  ∥y∥^{2} − |⟨y, e⟩|^{2} ≤ \frac{1}{4} · \begin{matrix}|B − b| & 2 \\ Re & (B\overline{b}) \end{matrix} |⟨y, e⟩|^{2} .

Utilizing (4.6) – (4.8), we deduce

|⟨x, y⟩ − ⟨x, e⟩ ⟨e, y⟩|^{2} ≤ \frac{1}{16} · \begin{matrix}|A − a| & 2 & |B − b| & 2 \\ Re & (A\overline{a}) & Re & (B\overline{b}) \end{matrix} |⟨x, e⟩ ⟨e, y⟩|^{2} ,

which is clearly equivalent to the desired inequality (4.5).

The sharpness of the constant follows from Corollary 11, and we omit the details.

Remark 29. With the assumptions of Theorem 22 and if ⟨x, e⟩ , ⟨y, e⟩ ≠ 0 (that is actually the interesting case), then one has the inequality

|\frac{⟨x, y⟩}{⟨x, e⟩ ⟨e, y⟩} − 1| ≤ \frac{1}{4} · \begin{matrix}|A − a| & |B − b| \\ √ & \overline{Re (A\overline{a}) Re (B\overline{b})} \end{matrix}.

The constant \frac{1}{4} is best possible.

We may state the following result [9].

Theorem 23. Let (H; ⟨·, ·⟩) be an inner product space over the real or complex number field 𝕂 and x, y, e ∈ H with ∥e∥ = 1. If r_{1}, r_{2} > 0 and

∥x − e∥ ≤ r_{1}, ∥y − e∥ ≤ r_{2},

then we have the inequalities

(4.9) |⟨x, y⟩ − ⟨x, e⟩ ⟨e, y⟩| ≤ \frac{1}{2}r_{1}r_{2}\sqrt{∥x∥ + |⟨x, e⟩|} · \sqrt{∥y∥ + |⟨y, e⟩|}
≤ r_{1}r_{2} ∥x∥ ∥y∥ .

The constant \frac{1}{2} is best possible in the sense that it cannot be replaced by a smaller constant.

Proof. Start with the inequality

(4.10) |⟨x, y⟩ − ⟨x, e⟩ ⟨e, y⟩|^{2} ≤ (∥x∥^{2} − |⟨x, e⟩|^{2}) (∥y∥^{2} − |⟨y, e⟩|^{2}) .

Using Theorem 7 for a = e, we have

(4.11)  0 ≤ ∥x∥^{2} − |⟨x, e⟩|^{2}

= (∥x∥ − |⟨x, e⟩|) (∥x∥ + |⟨x, e⟩|)

≤ \frac{1}{2}r_{1}^{2} (∥x∥ + |⟨x, e⟩|) ≤ r_{1}^{2} ∥x∥ ,

and, in a similar way

(4.12)  0 ≤ ∥y∥^{2} − |⟨y, e⟩|^{2}

≤ \frac{1}{2}r_{2}^{2} (∥y∥ + |⟨y, e⟩|) ≤ r_{2}^{2} ∥y∥ .

Utilising (4.10) – (4.12), we may state that

(4.13) |⟨x, y⟩ − ⟨x, e⟩ ⟨e, y⟩|^{2} ≤ \frac{1}{4}r_{1}^{2}r_{2}^{2} (∥x∥ + |⟨x, e⟩|) (∥y∥ + |⟨y, e⟩|)
≤ r_{1}^{2}r_{2}^{2} ∥x∥ ∥y∥ ,

giving the desired inequality (4.9).

To prove the sharpness of the constant \frac{1}{2}, let us assume that x = y in (4.9), to get

(4.14)  ∥x∥^{2} − |⟨x, e⟩|^{2} ≤ \frac{1}{2}r_{1}^{2} (∥x∥ + |⟨x, e⟩|) ,

provided ∥x − e∥ ≤ r_{1}. If x ≠ 0, then dividing (4.14) with ∥x∥ + |⟨x, e⟩| > 0 we get

(4.15)  ∥x∥ − |⟨x, e⟩| ≤ \frac{1}{2}r_{1}^{2}

provided ∥x − e∥ ≤ r_{1}, ∥e∥ = 1. However, (4.15) is in fact (5.2) for a = e, for which we have shown that \frac{1}{2} is the best possible constant.

The following result also holds [9].

Theorem 24. With the assumptions of Theorem 23, we have the inequality

(4.16) |⟨x, y⟩ − ⟨x, e⟩ ⟨e, y⟩| ≤ r_{1}r_{2}\sqrt{\frac{1}{4}r_{1}^{2} + |⟨x, e⟩|} · \sqrt{\frac{1}{4}r_{2}^{2} + |⟨y, e⟩|}.

Proof. Note that, from Theorem 8, we have

(4.17)  ∥x∥ ∥a∥ ≤ |⟨x, a⟩| + \frac{1}{2}r^{2},

provided ∥x − a∥ ≤ r.

Taking the square of (4.17) and re-arranging the terms, we obtain:

0 ≤ ∥x∥^{2} ∥a∥^{2} − |⟨x, a⟩|^{2} ≤ r^{2} (\frac{1}{4}r^{2} + |⟨x, a⟩|) ,

provided ∥x − a∥ ≤ r.

Using the assumption of the theorem, we then have

(4.18)  0 ≤ ∥x∥^{2} − |⟨x, e⟩|^{2} ≤ r_{1}^{2} (\frac{1}{4}r_{1}^{2} + |⟨x, e⟩|) ,

and

(4.19)  0 ≤ ∥y∥^{2} − |⟨y, e⟩|^{2} ≤ r_{2}^{2} (\frac{1}{4}r_{2}^{2} + |⟨y, e⟩|) .

Utilising (4.10), (4.18) and (4.19), we deduce the desired inequality (4.16).

The following result may be stated as well [9].

Theorem 25. Let (H; ⟨·, ·⟩) be an inner product space over 𝕂 and x, y, e ∈ H with ∥e∥ = 1. Suppose also that a, A, b, B ∈ 𝕂 such that A ≠ −a, B ≠ −b. If either

Re ⟨Ae − x, x − ae⟩ ≥ 0, Re ⟨Be − y, y − be⟩ ≥ 0,

or, equivalently,

∥x − \frac{a + A}{2}e∥ ≤ \frac{1}{2} |A − a| , ∥y − \frac{b + B}{2} e∥ ≤ \frac{1}{2} |B − b| ,

holds, then we have the inequality

|⟨x, y⟩ − ⟨x, e⟩ ⟨e, y⟩| (4.20)

≤ \frac{1}{4} · \begin{matrix}|A − a| & |B − b| \\ √ & \overline{|A + a| |B + b|} \end{matrix}\sqrt{∥x∥ + |⟨x, e⟩|} · \sqrt{∥y∥ + |⟨y, e⟩|}

≤ \frac{1}{2} · \begin{matrix}|A − a| & |B − b| \\ √ & \overline{|A + a| |B + b|} \end{matrix}\sqrt{∥x∥ ∥y∥}.

The constant \frac{1}{4} is best possible in (4.20).

Proof. From Theorem 8, we may state that

(4.21)  0 ≤ ∥x∥^{2} − |⟨x, e⟩|^{2}

= (∥x∥ − |⟨x, e⟩|) (∥x∥ + |⟨x, e⟩|)

≤ \frac{1}{4} · \frac{|A − a|^{2}}{|A + a|} (∥x∥ + |⟨x, e⟩|) ,

and

(4.22)  0 ≤ ∥y∥^{2} − |⟨y, e⟩|^{2} ≤ \frac{1}{4} · \frac{|B − b|^{2}}{|B + b|} (∥y∥ + |⟨y, e⟩|) .

Making use of (4.10) and (4.21), (4.22), we deduce the first inequality in (4.20).

The best constant follows by the use of Theorem 8, and we omit the details.

Finally, we may state the following theorem as well [9].

Theorem 26. With the assumptions of Theorem 25, we have the inequality

(4.23) |⟨x, y⟩ − ⟨x, e⟩ ⟨e, y⟩|

≤ \frac{1}{2} · \begin{matrix}|A − a| & |B − b| \\ √ & \overline{|A + a| |B + b|} \end{matrix}\sqrt{\frac{1}{8} · \frac{|A − a|^{2}}{|A + a|} + |⟨x, e⟩|}

× \sqrt{\frac{1}{8} · \frac{|B − b|^{2}}{|B + b|} + |⟨y, e⟩|}.

Proof. Using Theorem 8, we may state that

0 ≤ ∥x∥ − |⟨x, e⟩| ≤ \frac{1}{4} · \frac{|A − a|^{2}}{|A + a|} .

This inequality implies that

∥x∥^{2} ≤ |⟨x, e⟩|^{2} + \frac{1}{2} |⟨x, e⟩| · \frac{|A − a|^{2}}{|A + a|} + \frac{1}{16} · \begin{matrix}|A − a| & 4 \\ |A + a| & 2 \end{matrix} ,

giving

(4.24) 0 ≤ ∥x∥^{2} − |⟨x, e⟩|^{2} ≤ \frac{1}{2} · \frac{|A − a|^{2}}{|A + a|} [|⟨x, e⟩| + \frac{1}{8} · \frac{|A − a|^{2}}{|A + a|} ] .

Similarly, we have

(4.25) 0 ≤ ∥y∥^{2} − |⟨y, e⟩|^{2} ≤ \frac{1}{2} · \frac{|B − b|^{2}}{|B + b|} [|⟨y, e⟩| + \frac{1}{8} · \frac{|B − b|^{2}}{|B + b|} ] .

By making use of (4.10) and (4.24), (4.25), we deduce the desired inequality (4.23).

4.2. Integral Inequalities. The following Grüss type integral inequality for real or complex-valued functions also holds [10].

Proposition 23. Let f, g, h ∈ L_{ρ}^{2} (Ω, 𝕂) with ∫_{Ω} ρ (s) |h (s)|^{2} dμ (s) = 1 and a, A, b, B ∈ 𝕂 such that Re (A\overline{a}) , Re (B\overline{b}) > 0 and

Re [(Ah (s) − f (s)) (\overline{f (s)} − \overline{a}\overline{h (s)})] ≥ 0,
Re [(Bh (s) − g (s)) (\overline{g (s)} − \overline{b}\overline{h (s)})] ≥ 0,

4. OTHER GRÜSS TYPE INEQUALITIES 67

for μ−a.e. s ∈ Ω. Then we have the inequalities

|∫_{Ω} ρ (s) f (s) \overline{g (s)}dμ (s)

− ∫_{Ω} ρ (s) f (s) \overline{h (s)}dμ (s) ∫_{Ω} ρ (s) h (s) \overline{g (s)}dμ (s)|

≤ \frac{1}{4} · \begin{matrix}|A − a| & |B − b| \\ √ & \overline{Re (A\overline{a}) Re (B\overline{b})} \end{matrix}

× |∫_{Ω} ρ (s) f (s) \overline{h (s)}dμ (s) ∫_{Ω} ρ (s) h (s) \overline{g (s)}dμ (s)| .

The constant \frac{1}{4} is best possible.

The proof follows by Theorem 22.

By making use of Theorem 25, we may also state:

Proposition 24. Let f, g, h ∈ L_{ρ}^{2} (Ω, 𝕂) be such that

∫_{Ω} ρ (s) |h (s)|^{2} dμ (s) = 1.

Suppose also that a, A, b, B ∈ 𝕂 with A ≠ −a, B ≠ −b and

Re [(Ah (s) − f (s)) (\overline{f (s)} − \overline{a}\overline{h (s)})] ≥ 0,
Re [(Bh (s) − g (s)) (\overline{g (s)} − \overline{b}\overline{h (s)})] ≥ 0,

for μ−a.e. s ∈ Ω. Then we have the inequality

|∫_{Ω} ρ (s) f (s) \overline{g (s)}dμ (s)

− ∫_{Ω} ρ (s) f (s) \overline{h (s)}dμ (s) ∫_{Ω} ρ (s) h (s) \overline{g (s)}dμ (s)|

≤ \frac{1}{4} · \begin{matrix}|A − a| & |B − b| \\ √ & \overline{|A + a| |B + b|} \end{matrix}

× \sqrt{(∫_{Ω} ρ (s) |f (s)|^{2} dμ (s))^{\frac{1}{2}} + |∫_{Ω} ρ (s) f (s) \overline{h (s)}dμ (s)|}

× \sqrt{(∫_{Ω} ρ (s) |g (s)|^{2} dμ (s))^{\frac{1}{2}} + |∫_{Ω} ρ (s) g (s) \overline{h (s)}dμ (s)|}.

The constant \frac{1}{4} is best possible.

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CHAPTER 3 Reverses of Bessel’s Inequality

1. Introduction

Let (H, ⟨·, ·⟩) be an inner product space over 𝕂 (𝕂 = ℝ, ℂ) and {e_{i}}_{i∈I} a finite or infinite family of orthonormal vectors in H, i.e.,

⟨e_{i}, e_{j}⟩ = \{ \begin{matrix}0 & if i ≠ j \\ 1 & if i = j \end{matrix} ; i, j ∈ I.

It is well known that, the following inequality due to Bessel, holds

∑_{i∈I} |⟨x, e_{i}⟩|^{2} ≤ ∥x∥^{2} ,

for any x ∈ H, where the meaning of the sum is:

∑_{i∈I} |⟨x, e_{i}⟩|^{2} := sup_{F ⊂I} {∑_{i∈F} |⟨x, e_{i}⟩|^{2} , F is a finite part of I} .

If (H, ⟨·, ·⟩) is an infinite dimensional Hilbert space and {e_{i}}_{i∈ℕ} an orthonormal family in H, then we also have

∑_{i=0}^{∞} |⟨x, e_{i}⟩|^{2} ≤ ∥x∥^{2}

for any x ∈ H. Here the meaning of the series is the usual one.

In this chapter we establish reverses of the Bessel inequality and some Grüss type inequalities for orthonormal families, namely, upper bounds for the expressions

∥x∥^{2} − ∑_{i∈I} |⟨x, e_{i}⟩|^{2} , ∥x∥ − (∑_{i∈I} |⟨x, e_{i}⟩|^{2})^{\frac{1}{2}} , x ∈ X

and

|⟨x, y⟩ − ∑_{i∈I} ⟨x, e_{i}⟩ ⟨e_{i}, y⟩| , x, y ∈ H,

under various assumptions for the vectors x, y and the orthonormal family {e_{i}}_{i∈I} .

2. Reverses of Bessel’s Inequality

2.1. Introduction. In [3], the author has proved the following Grüss type inequality in real or complex inner product spaces.

Theorem 27. Let (H, ⟨·, ·⟩) be an inner product space over 𝕂 and e ∈ H, ∥e∥ = 1. If ϕ, Φ, γ, Γ are real or complex numbers and x, y are vectors in H such that the conditions

(2.1)  Re ⟨Φe − x, x − ϕe⟩ ≥ 0 and Re ⟨Γe − y, y − γe⟩ ≥ 0

hold, then we have the inequality

(2.2)  |⟨x, y⟩ − ⟨x, e⟩ ⟨e, y⟩| ≤ \frac{1}{4} |Φ − ϕ| |Γ − γ| .

The constant \frac{1}{4} is best possible in the sense that it cannot be replaced by a smaller constant.

In [8], the following refinement of (2.2) has been pointed out.

Theorem 28. Let H, 𝕂 and e be as in Theorem 27. If ϕ, Φ, γ, Γ, x, y satisfy (2.1) or, equivalently,

∥x − \frac{ϕ + Φ}{2}e∥ ≤ \frac{1}{2} |Φ − ϕ| , ∥y − \frac{γ + Γ}{2}e∥ ≤ \frac{1}{2} |Γ − γ| ,

then

(2.3) |⟨x, y⟩ − ⟨x, e⟩ ⟨e, y⟩| ≤ \frac{1}{4} |Φ − ϕ| |Γ − γ|

− [Re ⟨Φe − x, x − ϕe⟩]^{\frac{1}{2}} [Re ⟨Γe − y, y − γe⟩]^{\frac{1}{2}} .

In [12], N. Ujević has generalised Theorem 27 for the case of real inner product spaces as follows.

Theorem 29. Let (H, ⟨·, ·⟩) be an inner product space over the real number field ℝ, and {e_{i}}_{i∈{1,...,n}} an orthornormal family in H. If ϕ_{i}, γ_{i}, Φ_{i}, Γ_{i} ∈ ℝ, i ∈ {1, …, n} satisfy the condition

⟨∑_{i=1}^{n} Φ_{i}e_{i} − x, x − ∑_{i=1}^{n} ϕ_{i}e_{i}⟩ ≥ 0, ⟨∑_{i=1}^{n} Γ_{i}e_{i} − y, y − ∑_{i=1}^{n} γ_{i}e_{i}⟩ ≥ 0,

then one has the inequality:

(2.4)  |⟨x, y⟩ − ∑_{i=1}^{n} ⟨x, e_{i}⟩ ⟨e_{i}, y⟩|

≤ \frac{1}{4} [∑_{i=1}^{n} (Φ_{i} − ϕ_{i})^{2} · ∑_{i=1}^{n} (Γ_{i} − γ_{i})^{2}]^{\frac{1}{2}} .

The constant \frac{1}{4} is best possible in the sense that it cannot be replaced by a smaller constant.

We note that the key point in his proof is the following identity:

∑_{i=1}^{n} (⟨x, e_{i}⟩ − ϕ_{i}) (Φ_{i} − ⟨x, e_{i}⟩)

− ⟨x − ∑_{i=1}^{n} ϕ_{i}e_{i}, ∑_{i=1}^{n} Φ_{i}e_{i} − x⟩

= ∥x∥^{2} − ∑_{i=1}^{n} ⟨x, e_{i}⟩^{2} ,

holding for x ∈ H, ϕ_{i}, Φ_{i} ∈ ℝ, i ∈ {1, …, n} and {e_{i}}_{i∈{1,...,n}} an orthornormal family of vectors in the real inner product space H.

In this section, by following [2], we point out a reverse of Bessel’s inequality in both real and complex inner product spaces. This result will then be employed to provide a refinement of the Grüss type inequality (2.4) for real or complex inner products. Related results as well as integral inequalities for general measure spaces are also given.

2.2. A General Result. The following lemma holds [2].

Lemma 4. Let {e_{i}}_{i∈I} be a family of orthornormal vectors in H, F a finite part of I and ϕ_{i}, Φ_{i} (i ∈ F) , real or complex numbers. The following statements are equivalent for x ∈ H :

(i) Re ⟨∑_{i∈F} Φ_{i}e_{i} − x, x − ∑_{i∈F} ϕ_{i}e_{i}⟩ ≥ 0,

(ii) ∥x − ∑_{i∈F} \frac{ϕ_{i}+Φ_{i}}{2} e_{i}∥ ≤ \frac{1}{2} (∑_{i∈F} |Φ_{i} − ϕ_{i}|^{2})^{\frac{1}{2}} .

Proof. It is easy to see that for y, a, A ∈ H, the following are equivalent (see [8, Lemma 1])

(b) Re ⟨A − y, y − a⟩ ≥ 0 and

(bb) ∥y − \frac{a+A}{2} ∥ ≤ \frac{1}{2} ∥A − a∥ .

Now, for a = ∑_{i∈F} ϕ_{i}e_{i}, A = ∑_{i∈F} Φ_{i}e_{i}, we have

∥A − a∥ = ∥∑_{i∈F} (Φ_{i} − ϕ_{i}) e_{i}∥ = (∥∑_{i∈F} (Φ_{i} − ϕ_{i}) e_{i}∥^{2})^{\frac{1}{2}}

= (∑_{i∈F} |Φ_{i} − ϕ_{i}|^{2} ∥e_{i}∥^{2})^{\frac{1}{2}} = (∑_{i∈F} |Φ_{i} − ϕ_{i}|^{2})^{\frac{1}{2}} ,

giving, for y = x, the desired equivalence.

74 3. REVERSES OF BESSEL’S INEQUALITY

The following reverse of Bessel’s inequality holds [2].

Theorem 30. Let {e_{i}}_{i∈I} , F, ϕ_{i}, Φ_{i}, i ∈ F and x ∈ H such that either (i) or (ii) of Lemma 4 holds. Then we have the inequality:

(2.5)  0 ≤ ∥x∥^{2} − ∑_{i∈F} |⟨x, e_{i}⟩|^{2}

≤ \frac{1}{4} ∑_{i∈F} |Φ_{i} − ϕ_{i}|^{2} − Re ⟨∑_{i∈F} Φ_{i}e_{i} − x, x − ∑_{i∈F} ϕ_{i}e_{i}⟩

≤ \frac{1}{4} ∑_{i∈F} |Φ_{i} − ϕ_{i}|^{2} .

The constant \frac{1}{4} is best in both inequalities.

Proof. Define

I_{1} := ∑_{i∈H} Re [(Φ_{i} − ⟨x, e_{i}⟩) (\overline{⟨x, e_{i}⟩} − \overline{ϕ_{i}})]

and

I_{2} := Re [⟨∑_{i∈H} Φ_{i}e_{i} − x, x − ∑_{i∈H} ϕ_{i}e_{i}⟩] .

Observe that

I_{1} = ∑_{i∈H} Re [Φ_{i}\overline{⟨x, e_{i}⟩}] + ∑_{i∈H} Re [\overline{ϕ_{i}} ⟨x, e_{i}⟩]

− ∑_{i∈H} Re [Φ_{i}\overline{ϕ_{i}}] − ∑_{i∈H} |⟨x, e_{i}⟩|^{2}

and

I_{2} = Re [∑_{i∈H} Φ_{i}\overline{⟨x, e_{i}⟩} + ∑_{i∈H} \overline{ϕ_{i}} ⟨x, e_{i}⟩ − ∥x∥^{2} − ∑_{i∈H} ∑_{j∈H} Φ_{i}\overline{ϕ_{i}} ⟨e_{i}, e_{j}⟩]

= ∑_{i∈H} Re [Φ_{i}\overline{⟨x, e_{i}⟩}] + ∑_{i∈H} Re [\overline{ϕ_{i}} ⟨x, e_{i}⟩] − ∥x∥^{2} − ∑_{i∈H} Re [Φ_{i}\overline{ϕ_{i}}] .

Consequently, subtracting I_{2} from I_{1}, we deduce the following equality that is interesting in its turn

(2.6) ∥x∥^{2} − ∑_{i∈F} |⟨x, e_{i}⟩|^{2} = ∑_{i∈H} Re [(Φ_{i} − ⟨x, e_{i}⟩) (\overline{⟨x, e_{i}⟩} − \overline{ϕ_{i}})]

− Re [⟨∑_{i∈H} Φ_{i}e_{i} − x, x − ∑_{i∈H} ϕ_{i}e_{i}⟩] .

Using the following elementary inequality for complex numbers

Re [a\overline{b}] ≤ \frac{1}{4} |a + b|^{2} , a, b ∈ 𝕂,

for the choices a = Φ_{i} − ⟨x, e_{i}⟩ , b = ⟨x, e_{i}⟩ − ϕ_{i} (i ∈ F) , we deduce

(2.7)  ∑_{i∈H} Re [(Φ_{i} − ⟨x, e_{i}⟩) (\overline{⟨x, e_{i}⟩} − \overline{ϕ_{i}})] ≤ \frac{1}{4} ∑_{i∈H} |Φ_{i} − ϕ_{i}|^{2} .

Making use of (2.6), (2.7) and the assumption (i), we deduce (2.5).

The sharpness of the constant \frac{1}{4} was proved for a single element e, ∥e∥ = 1 in [3], or for the real case in [12].

We can give here a simple proof as follows.

Assume that there is a c > 0 such that

(2.8)  0 ≤ ∥x∥^{2} − ∑_{i∈F} |⟨x, e_{i}⟩|^{2}

≤ c ∑_{i∈F} |Φ_{i} − ϕ_{i}|^{2} − Re ⟨∑_{i∈F} Φ_{i}e_{i} − x, x − ∑_{i∈F} ϕ_{i}e_{i}⟩ ,

provided ϕ_{i}, Φ_{i}, x and F satisfy (i) or (ii).

We choose F = \{1\} , e_{1} = e_{2} = (\frac{1}{\sqrt{2}}, \frac{1}{\sqrt{2}}) ∈ ℝ^{2}, x = (x_{1}, x_{2}) ∈ ℝ^{2}, Φ_{1} = Φ = m > 0, ϕ_{1} = ϕ = −m, H = ℝ^{2} to get from (2.8) that

(2.9)  0 ≤ x_{1}^{2} + x_{2}^{2} − \frac{(x_{1} + x_{2})^{2}}{2}

≤ 4cm^{2} − (\frac{m}{\sqrt{2}} − x_{1}) (x_{1} + \frac{m}{\sqrt{2}})

− (\frac{m}{\sqrt{2}} − x_{2}) (x_{2} + \frac{m}{\sqrt{2}}) ,

provided

(2.10) 0 ≤ ⟨me − x, x + me⟩

= (\frac{m}{\sqrt{2}} − x_{1}) (x_{1} + \frac{m}{\sqrt{2}}) + (\frac{m}{\sqrt{2}} − x_{2}) (x_{2} + \frac{m}{\sqrt{2}}) .

If we choose x_{1} = \frac{m}{\sqrt{2}}, x_{2} = −\frac{m}{\sqrt{2}}, then (2.10) is fulfilled and by (2.9) we get m^{2} ≤ 4cm^{2}, giving c ≥ \frac{1}{4}.

2.3. A Refinement of the Grüss Inequality for Orthonormal Families. The following result holds [2].

Theorem 31. Let {e_{i}}_{i∈I} be a family of orthornormal vectors in H, F a finite part of I and ϕ_{i}, Φ_{i}, γ_{i}, Γ_{i} ∈ 𝕂, i ∈ F and x, y ∈ H. If either

Re ⟨∑_{i∈F} Φ_{i}e_{i} − x, x − ∑_{i∈F} ϕ_{i}e_{i}⟩ ≥ 0,

Re ⟨∑_{i∈F} Γ_{i}e_{i} − y, y − ∑_{i∈F} γ_{i}e_{i}⟩ ≥ 0,

or, equivalently,

∥x − ∑_{i∈F} \frac{Φ_{i} + ϕ_{i}}{2} e_{i}∥ ≤ \frac{1}{2} (∑_{i∈F} |Φ_{i} − ϕ_{i}|^{2})^{\frac{1}{2}} ,

∥y − ∑_{i∈F} \frac{Γ_{i} + γ_{i}}{2} e_{i}∥ ≤ \frac{1}{2} (∑_{i∈F} |Γ_{i} − γ_{i}|^{2})^{\frac{1}{2}} ,

hold, then we have the inequalities

|⟨x, y⟩ − ∑_{i∈F} ⟨x, e_{i}⟩ ⟨e_{i}, y⟩| (2.11)

≤ \frac{1}{4} (∑_{i∈F} |Φ_{i} − ϕ_{i}|^{2})^{\frac{1}{2}} · (∑_{i∈F} |Γ_{i} − γ_{i}|^{2})^{\frac{1}{2}}

− [Re ⟨∑_{i∈F} Φ_{i}e_{i} − x, x − ∑_{i∈F} ϕ_{i}e_{i}⟩]^{\frac{1}{2}}

× [Re ⟨∑_{i∈F} Γ_{i}e_{i} − y, y − ∑_{i∈F} γ_{i}e_{i}⟩]^{\frac{1}{2}}

≤ \frac{1}{4} (∑_{i∈F} |Φ_{i} − ϕ_{i}|^{2})^{\frac{1}{2}} · (∑_{i∈F} |Γ_{i} − γ_{i}|^{2})^{\frac{1}{2}} .

Proof. Using Schwarz’s inequality in the inner product space (H, ⟨·, ·⟩) one has

(2.12) |⟨x − ∑_{i∈F} ⟨x, e_{i}⟩ e_{i}, y − ∑_{i∈F} ⟨y, e_{i}⟩ e_{i}⟩|^{2}

≤ ∥x − ∑_{i∈F} ⟨x, e_{i}⟩ e_{i}∥^{2} ∥y − ∑_{i∈F} ⟨y, e_{i}⟩ e_{i}∥^{2}

and since a simple calculation shows that

⟨x − ∑_{i∈F} ⟨x, e_{i}⟩ e_{i}, y − ∑_{i∈F} ⟨y, e_{i}⟩ e_{i}⟩ = ⟨x, y⟩ − ∑_{i∈F} ⟨x, e_{i}⟩ ⟨e_{i}, y⟩

and

∥x − ∑_{i∈F} ⟨x, e_{i}⟩ e_{i}∥^{2} = ∥x∥^{2} − ∑_{i∈F} |⟨x, e_{i}⟩|^{2} ,

for any x, y ∈ H, then by (2.12) and by the reverse of Bessel’s inequality in Theorem 30, we have

|⟨x, y⟩ − ∑_{i∈F} ⟨x, e_{i}⟩ ⟨e_{i}, y⟩|^{2} (2.13)

≤ (∥x∥^{2} − ∑_{i∈F} |⟨x, e_{i}⟩|^{2}) (∥y∥^{2} − ∑_{i∈F} |⟨y, e_{i}⟩|^{2})

≤ [\frac{1}{4} ∑_{i∈F} |Φ_{i} − ϕ_{i}|^{2} − Re ⟨∑_{i∈F} Φ_{i}e_{i} − x, x − ∑_{i∈F} ϕ_{i}e_{i}⟩]

× [\frac{1}{4} ∑_{i∈F} |Γ_{i} − γ_{i}|^{2} − Re ⟨∑_{i∈F} Γ_{i}e_{i} − y, y − ∑_{i∈F} γ_{i}e_{i}⟩]

≤ [\frac{1}{4} (∑_{i∈F} |Φ_{i} − ϕ_{i}|^{2})^{\frac{1}{2}} · (∑_{i∈F} |Γ_{i} − γ_{i}|^{2})^{\frac{1}{2}}

− [Re ⟨∑_{i∈F} Φ_{i}e_{i} − x, x − ∑_{i∈F} ϕ_{i}e_{i}⟩]^{\frac{1}{2}}

× [Re ⟨∑_{i∈F} Γ_{i}e_{i} − y, y − ∑_{i∈F} γ_{i}e_{i}⟩]^{\frac{1}{2}} ]

where, for the last inequality, we have made use of the inequality

(m^{2} − n^{2}) (p^{2} − q^{2}) ≤ (mp − nq)^{2} ,

holding for any m, n, p, q > 0.

Taking the square root in (2.13) and observing that the quantity in the last square bracket is nonnegative (see for example (2.5)), we deduce the desired result (2.11).

The best constant has been proved in [3] for one element and we omit the details.

2.4. Some Companion Inequalities. The following companion of the Grüss inequality also holds [2].

Theorem 32. Let {e_{i}}_{i∈I} be a family of orthornormal vectors in H, F a finite part of I and ϕ_{i}, Φ_{i} ∈ 𝕂, i ∈ F and x, y ∈ H such that

(2.14)  Re ⟨∑_{i∈F} Φ_{i}e_{i} − \frac{x + y}{2} , \frac{x + y}{2} − ∑_{i∈F} ϕ_{i}e_{i}⟩ ≥ 0,

or, equivalently,

∥\frac{x + y}{2} − ∑_{i∈F} \frac{Φ_{i} + ϕ_{i}}{2} · e_{i}∥ ≤ \frac{1}{2} (∑_{i∈F} |Φ_{i} − ϕ_{i}|^{2})^{\frac{1}{2}} ,

holds, then we have the inequality

(2.15)  Re [⟨x, y⟩ − ∑_{i∈F} ⟨x, e_{i}⟩ ⟨e_{i}, y⟩] ≤ \frac{1}{4} ∑_{i∈F} |Φ_{i} − ϕ_{i}|^{2} .

The constant \frac{1}{4} is best possible.

Proof. Start with the well known inequality

(2.16)  Re ⟨z, u⟩ ≤ \frac{1}{4} ∥z + u∥^{2} , z, u ∈ H.

Since

⟨x, y⟩ − ∑_{i∈F} ⟨x, e_{i}⟩ ⟨e_{i}, y⟩ = ⟨x − ∑_{i∈F} ⟨x, e_{i}⟩ e_{i}, y − ∑_{i∈F} ⟨y, e_{i}⟩ e_{i}⟩ ,

for any x, y ∈ H, then, by (2.16), we get

Re [⟨x, y⟩ − ∑_{i∈F} ⟨x, e_{i}⟩ ⟨e_{i}, y⟩] (2.17)

= Re [⟨x − ∑_{i∈F} ⟨x, e_{i}⟩ e_{i}, y − ∑_{i∈F} ⟨y, e_{i}⟩ e_{i}⟩]

≤ \frac{1}{4} ∥x − ∑_{i∈F} ⟨x, e_{i}⟩ e_{i} + y − ∑_{i∈F} ⟨y, e_{i}⟩ e_{i}∥^{2}

= ∥\frac{x + y}{2} − ∑_{i∈F} ⟨\frac{x + y}{2} , e_{i}⟩ e_{i}∥^{2}

= ∥\frac{x + y}{2} ∥^{2} − ∑_{i∈F} |⟨\frac{x + y}{2} , e_{i}⟩|^{2} .

If we apply the reverse of Bessel’s inequality in Theorem 30 for \frac{x+y}{2}, we may state that

(2.18)  ∥\frac{x + y}{2} ∥^{2} − ∑_{i∈F} |⟨\frac{x + y}{2} , e_{i}⟩|^{2} ≤ \frac{1}{4} (∑_{i∈F} |Φ_{i} − ϕ_{i}|^{2})^{\frac{1}{2}} .

Now, by making use of (2.17) and (2.18), we deduce (2.15).

The fact that \frac{1}{4} is the best constant in (2.15) follows by the fact that if in (2.14) we choose x = y, then it becomes (i) of Lemma 4, implying (2.5), for which, we have shown that \frac{1}{4} was the best constant.

The following corollary may be of interest if we wish to evaluate the absolute value of

Re [⟨x, y⟩ − ∑_{i∈F} ⟨x, e_{i}⟩ ⟨e_{i}, y⟩] .

Corollary 23. With the assumptions of Theorem 32 and if

Re ⟨∑_{i∈F} Φ_{i}e_{i} − \frac{x ± y}{2} , \frac{x ± y}{2} − ∑_{i∈F} ϕ_{i}e_{i}⟩ ≥ 0,

or, equivalently,

∥\frac{x ± y}{2} − ∑_{i∈F} \frac{Φ_{i} + ϕ_{i}}{2} · e_{i}∥ ≤ \frac{1}{2} (∑_{i∈F} |Φ_{i} − ϕ_{i}|^{2})^{\frac{1}{2}} ,

holds, then we have the inequality

(2.19)  |Re [⟨x, y⟩ − ∑_{i∈F} ⟨x, e_{i}⟩ ⟨e_{i}, y⟩]| ≤ \frac{1}{4} ∑_{i∈F} |Φ_{i} − ϕ_{i}|^{2} .

Proof. We only remark that, if

Re ⟨∑_{i∈F} Φ_{i}e_{i} − \frac{x − y}{2} , \frac{x − y}{2} − ∑_{i∈F} ϕ_{i}e_{i}⟩ ≥ 0

holds, then by Theorem 32 for (−y) instead of y, we have

Re [− ⟨x, y⟩ + ∑_{i∈F} ⟨x, e_{i}⟩ ⟨e_{i}, y⟩] ≤ \frac{1}{4} ∑_{i∈F} |Φ_{i} − ϕ_{i}|^{2} ,

showing that

(2.20)  Re [⟨x, y⟩ − ∑_{i∈F} ⟨x, e_{i}⟩ ⟨e_{i}, y⟩] ≥ −\frac{1}{4} ∑_{i∈F} |Φ_{i} − ϕ_{i}|^{2} .

Making use of (2.15) and (2.20), we deduce the desired inequality (2.19).

Remark 30. If H is a real inner product space and m_{i}, M_{i} ∈ ℝ with the property that

⟨∑_{i∈F} M_{i}e_{i} − \frac{x ± y}{2} , \frac{x ± y}{2} − ∑_{i∈F} m_{i}e_{i}⟩ ≥ 0,

or, equivalently,

∥\frac{x ± y}{2} − ∑_{i∈F} \frac{M_{i} + m_{i}}{2} · e_{i}∥ ≤ \frac{1}{2} (∑_{i∈F} (M_{i} − m_{i})^{2})^{\frac{1}{2}} ,

then we have the Grüss type inequality

|⟨x, y⟩ − ∑_{i∈F} ⟨x, e_{i}⟩ ⟨e_{i}, y⟩| ≤ \frac{1}{4} ∑_{i∈F} (M_{i} − m_{i})^{2} .

2.5. Integral Inequalities. Let (Ω, Σ, μ) be a measure space consisting of a set Ω, Σ a σ−algebra of parts and μ a countably additive and positive measure on Σ with values in ℝ ∪ {∞} . Let ρ ≥ 0 be a μ−measurable function on Ω. Denote by L_{ρ}^{2} (Ω, 𝕂) the Hilbert space of all real or complex valued functions defined on Ω and 2 − ρ−integrable on Ω, i.e.,

∫_{Ω} ρ (s) |f (s)|^{2} dμ (s) < ∞.

Consider the family {f_{i}}_{i∈I} of functions in L_{ρ}^{2} (Ω, 𝕂) with the properties that

∫_{Ω} ρ (s) f_{i} (s) \overline{f_{j}} (s) dμ (s) = δ_{ij}, i, j ∈ I,

where δ_{ij} is 0 if i ≠ j and δ_{ij} = 1 if i = j. {f_{i}}_{i∈I} is an orthornormal family in L_{ρ}^{2} (Ω, 𝕂) .

The following proposition holds [2].

Proposition 25. Let {f_{i}}_{i∈I} be an orthornormal family of functions in L_{ρ}^{2} (Ω, 𝕂) , F a finite subset of I, ϕ_{i}, Φ_{i} ∈ 𝕂 (i ∈ F) and f ∈ L_{ρ}^{2} (Ω, 𝕂) , such that either

(2.21) ∫_{Ω} ρ (s) Re [(∑_{i∈F} Φ_{i}f_{i} (s) − f (s))

× (\overline{f} (s) − ∑_{i∈F} \overline{ϕ_{i}} \overline{f_{i}} (s))] dμ (s) ≥ 0,

or, equivalently,

∫_{Ω} ρ (s) |f (s) − ∑_{i∈F} \frac{Φ_{i} + ϕ_{i}}{2} f_{i} (s)|^{2} dμ (s) ≤ \frac{1}{4} ∑_{i∈F} |Φ_{i} − ϕ_{i}|^{2} .

Then we have the inequality

(2.22) 0 ≤ ∫_{Ω} ρ (s) |f (s)|^{2} dμ (s) − ∑_{i∈F} |∫_{Ω} ρ (s) f (s) \overline{f_{i}} (s) dμ (s)|^{2}

≤ \frac{1}{4} ∑_{i∈F} |Φ_{i} − ϕ_{i}|^{2}

− ∫_{Ω} ρ (s) Re [(∑_{i∈F} Φ_{i}f_{i} (s) − f (s))

× (\overline{f} (s) − ∑_{i∈F} \overline{ϕ_{i}} \overline{f_{i}} (s))] dμ (s)

≤ \frac{1}{4} ∑_{i∈F} |Φ_{i} − ϕ_{i}|^{2} .

The constant \frac{1}{4} is best possible in both inequalities.

The proof follows by Theorem 30 applied for the Hilbert space L_{ρ}^{2} (Ω, 𝕂) and the orthornormal family {f_{i}}_{i∈I} .

The following Grüss type inequality also holds [2].

82 3. REVERSES OF BESSEL’S INEQUALITY

Proposition 26. Let {f_{i}}_{i∈I} and F be as in Proposition 25. If ϕ_{i}, Φ_{i}, γ_{i}, Γ_{i} ∈ 𝕂 (i ∈ F) and f, g ∈ L_{ρ}^{2} (Ω, 𝕂) so that either

∫_{Ω} ρ (s) Re [(∑_{i∈F} Φ_{i}f_{i} (s) − f (s)) (\overline{f} (s) − ∑_{i∈F} \overline{ϕ_{i}} \overline{f_{i}} (s))] dμ (s) ≥ 0,
∫_{Ω} ρ (s) Re [(∑_{i∈F} Γ_{i}f_{i} (s) − g (s)) (\overline{g} (s) − ∑_{i∈F} \overline{γ_{i}} \overline{f_{i}} (s))] dμ (s) ≥ 0,

or, equivalently,

∫_{Ω} ρ (s) |f (s) − ∑_{i∈F} \frac{Φ_{i} + ϕ_{i}}{2} f_{i} (s)|^{2} dμ (s) ≤ \frac{1}{4} ∑_{i∈F} |Φ_{i} − ϕ_{i}|^{2} ,
∫_{Ω} ρ (s) |g (s) − ∑_{i∈F} \frac{Γ_{i} + γ_{i}}{2} f_{i} (s)|^{2} dμ (s) ≤ \frac{1}{4} ∑_{i∈F} |Γ_{i} − γ_{i}|^{2} ,

hold, then we have the inequalities

(2.23) |∫_{Ω} ρ (s) f (s) g (s) dμ (s)

− ∑_{i∈F} ∫_{Ω} ρ (s) f (s) \overline{f_{i}} (s) dμ (s) ∫_{Ω} ρ (s) f_{i} (s) \overline{g (s)}dμ (s)|

≤ \frac{1}{4} (∑_{i∈F} |Φ_{i} − ϕ_{i}|^{2})^{\frac{1}{2}} (∑_{i∈F} |Γ_{i} − γ_{i}|^{2})^{\frac{1}{2}}

− [∫_{Ω} ρ (s) Re [(∑_{i∈F} Φ_{i}f_{i} (s) − f (s)) (\overline{f} (s) − ∑_{i∈F} ϕ\overline{i}\overline{f_{i}} (s))] dμ (s)]^{\frac{1}{2}}

× [∫_{Ω} ρ (s) Re [(∑_{i∈F} Γ_{i}f_{i} (s) − g (s)) (\overline{g} (s) − ∑_{i∈F} \overline{γ_{i}}\overline{f_{i}} (s))] dμ (s)]^{\frac{1}{2}}

≤ \frac{1}{4} (∑_{i∈F} |Φ_{i} − ϕ_{i}|^{2})^{\frac{1}{2}} (∑_{i∈F} |Γ_{i} − γ_{i}|^{2})^{\frac{1}{2}} .

The constant \frac{1}{4} is the best possible.

The proof follows by Theorem 31 and we omit the details.

Remark 31. Similar results may be stated if we apply the other inequalities obtained above. We omit the details.

In the case of real spaces, the following corollaries provide much simpler sufficient conditions for the reverse of Bessel’s inequality (2.22) or for the Grüss type inequality (2.23) to hold.

Corollary 24. Let {f_{i}}_{i∈I} be an orthornormal family of functions in the real Hilbert space L_{ρ}^{2} (Ω) , F a finite part of I, M_{i}, m_{i} ∈ ℝ (i ∈ F) and f ∈ L_{ρ}^{2} (Ω) such that

∑_{i∈F} m_{i}f_{i} (s) ≤ f (s) ≤ ∑_{i∈F} M_{i}f_{i} (s) for μ − a.e. s ∈ Ω.

Then we have the inequalities

0 ≤ ∫_{Ω} ρ (s) f^{2} (s) dμ (s) − ∑_{i∈F} [∫_{Ω} ρ (s) f (s) f_{i} (s) dμ (s)]^{2}

≤ \frac{1}{4} ∑_{i∈F} (M_{i} − m_{i})^{2}

− ∫_{Ω} ρ (s) (∑_{i∈F} M_{i}f_{i} (s) − f (s)) (f (s) − ∑_{i∈F} m_{i}f_{i} (s)) dμ (s)

≤ \frac{1}{4} ∑_{i∈F} (M_{i} − m_{i})^{2} .

The constant \frac{1}{4} is best possible.

Corollary 25. Let {f_{i}}_{i∈I} and F be as in Corollary 24. If M_{i}, m_{i}, N_{i}, n_{i} ∈ ℝ (i ∈ F) and f, g ∈ L_{ρ}^{2} (Ω) are such that

∑_{i∈F} m_{i}f_{i} (s) ≤ f (s) ≤ ∑_{i∈F} M_{i}f_{i} (s)

and

∑_{i∈F} n_{i}f_{i} (s) ≤ g (s) ≤ ∑_{i∈F} N_{i}f_{i} (s) , for μ − a.e. s ∈ Ω,

then we have the inequalities

|∫_{Ω} ρ (s) f (s) g (s) dμ (s)

− ∑_{i∈F} ∫_{Ω} ρ (s) f (s) f_{i} (s) dμ (s) ∫_{Ω} ρ (s) g (s) f_{i} (s) dμ (s)|

84 3. REVERSES OF BESSEL’S INEQUALITY

≤ \frac{1}{4} (∑_{i∈F} (M_{i} − m_{i})^{2})^{\frac{1}{2}} (∑_{i∈F} (N_{i} − n_{i})^{2})^{\frac{1}{2}}

− [∫_{Ω} ρ (s) (∑_{i∈F} M_{i}f_{i} (s) − f (s)) (f (s) − ∑_{i∈F} m_{i}f_{i} (s)) dμ (s)]^{\frac{1}{2}}

× [∫_{Ω} ρ (s) (∑_{i∈F} N_{i}f_{i} (s) − g (s)) (g (s) − ∑_{i∈F} n_{i}f_{i} (s)) dμ (s)]^{\frac{1}{2}}

≤ \frac{1}{4} (∑_{i∈F} (M_{i} − m_{i})^{2})^{\frac{1}{2}} (∑_{i∈F} (N_{i} − n_{i})^{2})^{\frac{1}{2}} .

3. Another Reverse for Bessel’s Inequality 3.1. A General Result. The following lemma holds [6]. Lemma 5. Let {e_{i}}_{i∈I} be a family of orthornormal vectors in H, F

a finite part of I, λ_{i} ∈ 𝕂, i ∈ F, r > 0 and x ∈ H. If

∥x − ∑_{i∈F} λ_{i}e_{i}∥ ≤ r,

then we have the inequality

(3.1)  0 ≤ ∥x∥^{2} − ∑_{i∈F} |⟨x, e_{i}⟩|^{2} ≤ r^{2} − ∑_{i∈F} |λ_{i} − ⟨x, e_{i}⟩|^{2} .

Proof. Consider

I_{1} := ∥x − ∑_{i∈F} λ_{i}e_{i}∥^{2} = ⟨x − ∑_{i∈F} λ_{i}e_{i}, x − ∑_{j∈F} λ_{j}e_{j}⟩

= ∥x∥^{2} − ∑_{i∈F} λ_{i}\overline{⟨x, e_{i}⟩} − ∑_{i∈F} \overline{λ_{i}} ⟨x, e_{i}⟩ + ∑_{i∈F} ∑_{j∈F} λ_{i}\overline{λ_{j}} ⟨e_{i}, e_{j}⟩

= ∥x∥^{2} − ∑_{i∈F} λ_{i}\overline{⟨x, e_{i}⟩} − ∑_{i∈F} \overline{λ_{i}} ⟨x, e_{i}⟩ + ∑_{i∈F} |λ_{i}|^{2}

and

I_{2} := ∑_{i∈F} |λ_{i} − ⟨x, e_{i}⟩|^{2} = ∑_{i∈F} (λ_{i} − ⟨x, e_{i}⟩) (\overline{λ_{i}} − \overline{⟨x, e_{i}⟩})

= ∑_{i∈F} [|λ_{i}|^{2} + |⟨x, e_{i}⟩|^{2} − \overline{λ_{i}} ⟨x, e_{i}⟩ − λ_{i}\overline{⟨x, e_{i}⟩}]

= ∑_{i∈F} |λ_{i}|^{2} + ∑_{i∈F} |⟨x, e_{i}⟩|^{2} − ∑_{i∈F} \overline{λ_{i}} ⟨x, e_{i}⟩ − ∑_{i∈F} λ_{i}\overline{⟨x, e_{i}⟩}.

If we subtract I_{2} from I_{1} we deduce the following identity that is interesting in its own right

∥x − ∑_{i∈F} λ_{i}e_{i}∥^{2} − ∑_{i∈F} |λ_{i} − ⟨x, e_{i}⟩|^{2} = ∥x∥^{2} − ∑_{i∈F} |⟨x, e_{i}⟩|^{2} ,

from which we easily deduce (3.1).

The following reverse of Bessel’s inequality holds [6].

Theorem 33. Let {e_{i}}_{i∈I} be a family of orthornormal vectors in H, F a finite part of I, ϕ_{i}, Φ_{i}, i ∈ I real or complex numbers. For x ∈ H, if either

(i) Re ⟨∑_{i∈F} Φ_{i}e_{i} − x, x − ∑_{i∈F} ϕ_{i}e_{i}⟩ ≥ 0;

or, equivalently, (ii) ∥x − ∑_{i∈F} \frac{ϕ_{i}+Φ_{i}}{2} e_{i}∥ ≤ \frac{1}{2} (∑_{i∈F} |Φ_{i} − ϕ_{i}|^{2})^{\frac{1}{2}} ;

holds, then the following reverse of Bessel’s inequality

(3.2)  0 ≤ ∥x∥^{2} − ∑_{i∈F} |⟨x, e_{i}⟩|^{2}

≤ \frac{1}{4} ∑_{i∈F} |Φ_{i} − ϕ_{i}|^{2} − ∑_{i∈F} |\frac{ϕ_{i} + Φ_{i}}{2} − ⟨x, e_{i}⟩|^{2}

≤ \frac{1}{4} ∑_{i∈F} |Φ_{i} − ϕ_{i}|^{2} ,

is valid.

The constant \frac{1}{4} is best possible in both inequalities.

Proof. If we apply Lemma 5 for λ_{i} = \frac{ϕ_{i}+Φ_{i}}{2} and

r := \frac{1}{2} (∑_{i∈F} |Φ_{i} − ϕ_{i}|^{2})^{\frac{1}{2}} ,

we deduce the first inequality in (3.2).

Let us prove that \frac{1}{4} is best possible in the second inequality in (3.2). Assume that there is a c > 0 such that

(3.3)  0 ≤ ∥x∥^{2} − ∑_{i∈F} |⟨x, e_{i}⟩|^{2}

≤ c ∑_{i∈F} |Φ_{i} − ϕ_{i}|^{2} − ∑_{i∈F} |\frac{ϕ_{i} + Φ_{i}}{2} − ⟨x, e_{i}⟩|^{2} ,

provided that ϕ_{i}, Φ_{i}, x and F satisfy (i) and (ii).

We choose F = \{1\} , e_{1} = e = (\frac{1}{\sqrt{2}}, \frac{1}{\sqrt{2}}) ∈ ℝ^{2}, x = (x_{1}, x_{2}) ∈ ℝ^{2}, Φ_{1} = Φ = m > 0, ϕ_{1} = ϕ = −m, H = ℝ^{2} to get from (3.3) that

(3.4)  0 ≤ x_{1}^{2} + x_{2}^{2} − \frac{(x_{1} + x_{2})^{2}}{2}

≤ 4cm^{2} − \frac{(x_{1} + x_{2})^{2}}{2} ,

provided

(3.5) 0 ≤ ⟨me − x, x + me⟩

= (\frac{m}{\sqrt{2}} − x_{1}) (x_{1} + \frac{m}{\sqrt{2}}) + (\frac{m}{\sqrt{2}} − x_{2}) (x_{2} + \frac{m}{\sqrt{2}}) .

From (3.4) we get

(3.6)  x_{1}^{2} + x_{2}^{2} ≤ 4cm^{2}

provided (3.5) holds.

If we choose x_{1} = \frac{m}{\sqrt{2}}, x_{2} = −\frac{m}{\sqrt{2}}, then (3.5) is fulfilled and by (3.6) we get m^{2} ≤ 4cm^{2}, giving c ≥ \frac{1}{4}.

Remark 32. If F = {1} , e_{1} = 1, ∥e∥ = 1 and for ϕ, Φ ∈ 𝕂 and x ∈ H one has either

Re ⟨Φe − x, x − ϕe⟩ ≥ 0,

or, equivalently,

∥x − \frac{ϕ + Φ}{2}e∥ ≤ \frac{1}{2} |Φ − ϕ| ,

then

0 ≤ ∥x∥^{2} − |⟨x, e⟩|^{2}

≤ \frac{1}{4} |Φ − ϕ|^{2} − |\frac{ϕ + Φ}{2} − ⟨x, e⟩|^{2} ≤ \frac{1}{4} |Φ − ϕ|^{2} .

The constant \frac{1}{4} is best possible in both inequalities.

Remark 33. It is important to compare the bounds provided by Theorem 30 and Theorem 33.

For this purpose, consider

B_{1} (x, e, ϕ, Φ) := \frac{1}{4} (Φ − ϕ)^{2} − ⟨Φe − x, x − ϕe⟩

and

B_{2} (x, e, ϕ, Φ) := \frac{1}{4} (Φ − ϕ)^{2} − (\frac{ϕ + Φ}{2} − ⟨x, e⟩)^{2} ,

where H is a real inner product, e ∈ H, ∥e∥ = 1, x ∈ H, ϕ, Φ ∈ ℝ with

⟨Φe − x, x − ϕe⟩ ≥ 0,

or, equivalently,

∥x − \frac{ϕ + Φ}{2}e∥ ≤ \frac{1}{2} |Φ − ϕ| .

If we choose ϕ = −1, Φ = 1, then we have

B_{1} (x, e) = 1 − ⟨e − x, x + e⟩ = 1 − (∥e∥^{2} − ∥x∥^{2}) = ∥x∥^{2} ,
B_{2} (x, e) = 1 − ⟨x, e⟩^{2} ,

provided ∥x∥ ≤ 1.

Consider H = ℝ^{2}, ⟨x, y⟩ = x_{1}y_{1} + x_{2}y_{2}, x = (x_{1}, x_{2}) , y = (y_{1}, y_{2}) ∈ ℝ^{2} and e = (\frac{\sqrt{2}}{2}, \frac{\sqrt{2}}{2}) . Then ∥e∥ = 1 and we must compare

B_{1} (x) = x_{1}^{2} + x_{2}^{2}

with

B_{2} (x) = 1 − \frac{(x_{1} + x_{2})^{2}}{2} ,

provided x_{1}^{2} + x_{2}^{2} ≤ 1.

If we choose x_{0} = (1, 0) , then ∥x_{0}∥ = 1 and B_{1} (x_{0}) = 1, B_{2} (x_{0}) = \frac{1}{2} showing that B_{1} > B_{2}. If we choose x_{00} = (−\frac{1}{2}, \frac{1}{2}) , then B_{1} (x_{00}) = \frac{1}{2}, B_{2} (x_{00}) = 1, showing that B_{1} < B_{2}.

We may state the following proposition.

Proposition 27. Let {e_{i}}_{i∈I} be a family of orthornormal vectors in H, F a finite part of I, ϕ_{i}, Φ_{i} ∈ 𝕂 (i ∈ F). If x ∈ H either satisfies (i), or, equivalently, (ii) of Theorem 33, then the upper bounds

B_{1} (x, e, ϕ, Φ, F) := \frac{1}{4} ∑_{i∈F} |Φ_{i} − ϕ_{i}|^{2} − Re ⟨∑_{i∈F} Φ_{i}e_{i} − x, x − ∑_{i∈F} ϕ_{i}e_{i}⟩ ,

B_{2} (x, e, ϕ, Φ, F) := \frac{1}{4} ∑_{i∈F} |Φ_{i} − ϕ_{i}|^{2} − ∑_{i∈F} |\frac{ϕ_{i} + Φ_{i}}{2} − ⟨x, e_{i}⟩|^{2} ,

for the Bessel’s difference B_{s} (x, e, F) := ∥x∥^{2} −∑_{i∈F} |⟨x, e_{i}⟩|^{2} , cannot be compared in general.

3.2. A Refinement of the Grüss Inequality for Orthonormal Families. The following result holds [6].

Theorem 34. Let {e_{i}}_{i∈I} be a family of orthornormal vectors in H, F a finite part of I, ϕ_{i}, Φ_{i}, γ_{i}, Γ_{i} ∈ 𝕂, i ∈ F and x, y ∈ H. If either

Re ⟨∑_{i∈F} Φ_{i}e_{i} − x, x − ∑_{i∈F} ϕ_{i}e_{i}⟩ ≥ 0,

Re ⟨∑_{i∈F} Γ_{i}e_{i} − y, y − ∑_{i∈F} γ_{i}e_{i}⟩ ≥ 0,

or, equivalently,

∥x − ∑_{i∈F} \frac{Φ_{i} + ϕ_{i}}{2} e_{i}∥ ≤ \frac{1}{2} (∑_{i∈F} |Φ_{i} − ϕ_{i}|^{2})^{\frac{1}{2}} ,

∥y − ∑_{i∈F} \frac{Γ_{i} + γ_{i}}{2} e_{i}∥ ≤ \frac{1}{2} (∑_{i∈F} |Γ_{i} − γ_{i}|^{2})^{\frac{1}{2}} ,

hold, then we have the inequalities

(3.7) 0 ≤ |⟨x, y⟩ − ∑_{i∈F} ⟨x, e_{i}⟩ ⟨e_{i}, y⟩|

≤ \frac{1}{4} (∑_{i∈F} |Φ_{i} − ϕ_{i}|^{2})^{\frac{1}{2}} · (∑_{i∈F} |Γ_{i} − γ_{i}|^{2})^{\frac{1}{2}}

− ∑_{i∈F} |\frac{Φ_{i} + ϕ_{i}}{2} − ⟨x, e_{i}⟩| |\frac{Γ_{i} + γ_{i}}{2} − ⟨y, e_{i}⟩|

≤ \frac{1}{4} (∑_{i∈F} |Φ_{i} − ϕ_{i}|^{2})^{\frac{1}{2}} · (∑_{i∈F} |Γ_{i} − γ_{i}|^{2})^{\frac{1}{2}} .

Proof. Using Schwarz’s inequality in the inner product space (H, ⟨·, ·⟩) one has

(3.8) |⟨x − ∑_{i∈F} ⟨x, e_{i}⟩ e_{i}, y − ∑_{i∈F} ⟨y, e_{i}⟩ e_{i}⟩|^{2}

≤ ∥x − ∑_{i∈F} ⟨x, e_{i}⟩ e_{i}∥^{2} ∥y − ∑_{i∈F} ⟨y, e_{i}⟩ e_{i}∥^{2}

and since a simple calculation shows that

⟨x − ∑_{i∈F} ⟨x, e_{i}⟩ e_{i}, y − ∑_{i∈F} ⟨y, e_{i}⟩ e_{i}⟩ = ⟨x, y⟩ − ∑_{i∈F} ⟨x, e_{i}⟩ ⟨e_{i}, y⟩

and

∥x − ∑_{i∈F} ⟨x, e_{i}⟩ e_{i}∥^{2} = ∥x∥^{2} − ∑_{i∈F} |⟨x, e_{i}⟩|^{2}

for any x, y ∈ H, then by (3.8) and by the reverse of Bessel’s inequality in Theorem 33, we have

|⟨x, y⟩ − ∑_{i∈F} ⟨x, e_{i}⟩ ⟨e_{i}, y⟩|^{2} (3.9)

≤ (∥x∥^{2} − ∑_{i∈F} |⟨x, e_{i}⟩|^{2}) (∥y∥^{2} − ∑_{i∈F} |⟨y, e_{i}⟩|^{2})

≤ [\frac{1}{4} ∑_{i∈F} |Φ_{i} − ϕ_{i}|^{2} − ∑_{i∈F} |\frac{Φ_{i} + ϕ_{i}}{2} − ⟨x, e_{i}⟩|^{2}]

× [\frac{1}{4} ∑_{i∈F} |Γ_{i} − γ_{i}|^{2} − ∑_{i∈F} |\frac{Γ_{i} + γ_{i}}{2} − ⟨y, e_{i}⟩|^{2}]

:= K.

Using Aczél’s inequality for real numbers, i.e., we recall that

(3.10)  (a^{2} − ∑_{i∈F} a_{i}^{2}) (b^{2} − ∑_{i∈F} b_{i}^{2}) ≤ (ab − ∑_{i∈F} a_{i}b_{i})^{2} ,

provided that a, b, a_{i}, b_{i} > 0, i ∈ F, we may state that

(3.11) K ≤ [\frac{1}{4} (∑_{i∈F} |Φ_{i} − ϕ_{i}|^{2})^{\frac{1}{2}} · (∑_{i∈F} |Γ_{i} − γ_{i}|^{2})^{\frac{1}{2}}

− ∑_{i∈F} |\frac{Φ_{i} + ϕ_{i}}{2} − ⟨x, e_{i}⟩| |\frac{Γ_{i} + γ_{i}}{2} − ⟨y, e_{i}⟩|]^{2} .

Using (3.9) and (3.11) we conclude that

(3.12) |⟨x, y⟩ − ∑_{i∈F} ⟨x, e_{i}⟩ ⟨e_{i}, y⟩|^{2}

≤ [\frac{1}{4} (∑_{i∈F} |Φ_{i} − ϕ_{i}|^{2})^{\frac{1}{2}} · (∑_{i∈F} |Γ_{i} − γ_{i}|^{2})^{\frac{1}{2}}

− ∑_{i∈F} |\frac{Φ_{i} + ϕ_{i}}{2} − ⟨x, e_{i}⟩| |\frac{Γ_{i} + γ_{i}}{2} − ⟨y, e_{i}⟩|]^{2} .

Taking the square root in (3.12) and taking into account that the quantity in the last square bracket is nonnegative (this follows by (3.2) and by the Cauchy-Bunyakovsky-Schwarz inequality), we deduce the second inequality in (3.7).

The fact that \frac{1}{4} is the best possible constant follows by Theorem 33 and we omit the details.

The following corollary may be stated [6].

Corollary 26. Let e ∈ H, ∥e∥ = 1, ϕ, Φ, γ, Γ ∈ 𝕂 and x, y ∈ H are such that either

Re ⟨Φe − x, x − ϕe⟩ ≥ 0 and Re ⟨Γe − y, y − γe⟩ ≥ 0,

or, equivalently,

∥x − \frac{ϕ + Φ}{2}e∥ ≤ \frac{1}{2} |Φ − ϕ| , ∥y − \frac{γ + Γ}{2}e∥ ≤ \frac{1}{2} |Γ − γ| ,

hold. Then we have the following refinement of Grüss’ inequality

0 ≤ |⟨x, y⟩ − ⟨x, e⟩ ⟨e, y⟩|

≤ \frac{1}{4} |Φ − ϕ| |Γ − γ| − |\frac{ϕ + Φ}{2} − ⟨x, e⟩| |\frac{γ + Γ}{2} − ⟨y, e⟩|

≤ \frac{1}{4} |Φ − ϕ| |Γ − γ| .

The constant \frac{1}{4} is best possible in both inequalities.

3.3. Some Companion Inequalities. The following companion of the Grüss inequality also holds [6].

Theorem 35. Let {e_{i}}_{i∈I} be a family of orthornormal vectors in H, F a finite part of I and ϕ_{i}, Φ_{i} ∈ 𝕂, i ∈ F, x, y ∈ H and λ ∈ (0, 1) , such that either

(3.13)  Re ⟨∑_{i∈F} Φ_{i}e_{i} − (λx + (1 − λ) y) ,

λx + (1 − λ) y − ∑_{i∈F} ϕ_{i}e_{i}⟩ ≥ 0

or, equivalently,

∥λx + (1 − λ) y − ∑_{i∈F} \frac{Φ_{i} + ϕ_{i}}{2} · e_{i}∥ ≤ \frac{1}{2} (∑_{i∈F} |Φ_{i} − ϕ_{i}|^{2})^{\frac{1}{2}} ,

holds. Then we have the inequality

Re [⟨x, y⟩ − ∑_{i∈F} ⟨x, e_{i}⟩ ⟨e_{i}, y⟩] (3.14)

≤ \frac{1}{16} · \frac{1}{λ (1 − λ)} ∑_{i∈F} |Φ_{i} − ϕ_{i}|^{2}

− \frac{1}{4} \frac{1}{λ (1 − λ)} ∑_{i∈F} |\frac{Φ_{i} + ϕ_{i}}{2} − ⟨λx + (1 − λ) y, e_{i}⟩|^{2}

≤ \frac{1}{16} · \frac{1}{λ (1 − λ)} ∑_{i∈F} |Φ_{i} − ϕ_{i}|^{2} .

The constant \frac{1}{16} is the best possible constant in (3.14) in the sense that it cannot be replaced by a smaller constant.

Proof. We know that for any z, u ∈ H, one has

Re ⟨z, u⟩ ≤ \frac{1}{4} ∥z + u∥^{2} .

Then for any a, b ∈ H and λ ∈ (0, 1) one has

(3.15)  Re ⟨a, b⟩ ≤ \frac{1}{4λ (1 − λ)} ∥λa + (1 − λ) b∥^{2} .

Since

⟨x, y⟩ − ∑_{i∈F} ⟨x, e_{i}⟩ ⟨e_{i}, y⟩ = ⟨x − ∑_{i∈F} ⟨x, e_{i}⟩ e_{i}, y − ∑_{i∈F} ⟨y, e_{i}⟩ e_{i}⟩ ,

for any x, y ∈ H, then, by (3.15), we get

Re [⟨x, y⟩ − ∑_{i∈F} ⟨x, e_{i}⟩ ⟨e_{i}, y⟩] (3.16)

= Re [⟨x − ∑_{i∈F} ⟨x, e_{i}⟩ e_{i}, y − ∑_{i∈F} ⟨y, e_{i}⟩ e_{i}⟩]

≤ \frac{1}{4λ (1 − λ)} ∥λ (x − ∑_{i∈F} ⟨x, e_{i}⟩ e_{i}) + (1 − λ) (y − ∑_{i∈F} ⟨y, e_{i}⟩ e_{i})∥^{2}

= \frac{1}{4λ (1 − λ)} ∥λx + (1 − λ) y − ∑_{i∈F} ⟨λx + (1 − λ) y, e_{i}⟩ e_{i}∥^{2}

= \frac{1}{4λ (1 − λ)} [∥λx + (1 − λ) y∥^{2} − ∑_{i∈F} |⟨λx + (1 − λ) y, e_{i}⟩|^{2}] .

If we apply the reverse of Bessel’s inequality in Theorem 33 for λx + (1 − λ) y, we may state that

∥λx + (1 − λ) y∥^{2} − ∑_{i∈F} |⟨λx + (1 − λ) y, e_{i}⟩|^{2} (3.17)

≤ \frac{1}{4} ∑_{i∈F} |Φ_{i} − ϕ_{i}|^{2} − ∑_{i∈F} |\frac{Φ_{i} + ϕ_{i}}{2} − ⟨λx + (1 − λ) y, e_{i}⟩|^{2}

≤ \frac{1}{4} ∑_{i∈F} |Φ_{i} − ϕ_{i}|^{2} .

Now, by making use of (3.16) and (3.17), we deduce (3.14).

The fact that \frac{1}{16} is the best possible constant in (3.14) follows by the fact that, if in (3.13) we choose x = y, then it becomes (i) of Theorem 33, implying for λ = \frac{1}{2} (3.2), for which, we have shown that \frac{1}{4} was the best constant.

Remark 34. In practical applications we may use only the inequality between the first and the last terms in (3.14).

3. ANOTHER REVERSE FOR BESSEL’S INEQUALITY 93

Remark 35. If in Theorem 35, we choose λ = \frac{1}{2}, then we get

Re [⟨x, y⟩ − ∑_{i∈F} ⟨x, e_{i}⟩ ⟨e_{i}, y⟩]

≤ \frac{1}{4} ∑_{i∈F} |Φ_{i} − ϕ_{i}|^{2} − ∑_{i∈F} |\frac{Φ_{i} + ϕ_{i}}{2} − ⟨\frac{x + y}{2} , e_{i}⟩|^{2}

≤ \frac{1}{4} ∑_{i∈F} |Φ_{i} − ϕ_{i}|^{2} ,

provided

Re ⟨∑_{i∈F} Φ_{i}e_{i} − \frac{x + y}{2} , \frac{x + y}{2} − ∑_{i∈F} ϕ_{i}e_{i}⟩ ≥ 0

or, equivalently,

∥\frac{x + y}{2} − ∑_{i∈F} \frac{Φ_{i} + ϕ_{i}}{2} · e_{i}∥ ≤ \frac{1}{2} (∑_{i∈F} |Φ_{i} − ϕ_{i}|^{2})^{\frac{1}{2}} .

Corollary 27. With the assumptions of Theorem 35 and if

Re ⟨∑_{i∈F} Φ_{i}e_{i} − (λx ± (1 − λ) y) , λx ± (1 − λ) y − ∑_{i∈F} ϕ_{i}e_{i}⟩ ≥ 0

or, equivalently,

∥λx ± (1 − λ) y − ∑_{i∈F} \frac{Φ_{i} + ϕ_{i}}{2} · e_{i}∥ ≤ \frac{1}{2} (∑_{i∈F} |Φ_{i} − ϕ_{i}|^{2})^{\frac{1}{2}} ,

then we have the inequality

(3.18) |Re [⟨x, y⟩ − ∑_{i∈F} ⟨x, e_{i}⟩ ⟨e_{i}, y⟩]| ≤ \frac{1}{16} · \frac{1}{λ (1 − λ)} ∑_{i∈F} |Φ_{i} − ϕ_{i}|^{2} .

The constant \frac{1}{16} is best possible in (3.18).

Remark 36. If H is a real inner product space and m_{i}, M_{i} ∈ ℝ with the property

⟨∑_{i∈F} M_{i}e_{i} − (λx ± (1 − λ) y) , λx ± (1 − λ) y − ∑_{i∈F} m_{i}e_{i}⟩ ≥ 0

or, equivalently,

∥λx ± (1 − λ) y − ∑_{i∈F} \frac{M_{i} + m_{i}}{2} · e_{i}∥ ≤ \frac{1}{2} [∑_{i∈F} (M_{i} − m_{i})^{2}]^{\frac{1}{2}} ,

then we have the Grüss type inequality

|⟨x, y⟩ − ∑_{i∈F} ⟨x, e_{i}⟩ ⟨e_{i}, y⟩| ≤ \frac{1}{16} · \frac{1}{λ (1 − λ)} ∑_{i∈F} (M_{i} − m_{i})^{2} .

3.4. Integral Inequalities. The following proposition holds [6]. Proposition 28. Let {f_{i}}_{i∈I} be an orthornormal family of func-

tions in L_{ρ}^{2} (Ω, 𝕂) , F a finite subset of I, ϕ_{i}, Φ_{i} ∈ 𝕂 (i ∈ F) and f ∈ L_{ρ}^{2} (Ω, 𝕂) , so that either

∫_{Ω} ρ (s) Re [(∑_{i∈F} Φ_{i}f_{i} (s) − f (s)) (\overline{f} (s) − ∑_{i∈F} \overline{ϕ_{i}} \overline{f_{i}} (s))] dμ (s) ≥ 0

or, equivalently,

∫_{Ω} ρ (s) |f (s) − ∑_{i∈F} \frac{Φ_{i} + ϕ_{i}}{2} · f_{i} (s)|^{2} dμ (s) ≤ \frac{1}{4} ∑_{i∈F} |Φ_{i} − ϕ_{i}|^{2} .

Then we have the inequality

(3.19) 0 ≤ ∫_{Ω} ρ (s) |f (s)|^{2} dμ (s) − ∑_{i∈F} |∫_{Ω} ρ (s) f (s) \overline{f_{i}} (s) dμ (s)|^{2}

≤ \frac{1}{4} ∑_{i∈F} |Φ_{i} − ϕ_{i}|^{2}

− ∑_{i∈F} |\frac{Φ_{i} + ϕ_{i}}{2} − ∫_{Ω} ρ (s) f (s) \overline{f_{i}} (s) dμ (s)|^{2}

≤ \frac{1}{4} ∑_{i∈F} |Φ_{i} − ϕ_{i}|^{2} .

The constant \frac{1}{4} is best possible in both inequalities.

The proof follows by Theorem 33 applied for the Hilbert space L_{ρ}^{2} (Ω, 𝕂) and the orthornormal family {f_{i}}_{i∈I} .

The following Grüss type inequality also holds [6].

Proposition 29. Let {f_{i}}_{i∈I} and F be as in Proposition 28. If ϕ_{i}, Φ_{i}, γ_{i}, Γ_{i} ∈ 𝕂 (i ∈ F) and f, g ∈ L_{ρ}^{2} (Ω, 𝕂) so that either

∫_{Ω} ρ (s) Re [(∑_{i∈F} Φ_{i}f_{i} (s) − f (s)) (\overline{f} (s) − ∑_{i∈F} \overline{ϕ_{i}} \overline{f_{i}} (s))] dμ (s) ≥ 0,
∫_{Ω} ρ (s) Re [(∑_{i∈F} Γ_{i}f_{i} (s) − g (s)) (\overline{g} (s) − ∑_{i∈F} \overline{γ_{i}} \overline{f_{i}} (s))] dμ (s) ≥ 0,

or, equivalently,

∫_{Ω} ρ (s) |f (s) − ∑_{i∈F} \frac{Φ_{i} + ϕ_{i}}{2} f_{i} (s)|^{2} dμ (s) ≤ \frac{1}{4} ∑_{i∈F} |Φ_{i} − ϕ_{i}|^{2} ,
∫_{Ω} ρ (s) |g (s) − ∑_{i∈F} \frac{Γ_{i} + γ_{i}}{2} f_{i} (s)|^{2} dμ (s) ≤ \frac{1}{4} ∑_{i∈F} |Γ_{i} − γ_{i}|^{2} ,

hold, then we have the inequalities

|∫_{Ω} ρ (s) f (s) \overline{g (s)}dμ (s) (3.20)

− ∑_{i∈F} ∫_{Ω} ρ (s) f (s) \overline{f_{i}} (s) dμ (s) ∫_{Ω} ρ (s) f_{i} (s) \overline{g (s)}dμ (s)|

≤ \frac{1}{4} (∑_{i∈F} |Φ_{i} − ϕ_{i}|^{2})^{\frac{1}{2}} (∑_{i∈F} |Γ_{i} − γ_{i}|^{2})^{\frac{1}{2}}

− ∑_{i∈F} |\frac{Φ_{i} + ϕ_{i}}{2} − ∫_{Ω} ρ (s) f (s) \overline{f_{i}} (s) dμ (s)|

× |\frac{Γ_{i} + γ_{i}}{2} − ∫_{Ω} ρ (s) g (s) \overline{f_{i}} (s) dμ (s)|

≤ \frac{1}{4} (∑_{i∈F} |Φ_{i} − ϕ_{i}|^{2})^{\frac{1}{2}} (∑_{i∈F} |Γ_{i} − γ_{i}|^{2})^{\frac{1}{2}} .

The constant \frac{1}{4} is the best possible.

The proof follows by Theorem 34 and we omit the details.

Remark 37. Similar results may be stated if one applies the inequalities in the above subsections. We omit the details.

In the case of real spaces, the following corollaries provide much simpler sufficient conditions for the reverse of Bessel’s inequality (3.19) or for the Grüss type inequality (3.20) to hold.

Corollary 28. Let {f_{i}}_{i∈I} be an orthornormal family of functions in the real Hilbert space L_{ρ}^{2} (Ω) , F a finite part of I, M_{i}, m_{i} ∈ ℝ (i ∈ F) and f ∈ L_{ρ}^{2} (Ω) so that

∑_{i∈F} m_{i}f_{i} (s) ≤ f (s) ≤ ∑_{i∈F} M_{i}f_{i} (s) for μ − a.e. s ∈ Ω.

Then we have the inequalities

0 ≤ ∫_{Ω} ρ (s) f^{2} (s) dμ (s) − ∑_{i∈F} [∫_{Ω} ρ (s) f (s) f_{i} (s) dμ (s)]^{2}

≤ \frac{1}{4} ∑_{i∈F} (M_{i} − m_{i})^{2} − ∑_{i∈F} [\frac{M_{i} + m_{i}}{2} − ∫_{Ω} ρ (s) f (s) f_{i} (s) dμ (s)]^{2}

≤ \frac{1}{4} ∑_{i∈F} (M_{i} − m_{i})^{2} .

The constant \frac{1}{4} is best possible.

Corollary 29. Let {f_{i}}_{i∈I} and F be as in Corollary 28. If M_{i}, m_{i}, N_{i}, n_{i} ∈ ℝ (i ∈ F) and f, g ∈ L_{ρ}^{2} (Ω) are such that

∑_{i∈F} m_{i}f_{i} (s) ≤ f (s) ≤ ∑_{i∈F} M_{i}f_{i} (s)

and

∑_{i∈F} n_{i}f_{i} (s) ≤ g (s) ≤ ∑_{i∈F} N_{i}f_{i} (s) for μ − a.e. s ∈ Ω,

hold, then we have the inequalities

|∫_{Ω} ρ (s) f (s) g (s) dμ (s)

− ∑_{i∈F} ∫_{Ω} ρ (s) f (s) f_{i} (s) dμ (s) ∫_{Ω} ρ (s) g (s) f_{i} (s) dμ (s)|

≤ \frac{1}{4} (∑_{i∈F} (M_{i} − m_{i})^{2})^{\frac{1}{2}} (∑_{i∈F} (N_{i} − n_{i})^{2})^{\frac{1}{2}}

− ∑_{i∈F} |\frac{M_{i} + m_{i}}{2} − ∫_{Ω} ρ (s) f (s) f_{i} (s) dμ (s)|

× |\frac{N_{i} + n_{i}}{2} − ∫_{Ω} ρ (s) g (s) f_{i} (s) dμ (s)|

≤ \frac{1}{4} (∑_{i∈F} (M_{i} − m_{i})^{2})^{\frac{1}{2}} (∑_{i∈F} (N_{i} − n_{i})^{2})^{\frac{1}{2}} .

4. More Reverses of Bessel’s Inequality

4.1. A General Result. The following reverse of Bessel’s inequality holds [9].

Theorem 36. Let {e_{i}}_{i∈I} be a family of orthornormal vectors in H, F a finite part of I, and ϕ_{i}, Φ_{i} (i ∈ F) , real or complex numbers such that ∑_{i∈F} Re (Φ_{i}\overline{ϕ_{i}}) > 0. If x ∈ H is such that either

(i) Re ⟨∑_{i∈F} Φ_{i}e_{i} − x, x − ∑_{i∈F} ϕ_{i}e_{i}⟩ ≥ 0; or, equivalently,

(ii) ∥x − ∑_{i∈F} \frac{ϕ_{i}+Φ_{i}}{2} e_{i}∥ ≤ \frac{1}{2} (∑_{i∈F} |Φ_{i} − ϕ_{i}|^{2})^{\frac{1}{2}} ;

holds, then one has the inequality

(4.1)  ∥x∥^{2} ≤ \frac{1}{4} · \begin{matrix}∑ & i∈F & (|Φ_{i}| + |ϕ_{i}|) & 2 \\ ∑ & i∈F & Re & (Φ_{i}\overline{ϕ_{i}}) \end{matrix} ∑_{i∈F} |⟨x, e_{i}⟩|^{2} .

The constant \frac{1}{4} is best possible in the sense that it cannot be replaced by a smaller constant.

Proof. Observe that

Re ⟨∑_{i∈F} Φ_{i}e_{i} − x, x − ∑_{i∈F} ϕ_{i}e_{i}⟩

= ∑_{i∈F} Re [Φ_{i}\overline{⟨x, e_{i}⟩} + \overline{ϕ_{i}} ⟨x, e_{i}⟩] − ∥x∥^{2} − ∑_{i∈F} Re (Φ_{i}\overline{ϕ_{i}}) ,

giving, from (i), that

(4.2)  ∥x∥^{2} + ∑_{i∈F} Re (Φ_{i}\overline{ϕ_{i}}) ≤ ∑_{i∈F} Re [Φ_{i}\overline{⟨x, e_{i}⟩} + \overline{ϕ_{i}} ⟨x, e_{i}⟩] .

On the other hand, by the elementary inequality

αp^{2} + \frac{1}{α}q^{2} ≥ 2pq, α > 0, p, q ≥ 0;

we deduce

(4.3)  2 ∥x∥ ≤ \begin{matrix}∥x∥ & 2 \\ [∑_{i∈F} Re (Φ_{i}\overline{ϕ_{i}})] & \frac{1}{2} \end{matrix} + [∑_{i∈F} Re (Φ_{i}\overline{ϕ_{i}})]^{\frac{1}{2}} .

Dividing (4.2) by [∑_{i∈F} Re (Φ_{i}\overline{ϕ_{i}})]^{\frac{1}{2}} > 0 and using (4.3), we obtain

(4.4)  ∥x∥ ≤ \frac{1}{2} \begin{matrix}∑ & i∈F & Re & [Φ_{i}\overline{⟨x, e_{i}⟩} + \overline{ϕ_{i}} ⟨x, e_{i}⟩] \\ [∑_{i∈F} Re (Φ_{i}\overline{ϕ_{i}})] & \frac{1}{2} \end{matrix},

which is also an interesting inequality in itself.

Using the Cauchy-Bunyakovsky-Schwarz inequality for real numbers, we get

∑_{i∈F} Re [Φ_{i}\overline{⟨x, e_{i}⟩} + \overline{ϕ_{i}} ⟨x, e_{i}⟩] (4.5)

≤ ∑_{i∈F} |Φ_{i}\overline{⟨x, e_{i}⟩} + \overline{ϕ_{i}} ⟨x, e_{i}⟩|

≤ ∑_{i∈F} (|Φ_{i}| + |ϕ_{i}|) |⟨x, e_{i}⟩|

≤ [∑_{i∈F} (|Φ_{i}| + |ϕ_{i}|)^{2}]^{\frac{1}{2}} [∑_{i∈F} |⟨x, e_{i}⟩|^{2}]^{\frac{1}{2}} .

Making use of (4.4) and (4.5), we deduce the desired result (4.1).

To prove the sharpness of the constant \frac{1}{4}, let us assume that (4.1) holds with a constant c > 0, i.e.,

(4.6)  ∥x∥^{2} ≤ c · \begin{matrix}∑ & i∈F & (|Φ_{i}| + |ϕ_{i}|) & 2 \\ ∑ & i∈F & Re & (Φ_{i}\overline{ϕ_{i}}) \end{matrix} ∑_{i∈F} |⟨x, e_{i}⟩|^{2} ,

provided x, ϕ_{i}, Φ_{i}, i ∈ F satisfies (i).

Choose F = {1} , e_{1} = e, ∥e∥ = 1, ϕ_{i} = m, Φ_{i} = M with m, M > 0, then, by (4.6), we get

(4.7)  ∥x∥^{2} ≤ c\frac{(M + m)^{2}}{mM} |⟨x, e⟩|^{2}

provided

(4.8)  Re ⟨Me − x, x − me⟩ ≥ 0.

If x = me, then obviously (4.8) holds, and by (4.7) we get

m^{2} ≤ c\frac{(M + m)^{2}}{mM} m^{2}

giving mM ≤ c (M + m)^{2} for m, M > 0. Now, if in this inequality we choose m = 1 − ε, M = 1 + ε (ε ∈ (0, 1)) , then we get 1 − ε^{2} ≤ 4c for ε ∈ (0, 1) , from where we deduce c ≥ \frac{1}{4}.

Remark 38. By the use of (4.4), the second inequality in (4.5) and the Hölder inequality, we may state the following reverses of Bessel’s inequality as well:

∥x∥^{2} ≤ \frac{1}{2} · \frac{1}{[∑_{i∈F} Re (Φ_{i}\overline{ϕ_{i}})]^{\frac{1}{2}}}

× \{ \begin{matrix}max_{i∈F} \{|Φ_{i}| + |ϕ_{i}|\} ∑_{i∈F} |⟨x, e_{i}⟩|; \\ [∑_{i∈F} (|Φ_{i}| + |ϕ_{i}|)^{p}]^{\frac{1}{p}} (∑_{i∈F} |⟨x, e_{i}⟩|^{q})^{\frac{1}{q}} , \\ for p > 1, \frac{1}{p} + \frac{1}{q} = 1; \\ max_{i∈F} |⟨x, e_{i}⟩| ∑_{i∈F} [|Φ_{i}| + |ϕ_{i}|] . \end{matrix}

The following corollary holds [9].

Corollary 30. With the assumption of Theorem 36 and if either (i) or (ii) holds, then

(4.9)  0 ≤ ∥x∥^{2} − ∑_{i∈F} |⟨x, e_{i}⟩|^{2} ≤ \frac{1}{4}M^{2} (Φ, ϕ, F) ∑_{i∈F} |⟨x, e_{i}⟩|^{2} ,

where

M (Φ, ϕ, F) := [\begin{matrix}∑ & i∈F & \{(|Φ_{i}| − |ϕ_{i}|)^{2} + 4 [|Φ_{i}\overline{ϕ_{i}}| − Re (Φ_{i}\overline{ϕ_{i}})]\} \\ ∑ & i∈F & Re & (Φ_{i}\overline{ϕ_{i}}) \end{matrix}]^{\frac{1}{2}} .

The constant \frac{1}{4} is best possible.

Proof. The inequality (4.9) follows by (4.1) on subtracting the same quantity ∑_{i∈F} |⟨x, e_{i}⟩|^{2} from both sides.

To prove the sharpness of the constant \frac{1}{4}, assume that (4.9) holds with c > 0, i.e.,

(4.10)  0 ≤ ∥x∥^{2} − ∑_{i∈F} |⟨x, e_{i}⟩|^{2} ≤ cM^{2} (Φ, ϕ, F) ∑_{i∈F} |⟨x, e_{i}⟩|^{2}

provided the condition (i) holds.

Choose F = {1} , e_{1} = e, ∥e∥ = 1, ϕ_{i} = ϕ, Φ_{i} = Φ, ϕ, Φ > 0 in (4.10) to get

(4.11)  0 ≤ ∥x∥^{2} − |⟨x, e⟩|^{2} ≤ c\frac{(Φ − ϕ)^{2}}{ϕΦ} |⟨x, e⟩|^{2} ,

provided

(4.12)  ⟨Φe − x, x − ϕe⟩ ≥ 0.

If H = ℝ^{2}, x = (x_{1}, x_{2}) ∈ ℝ^{2}, e = (\frac{1}{\sqrt{2}}, \frac{1}{\sqrt{2}}) then we have

∥x∥^{2} − |⟨x, e⟩|^{2} = x_{1}^{2} + x_{2}^{2} − \frac{(x_{1} + x_{2})^{2}}{2} = \frac{1}{2} (x_{1} − x_{2})^{2} ,
|⟨x, e⟩|^{2} = \frac{(x_{1} + x_{2})^{2}}{2}

and by (4.11) we get

(4.13)  \frac{(x_{1} − x_{2})^{2}}{2} ≤ c\frac{(Φ − ϕ)^{2}}{ϕΦ} · \frac{(x_{1} + x_{2})^{2}}{2} .

Now, if we let x_{1} = \frac{ϕ}{\sqrt{2}}, x_{2} = \frac{Φ}{\sqrt{2}} (ϕ, Φ > 0) then obviously

⟨Φe − x, x − ϕe⟩ = ∑_{i=1}^{2} (\frac{Φ}{\sqrt{2}} − x_{i}) (x_{i} − \frac{ϕ}{\sqrt{2}}) = 0,

which shows that (4.12) is fulfilled, and thus by (4.13) we obtain

\frac{(Φ − ϕ)^{2}}{4} ≤ c\frac{(Φ − ϕ)^{2}}{ϕΦ} · \frac{(Φ + ϕ)^{2}}{4}

for any Φ > ϕ > 0. This implies

(4.14)  c (Φ + ϕ)^{2} ≥ ϕΦ

for any Φ > ϕ > 0.

Finally, let ϕ = 1 − ε, Φ = 1 + ε, ε ∈ (0, 1). Then from (4.14) we get 4c ≥ 1 − ε^{2} for any ε ∈ (0, 1) which produces c ≥ \frac{1}{4}.

Remark 39. If {e_{i}}_{i∈I} is an orthornormal family in the real inner product space (H; ⟨·, ·⟩) and M_{i}, m_{i} ∈ ℝ, i ∈ F (F is a finite part of I) and x ∈ H are such that M_{i}, m_{i} ≥ 0 for i ∈ F with ∑_{i∈F} M_{i}m_{i} ≥ 0 and

⟨∑_{i∈F} M_{i}e_{i} − x, x − ∑_{i∈F} m_{i}e_{i}⟩ ≥ 0,

4. MORE REVERSES OF BESSEL’S INEQUALITY 101

then we have the inequality

0 ≤ ∥x∥^{2} − ∑_{i∈F} [⟨x, e_{i}⟩]^{2} ≤ \frac{1}{4} · \frac{∑_{i∈F} (M_{i} − m_{i})^{2}}{∑_{i∈F} M_{i}m_{i}} · ∑_{i∈F} [⟨x, e_{i}⟩]^{2} .

The constant \frac{1}{4} is best possible.

The following reverse of the Schwarz’s inequality in inner product spaces holds.

Corollary 31. Let x, y ∈ H and δ, Δ ∈ 𝕂 (𝕂 = ℂ, ℝ) with the property that Re (Δ\overline{δ}) > 0. If either

Re ⟨Δy − x, x − δy⟩ ≥ 0,

or, equivalently,

∥x − \frac{δ + Δ}{2} · y∥ ≤ \frac{1}{2} |Δ − δ| ∥y∥ ,

holds, then we have the inequalities

(4.15)  ∥x∥ ∥y∥ ≤ \frac{1}{2} · \begin{matrix}Re & [Δ\overline{⟨x, y⟩} + \overline{δ} ⟨x, y⟩] \\ √ & \overline{Δ\overline{δ}} \end{matrix}

≤ \frac{1}{2} · \frac{|Δ| + |δ|}{\sqrt{Δ\overline{δ}}} |⟨x, y⟩| ,

(4.16) 0 ≤ ∥x∥ ∥y∥ − |⟨x, y⟩|

≤ \frac{1}{2} · \frac{(\sqrt{|Δ|} − \sqrt{|δ|})^{2} + 2 (\sqrt{Δ\overline{δ}} − \sqrt{Re (Δ\overline{δ})})}{\sqrt{Δ\overline{δ}}} |⟨x, y⟩| ,

(4.17)  ∥x∥^{2} ∥y∥^{2} ≤ \frac{1}{4} · \begin{matrix}(|Δ| + |δ|) & 2 \\ Re & (Δ\overline{δ}) \end{matrix} |⟨x, y⟩|^{2} ,

and

(4.18)  0 ≤ ∥x∥^{2} ∥y∥^{2} − |⟨x, y⟩|^{2}

≤ \frac{1}{4} · \frac{(|Δ| + |δ|)^{2} + 4 (|Δ\overline{δ}| − Re (Δ\overline{δ}))}{Re (Δ\overline{δ})} |⟨x, y⟩|^{2} .

The constants \frac{1}{2} and \frac{1}{4} are best possible.

Proof. The inequality (4.15) follows from (4.4) on choosing F = \{1\} , e_{1} = e = \frac{y}{∥y∥}, Φ_{1} = Φ = Δ ∥y∥ , ϕ_{1} = ϕ = δ ∥y∥ (y ≠ 0) . The inequality (4.16) is equivalent with (4.15). The inequality (4.17) follows from (4.1) for F = {1} and the same choices as above. Finally, (4.18) is obviously equivalent with (4.17).

4.2. Some Grüss Type Inequalities for Orthonormal Families. The following result holds [9].

Theorem 37. Let {e_{i}}_{i∈I} be a family of orthornormal vectors in H, F a finite part of I, ϕ_{i}, Φ_{i}, γ_{i}, Γ_{i} ∈ 𝕂, i ∈ F and x, y ∈ H. If either

Re ⟨∑_{i∈F} Φ_{i}e_{i} − x, x − ∑_{i∈F} ϕ_{i}e_{i}⟩ ≥ 0,

Re ⟨∑_{i∈F} Γ_{i}e_{i} − y, y − ∑_{i∈F} γ_{i}e_{i}⟩ ≥ 0,

or, equivalently,

∥x − ∑_{i∈F} \frac{Φ_{i} + ϕ_{i}}{2} e_{i}∥ ≤ \frac{1}{2} (∑_{i∈F} |Φ_{i} − ϕ_{i}|^{2})^{\frac{1}{2}} ,

∥y − ∑_{i∈F} \frac{Γ_{i} + γ_{i}}{2} e_{i}∥ ≤ \frac{1}{2} (∑_{i∈F} |Γ_{i} − γ_{i}|^{2})^{\frac{1}{2}} ,

hold, then we have the inequality

(4.19)  0 ≤ |⟨x, y⟩ − ∑_{i∈F} ⟨x, e_{i}⟩ ⟨e_{i}, y⟩|

≤ \frac{1}{4}M (Φ, ϕ, F) M (Γ, γ, F)

× (∑_{i∈F} |⟨x, e_{i}⟩|^{2})^{\frac{1}{2}} (∑_{i∈F} |⟨y, e_{i}⟩|^{2})^{\frac{1}{2}} ,

where M (Φ, ϕ, F) is defined in Corollary 30.

The constant \frac{1}{4} is best possible.

Proof. By the reverse of Bessel’s inequality in Corollary 30, we have

|⟨x, y⟩ − ∑_{i∈F} ⟨x, e_{i}⟩ ⟨e_{i}, y⟩|^{2} (4.20)

≤ (∥x∥^{2} − ∑_{i∈F} |⟨x, e_{i}⟩|^{2}) (∥y∥^{2} − ∑_{i∈F} |⟨y, e_{i}⟩|^{2})

≤ \frac{1}{4}M^{2} (Φ, ϕ, F) ∑_{i∈F} |⟨x, e_{i}⟩|^{2} · \frac{1}{4}M^{2} (Γ, γ, F) ∑_{i∈F} |⟨y, e_{i}⟩|^{2} .

Taking the square root in (4.20), we deduce (4.19).

The fact that \frac{1}{4} is the best possible constant follows by Corollary 30 and we omit the details.

The following corollary for real inner product spaces holds [9].

Corollary 32. Let {e_{i}}_{i∈I} be a family of orthornormal vectors in H, F a finite part of I, M_{i}, m_{i}, N_{i}, n_{i} ≥ 0, i ∈ F and x, y ∈ H such that ∑_{i∈F} M_{i}m_{i} > 0, ∑_{i∈F} N_{i}n_{i} > 0 and

⟨∑_{i∈F} M_{i}e_{i} − x, x − ∑_{i∈F} m_{i}e_{i}⟩ ≥ 0, ⟨∑_{i∈F} N_{i}e_{i} − y, y − ∑_{i∈F} n_{i}e_{i}⟩ ≥ 0.

Then we have the inequality

0 ≤ |⟨x, y⟩ − ∑_{i∈F} ⟨x, e_{i}⟩ ⟨y, e_{i}⟩|^{2}

≤ \frac{1}{16} · \frac{∑_{i∈F} (M_{i} − m_{i})^{2} ∑_{i∈F} (N_{i} − n_{i})^{2} ∑_{i∈F} |⟨x, e_{i}⟩|^{2} ∑_{i∈F} |⟨y, e_{i}⟩|^{2}}{∑_{i∈F} M_{i}m_{i} ∑_{i∈F} N_{i}n_{i}} .

The constant \frac{1}{16} is best possible.

In the case where the family {e_{i}}_{i∈I} reduces to a single vector, we may deduce from Theorem 37 the following particular case.

Corollary 33. Let e ∈ H, ∥e∥ = 1, ϕ, Φ, γ, Γ ∈ 𝕂 with Re (Φ\overline{ϕ}) , Re (Γ\overline{γ}) > 0 and x, y ∈ H such that either

Re ⟨Φe − x, x − ϕe⟩ ≥ 0, Re ⟨Γe − y, y − γe⟩ ≥ 0,

or, equivalently,

∥x − \frac{ϕ + Φ}{2}e∥ ≤ \frac{1}{2} |Φ − ϕ| , ∥y − \frac{γ + Γ}{2}e∥ ≤ \frac{1}{2} |Γ − γ|

hold, then

0 ≤ |⟨x, y⟩ − ⟨x, e⟩ ⟨e, y⟩| ≤ \frac{1}{4}M (Φ, ϕ) M (Γ, γ) |⟨x, e⟩ ⟨e, y⟩| ,

where

M (Φ, ϕ) := [\frac{(|Φ| − |ϕ|)^{2} + 4 [|ϕΦ| − Re (Φ\overline{ϕ})]}{Re (Φ\overline{ϕ})}]^{\frac{1}{2}} .

The constant \frac{1}{4} is best possible.

Remark 40. If H is real, e ∈ H, ∥e∥ = 1 and a, b, A, B ∈ ℝ are such that A > a > 0, B > b > 0 and

∥x − \frac{a + A}{2}e∥ ≤ \frac{1}{2} (A − a) , ∥y − \frac{b + B}{2} e∥ ≤ \frac{1}{2} (B − b) ,

then

(4.21) |⟨x, y⟩ − ⟨x, e⟩ ⟨e, y⟩| ≤ \frac{1}{4} · \begin{matrix}(A − a) & (B − b) \\ √ & \overline{abAB} \end{matrix} |⟨x, e⟩ ⟨e, y⟩| .

The constant \frac{1}{4} is best possible.

If ⟨x, e⟩ , ⟨y, e⟩ ≠ 0, then the following equivalent form of (4.21) also holds

|\frac{⟨x, y⟩}{⟨x, e⟩ ⟨e, y⟩} − 1| ≤ \frac{1}{4} · \begin{matrix}(A − a) & (B − b) \\ √ & \overline{abAB} \end{matrix}.

4.3. Some Companion Inequalities. The following companion of the Grüss inequality also holds [9].

Theorem 38. Let {e_{i}}_{i∈I} be a family of orthornormal vectors in H, F a finite part of I, ϕ_{i}, Φ_{i} ∈ 𝕂, (i ∈ F), x, y ∈ H and λ ∈ (0, 1) , such that either (4.22)

Re ⟨∑_{i∈F} Φ_{i}e_{i} − (λx + (1 − λ) y) , λx + (1 − λ) y − ∑_{i∈F} ϕ_{i}e_{i}⟩ ≥ 0,

or, equivalently,

∥λx + (1 − λ) y − ∑_{i∈F} \frac{Φ_{i} + ϕ_{i}}{2} · e_{i}∥ ≤ \frac{1}{2} (∑_{i∈F} |Φ_{i} − ϕ_{i}|^{2})^{\frac{1}{2}} ,

4. MORE REVERSES OF BESSEL’S INEQUALITY 105

holds. Then we have the inequality

(4.23) Re [⟨x, y⟩ − ∑_{i∈F} ⟨x, e_{i}⟩ ⟨e_{i}, y⟩]

≤ \frac{1}{16} · \frac{1}{λ (1 − λ)} ∑_{i∈F} M^{2} (Φ, ϕ, F) ∑_{i∈F} |⟨λx + (1 − λ) y, e_{i}⟩|^{2} .

The constant \frac{1}{16} is the best possible constant in (4.23) in the sense that it cannot be replaced by a smaller constant.

Proof. Using the known inequality

Re ⟨z, u⟩ ≤ \frac{1}{4} ∥z + u∥^{2}

we may state that for any a, b ∈ H and λ ∈ (0, 1)

(4.24)  Re ⟨a, b⟩ ≤ \frac{1}{4λ (1 − λ)} ∥λa + (1 − λ) b∥^{2} .

Since

⟨x, y⟩ − ∑_{i∈F} ⟨x, e_{i}⟩ ⟨e_{i}, y⟩ = ⟨x − ∑_{i∈F} ⟨x, e_{i}⟩ e_{i}, y − ∑_{i∈F} ⟨y, e_{i}⟩ e_{i}⟩ ,

for any x, y ∈ H, then, by (4.24), we get

Re [⟨x, y⟩ − ∑_{i∈F} ⟨x, e_{i}⟩ ⟨e_{i}, y⟩] (4.25)

= Re [⟨x − ∑_{i∈F} ⟨x, e_{i}⟩ e_{i}, y − ∑_{i∈F} ⟨y, e_{i}⟩ e_{i}⟩]

≤ \frac{1}{4λ (1 − λ)} ∥λ (x − ∑_{i∈F} ⟨x, e_{i}⟩ e_{i})

+ (1 − λ) (y − ∑_{i∈F} ⟨y, e_{i}⟩ e_{i})∥^{2}

= \frac{1}{4λ (1 − λ)} ∥λx + (1 − λ) y − ∑_{i∈F} ⟨λx + (1 − λ) y, e_{i}⟩ e_{i}∥^{2}

= \frac{1}{4λ (1 − λ)} [∥λx + (1−λ) y∥^{2}−∑_{i∈F} |⟨λx + (1−λ) y, e_{i}⟩|^{2}] .

If we apply the reverse of Bessel’s inequality from Corollary 30 for λx + (1 − λ) y, we may state that

(4.26) ∥λx + (1 − λ) y∥^{2} − ∑_{i∈F} |⟨λx + (1 − λ) y, e_{i}⟩|^{2}

≤ \frac{1}{4}M^{2} (Φ, ϕ, F) ∑_{i∈F} |⟨λx + (1 − λ) y, e_{i}⟩|^{2} .

Now, by making use of (4.25) and (4.26), we deduce (4.23).

The fact that \frac{1}{16} is the best possible constant in (4.23) follows by the fact that if in (4.22) we choose x = y, then it becomes (i) of Theorem 36, implying for λ = \frac{1}{2} the inequality (4.9), for which, we have shown that \frac{1}{4} is the best constant.

Remark 41. If in Theorem 38, we choose λ = \frac{1}{2}, then we get

Re [⟨x, y⟩ − ∑_{i∈F} ⟨x, e_{i}⟩ ⟨e_{i}, y⟩] ≤ \frac{1}{4}M^{2} (Φ, ϕ, F) ∑_{i∈F} |⟨\frac{x + y}{2} , e_{i}⟩|^{2} ,

provided

Re ⟨∑_{i∈F} Φ_{i}e_{i} − \frac{x + y}{2} , \frac{x + y}{2} − ∑_{i∈F} ϕ_{i}e_{i}⟩ ≥ 0

or, equivalently,

∥\frac{x + y}{2} − ∑_{i∈F} \frac{Φ_{i} + ϕ_{i}}{2} · e_{i}∥ ≤ \frac{1}{2} (∑_{i∈F} |Φ_{i} − ϕ_{i}|^{2})^{\frac{1}{2}} .

4.4. Integral Inequalities. The following proposition holds [9].

Proposition 30. Let {f_{i}}_{i∈I} be an orthornormal family of functions in L_{ρ}^{2} (Ω, 𝕂) , F a finite subset of I, ϕ_{i}, Φ_{i} ∈ 𝕂 (i ∈ F) such that ∑_{i∈F} Re (Φ_{i}\overline{ϕ_{i}}) > 0 and f ∈ L_{ρ}^{2} (Ω, 𝕂) , so that either

∫_{Ω} ρ (s) Re [(∑_{i∈F} Φ_{i}f_{i} (s) − f (s)) (\overline{f} (s) − ∑_{i∈F} \overline{ϕ_{i}} \overline{f_{i}} (s))] dμ (s) ≥ 0,

or, equivalently,

∫_{Ω} ρ (s) |f (s) − ∑_{i∈F} \frac{Φ_{i} + ϕ_{i}}{2} f_{i} (s)|^{2} dμ (s) ≤ \frac{1}{4} ∑_{i∈F} |Φ_{i} − ϕ_{i}|^{2} .

Then we have the inequality

(∫_{Ω} ρ (s) |f (s)|^{2} dμ (s))^{\frac{1}{2}} ≤ \frac{1}{2} · \frac{1}{[∑_{i∈F} Re (Φ_{i}\overline{ϕ_{i}})]^{\frac{1}{2}}}

× \{ \begin{matrix}max_{i∈F} & \{|Φ_{i}| + |ϕ_{i}|\} & ∑_{i∈F} & |∫ _{Ω} ρ (s) f (s) \overline{f_{i}} (s) dμ (s)| \\ [∑_{i∈F} (|Φ_{i}| + |ϕ_{i}|)^{p}]^{\frac{1}{p}} (∑_{i∈F} |∫ _{Ω} ρ (s) f (s) \overline{f_{i}} (s) dμ (s)|^{q})^{\frac{1}{q}} , \\ for p > 1, \frac{1}{p} + \frac{1}{q} = 1 \\ max_{i∈F} |∫ _{Ω} ρ (s) f (s) \overline{f_{i}} (s) dμ (s)| ∑_{i∈F} [|Φ_{i}| + |ϕ_{i}|] . \end{matrix}

In particular, we have

(4.27) ∫_{Ω} ρ (s) |f (s)|^{2} dμ (s)

≤ \frac{1}{4} · \begin{matrix}∑ & i∈F & (|Φ_{i}| + |ϕ_{i}|) & 2 \\ ∑ & i∈F & Re & (Φ_{i}\overline{ϕ_{i}}) \end{matrix} ∑_{i∈F} |∫_{Ω} ρ (s) f (s) \overline{f_{i}} (s) dμ (s)|^{2} .

The constant \frac{1}{4} is best possible.

The proof is obvious by Theorem 36 and Remark 38. We omit the details.

The following proposition also holds.

Proposition 31. Assume that f_{i}, f, ϕ_{i}, Φ_{i} and F satisfy the assumptions of Proposition 30. Then we have the following reverse of Bessel’s inequality:

(4.28) 0 ≤ ∫_{Ω} ρ (s) f^{2} (s) dμ (s) − ∑_{i∈F} |∫_{Ω} ρ (s) f (s) \overline{f_{i}} (s) dμ (s)|^{2}

≤ \frac{1}{4}M^{2} (Φ, ϕ, F) · ∑_{i∈F} |∫_{Ω} ρ (s) f (s) \overline{f_{i}} (s) dμ (s)|^{2} ,

where, as above,

(4.29) M (Φ, ϕ, F)

:= [\begin{matrix}∑_{i∈F} & \{(|Φ_{i}| − |ϕ_{i}|)^{2} + 4 [|ϕ_{i}Φ_{i}| − Re (Φ_{i}\overline{ϕ_{i}})]\} \\ Re & (Φ_{i}\overline{ϕ_{i}}) \end{matrix}]^{\frac{1}{2}} .

108 3. REVERSES OF BESSEL’S INEQUALITY

The constant \frac{1}{4} is the best possible.

The following Grüss type inequality also holds.

Proposition 32. Let {f_{i}}_{i∈I} and F be as in Proposition 30. If ϕ_{i}, Φ_{i}, γ_{i}, Γ_{i} ∈ 𝕂 (i ∈ F) and f, g ∈ L_{ρ}^{2} (Ω, 𝕂) such that either

∫_{Ω} ρ (s) Re [(∑_{i∈F} Φ_{i}f_{i} (s) − f (s)) (\overline{f} (s) − ∑_{i∈F} \overline{ϕ_{i}} \overline{f_{i}} (s))] dμ (s) ≥ 0,
∫_{Ω} ρ (s) Re [(∑_{i∈F} Γ_{i}f_{i} (s) − g (s)) (\overline{g} (s) − ∑_{i∈F} \overline{γ_{i}} \overline{f_{i}} (s))] dμ (s) ≥ 0,

or, equivalently,

∫_{Ω} ρ (s) |f (s) − ∑_{i∈F} \frac{Φ_{i} + ϕ_{i}}{2} · f_{i} (s)|^{2} dμ (s) ≤ \frac{1}{4} ∑_{i∈F} |Φ_{i} − ϕ_{i}|^{2} ,
∫_{Ω} ρ (s) |g (s) − ∑_{i∈F} \frac{Γ_{i} + γ_{i}}{2} · f_{i} (s)|^{2} dμ (s) ≤ \frac{1}{4} ∑_{i∈F} |Γ_{i} − γ_{i}|^{2} ,

hold, then we have the inequality

(4.30) |∫_{Ω} ρ (s) f (s) \overline{g (s)}dμ (s)

− ∑_{i∈F} ∫_{Ω} ρ (s) f (s) \overline{f_{i}} (s) dμ (s) ∫_{Ω} ρ (s) f_{i} (s) \overline{g (s)}dμ (s)|

≤ \frac{1}{4}M (Φ, ϕ, F) M (Γ, γ, F) (∑_{i∈F} |∫_{Ω} ρ (s) f (s) \overline{f_{i}} (s) dμ (s)|^{2})^{\frac{1}{2}}

× (∑_{i∈F} |ρ (s) f_{i} (s) \overline{g (s)}dμ (s)|^{2})^{\frac{1}{2}} ,

where M (Φ, ϕ, F) is as defined in (4.29).

The constant \frac{1}{4} is the best possible.

The proof follows by Theorem 37 and we omit the details.

In the case of real spaces, the following corollaries provide much simpler sufficient conditions for the reverse of Bessel’s inequality (4.28) or for the Grüss type inequality (4.30) to hold.

Corollary 34. Let {f_{i}}_{i∈I} be an orthornormal family of functions in the real Hilbert space L_{ρ}^{2} (Ω) , F a finite part of I, M_{i}, m_{i} ≥ 0 (i ∈ F) , with ∑_{i∈F} M_{i}m_{i} > 0 and f ∈ L_{ρ}^{2} (Ω) so that

∑_{i∈F} m_{i}f_{i} (s) ≤ f (s) ≤ ∑_{i∈F} M_{i}f_{i} (s) for μ − a.e. s ∈ Ω.

Then we have the inequality

0 ≤ ∫_{Ω} ρ (s) f^{2} (s) dμ (s) − ∑_{i∈F} [∫_{Ω} ρ (s) f (s) f_{i} (s) dμ (s)]^{2}

≤ \frac{1}{4} · \frac{∑_{i∈F} (M_{i} − m_{i})^{2}}{∑_{i∈F} M_{i}m_{i}} · ∑_{i∈F} [∫_{Ω} ρ (s) f (s) f_{i} (s) dμ (s)]^{2} .

The constant \frac{1}{4} is best possible.

Corollary 35. Let {f_{i}}_{i∈I} and F be as above. If M_{i}, m_{i}, N_{i}, n_{i} ≥ 0 (i ∈ F) with ∑_{i∈F} M_{i}m_{i}, ∑_{i∈F} N_{i}n_{i} > 0 and f, g ∈ L_{ρ}^{2} (Ω) such that

∑_{i∈F} m_{i}f_{i} (s) ≤ f (s) ≤ ∑_{i∈F} M_{i}f_{i} (s)

and

∑_{i∈F} n_{i}f_{i} (s) ≤ g (s) ≤ ∑_{i∈F} N_{i}f_{i} (s) for μ − a.e. s ∈ Ω,

then we have the inequality

|∫_{Ω} ρ (s) f (s) g (s) dμ (s)

− ∑_{i∈F} ∫_{Ω} ρ (s) f (s) f_{i} (s) dμ (s) ∫_{Ω} ρ (s) g (s) f_{i} (s) dμ (s)|

≤ \frac{1}{4} (\frac{∑_{i∈F} (M_{i} − m_{i})^{2}}{∑_{i∈F} M_{i}m_{i}} )^{\frac{1}{2}} (\frac{∑_{i∈F} (N_{i} − n_{i})^{2}}{∑_{i∈F} N_{i}n_{i}} )^{\frac{1}{2}}

× [∑_{i∈F} (∫_{Ω} ρ (s) f (s) f_{i} (s) dμ (s))^{2} ∑_{i∈F} (∫_{Ω} ρ (s) g (s) f_{i} (s) dμ (s))^{2}]^{\frac{1}{2}} .

5. General Reverses of Bessel’s Inequality

5.1. Some Reverses of Bessel’s Inequality. Let (H; ⟨·, ·⟩) be a real or complex infinite dimensional Hilbert space and (e_{i})_{i∈ℕ} an orthornormal family in H, i.e., we recall that ⟨e_{i}, e_{j}⟩ = 0 if i, j ∈ ℕ, i ≠ j and ∥e_{i}∥ = 1 for i ∈ ℕ.

It is well known that, if x ∈ H, then the sum ∑_{i=1}^{∞} |⟨x, e_{i}⟩|^{2} is convergent and the following inequality, called Bessel’s inequality

(5.1)  ∑_{i=1}^{∞} |⟨x, e_{i}⟩|^{2} ≤ ∥x∥^{2} ,

holds.

If ℓ^{2} (𝕂) := {a = (a_{i})_{i∈ℕ} ⊂ 𝕂 |∑_{i=1}^{∞} |a_{i}|^{2} < ∞} , where 𝕂 = ℂ or 𝕂 = ℝ, is the Hilbert space of all complex or real sequences that are 2-summable and λ = (λ_{i})_{i∈ℕ} ∈ ℓ^{2} (𝕂) , then the sum ∑_{i=1}^{∞} λ_{i}e_{i} is

convergent in H and if y := ∑_{i=1}^{∞} λ_{i}e_{i} ∈ H, then ∥y∥ = (∑_{i=1}^{∞} |λ_{i}|^{2})^{\frac{1}{2}} . We may state the following result [7].

Theorem 39. Let (H; ⟨·, ·⟩) be an infinite dimensional Hilbert space over the real or complex number field 𝕂, (e_{i})_{i∈ℕ} an orthornormal family in H, λ = (λ_{i})_{i∈ℕ} ∈ ℓ^{2} (𝕂) and r > 0 with the property that

∑_{i=1}^{∞} |λ_{i}|^{2} > r^{2}.

If x ∈ H is such that

∥x − ∑_{i=1}^{∞} λ_{i}e_{i}∥ ≤ r,

then we have the inequality

(5.2)  ∥x∥^{2} ≤ \frac{(∑_{i=1}^{∞} Re [\overline{λ_{i}} ⟨x, e_{i}⟩])^{2}}{∑_{i=1}^{∞} |λ_{i}|^{2} − r^{2}}

≤ \frac{|∑_{i=1}^{∞} \overline{λ_{i}} ⟨x, e_{i}⟩|^{2}}{∑_{i=1}^{∞} |λ_{i}|^{2} − r^{2}}

≤ \frac{∑_{i=1}^{∞} |λ_{i}|^{2}}{∑_{i=1}^{∞} |λ_{i}|^{2} − r^{2}} ∑_{i=1}^{∞} |⟨x, e_{i}⟩|^{2}

and

(5.3)  0 ≤ ∥x∥^{2} − ∑_{i=1}^{∞} |⟨x, e_{i}⟩|^{2}

≤ \frac{r^{2}}{∑_{i=1}^{∞} |λ_{i}|^{2} − r^{2}} ∑_{i=1}^{∞} |⟨x, e_{i}⟩|^{2} .

Proof. Applying the third inequality in (4.9) for a = ∑_{i=1}^{∞} λ_{i}e_{i} ∈ H, we have

(5.4)  ∥x∥^{2} ∥∑_{i=1}^{∞} λ_{i}e_{i}∥^{2} − [Re ⟨x, ∑_{i=1}^{∞} λ_{i}e_{i}⟩]^{2} ≤ r^{2} ∥x∥^{2}

and since

∥∑_{i=1}^{∞} λ_{i}e_{i}∥^{2} = ∑_{i=1}^{∞} |λ_{i}|^{2} ,

Re ⟨x, ∑_{i=1}^{∞} λ_{i}e_{i}⟩ = ∑_{i=1}^{∞} Re [\overline{λ_{i}} ⟨x, e_{i}⟩] ,

then, by (5.4), we deduce

∥x∥^{2} ∑_{i=1}^{∞} |λ_{i}|^{2} − [Re ⟨x, ∑_{i=1}^{∞} λ_{i}e_{i}⟩]^{2} ≤ r^{2} ∥x∥^{2} ,

giving the first inequality in (5.2).

The second inequality is obvious by the modulus property.

The last inequality follows by the Cauchy-Bunyakovsky-Schwarz inequality

|∑_{i=1}^{∞} \overline{λ_{i}} ⟨x, e_{i}⟩|^{2} ≤ ∑_{i=1}^{∞} |λ_{i}|^{2} ∑_{i=1}^{∞} |⟨x, e_{i}⟩|^{2} .

The inequality (5.3) follows by the last inequality in (5.2) on subtracting from both sides the quantity ∑_{i=1}^{∞} |⟨x, e_{i}⟩|^{2} < ∞.

The following result provides a generalization for the reverse of Bessel’s inequality obtained in [9].

Theorem 40. Let (H; ⟨·, ·⟩) and (e_{i})_{i∈ℕ} be as in Theorem 39. Suppose that Γ = (Γ_{i})_{i∈ℕ} ∈ ℓ^{2} (𝕂) , γ = (γ_{i})_{i∈ℕ} ∈ ℓ^{2} (𝕂) are sequences of real or complex numbers such that

∑_{i=1}^{∞} Re (Γ_{i}\overline{γ_{i}}) > 0.

If x ∈ H is such that either

(5.5)  ∥x − ∑_{i=1}^{∞} \frac{Γ_{i} + γ_{i}}{2} e_{i}∥ ≤ \frac{1}{2} (∑_{i=1}^{∞} |Γ_{i} − γ_{i}|^{2})^{\frac{1}{2}}

or, equivalently,

(5.6)  Re ⟨∑_{i=1}^{∞} Γ_{i}e_{i} − x, x − ∑_{i=1}^{∞} γ_{i}e_{i}⟩ ≥ 0

holds, then we have the inequalities

(5.7)  ∥x∥^{2} ≤ \frac{1}{4} · \begin{matrix}(∑_{i=1}^{∞} Re [(\overline{Γ_{i}} + \overline{γ_{i}}) ⟨x, e_{i}⟩]) & 2 \\ ∑ & {∞ \atop i=1} & Re & (Γ_{i}\overline{γ_{i}}) \end{matrix}

≤ \frac{1}{4} · \begin{matrix}|∑_{i=1}^{∞} (\overline{Γ_{i}} + \overline{γ_{i}}) ⟨x, e_{i}⟩| & 2 \\ ∑ & {∞ \atop i=1} & Re & (Γ_{i}\overline{γ_{i}}) \end{matrix}

≤ \frac{1}{4} · \begin{matrix}∑ & {∞ \atop i=1} & |Γ_{i} + γ_{i}| & 2 \\ ∑ & {∞ \atop i=1} & Re & (Γ_{i}\overline{γ_{i}}) \end{matrix} ∑_{i=1}^{∞} |⟨x, e_{i}⟩|^{2} .

The constant \frac{1}{4} is best possible in all inequalities in (5.7).

We also have the inequalities:

(5.8) 0 ≤ ∥x∥^{2} − ∑_{i=1}^{∞} |⟨x, e_{i}⟩|^{2} ≤ \frac{1}{4} · \begin{matrix}∑ & {∞ \atop i=1} & |Γ_{i} − γ_{i}| & 2 \\ ∑ & {∞ \atop i=1} & Re & (Γ_{i}\overline{γ_{i}}) \end{matrix} ∑_{i=1}^{∞} |⟨x, e_{i}⟩|^{2} .

Here the constant \frac{1}{4} is also best possible.

Proof. Since Γ, γ ∈ ℓ^{2} (𝕂) , then also \frac{1}{2} (Γ ± γ) ∈ ℓ^{2} (𝕂) , showing that the series

∑_{i=1}^{∞} |\frac{Γ_{i} + γ_{i}}{2} |^{2} , ∑_{i=1}^{∞} |\frac{Γ_{i} − γ_{i}}{2} |^{2} and ∑_{i=1}^{∞} Re (Γ_{i}\overline{γ_{i}})

are convergent. Also, the series

∑_{i=1}^{∞} Γ_{i}e_{i}, ∑_{i=1}^{∞} γ_{i}e_{i} and ∑_{i=1}^{∞} \frac{γ_{i} + Γ_{i}}{2} e_{i}

are convergent in the Hilbert space H.

Now, we observe that the inequalities (5.7) and (5.8) follow from

Theorem 39 on choosing λ_{i} = \frac{γ_{i}+Γ_{i}}{2} , i ∈ ℕ and r = \frac{1}{2} (∑_{i=1}^{∞} |Γ_{i} − γ_{i}|^{2})^{\frac{1}{2}} .

The fact that \frac{1}{4} is the best constant in both (5.7) and (5.8) follows from Theorem 6 and Corollary 11, and we omit the details.

For some recent results related to the Bessel inequality, see [1], [4], [10], and [11].

5.2. Some Grüss Type Inequalities for Orthonormal Families. The following result related to the Grüss inequality in inner product spaces holds [7].

Theorem 41. Let (H; ⟨·, ·⟩) be an infinite dimensional Hilbert space over the real or complex number field 𝕂, and (e_{i})_{i∈ℕ} an orthornormal family in H. Assume that λ = (λ_{i})_{i∈ℕ} , μ = (μ_{i})_{i∈ℕ} ∈ ℓ^{2} (𝕂) and r_{1}, r_{2} > 0 with the properties that

∑_{i=1}^{∞} |λ_{i}|^{2} > r_{1}^{2}, ∑_{i=1}^{∞} |μ_{i}|^{2} > r_{2}^{2}.

If x, y ∈ H are such that

∥x − ∑_{i=1}^{∞} λ_{i}e_{i}∥ ≤ r_{1},  ∥y − ∑_{i=1}^{∞} μ_{i}e_{i}∥ ≤ r_{2},

then we have the inequalities

|⟨x, y⟩ − ∑_{i=1}^{∞} ⟨x, e_{i}⟩ ⟨e_{i}, y⟩| (5.9)

≤ \frac{r_{1}r_{2}}{\sqrt{∑_{i=1}^{∞} |λ_{i}|^{2} − r_{1}^{2}}\sqrt{∑_{i=1}^{∞} |μ_{i}|^{2} − r_{2}^{2}}}

× \sqrt{∑_{i=1}^{∞} |⟨x, e_{i}⟩|^{2} ∑_{i=1}^{∞} |⟨y, e_{i}⟩|^{2}}

≤ \frac{r_{1}r_{2} ∥x∥ ∥y∥}{\sqrt{∑_{i=1}^{∞} |λ_{i}|^{2} − r_{1}^{2}}\sqrt{∑_{i=1}^{∞} |μ_{i}|^{2} − r_{2}^{2}}} .

Proof. Applying Schwarz’s inequality for the vectors

x − ∑_{i=1}^{∞} ⟨x, e_{i}⟩ e_{i}, y − ∑_{i=1}^{∞} ⟨y, e_{i}⟩ e_{i},

we have

(5.10) |⟨x − ∑_{i=1}^{∞} ⟨x, e_{i}⟩ e_{i}, y − ∑_{i=1}^{∞} ⟨y, e_{i}⟩ e_{i}⟩|^{2}

≤ ∥x − ∑_{i=1}^{∞} ⟨x, e_{i}⟩ e_{i}∥^{2} ∥y − ∑_{i=1}^{∞} ⟨y, e_{i}⟩ e_{i}∥^{2} .

Since

⟨x − ∑_{i=1}^{∞} ⟨x, e_{i}⟩ e_{i}, y − ∑_{i=1}^{∞} ⟨y, e_{i}⟩ e_{i}⟩ = ⟨x, y⟩ − ∑_{i=1}^{∞} ⟨x, e_{i}⟩ ⟨e_{i}, y⟩

and

∥x − ∑_{i=1}^{∞} ⟨x, e_{i}⟩ e_{i}∥^{2} = ∥x∥^{2} − ∑_{i=1}^{∞} |⟨x, e_{i}⟩|^{2} ,

then by (5.3) applied for x and y, and from (5.10), we deduce the first part of (5.9).

The second part follows by Bessel’s inequality.

The following Grüss type inequality may be stated as well.

Theorem 42. Let (H; ⟨·, ·⟩) be an infinite dimensional Hilbert space and (e_{i})_{i∈ℕ} an orthornormal family in H. Suppose that (Γ_{i})_{i∈ℕ} , (γ_{i})_{i∈ℕ} , (ϕ_{i})_{i∈ℕ} , (Φ_{i})_{i∈ℕ} ∈ ℓ^{2} (𝕂) are sequences of real and complex numbers such that

∑_{i=1}^{∞} Re (Γ_{i}\overline{γ_{i}}) > 0, ∑_{i=1}^{∞} Re (Φ_{i}\overline{ϕ_{i}}) > 0.

If x, y ∈ H are such that either

∥x − ∑_{i=1}^{∞} \frac{Γ_{i} + γ_{i}}{2} · e_{i}∥ ≤ \frac{1}{2} (∑_{i=1}^{∞} |Γ_{i} − γ_{i}|^{2})^{\frac{1}{2}} ,

∥y − ∑_{i=1}^{∞} \frac{Φ_{i} + ϕ_{i}}{2} · e_{i}∥ ≤ \frac{1}{2} (∑_{i=1}^{∞} |Φ_{i} − ϕ_{i}|^{2})^{\frac{1}{2}} ,

or, equivalently,

Re ⟨∑_{i=1}^{∞} Γ_{i}e_{i} − x, x − ∑_{i=1}^{∞} γ_{i}e_{i}⟩ ≥ 0,

Re ⟨∑_{i=1}^{∞} Φ_{i}e_{i} − y, y − ∑_{i=1}^{∞} ϕ_{i}e_{i}⟩ ≥ 0,

holds, then we have the inequality

|⟨x, y⟩ − ∑_{i=1}^{∞} ⟨x, e_{i}⟩ ⟨e_{i}, y⟩|

≤ \frac{1}{4} · (∑_{i=1}^{∞} |Γ_{i} − γ_{i}|^{2})^{\frac{1}{2}} (∑_{i=1}^{∞} |Φ_{i} − ϕ_{i}|^{2})^{\frac{1}{2}} ; (∑_{i=1}^{∞} Re (Γ_{i}\overline{γ_{i}}))^{\frac{1}{2}} (∑_{i=1}^{∞} Re (Φ_{i}\overline{ϕ_{i}}))^{\frac{1}{2}}

× (∑_{i=1}^{∞} |⟨x, e_{i}⟩|^{2})^{\frac{1}{2}} (∑_{i=1}^{∞} |⟨y, e_{i}⟩|^{2})^{\frac{1}{2}}

≤ \frac{1}{4} · (∑_{i=1}^{∞} |Γ_{i} − γ_{i}|^{2})^{\frac{1}{2}} (∑_{i=1}^{∞} |Φ_{i} − ϕ_{i}|^{2})^{\frac{1}{2}} ; [∑_{i=1}^{∞} Re (Γ_{i}\overline{γ_{i}})]^{\frac{1}{2}} [∑_{i=1}^{∞} Re (Φ_{i}\overline{ϕ_{i}})]^{\frac{1}{2}} ∥x∥ ∥y∥ .

The constant \frac{1}{4} is best possible in the first inequality.

Proof. Follows by (5.8) and (5.10). The best constant follows from Theorem 22, and we omit the details.

5.3. Other Reverses of Bessel’s Inequality. We may state the following result [5].

Theorem 43. Let (H; ⟨·, ·⟩) be an infinite dimensional Hilbert space over the real or complex number field 𝕂, (e_{i})_{i∈ℕ} an orthornormal family in H, λ = (λ_{i})_{i∈ℕ} ∈ ℓ^{2} (𝕂) , λ ≠ 0 and r > 0. If x ∈ H is such that

∥x − ∑_{i=1}^{∞} λ_{i}e_{i}∥ ≤ r,

then we have the inequality

(5.11)  0 ≤ ∥x∥ − (∑_{i=1}^{∞} |⟨x, e_{i}⟩|^{2})^{\frac{1}{2}} ≤ \frac{1}{2} · \frac{r^{2}}{(∑_{i=1}^{∞} |λ_{i}|^{2})^{\frac{1}{2}}} .

The constant \frac{1}{2} is best possible in (5.11) in the sense that it cannot be replaced by a smaller constant.

Proof. Let a := ∑_{i=1}^{∞} λ_{i}e_{i} ∈ H. Then by Theorem 7, we have

∥x∥ ∥∑_{i=1}^{∞} λ_{i}e_{i}∥ − |∑_{i=1}^{∞} λ̄_{i} ⟨x, e_{i}⟩| ≤ \frac{1}{2}r^{2},

giving

(5.12)  ∥x∥ (∑_{i=1}^{∞} |λ_{i}|^{2})^{\frac{1}{2}} ≤ \frac{1}{2}r^{2} + |∑_{i=1}^{∞} λ̄_{i} ⟨x, e_{i}⟩| ,

since

∥∑_{i=1}^{∞} λ_{i}e_{i}∥ = (∑_{i=1}^{∞} |λ_{i}|^{2})^{\frac{1}{2}} .

Using the Cauchy-Bunyakovsky-Schwarz inequality, we may state that

(5.13)  |∑_{i=1}^{∞} λ̄_{i} ⟨x, e_{i}⟩| ≤ (∑_{i=1}^{∞} |λ_{i}|^{2})^{\frac{1}{2}} (∑_{i=1}^{∞} |⟨x, e_{i}⟩|^{2})^{\frac{1}{2}} ,

and thus, by (5.12) and (5.13), we may state that

∥x∥ (∑_{i=1}^{∞} |λ_{i}|^{2})^{\frac{1}{2}} ≤ \frac{1}{2}r^{2} + (∑_{i=1}^{∞} |λ_{i}|^{2})^{\frac{1}{2}} (∑_{i=1}^{∞} |⟨x, e_{i}⟩|^{2})^{\frac{1}{2}} ,

from where we get the desired inequality in (5.11).

The best constant, follows by Theorem 7 on choosing (e_{i})_{i∈ℕ} = {e} , with ∥e∥ = 1 and we omit the details.

Remark 42. Under the assumptions of Theorem 43, and if we multiply by ∥x∥ + (∑_{i=1}^{∞} |⟨x, e_{i}⟩|^{2})^{\frac{1}{2}} > 0, then we deduce, from (5.11), that

(5.14)  0 ≤ ∥x∥^{2} − ∑_{i=1}^{∞} |⟨x, e_{i}⟩|^{2}

≤ \frac{1}{2} · \frac{r^{2} (∥x∥ + (∑_{i=1}^{∞} |⟨x, e_{i}⟩|^{2})^{\frac{1}{2}} )}{(∑_{i=1}^{∞} |λ_{i}|^{2})^{\frac{1}{2}}}

≤ \frac{r^{2} ∥x∥}{(∑_{i=1}^{∞} |λ_{i}|^{2})^{\frac{1}{2}}} ,

where, for the last inequality, we have used Bessel’s inequality

(∑_{i=1}^{∞} |⟨x, e_{i}⟩|^{2})^{\frac{1}{2}} ≤ ∥x∥ , x ∈ H.

The following result also holds [5].

Theorem 44. Assume that (H; ⟨·, ·⟩) and (e_{i})_{i∈ℕ} are as in Theorem 43. If Γ = (Γ_{i})_{i∈ℕ} , γ = (γ_{i})_{i∈ℕ} ∈ ℓ^{2} (𝕂) , with Γ ≠ −γ, and x ∈ H with the property that, either

∥x − ∑_{i=1}^{∞} \frac{Γ_{i} + γ_{i}}{2} · e_{i}∥ ≤ \frac{1}{2} (∑_{i=1}^{∞} |Γ_{i} − γ_{i}|^{2})^{\frac{1}{2}} ,

or, equivalently,

Re ⟨∑_{i=1}^{∞} Γ_{i}e_{i} − x, x − ∑_{i=1}^{∞} γ_{i}e_{i}⟩ ≥ 0,

holds, then we have the inequality

(5.15)  0 ≤ ∥x∥ − (∑_{i=1}^{∞} |⟨x, e_{i}⟩|^{2})^{\frac{1}{2}} ≤ \frac{1}{4} · \begin{matrix}∑ & {∞ \atop i=1} & |Γ_{i} − γ_{i}| & 2 \\ (∑_{i=1}^{∞} |Γ_{i} + γ_{i}|^{2}) & \frac{1}{2} \end{matrix} .

The constant \frac{1}{4} is best possible in the sense that it cannot be replaced by a smaller constant.

Proof. Since Γ, γ ∈ ℓ^{2} (𝕂) , then we have that \frac{1}{2} (Γ ± γ) ∈ ℓ^{2} (𝕂) , showing that the series

∑_{i=1}^{∞} |\frac{Γ_{i} + γ_{i}}{2} |^{2} , ∑_{i=1}^{∞} |\frac{Γ_{i} − γ_{i}}{2} |^{2}

are convergent. In addition, the series ∑_{i=1}^{∞} Γ_{i}e_{i}, ∑_{i=1}^{∞} γ_{i}e_{i} and ∑_{i=1}^{∞} \frac{Γ_{i}+γ_{i}}{2} e_{i} are also convergent in the Hilbert space H.

Now, we observe that the inequality (5.15) follows from Theorem

43 on choosing λ_{i} = \frac{Γ_{i}+γ_{i}}{2} , i ∈ ℕ and r = \frac{1}{2} (∑_{i=1}^{∞} |Γ_{i} − γ_{i}|^{2})^{\frac{1}{2}} .

The fact that \frac{1}{4} is the best possible constant in (5.15) follows from Theorem 8, and we omit the details.

Remark 43. With the assumptions of Theorem 44, we have

(5.16)  0 ≤ ∥x∥^{2} − ∑_{i=1}^{∞} |⟨x, e_{i}⟩|^{2}

≤ \frac{1}{4} · \begin{matrix}∑ & {∞ \atop i=1} & |Γ_{i} − γ_{i}| & 2 \\ (∑_{i=1}^{∞} |Γ_{i} + γ_{i}|^{2}) & \frac{1}{2} \end{matrix} [∥x∥ + (∑_{i=1}^{∞} |⟨x, e_{i}⟩|^{2})^{\frac{1}{2}} ]

≤ \frac{1}{2} · \begin{matrix}∑ & {∞ \atop i=1} & |Γ_{i} − γ_{i}| & 2 \\ (∑_{i=1}^{∞} |Γ_{i} + γ_{i}|^{2}) & \frac{1}{2} \end{matrix} ∥x∥ .

5.4. More Grüss Type Inequalities for Orthonormal Families. The following result holds [5].

Theorem 45. Let (H; ⟨·, ·⟩) be an infinite dimensional Hilbert space over the real or complex number field 𝕂 and (e_{i})_{i∈ℕ} an orthornormal family in H. If λ = (λ_{i})_{i∈ℕ} , μ = (μ_{i})_{i∈ℕ} ∈ ℓ^{2} (𝕂) , λ, μ ≠ 0, r_{1}, r_{2} > 0 and x, y ∈ H are such that

∥x − ∑_{i=1}^{∞} λ_{i}e_{i}∥ ≤ r_{1}, ∥y − ∑_{i=1}^{∞} μ_{i}e_{i}∥ ≤ r_{2},

then we have the inequality

|⟨x, y⟩ − ∑_{i=1}^{∞} ⟨x, e_{i}⟩ ⟨e_{i}, y⟩|

≤ \frac{1}{2}r_{1}r_{2} \begin{matrix}[∥x∥ + (∑_{i=1}^{∞} |⟨x, e_{i}⟩|^{2})^{\frac{1}{2}} ] & \frac{1}{2} & [∥y∥ + (∑_{i=1}^{∞} |⟨y, e_{i}⟩|^{2})^{\frac{1}{2}} ] & \frac{1}{2} \\ (∑_{i=1}^{∞} |λ_{i}|^{2}) & \frac{1}{4} & (∑_{i=1}^{∞} |μ_{i}|^{2}) & \frac{1}{4} \end{matrix}

≤ r_{1}r_{2} \begin{matrix}∥x∥ & \frac{1}{2} & ∥y∥ & \frac{1}{2} \\ (∑_{i=1}^{∞} |λ_{i}|^{2}) & \frac{1}{4} & (∑_{i=1}^{∞} |μ_{i}|^{2}) & \frac{1}{4} \end{matrix} .

Proof. It follows by (5.14) applied for x and y. We omit the details.

Finally we may state the following theorem [5].

5. GENERAL REVERSES OF BESSEL’S INEQUALITY 119

Theorem 46. Assume that (H; ⟨·, ·⟩) and (e_{i})_{i∈ℕ} are as in Theorem 45. If Γ = (Γ_{i})_{i∈ℕ} , Γ = (Γ_{i})_{i∈ℕ} , ϕ = (ϕ_{i})_{i∈ℕ} , Φ = (Φ_{i})_{i∈ℕ} ∈ ℓ^{2} (𝕂) , with Γ ≠ −γ, Φ ≠ −ϕ, and x, y ∈ H are such that, either

∥x − ∑_{i=1}^{∞} \frac{Γ_{i} + γ_{i}}{2} · e_{i}∥ ≤ \frac{1}{2} (∑_{i=1}^{∞} |Γ_{i} − γ_{i}|^{2})^{\frac{1}{2}} ,
∥y − ∑_{i=1}^{∞} \frac{Φ_{i} + ϕ_{i}}{2} · e_{i}∥ ≤ \frac{1}{2} (∑_{i=1}^{∞} |Φ_{i} − ϕ_{i}|^{2})^{\frac{1}{2}} ,

or, equivalently,

Re ⟨∑_{i=1}^{∞} Γ_{i}e_{i} − x, x − ∑_{i=1}^{∞} γ_{i}e_{i}⟩ ≥ 0,
Re ⟨∑_{i=1}^{∞} Φ_{i}e_{i} − y, y − ∑_{i=1}^{∞} ϕ_{i}e_{i}⟩ ≥ 0,

holds, then we have the inequality

|⟨x, y⟩ − ∑_{i=1}^{∞} ⟨x, e_{i}⟩ ⟨e_{i}, y⟩|

≤ \frac{1}{4} · (∑_{i=1}^{∞} |Φ_{i} − ϕ_{i}|^{2})^{\frac{1}{2}} (∑_{i=1}^{∞} |Γ_{i} − γ_{i}|^{2})^{\frac{1}{2}}

× \begin{matrix}[∥x∥ + (∑_{i=1}^{∞} |⟨x, e_{i}⟩|^{2})^{\frac{1}{2}} ] & \frac{1}{2} & [∥y∥ + (∑_{i=1}^{∞} |⟨y, e_{i}⟩|^{2})^{\frac{1}{2}} ] & \frac{1}{2} \\ (∑_{i=1}^{∞} |Φ_{i} + ϕ_{i}|^{2}) & \frac{1}{4} & (∑_{i=1}^{∞} |Γ_{i} + γ_{i}|^{2}) & \frac{1}{4} \end{matrix}

≤ \frac{1}{2} · \begin{matrix}(∑_{i=1}^{∞} |Φ_{i} − ϕ_{i}|^{2}) & \frac{1}{2} & (∑_{i=1}^{∞} |Γ_{i} − γ_{i}|^{2}) & \frac{1}{2} \\ (∑_{i=1}^{∞} |Φ_{i} + ϕ_{i}|^{2}) & \frac{1}{4} & (∑_{i=1}^{∞} |Γ_{i} + γ_{i}|^{2}) & \frac{1}{4} \end{matrix} ∥x∥^{\frac{1}{2}} ∥y∥^{\frac{1}{2}} .

The proof follow by (5.16) applied for x and y. We omit the details.

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[2] S.S. DRAGOMIR, A counterpart of Bessel’s inequality in inner product spaces and some Grüss type related results, RGMIA Res. Rep. Coll. 6(2003), Supplement,

Article 10, [ON LINE: http://rgmia.vu.edu.au/v6(E).html].

[3] S.S. DRAGOMIR, A generalisation of Grüss’ inequality in inner product spaces and applications, J. Math. Anal. Appl., 237 (1999), 74-82.

[4] S.S. DRAGOMIR, A note on Bessel’s inequality, Austral. Math. Soc. Gaz. 28 (2001), no. 5, 246–248.

[5] S.S. DRAGOMIR, New reverses of Schwarz, triangle and Bessel inequalities in inner product spaces, RGMIA Res. Rep. Coll., 6(2003), Supplement, Article 20,

[ONLINE: http://rgmia.vu.edu.au/v6(E).html].

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[ONLINE: http://rgmia.vu.edu.au/v6(E).html].

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Preprint/Sever/20030828.2/1/].

[8] S.S. DRAGOMIR, Some Grüss type inequalities in inner product spaces, J. Inequal. Pure & Appl. Math., 4(2003), No. 2, Article 42, [ON LINE:

http://jipam.vu.edu.au/v4n2/032 03.html].

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Supplement, Article 13, [ONLINE: http://rgmia.vu.edu.au/v6(E).html].

[10] S.S. DRAGOMIR and J. SÁNDOR, On Bessel’s and Gram’s inequalities in pre-Hilbertian spaces. Period. Math. Hungar., 29 (1994), no. 3, 197–205.

[11] H. GUNAWAN, A generalization of Bessel’s inequality and Parseval’s identity, Period. Math. Hungar., 44 (2002), no. 2, 177–181.

[12] N. UJEVIĆ, A new generalisation of Grüss inequality in inner product spaces, Math. Ineq. & Appl., 6(2003), No. 4, pp. 617-623.

Part 2

Other Inequalities in Inner Product Spaces

CHAPTER 4 Generalisations of Bessel’s Inequality

1. Boas-Bellman Type Inequalities

1.1. Introduction. Let (H; (·, ·)) be an inner product space over the real or complex number field 𝕂. If (e_{i})_{1≤i≤n} are orthonormal vectors in the inner product space H, i.e., (e_{i}, e_{j}) = δ_{ij} for all i, j ∈ {1, …, n} , where δ_{ij} is the Kronecker delta, then we have the following inequality that is well known in the literature as Bessel’s inequality (see for example [11, p. 391]):

∑_{i=1}^{n} |(x, e_{i})|^{2} ≤ ∥x∥^{2} for any x ∈ H.

For other results related to Bessel’s inequality, see [8]-[9]and Chapter XV in the book [11].

In 1941, R.P. Boas [2] and in 1944, independently, R. Bellman [1] proved the following generalisation of Bessel’s inequality (see also [11, p. 392]).

Theorem 47. If x, y_{1}, …, y_{n} are vectors in an inner product space (H; (·, ·)) , then the following inequality:

(1.1) ∑_{i=1}^{n} |(x, y_{i})|^{2} ≤ ∥x∥^{2} [ max_{1≤i≤n} ∥y_{i}∥^{2} + ( ∑_{1≤i≠j≤n} |(y_{i}, y_{j})|^{2})^{\frac{1}{2}} ] ,

holds.

A recent generalisation of the Boas-Bellman result was given in Mitrinović-Pečarić-Fink [11, p. 392] where they proved the following:

Theorem 48. If x, y_{1}, …, y_{n} are as in Theorem 47 and c_{1}, …, c_{n} ∈ 𝕂, then one has the inequality:

(1.2) |∑_{i=1}^{n} c_{i} (x, y_{i})|^{2}

≤ ∥x∥^{2} ∑_{i=1}^{n} |c_{i}|^{2} [ max_{1≤i≤n} ∥y_{i}∥^{2} + ( ∑_{1≤i≠j≤n} |(y_{i}, y_{j})|^{2})^{\frac{1}{2}} ] .

They also noted that if in (1.2) one chooses c_{i} = \overline{(x, y_{i})}, then this inequality becomes (1.1).

For other results related to the Boas-Bellman inequality, see [7]. In this section, by following [5], we point out some new results that

may be related to both the Mitrinović-Pečarić-Fink and Boas-Bellman inequalities.

1.2. Preliminary Results. We start with the following lemma which is also interesting in itself [5].

Lemma 6. Let z_{1}, …, z_{n} ∈ H and α_{1}, …, α_{n} ∈ 𝕂. Then one has the inequality:

(1.3) ∥∑_{i=1}^{n} α_{i}z_{i}∥^{2}

≤ \{ \begin{matrix}max_{1≤i≤n} |α_{i}|^{2} ∑_{i=1}^{n} ∥z_{i}∥^{2} ; \\ (∑_{i=1}^{n} |α_{i}|^{2α})^{\frac{1}{α}} (∑_{i=1}^{n} ∥z_{i}∥^{2β})^{\frac{1}{β}} , where α > 1, \\ \frac{1}{α} + \frac{1}{β} = 1; \\ ∑_{i=1}^{n} |α_{i}|^{2} max_{1≤i≤n} ∥z_{i}∥^{2} , \end{matrix}

+ \{ \begin{matrix}max_{1≤i≠j≤n} \{|α_{i}α_{j}|\} ∑_{1≤i≠j≤n} |(z_{i}, z_{j})|; \\ [(∑_{i=1}^{n} |α_{i}|^{γ})^{2} − (∑_{i=1}^{n} |α_{i}|^{2γ})]^{\frac{1}{γ}} ( ∑_{1≤i≠j≤n} |(z_{i}, z_{j})|^{δ})^{\frac{1}{δ}} , \\ where γ > 1, \frac{1}{γ} + \frac{1}{δ} = 1; \\ [(∑_{i=1}^{n} |α_{i}|)^{2} − ∑_{i=1}^{n} |α_{i}|^{2}] max_{1≤i≠j≤n} |(z_{i}, z_{j})| . \end{matrix}

Proof. We observe that

(1.4) ∥∑_{i=1}^{n} α_{i}z_{i}∥^{2} = (∑_{i=1}^{n} α_{i}z_{i}, ∑_{j=1}^{n} α_{j}z_{j})

= ∑_{i=1}^{n} ∑_{j=1}^{n} α_{i}\overline{α_{j}} (z_{i}, z_{j}) = |∑_{i=1}^{n} ∑_{j=1}^{n} α_{i}\overline{α_{j}} (z_{i}, z_{j})|

≤ ∑_{i=1}^{n} ∑_{j=1}^{n} |α_{i}| |\overline{α_{j}}| |(z_{i}, z_{j})|

= ∑_{i=1}^{n} |α_{i}|^{2} ∥z_{i}∥^{2} + ∑_{1≤i≠j≤n} |α_{i}| |α_{j}| |(z_{i}, z_{j})| .

Using Hölder’s inequality, we may write that

(1.5) ∑_{i=1}^{n} |α_{i}|^{2} ∥z_{i}∥^{2}

≤ \{ \begin{matrix}max_{1≤i≤n} |α_{i}|^{2} ∑_{i=1}^{n} ∥z_{i}∥^{2} ; \\ (∑_{i=1}^{n} |α_{i}|^{2α})^{\frac{1}{α}} (∑_{i=1}^{n} ∥z_{i}∥^{2β})^{\frac{1}{β}} , where α > 1, \frac{1}{α} + \frac{1}{β} = 1; \\ ∑_{i=1}^{n} |α_{i}|^{2} max_{1≤i≤n} ∥z_{i}∥^{2} . \end{matrix}

By Hölder’s inequality for double sums we also have

∑_{1≤i≠j≤n} |α_{i}| |α_{j}| |(z_{i}, z_{j})| (1.6)

≤ \{ \begin{matrix}max_{1≤i≠j≤n} |α_{i}α_{j}| ∑_{1≤i≠j≤n} |(z_{i}, z_{j})|; \\ ( ∑_{1≤i≠j≤n} |α_{i}|^{γ} |α_{j}|^{γ})^{\frac{1}{γ}} ( ∑_{1≤i≠j≤n} |(z_{i}, z_{j})|^{δ})^{\frac{1}{δ}} , \\ where γ > 1, \frac{1}{γ} + \frac{1}{δ} = 1; \\ ∑_{1≤i≠j≤n} |α_{i}| |α_{j}| max_{1≤i≠j≤n} |(z_{i}, z_{j})| , \end{matrix}

= \{ \begin{matrix}max_{1≤i≠j≤n} \{|α_{i}α_{j}|\} ∑_{1≤i≠j≤n} |(z_{i}, z_{j})|; \\ [(∑_{i=1}^{n} |α_{i}|^{γ})^{2} − (∑_{i=1}^{n} |α_{i}|^{2γ})]^{\frac{1}{γ}} ( ∑_{1≤i≠j≤n} |(z_{i}, z_{j})|^{δ})^{\frac{1}{δ}} , \\ where γ > 1, \frac{1}{γ} + \frac{1}{δ} = 1; \\ [(∑_{i=1}^{n} |α_{i}|)^{2} − ∑_{i=1}^{n} |α_{i}|^{2}] max_{1≤i≠j≤n} |(z_{i}, z_{j})| . \end{matrix}

Utilising (1.5) and (1.6) in (1.4), we may deduce the desired result (1.3).

Remark 44. Inequality (1.3) contains in fact 9 different inequalities which may be obtained by combining the first 3 ones with the last 3 ones.

A particular case that may be related to the Boas-Bellman result is embodied in the following inequality [5].

Corollary 36. With the assumptions in Lemma 6, we have

(1.7)  ∥∑_{i=1}^{n} α_{i}z_{i}∥^{2}

≤ ∑_{i=1}^{n} |α_{i}|^{2}

· \{ max_{1≤i≤n} ∥z_{i}∥^{2} + \begin{matrix}[(∑_{i=1}^{n} |α_{i}|^{2})^{2} − ∑_{i=1}^{n} |α_{i}|^{4}] & \frac{1}{2} \\ ∑ & {n \atop i=1} & |α_{i}| & 2 \end{matrix} ( ∑_{1≤i≠j≤n} |(z_{i}, z_{j})|^{2})^{\frac{1}{2}} \}

≤ ∑_{i=1}^{n} |α_{i}|^{2} \{ max_{1≤i≤n} ∥z_{i}∥^{2} + ( ∑_{1≤i≠j≤n} |(z_{i}, z_{j})|^{2})^{\frac{1}{2}} \} .

The first inequality follows by taking the third branch in the first curly bracket with the second branch in the second curly bracket for γ = δ = 2.

The second inequality in (1.7) follows by the fact that

[(∑_{i=1}^{n} |α_{i}|^{2})^{2} − ∑_{i=1}^{n} |α_{i}|^{4}]^{\frac{1}{2}} ≤ ∑_{i=1}^{n} |α_{i}|^{2} .

Applying the following Cauchy-Bunyakovsky-Schwarz type inequality

(∑_{i=1}^{n} a_{i})^{2} ≤ n ∑_{i=1}^{n} a_{i}^{2}, a_{i} ∈ ℝ_{+}, 1 ≤ i ≤ n,

we may write that

(1.8) (∑_{i=1}^{n} |α_{i}|^{γ})^{2} − ∑_{i=1}^{n} |α_{i}|^{2γ} ≤ (n − 1) ∑_{i=1}^{n} |α_{i}|^{2γ}  (n ≥ 1)

and

(1.9)  (∑_{i=1}^{n} |α_{i}|)^{2} − ∑_{i=1}^{n} |α_{i}|^{2} ≤ (n − 1) ∑_{i=1}^{n} |α_{i}|^{2}  (n ≥ 1) .

Also, it is obvious that:

(1.10)  max_{1≤i≠j≤n} {|α_{i}α_{j}|} ≤ max_{1≤i≤n} |α_{i}|^{2} .

Consequently, we may state the following coarser upper bounds for ∥∑_{i=1}^{n} α_{i}z_{i}∥^{2} that may be useful in applications [5].

Corollary 37. With the assumptions in Lemma 6, we have the inequalities:

(1.11) ∥∑_{i=1}^{n} α_{i}z_{i}∥^{2}

≤ \{ \begin{matrix}max_{1≤i≤n} |α_{i}|^{2} ∑_{i=1}^{n} ∥z_{i}∥^{2} ; \\ (∑_{i=1}^{n} |α_{i}|^{2α})^{\frac{1}{α}} (∑_{i=1}^{n} ∥z_{i}∥^{2β})^{\frac{1}{β}} , where α > 1, \frac{1}{α} + \frac{1}{β} = 1; \\ ∑_{i=1}^{n} |α_{i}|^{2} max_{1≤i≤n} ∥z_{i}∥^{2} , \end{matrix}

+ \{ \begin{matrix}max_{1≤i≤n} |α_{i}|^{2} ∑_{1≤i≠j≤n} |(z_{i}, z_{j})|; \\ (n − 1)^{\frac{1}{γ}} (∑_{i=1}^{n} |α_{i}|^{2γ})^{\frac{1}{γ}} ( ∑_{1≤i≠j≤n} |(z_{i}, z_{j})|^{δ})^{\frac{1}{δ}} , \\ where γ > 1, \frac{1}{γ} + \frac{1}{δ} = 1; \\ (n − 1) ∑_{i=1}^{n} |α_{i}|^{2} max_{1≤i≠j≤n} |(z_{i}, z_{j})| . \end{matrix}

The proof is obvious by Lemma 6 in applying the inequalities (1.8)

– (1.10).

Remark 45. The following inequalities which are incorporated in (1.11) are of special interest:

(1.12)  ∥∑_{i=1}^{n} α_{i}z_{i}∥^{2} ≤ max_{1≤i≤n} |α_{i}|^{2} [∑_{i=1}^{n} ∥z_{i}∥^{2} + ∑_{1≤i≠j≤n} |(z_{i}, z_{j})|];

(1.13) ∥∑_{i=1}^{n} α_{i}z_{i}∥^{2} ≤ (∑_{i=1}^{n} |α_{i}|^{2p})^{\frac{1}{p}} [(∑_{i=1}^{n} ∥z_{i}∥^{2q})^{\frac{1}{q}}

+ (n − 1) ^{\frac{1}{p}} ( ∑_{1≤i≠j≤n} |(z_{i}, z_{j})|^{q})^{\frac{1}{q}} ] ,

where p > 1, \frac{1}{p} + \frac{1}{q} = 1; and

(1.14)  ∥∑_{i=1}^{n} α_{i}z_{i}∥^{2}

≤ ∑_{i=1}^{n} |α_{i}|^{2} [ max_{1≤i≤n} ∥z_{i}∥^{2} + (n − 1) max_{1≤i≠j≤n} |(z_{i}, z_{j})|] .

1.3. Mitrinović-Pečarić-Fink Type Inequalities. We are now able to present the following result obtained in [5], which complements the inequality (1.2) due to Mitrinović, Pečarić and Fink [11, p. 392].

Theorem 49. Let x, y_{1}, …, y_{n} be vectors of an inner product space (H; (·, ·)) and c_{1}, …, c_{n} ∈ 𝕂 (𝕂 = ℂ, ℝ) . Then one has the inequalities:

|∑_{i=1}^{n} c_{i} (x, y_{i})|^{2}

≤ ∥x∥^{2} × \{ \begin{matrix}max_{1≤i≤n} |c_{i}|^{2} ∑_{i=1}^{n} ∥y_{i}∥^{2} ; \\ (∑_{i=1}^{n} |c_{i}|^{2α})^{\frac{1}{α}} (∑_{i=1}^{n} ∥y_{i}∥^{2β})^{\frac{1}{β}} , where α > 1, \\ \frac{1}{α} + \frac{1}{β} = 1; \\ ∑_{i=1}^{n} |c_{i}|^{2} max_{1≤i≤n} ∥y_{i}∥^{2} , \end{matrix}

(1.15) + ∥x∥^{2} × \{ \begin{matrix}max_{1≤i≠j≤n} \{|c_{i}c_{j}|\} ∑_{1≤i≠j≤n} |(y_{i}, y_{j})|; \\ [(∑_{i=1}^{n} |c_{i}|^{γ})^{2} − (∑_{i=1}^{n} |c_{i}|^{2γ})] & \frac{1}{γ} \\ × (∑_{1≤i≠j≤n} |(y_{i}, y_{j})|^{δ})^{\frac{1}{δ}} , \\ where γ > 1, \frac{1}{γ} + \frac{1}{δ} = 1; \\ [(∑_{i=1}^{n} |c_{i}|)^{2} − ∑_{i=1}^{n} |c_{i}|^{2}] max_{1≤i≠j≤n} |(y_{i}, y_{j})| . \end{matrix}

Proof. We note that

∑_{i=1}^{n} c_{i} (x, y_{i}) = (x, ∑_{i=1}^{n} \overline{c_{i}}y_{i}) .

Using Schwarz’s inequality in inner product spaces, we have:

|∑_{i=1}^{n} c_{i} (x, y_{i})|^{2} ≤ ∥x∥^{2} ∥∑_{i=1}^{n} \overline{c_{i}}y_{i}∥^{2} .

Now using Lemma 6 with α_{i} = \overline{c_{i}}, z_{i} = y_{i} (i = 1, …, n) , we deduce the desired inequality (1.15).

The following particular inequalities that may be obtained by the Corollaries 36 and 37 and Remark 45 hold [5].

Corollary 38. With the assumptions in Theorem 49, one has the inequalities:

(1.16) |∑_{i=1}^{n} c_{i} (x, y_{i})|^{2}

≤ ∥x∥^{2} × \{ \begin{matrix}∑_{i=1}^{n} |c_{i}|^{2} \{ max_{1≤i≤n} ∥y_{i}∥^{2} + (∑_{1≤i≠j≤n} |(y_{i}, y_{j})|^{2})^{\frac{1}{2}} \}; \\ max_{1≤i≤n} |c_{i}|^{2} \{∑_{i=1}^{n} ∥y_{i}∥^{2} + ∑_{1≤i≠j≤n} |(y_{i}, y_{j})|\}; \\ (∑_{i=1}^{n} |c_{i}|^{2p}) & \frac{1}{p} & \{(∑_{i=1}^{n} ∥y_{i}∥^{2q})^{\frac{1}{q}} \\ + (n − 1)^{\frac{1}{p}} (∑_{1≤i≠j≤n} |(y_{i}, y_{j})|^{q})^{\frac{1}{q}} \} , \\ where p > 1, \frac{1}{p} + \frac{1}{q} = 1; \\ ∑_{i=1}^{n} |c_{i}|^{2} \{ max_{1≤i≤n} ∥y_{i}∥^{2} + (n − 1) max_{1≤i≠j≤n} |(y_{i}, y_{j})|\} . \end{matrix}

Remark 46. Note that the first inequality in (1.16) is the result obtained by Mitrinović-Pečarić-Fink in [11]. The other 3 provide similar bounds in terms of the p−norms of the vector (|c_{1}|^{2} , …, |c_{n}|^{2}) .

1.4. Boas-Bellman Type Inequalities. If one chooses c_{i} = \overline{(x, y_{i})} (i = 1, …, n) in (1.15), then it is possible to obtain 9 different inequalities between the Fourier coefficients (x, y_{i}) and the norms and inner products of the vectors y_{i} (i = 1, …, n) . We restrict ourselves only to those inequalities that may be obtained from (1.16).

As Mitrinović, Pečarić and Fink noted in [11, p. 392], the first inequality in (1.16) for the above selection of c_{i} will produce the Boas-Bellman inequality (1.1).

From the second inequality in (1.16) for c_{i} = \overline{(x, y_{i})} we get

(∑_{i=1}^{n} |(x, y_{i})|^{2})^{2} ≤ ∥x∥^{2} max_{1≤i≤n} |(x, y_{i})|^{2} {∑_{i=1}^{n} ∥y_{i}∥^{2} + ∑_{1≤i≠j≤n} |(y_{i}, y_{j})|} .

Taking the square root in this inequality we obtain:

(1.17) ∑_{i=1}^{n} |(x, y_{i})|^{2}

≤ ∥x∥ max_{1≤i≤n} |(x, y_{i})| \{∑_{i=1}^{n} ∥y_{i}∥^{2} + ∑_{1≤i≠j≤n} |(y_{i}, y_{j})|\}^{\frac{1}{2}} ,

for any x, y_{1}, …, y_{n} vectors in the inner product space (H; (·, ·)) .

If we assume that (e_{i})_{1≤i≤n} is an orthonormal family in H, then by (1.17) we have

∑_{i=1}^{n} |(x, e_{i})|^{2} ≤ \sqrt{n} ∥x∥ max_{1≤i≤n} |(x, e_{i})| , x ∈ H.

From the third inequality in (1.16) for c_{i} = \overline{(x, y_{i})} we deduce

(∑_{i=1}^{n} |(x, y_{i})|^{2})^{2} ≤ ∥x∥^{2} (∑_{i=1}^{n} |(x, y_{i})|^{2p})^{\frac{1}{p}}

× \{(∑_{i=1}^{n} ∥y_{i}∥^{2q})^{\frac{1}{q}} + (n − 1)^{\frac{1}{p}} ( ∑_{1≤i≠j≤n} |(y_{i}, y_{j})|^{q})^{\frac{1}{q}} \} ,

for p > 1, \frac{1}{p} + \frac{1}{q} = 1.

Taking the square root in this inequality we get

(1.18) ∑_{i=1}^{n} |(x, y_{i})|^{2} ≤ ∥x∥ (∑_{i=1}^{n} |(x, y_{i})|^{2p})^{\frac{1}{2p}}

× \{(∑_{i=1}^{n} ∥y_{i}∥^{2q})^{\frac{1}{q}} + (n − 1)^{\frac{1}{p}} ( ∑_{1≤i≠j≤n} |(y_{i}, y_{j})|^{q})^{\frac{1}{q}} \}^{\frac{1}{2}} ,

for any x, y_{1}, …, y_{n} ∈ H, p > 1, \frac{1}{p} + \frac{1}{q} = 1.

The above inequality (1.18) becomes, for an orthornormal family (e_{i})_{1≤i≤n} ,

∑_{i=1}^{n} |(x, e_{i})|^{2} ≤ n^{\frac{1}{q}} ∥x∥ (∑_{i=1}^{n} |(x, e_{i})|^{2p})^{\frac{1}{2p}} , x ∈ H.

Finally, the choice c_{i} = \overline{(x, y_{i})} (i = 1, …, n) will produce in the last inequality in (1.16)

(∑_{i=1}^{n} |(x, y_{i})|^{2})^{2}

≤ ∥x∥^{2} ∑_{i=1}^{n} |(x, y_{i})|^{2} { max_{1≤i≤n} ∥y_{i}∥^{2} + (n − 1) max_{1≤i≠j≤n} |(y_{i}, y_{j})|}

giving the following Boas-Bellman type inequality

(1.19) ∑_{i=1}^{n} |(x, y_{i})|^{2} ≤ ∥x∥^{2} { max_{1≤i≤n} ∥y_{i}∥^{2} + (n − 1) max_{1≤i≠j≤n} |(y_{i}, y_{j})|} ,

for any x, y_{1}, …, y_{n} ∈ H.

It is obvious that (1.19) will give for orthonormal families the well known Bessel inequality.

Remark 47. In order the compare the Boas-Bellman result with our result (1.19), it is enough to compare the quantities

A := ( ∑_{1≤i≠j≤n} |(y_{i}, y_{j})|^{2})^{\frac{1}{2}}

and

B := (n − 1) max_{1≤i≠j≤n} |(y_{i}, y_{j})| .

Consider the inner product space H = ℝ with (x, y) = xy, and choose n = 3, y_{1} = a > 0, y_{2} = b > 0, y_{3} = c > 0. Then

A = \sqrt{2} (a^{2}b^{2} + b^{2}c^{2} + c^{2}a^{2})^{\frac{1}{2}} , B = 2 max (ab, ac, bc) .

Denote ab = p, bc = q, ca = r. Then

A = \sqrt{2} (p^{2} + q^{2} + r^{2})^{\frac{1}{2}} , B = 2 max (p, q, r) .

Firstly, if we assume that p = q = r, then A = \sqrt{6}p, B = 2p which shows that A > B.

Now choose r = 1 and p, q = \frac{1}{2}. Then A = \sqrt{3} and B = 2 showing that B > A.

Consequently, in general, the Boas-Bellman inequality and our inequality (1.19) cannot be compared.

2. Bombieri Type Inequalities

2.1. Introduction. In 1971, E. Bombieri [3] (see also [11, p. 394]) gave the following generalisation of Bessel’s inequality.

Theorem 50. If x, y_{1}, …, y_{n} are vectors in the inner product space (H; (·, ·)) , then the following inequality:

(2.1)  ∑_{i=1}^{n} |(x, y_{i})|^{2} ≤ ∥x∥^{2} max_{1≤i≤n} {∑_{j=1}^{n} |(y_{i}, y_{j})|} ,

holds.

It is obvious that if (y_{i})_{1≤i≤n} are orthonormal, then from (2.1) one can deduce Bessel’s inequality.

Another generalisation of Bessel’s inequality was obtained by A. Selberg (see for example [11, p. 394]):

Theorem 51. Let x, y_{1}, …, y_{n} be vectors in H with y_{i} ≠ 0 (i = 1, …, n) . Then one has the inequality:

(2.2)  ∑_{i=1}^{n} \begin{matrix}|(x, y_{i})| & 2 \\ ∑ & {n \atop j=1} & |(y_{i}, y_{j})| \end{matrix} ≤ ∥x∥^{2} .

In this case, also, if (y_{i})_{1≤i≤n} are orthonormal, then from (2.2) one may deduce Bessel’s inequality.

Another type of inequality related to Bessel’s result, was discovered in 1958 by H. Heilbronn [10] (see also [11, p. 395]).

Theorem 52. With the assumptions in Theorem 50, one has

(2.3)  ∑_{i=1}^{n} |(x, y_{i})| ≤ ∥x∥ (∑_{i,j=1}^{n} |(y_{i}, y_{j})|)^{\frac{1}{2}} .

If in (2.3) one chooses y_{i} = e_{i} (i = 1, …, n) , where (e_{i})_{1≤i≤n} are orthonormal vectors in H, then

∑_{i=1}^{n} |(x, e_{i})| ≤ \sqrt{n} ∥x∥ , for any x ∈ H.

In 1992, J.E. Pečarić [12] (see also [11, p. 394]) proved the following general inequality in inner product spaces.

Theorem 53. Let x, y_{1}, …, y_{n} ∈ H and c_{1}, …, c_{n} ∈ 𝕂. Then

(2.4)  |∑_{i=1}^{n} c_{i} (x, y_{i})|^{2} ≤ ∥x∥^{2} ∑_{i=1}^{n} |c_{i}|^{2} (∑_{j=1}^{n} |(y_{i}, y_{j})|)

≤ ∥x∥^{2} ∑_{i=1}^{n} |c_{i}|^{2} max_{1≤i≤n} {∑_{j=1}^{n} |(y_{i}, y_{j})|} .

He showed that the Bombieri inequality (2.1) may be obtained from (2.4) for the choice c_{i} = \overline{(x, y_{i})} (using the second inequality), the Selberg inequality (2.2) may be obtained from the first part of (2.4) for the choice

c_{i} = \frac{\overline{(x, y_{i})}}{∑_{j=1}^{n} |(y_{i}, y_{j})|}, i ∈ \{1, …, n\};

while the Heilbronn inequality (2.3) may be obtained from the first part of (2.4) if one chooses c_{i} = \frac{\overline{(x,y_{i})}}{|(x,y_{i})|}, for any i ∈ \{1, …, n\} .

For other results connected with the above ones, see [7] and [8].

2.2. Some Norm Inequalities. We start with the following lemma which is also interesting in itself [6].

Lemma 7. Let z_{1}, …, z_{n} ∈ H and α_{1}, …, α_{n} ∈ 𝕂. Then one has the inequality:

(2.5)  ∥∑_{i=1}^{n} α_{i}z_{i}∥^{2} ≤ \{ \begin{matrix}A \\ B \\ C \end{matrix} ,

where

A := \{ \begin{matrix}max_{1≤k≤n} |α_{k}|^{2} ∑_{i,j=1}^{n} |(z_{i}, z_{j})|; \\ max_{1≤k≤n} |α_{k}| (∑_{i=1}^{n} |α_{i}|^{r})^{\frac{1}{r}} (∑_{i=1}^{n} (∑_{j=1}^{n} |(z_{i}, z_{j})|)^{s})^{\frac{1}{s}} , \\ r > 1, \frac{1}{r} + \frac{1}{s} = 1; \\ max_{1≤k≤n} |α_{k}| ∑_{k=1}^{n} |α_{k}| max_{1≤i≤n} (∑_{j=1}^{n} |(z_{i}, z_{j})|); \end{matrix}

136 4. GENERALISATIONS OF BESSEL’S INEQUALITY

B := \{ \begin{matrix}(∑_{k=1}^{n} |α_{k}|^{p})^{\frac{1}{p}} max_{1≤i≤n} |α_{i}| (∑_{i=1}^{n} (∑_{j=1}^{n} |(z_{i}, z_{j})|)^{q})^{\frac{1}{q}} , \\ p > 1, \frac{1}{p} + \frac{1}{q} = 1; \\ (∑_{k=1}^{n} |α_{k}|^{p})^{\frac{1}{p}} (∑_{i=1}^{n} |α_{i}|^{t})^{\frac{1}{t}} [∑_{i=1}^{n} (∑_{j=1}^{n} |(z_{i}, z_{j})|^{q})^{\frac{u}{q}} ]^{\frac{1}{u}} , \\ p > 1, \frac{1}{p} + \frac{1}{q} = 1; t > 1, \frac{1}{t} + \frac{1}{u} = 1; \\ (∑_{k=1}^{n} |α_{k}|^{p})^{\frac{1}{p}} ∑_{i=1}^{n} |α_{i}| max_{1≤i≤n} \{(∑_{j=1}^{n} |(z_{i}, z_{j})|^{q})^{\frac{1}{q}} \} , \\ p > 1, \frac{1}{p} + \frac{1}{q} = 1; \end{matrix}

and

C := \{ \begin{matrix}∑_{k=1}^{n} |α_{k}| max_{1≤i≤n} |α_{i}| ∑_{i=1}^{n} [ max_{1≤j≤n} |(z_{i}, z_{j})|]; \\ ∑_{k=1}^{n} |α_{k}| (∑_{i=1}^{n} |α_{i}|^{m})^{\frac{1}{m}} (∑_{i=1}^{n} [ max_{1≤j≤n} |(z_{i}, z_{j})|]^{l})^{\frac{1}{l}} , \\ m > 1, \frac{1}{m} + \frac{1}{l} = 1; \\ (∑_{k=1}^{n} |α_{k}|)^{2} max_{1≤j≤n} |(z_{i}, z_{j})| . \end{matrix}

Proof. We observe that

∥∑_{i=1}^{n} α_{i}z_{i}∥^{2} = (∑_{i=1}^{n} α_{i}z_{i}, ∑_{j=1}^{n} α_{j}z_{j})

= ∑_{i=1}^{n} ∑_{j=1}^{n} α_{i}\overline{α_{j}} (z_{i}, z_{j}) = |∑_{i=1}^{n} ∑_{j=1}^{n} α_{i}\overline{α_{j}} (z_{i}, z_{j})|

≤ ∑_{i=1}^{n} ∑_{j=1}^{n} |α_{i}| |α_{j}| |(z_{i}, z_{j})| = ∑_{i=1}^{n} |α_{i}| (∑_{j=1}^{n} |α_{j}| |(z_{i}, z_{j})|)

:= M.

Using Hölder’s inequality, we may write that

∑_{j=1}^{n} |α_{j}| |(z_{i}, z_{j})| ≤ \{ \begin{matrix}max_{1≤k≤n} & |α_{k}| & ∑_{j=1}^{n} & |(z_{i}, z_{j})| \\ (∑_{k=1}^{n} |α_{k}|^{p})^{\frac{1}{p}} (∑_{j=1}^{n} |(z_{i}, z_{j})|^{q})^{\frac{1}{q}} , \\ p > 1, \frac{1}{p} + \frac{1}{q} = 1; \\ ∑_{k=1}^{n} & |α_{k}| & max_{1≤j≤n} & |(z_{i}, z_{j})| \end{matrix}

for any i ∈ {1, …, n} , giving

M ≤ \{ \begin{matrix}max_{1≤k≤n} |α_{k}| ∑_{i=1}^{n} |α_{i}| ∑_{j=1}^{n} |(z_{i}, z_{j})| =: M_{1}; \\ (∑_{k=1}^{n} |α_{k}|^{p})^{\frac{1}{p}} ∑_{i=1}^{n} |α_{i}| (∑_{j=1}^{n} |(z_{i}, z_{j})|^{q})^{\frac{1}{q}} := M_{p}, \\ p > 1, \frac{1}{p} + \frac{1}{q} = 1; \\ ∑_{k=1}^{n} |α_{k}| ∑_{i=1}^{n} |α_{i}| max_{1≤j≤n} |(z_{i}, z_{j})| =: M_{∞}. \end{matrix}

By Hölder’s inequality we also have:

∑_{i=1}^{n} |α_{i}| (∑_{j=1}^{n} |(z_{i}, z_{j})|) ≤ \{ \begin{matrix}max_{1≤i≤n} |α_{i}| ∑_{i,j=1}^{n} |(z_{i}, z_{j})|; \\ (∑_{i=1}^{n} |α_{i}|^{r})^{\frac{1}{r}} (∑_{i=1}^{n} (∑_{j=1}^{n} |(z_{i}, z_{j})|)^{s})^{\frac{1}{s}} , \\ r > 1, \frac{1}{r} + \frac{1}{s} = 1; \\ ∑_{i=1}^{n} |α_{i}| max_{1≤i≤n} (∑_{j=1}^{n} |(z_{i}, z_{j})|); \end{matrix}

and thus

M_{1} ≤ \{ \begin{matrix}max_{1≤k≤n} |α_{k}|^{2} ∑_{i,j=1}^{n} |(z_{i}, z_{j})|; \\ max_{1≤k≤n} |α_{k}| (∑_{i=1}^{n} |α_{i}|^{r})^{\frac{1}{r}} (∑_{i=1}^{n} (∑_{j=1}^{n} |(z_{i}, z_{j})|)^{s})^{\frac{1}{s}} , \\ r > 1, \frac{1}{r} + \frac{1}{s} = 1; \\ max_{1≤k≤n} |α_{k}| ∑_{i=1}^{n} |α_{i}| max_{1≤i≤n} (∑_{j=1}^{n} |(z_{i}, z_{j})|); \end{matrix}

and the first 3 inequalities in (2.5) are obtained.

By Hölder’s inequality we also have:

M_{p} ≤ (∑_{k=1}^{n} |α_{k}|^{p})^{\frac{1}{p}} × \{ \begin{matrix}max_{1≤i≤n} |α_{i}| ∑_{i=1}^{n} (∑_{j=1}^{n} |(z_{i}, z_{j})|^{q})^{\frac{1}{q}} ; \\ (∑_{i=1}^{n} |α_{i}|^{t})^{\frac{1}{t}} (∑_{i=1}^{n} (∑_{j=1}^{n} |(z_{i}, z_{j})|^{q})^{\frac{u}{q}} )^{\frac{1}{u}} , \\ t > 1, \frac{1}{t} + \frac{1}{u} = 1; \\ ∑_{i=1}^{n} |α_{i}| max_{1≤i≤n} \{(∑_{j=1}^{n} |(z_{i}, z_{j})|^{q})^{\frac{1}{q}} \}; \end{matrix}

and the next 3 inequalities in (2.5) are proved.

Finally, by the same Hölder inequality we may state that:

M_{∞} ≤ ∑_{k=1}^{n} |α_{k}| × \{ \begin{matrix}max_{1≤i≤n} |α_{i}| ∑_{i=1}^{n} ( max_{1≤j≤n} |(z_{i}, z_{j})|); \\ (∑_{i=1}^{n} |α_{i}|^{m})^{\frac{1}{m}} (∑_{i=1}^{n} ( max_{1≤j≤n} |(z_{i}, z_{j})|)^{l})^{\frac{1}{l}} , \\ m > 1, \frac{1}{m} + \frac{1}{l} = 1; \\ ∑_{i=1}^{n} |α_{i}| max_{1≤i,j≤n} |(z_{i}, z_{j})|; \end{matrix}

and the last 3 inequalities in (2.5) are proved.

If we would like to have some bounds for ∥∑_{i=1}^{n} α_{i}z_{i}∥^{2} in terms of ∑_{i=1}^{n} |α_{i}|^{2} , then the following corollaries may be used.

Corollary 39. Let z_{1}, …, z_{n} and α_{1}, …, α_{n} be as in Lemma 7. If 1 < p ≤ 2, 1 < t ≤ 2, then one has the inequality

(2.6) ∥∑_{i=1}^{n} α_{i}z_{i}∥^{2} ≤ n^{\frac{1}{p} + \frac{1}{t} −1} ∑_{k=1}^{n} |α_{k}|^{2} [∑_{i=1}^{n} (∑_{j=1}^{n} |(z_{i}, z_{j})|^{q})^{\frac{u}{q}} ]^{\frac{1}{u}} ,

where \frac{1}{p} + \frac{1}{q} = 1, \frac{1}{t} + \frac{1}{u} = 1.

Proof. Observe, by the monotonicity of power means, we may write that

(\frac{∑_{k=1}^{n} |α_{k}|^{p}}{n} )^{\frac{1}{p}} ≤ (\frac{∑_{k=1}^{n} |α_{k}|^{2}}{n} )^{\frac{1}{2}} ; 1 < p ≤ 2,

(\frac{∑_{k=1}^{n} |α_{k}|^{t}}{n} )^{\frac{1}{t}} ≤ (\frac{∑_{k=1}^{n} |α_{k}|^{2}}{n} )^{\frac{1}{2}} ; 1 < t ≤ 2,

from where we get

(∑_{k=1}^{n} |α_{k}|^{p})^{\frac{1}{p}} ≤ n^{\frac{1}{p} − \frac{1}{2}} (∑_{k=1}^{n} |α_{k}|^{2})^{\frac{1}{2}} ,

(∑_{k=1}^{n} |α_{k}|^{t})^{\frac{1}{t}} ≤ n^{\frac{1}{t} − \frac{1}{2}} (∑_{k=1}^{n} |α_{k}|^{2})^{\frac{1}{2}} .

Using the fifth inequality in (2.5), we then deduce (2.6).

Remark 48. An interesting particular case is the one for p = q = t = u = 2, giving

∥∑_{i=1}^{n} α_{i}z_{i}∥^{2} ≤ ∑_{k=1}^{n} |α_{k}|^{2} (∑_{i,j=1}^{n} |(z_{i}, z_{j})|^{2})^{\frac{1}{2}} .

Corollary 40. With the assumptions of Lemma 7 and if 1 < p ≤ 2, then

(2.7)  ∥∑_{i=1}^{n} α_{i}z_{i}∥^{2} ≤ n^{\frac{1}{p}} ∑_{k=1}^{n} |α_{k}|^{2} max_{1≤i≤n} [(∑_{j=1}^{n} |(z_{i}, z_{j})|^{q})^{\frac{1}{q}} ] ,

where \frac{1}{p} + \frac{1}{q} = 1.

Proof. Since

(∑_{k=1}^{n} |α_{k}|^{p})^{\frac{1}{p}} ≤ n^{\frac{1}{p} − \frac{1}{2}} (∑_{k=1}^{n} |α_{k}|^{2})^{\frac{1}{2}} ,

and

∑_{k=1}^{n} |α_{k}| ≤ n^{\frac{1}{2}} (∑_{k=1}^{n} |α_{k}|^{2})^{\frac{1}{2}} ,

then by the sixth inequality in (2.5) we deduce (2.7).

In a similar fashion, one may prove the following two corollaries.

Corollary 41. With the assumptions of Lemma 7 and if 1 < m ≤ 2, then

∥∑_{i=1}^{n} α_{i}z_{i}∥^{2} ≤ n^{\frac{1}{m}} ∑_{k=1}^{n} |α_{k}|^{2} (∑_{i=1}^{n} [ max_{1≤j≤n} |(z_{i}, z_{j})|]^{l})^{\frac{1}{l}} ,

where \frac{1}{m} + \frac{1}{l} = 1.

Corollary 42. With the assumptions of Lemma 7, we have:

∥∑_{i=1}^{n} α_{i}z_{i}∥^{2} ≤ n ∑_{k=1}^{n} |α_{k}|^{2} max_{1≤i,j≤n} |(z_{i}, z_{j})| .

The following lemma may be of interest as well [6].

Lemma 8. With the assumptions of Lemma 7, one has the inequalities

(2.8) ∥∑_{i=1}^{n} α_{i}z_{i}∥^{2} ≤ ∑_{i=1}^{n} |α_{i}|^{2} ∑_{j=1}^{n} |(z_{i}, z_{j})|

≤ \{ \begin{matrix}∑_{i=1}^{n} |α_{i}|^{2} max_{1≤i≤n} [∑_{j=1}^{n} |(z_{i}, z_{j})|]; \\ (∑_{i=1}^{n} |α_{i}|^{2p})^{\frac{1}{p}} ((∑_{j=1}^{n} |(z_{i}, z_{j})|)^{q})^{\frac{1}{q}} , \\ p > 1, \frac{1}{p} + \frac{1}{q} = 1; \\ max_{1≤i≤n} |α_{i}|^{2} ∑_{i,j=1}^{n} |(z_{i}, z_{j})| . \end{matrix}

Proof. As in Lemma 7, we know that

∥∑_{i=1}^{n} α_{i}z_{i}∥^{2} ≤ ∑_{i=1}^{n} ∑_{j=1}^{n} |α_{i}| |α_{j}| |(z_{i}, z_{j})| .

Using the simple observation that (see also [11, p. 394])

|α_{i}| |α_{j}| ≤ \frac{1}{2} (|α_{i}|^{2} + |α_{j}|^{2}) , i, j ∈ \{1, …, n\} ,

we have

∑_{i=1}^{n} ∑_{j=1}^{n} |α_{i}| |α_{j}| |(z_{i}, z_{j})| ≤ \frac{1}{2} ∑_{i,j=1}^{n} (|α_{i}|^{2} + |α_{j}|^{2}) |(z_{i}, z_{j})|

= \frac{1}{2} [∑_{i,j=1}^{n} |α_{i}|^{2} |(z_{i}, z_{j})| + ∑_{i,j=1}^{n} |α_{j}|^{2} |(z_{i}, z_{j})|]

= ∑_{i,j=1}^{n} |α_{i}|^{2} |(z_{i}, z_{j})| ,

which proves the first inequality in (2.8).

The second part follows by Hölder’s inequality and we omit the details.

Remark 49. The first part in (2.8) is the inequality obtained by Pečarić in [12].

2.3. Pečarić Type Inequalities. We are now able to present the following result obtainend in [6], which complements the inequality (2.4) due to J.E. Pečarić [12] (see also [11, p. 394]).

Theorem 54. Let x, y_{1}, …, y_{n} be vectors of an inner product space (H; (·, ·)) and c_{1}, …, c_{n} ∈ 𝕂. Then one has the inequalities:

(2.9)  |∑_{i=1}^{n} c_{i} (x, y_{i})|^{2} ≤ ∥x∥^{2} × \{ \begin{matrix}D \\ E \\ F \end{matrix} ,

where

D := \{ \begin{matrix}max_{1≤k≤n} |c_{k}|^{2} ∑_{i,j=1}^{n} |(y_{i}, y_{j})|; \\ max_{1≤k≤n} |c_{k}| (∑_{i=1}^{n} |c_{i}|^{r})^{\frac{1}{r}} [∑_{i=1}^{n} (∑_{j=1}^{n} |(y_{i}, y_{j})|)^{s}]^{\frac{1}{s}} , \\ r > 1, \frac{1}{r} + \frac{1}{s} = 1; \\ max_{1≤k≤n} |c_{k}| ∑_{i=1}^{n} |c_{i}| max_{1≤i≤n} (∑_{j=1}^{n} |(y_{i}, y_{j})|); \end{matrix}

E := \{ \begin{matrix}(∑_{k=1}^{n} |c_{k}|^{p})^{\frac{1}{p}} max_{1≤i≤n} |c_{i}| (∑_{i=1}^{n} (∑_{j=1}^{n} |(y_{i}, y_{j})|)^{q})^{\frac{1}{q}} , \\ p > 1, \frac{1}{p} + \frac{1}{q} = 1; \\ (∑_{k=1}^{n} |c_{k}|^{p})^{\frac{1}{p}} (∑_{i=1}^{n} |c_{i}|^{t})^{\frac{1}{t}} [∑_{i=1}^{n} (∑_{j=1}^{n} |(y_{i}, y_{j})|^{q})^{\frac{u}{q}} ]^{\frac{1}{u}} , \\ p > 1, \frac{1}{p} + \frac{1}{q} = 1; t > 1, \frac{1}{t} + \frac{1}{u} = 1; \\ (∑_{k=1}^{n} |c_{k}|^{p})^{\frac{1}{p}} ∑_{i=1}^{n} |c_{i}| max_{1≤i≤n} \{(∑_{j=1}^{n} |(y_{i}, y_{j})|^{q})^{\frac{1}{q}} \} , \\ p > 1, \frac{1}{p} + \frac{1}{q} = 1; \end{matrix}

and

F := \{ \begin{matrix}∑_{k=1}^{n} |c_{k}| max_{1≤i≤n} |c_{i}| ∑_{i=1}^{n} [ max_{1≤j≤n} |(y_{i}, y_{j})|]; \\ ∑_{k=1}^{n} |c_{k}| (∑_{i=1}^{n} |c_{i}|^{m})^{\frac{1}{m}} (∑_{i=1}^{n} [ max_{1≤j≤n} |(y_{i}, y_{j})|]^{l})^{\frac{1}{l}} , \\ m > 1, \frac{1}{m} + \frac{1}{l} = 1; \\ (∑_{k=1}^{n} |c_{k}|)^{2} max_{1≤j≤n} |(y_{i}, y_{j})| . \end{matrix}

Proof. We note that

∑_{i=1}^{n} c_{i} (x, y_{i}) = (x, ∑_{i=1}^{n} \overline{c_{i}}y_{i}) .

Using Schwarz’s inequality in inner product spaces, we have

|∑_{i=1}^{n} c_{i} (x, y_{i})|^{2} ≤ ∥x∥^{2} ∥∑_{i=1}^{n} \overline{c_{i}}y_{i}∥^{2} .

Finally, using Lemma 7 with α_{i} = \overline{c_{i}}, z_{i} = y_{i} (i = 1, …, n) , we deduce the desired inequality (2.9). We omit the details.

The following corollaries may be useful if one needs bounds in terms of ∑_{i=1}^{n} |c_{i}|^{2} .

Corollary 43. With the assumptions in Theorem 54 and if 1 < p ≤ 2, 1 < t ≤ 2, \frac{1}{p} + \frac{1}{q} = 1, \frac{1}{t} + \frac{1}{u} = 1, then one has the inequality:

(2.10) |∑_{i=1}^{n} c_{i} (x, y_{i})|^{2}

≤ ∥x∥^{2} n^{\frac{1}{p} + \frac{1}{t} −1} ∑_{i=1}^{n} |c_{i}|^{2} [∑_{i=1}^{n} (∑_{j=1}^{n} |(y_{i}, y_{j})|^{q})^{\frac{u}{q}} ]^{\frac{1}{u}} ,

and, in particular, for p = q = t = u = 2,

|∑_{i=1}^{n} c_{i} (x, y_{i})|^{2} ≤ ∥x∥^{2} ∑_{i=1}^{n} |c_{i}|^{2} (∑_{i,j=1}^{n} |(y_{i}, y_{j})|^{2})^{\frac{1}{2}} .

The proof is similar to the one in Corollary 39.

Corollary 44. With the assumptions in Theorem 54 and if 1 < p ≤ 2, then

|∑_{i=1}^{n} c_{i} (x, y_{i})|^{2} ≤ ∥x∥^{2} n^{\frac{1}{p}} ∑_{k=1}^{n} |c_{k}|^{2} max_{1≤i≤n} [∑_{j=1}^{n} |(y_{i}, y_{j})|^{q}]^{\frac{1}{q}} ,

where \frac{1}{p} + \frac{1}{q} = 1.

The proof is similar to the one in Corollary 40. The following two inequalities also hold.

Corollary 45. With the above assumptions for x, y_{i}, c_{i} and if 1 < m ≤ 2, then

(2.11) |∑_{i=1}^{n} c_{i} (x, y_{i})|^{2} ≤ ∥x∥^{2} n^{\frac{1}{m}} ∑_{k=1}^{n} |c_{k}|^{2} (∑_{i=1}^{n} [ max_{1≤j≤n} |(y_{i}, y_{j})|]^{l})^{\frac{1}{l}} ,

where \frac{1}{m} + \frac{1}{l} = 1.

Corollary 46. With the above assumptions for x, y_{i}, c_{i}, one has

(2.12)  |∑_{i=1}^{n} c_{i} (x, y_{i})|^{2} ≤ ∥x∥^{2} n ∑_{k=1}^{n} |c_{k}|^{2} max_{1≤j≤n} |(y_{i}, y_{j})| .

Using Lemma 8, we may state the following result as well.

Remark 50. With the assumptions of Theorem 54, one has the inequalities:

|∑_{i=1}^{n} c_{i} (x, y_{i})|^{2} ≤ ∥x∥^{2} ∑_{i=1}^{n} |c_{i}|^{2} ∑_{j=1}^{n} |(y_{i}, y_{j})|

≤ ∥x∥^{2} × \{ \begin{matrix}∑_{i=1}^{n} |c_{i}|^{2} max_{1≤i≤n} [∑_{j=1}^{n} |(y_{i}, y_{j})|]; \\ (∑_{i=1}^{n} |c_{i}|^{2p})^{\frac{1}{p}} (∑_{i=1}^{n} (∑_{j=1}^{n} |(y_{i}, y_{j})|)^{q})^{\frac{1}{q}} , \\ p > 1, \frac{1}{p} + \frac{1}{q} = 1; \\ max_{1≤i≤n} |c_{i}|^{2} ∑_{i,j=1}^{n} |(y_{i}, y_{j})|; \end{matrix}

that provide some alternatives to Pečarić’s result (2.4).

2.4. Inequalities of Bombieri Type. In this section we point out some inequalities of Bombieri type that may be obtained from (2.9) on choosing c_{i} = \overline{(x, y_{i})} (i = 1, …, n) .

If the above choice was made in the first inequality in (2.9), then one can obtain:

(∑_{i=1}^{n} |(x, y_{i})|^{2})^{2} ≤ ∥x∥^{2} max_{1≤i≤n} |(x, y_{i})|^{2} ∑_{i,j=1}^{n} |(y_{i}, y_{j})|

giving, by taking the square root,

(2.13) ∑_{i=1}^{n} |(x, y_{i})|^{2} ≤ ∥x∥ max_{1≤i≤n} |(x, y_{i})| (∑_{i,j=1}^{n} |(y_{i}, y_{j})|)^{\frac{1}{2}} , x ∈ H.

If the same choice for c_{i} is made in the second inequality in (2.9), then one can get

(∑_{i=1}^{n} |(x, y_{i})|^{2})^{2} ≤ ∥x∥^{2} max_{1≤i≤n} |(x, y_{i})| (∑_{i=1}^{n} |(x, y_{i})|^{r})^{\frac{1}{r}}

× [∑_{i=1}^{n} (∑_{j=1}^{n} |(y_{i}, y_{j})|)^{s}]^{\frac{1}{s}} ,

implying

(2.14) ∑_{i=1}^{n} |(x, y_{i})|^{2} ≤ ∥x∥ max_{1≤i≤n} |(x, y_{i})|^{\frac{1}{2}} (∑_{i=1}^{n} |(x, y_{i})|^{r})^{\frac{1}{2r}}

× [∑_{i=1}^{n} (∑_{j=1}^{n} |(y_{i}, y_{j})|)^{s}]^{\frac{1}{2s}} ,

where \frac{1}{r} + \frac{1}{s} = 1, s > 1.

The other inequalities in (2.9) will produce the following results, respectively

(2.15) ∑_{i=1}^{n} |(x, y_{i})|^{2} ≤ ∥x∥ max_{1≤i≤n} |(x, y_{i})|^{\frac{1}{2}} (∑_{i=1}^{n} |(x, y_{i})|)^{\frac{1}{2}}

× [ max_{1≤i≤n} (∑_{j=1}^{n} |(y_{i}, y_{j})|)];

(2.16) ∑_{i=1}^{n} |(x, y_{i})|^{2} ≤ ∥x∥ max_{1≤i≤n} |(x, y_{i})|^{\frac{1}{2}} (∑_{i=1}^{n} |(x, y_{i})|^{p})^{\frac{1}{2p}}

× [∑_{i=1}^{n} (∑_{j=1}^{n} |(y_{i}, y_{j})|^{q})^{\frac{1}{q}} ]^{\frac{1}{2}} ,

where p > 1, \frac{1}{p} + \frac{1}{q} = 1;

(2.17) ∑_{i=1}^{n} |(x, y_{i})|^{2} ≤ ∥x∥ (∑_{i=1}^{n} |(x, y_{i})|^{p})^{\frac{1}{2p}} (∑_{i=1}^{n} |(x, y_{i})|^{t})^{\frac{1}{2t}}

× [∑_{i=1}^{n} (∑_{j=1}^{n} |(y_{i}, y_{j})|^{q})^{\frac{u}{q}} ]^{\frac{1}{2u}} ,

where p > 1, \frac{1}{p} + \frac{1}{q} = 1, t > 1, \frac{1}{t} + \frac{1}{u} = 1;

(2.18) ∑_{i=1}^{n} |(x, y_{i})|^{2} ≤ ∥x∥ (∑_{i=1}^{n} |(x, y_{i})|^{p})^{\frac{1}{2p}} (∑_{i=1}^{n} |(x, y_{i})|)^{\frac{1}{2}}

× max_{1≤i≤n} \{(∑_{j=1}^{n} |(y_{i}, y_{j})|^{q})^{\frac{1}{2q}} \} ,

where p > 1, \frac{1}{p} + \frac{1}{q} = 1;

(2.19) ∑_{i=1}^{n} |(x, y_{i})|^{2} ≤ ∥x∥ [∑_{i=1}^{n} |(x, y_{i})|]^{\frac{1}{2}} max_{1≤i≤n} |(x, y_{i})|^{\frac{1}{2}}

× (∑_{i=1}^{n} [ max_{1≤j≤n} |(y_{i}, y_{j})|])^{\frac{1}{2}} ;

(2.20) ∑_{i=1}^{n} |(x, y_{i})|^{2}

≤ ∥x∥ [∑_{i=1}^{n} |(x, y_{i})|^{m}]^{\frac{1}{2m}} [∑_{i=1}^{n} [ max_{1≤j≤n} |(y_{i}, y_{j})|^{l}]]^{\frac{1}{2l}} ,

where m > 1, \frac{1}{m} + \frac{1}{l} = 1; and

(2.21)  ∑_{i=1}^{n} |(x, y_{i})|^{2} ≤ ∥x∥ ∑_{i=1}^{n} |(x, y_{i})| max_{1≤j≤n} |(y_{i}, y_{j})|^{\frac{1}{2}} .

If in the above inequalities we assume that (y_{i})_{1≤i≤n} = (e_{i})_{1≤i≤n} , where (e_{i})_{1≤i≤n} are orthonormal vectors in the inner product space (H, (·, ·)) , then from (2.13) – (2.21) we may deduce the following inequalities similar in a sense to Bessel’s inequality:

∑_{i=1}^{n} |(x, e_{i})|^{2} ≤ \sqrt{n} ∥x∥ max_{1≤i≤n} \{|(x, e_{i})|\};

∑_{i=1}^{n} |(x, e_{i})|^{2} ≤ n^{\frac{1}{2s}} ∥x∥ max_{1≤i≤n} \{|(x, e_{i})|^{\frac{1}{2}} \} (∑_{i=1}^{n} |(x, e_{i})|^{r})^{\frac{1}{2r}} ,

where r > 1, \frac{1}{r} + \frac{1}{s} = 1;

∑_{i=1}^{n} |(x, e_{i})|^{2} ≤ ∥x∥ max_{1≤i≤n} \{|(x, e_{i})|^{\frac{1}{2}} \} (∑_{i=1}^{n} |(x, e_{i})|)^{\frac{1}{2}} ,

∑_{i=1}^{n} |(x, e_{i})|^{2} ≤ \sqrt{n} ∥x∥ max_{1≤i≤n} \{|(x, e_{i})|^{\frac{1}{2}} \} (∑_{i=1}^{n} |(x, e_{i})|^{p})^{\frac{1}{2p}} ,

where p > 1;

∑_{i=1}^{n} |(x, e_{i})|^{2} ≤ n^{\frac{1}{2u}} ∥x∥ (∑_{i=1}^{n} |(x, e_{i})|^{p})^{\frac{1}{2p}} (∑_{i=1}^{n} |(x, e_{i})|^{t})^{\frac{1}{2t}} ,

where p > 1, t > 1, \frac{1}{t} + \frac{1}{u} = 1;

∑_{i=1}^{n} |(x, e_{i})|^{2} ≤ ∥x∥ (∑_{i=1}^{n} |(x, e_{i})|^{p})^{\frac{1}{2p}} (∑_{i=1}^{n} |(x, e_{i})|)^{\frac{1}{2}} , p > 1;

∑_{i=1}^{n} |(x, e_{i})|^{2} ≤ \sqrt{n} ∥x∥ (∑_{i=1}^{n} |(x, e_{i})|)^{\frac{1}{2}} max_{1≤i≤n} \{|(x, e_{i})|^{\frac{1}{2}} \};

∑_{i=1}^{n} |(x, e_{i})|^{2} ≤ n^{\frac{1}{2l}} ∥x∥ [∑_{i=1}^{n} |(x, e_{i})|^{m}]^{\frac{1}{m}} , m > 1, \frac{1}{m} + \frac{1}{l} = 1;

∑_{i=1}^{n} |(x, e_{i})|^{2} ≤ ∥x∥ ∑_{i=1}^{n} |(x, e_{i})| .

2. BOMBIERI TYPE INEQUALITIES 147

Corollaries 43 – 46 will produce the following results which do not contain the Fourier coefficients in the right side of the inequality.

Indeed, if one chooses c_{i} = \overline{(x, y_{i})} in (2.10), then

(∑_{i=1}^{n} |(x, y_{i})|^{2})^{2} ≤ ∥x∥^{2} n^{\frac{1}{p} + \frac{1}{t} −1} ∑_{i=1}^{n} |(x, y_{i})|^{2} [∑_{i=1}^{n} (∑_{j=1}^{n} |(y_{i}, y_{j})|^{q})^{\frac{u}{q}} ]^{\frac{1}{u}} ,

giving the following Bombieri type inequality:

∑_{i=1}^{n} |(x, y_{i})|^{2} ≤ n^{\frac{1}{p} + \frac{1}{t} −1} ∥x∥^{2} [∑_{i=1}^{n} (∑_{j=1}^{n} |(y_{i}, y_{j})|^{q})^{\frac{u}{q}} ]^{\frac{1}{u}} ,

where 1 < p ≤ 2, 1 < t ≤ 2, \frac{1}{p} + \frac{1}{q} = 1, \frac{1}{t} + \frac{1}{u} = 1.

If in this inequality we consider p = q = t = u = 2, then

∑_{i=1}^{n} |(x, y_{i})|^{2} ≤ ∥x∥^{2} (∑_{i,j=1}^{n} |(y_{i}, y_{j})|^{2})^{\frac{1}{2}} .

For a different proof of this result see also [8].

In a similar way, if c_{i} = \overline{(x, y_{i})} in (2.11), then

∑_{i=1}^{n} |(x, y_{i})|^{2} ≤ n^{\frac{1}{m}} ∥x∥^{2} (∑_{i=1}^{n} [ max_{1≤j≤n} |(y_{i}, y_{j})|]^{l})^{\frac{1}{l}} ,

where m > 1, \frac{1}{m} + \frac{1}{l} = 1.

Finally, if c_{i} = \overline{(x, y_{i})} (i = 1, …, n) , is taken in (2.12), then

∑_{i=1}^{n} |(x, y_{i})|^{2} ≤ n ∥x∥^{2} max_{1≤i,j≤n} |(y_{i}, y_{j})| .

Remark 51. Let us compare Bombieri’s result

(2.22)  ∑_{i=1}^{n} |(x, y_{i})|^{2} ≤ ∥x∥^{2} max_{1≤i≤n} {∑_{j=1}^{n} |(y_{i}, y_{j})|}

with our result

(2.23)  ∑_{i=1}^{n} |(x, y_{i})|^{2} ≤ ∥x∥^{2} \{∑_{i,j=1}^{n} |(y_{i}, y_{j})|^{2}\}^{\frac{1}{2}} .

Denote

M_{1} := max_{1≤i≤n} {∑_{j=1}^{n} |(y_{i}, y_{j})|}

and

M_{2} := [∑_{i,j=1}^{n} |(y_{i}, y_{j})|^{2}]^{\frac{1}{2}} .

If we choose the inner product space H = ℝ, (x, y) := xy and n = 2, then for y_{1} = a, y_{2} = b, a, b > 0, we have

M_{1} = max {a^{2} + ab, ab + b^{2}} = (a + b) max (a, b) ,
M_{2} = (a^{4} + a^{2}b^{2} + a^{2}b^{2} + b^{4})^{\frac{1}{2}} = a^{2} + b^{2}.

Assume that a ≥ b. Then M_{1} = a^{2} + ab ≥ a^{2} + b^{2} = M_{2}, showing that, in this case, the bound provided by (2.23) is better than the bound provided by (2.22). If (y_{i})_{1≤i≤n} are orthonormal vectors, then M_{1} = 1, M_{2} = \sqrt{n}, showing that in this case the Bombieri inequality (which becomes Bessel’s inequality) provides a better bound than (2.23).

3. Pečarić Type Inequalities

3.1. Introduction. In 1992, J.E. Pečarić [12] proved the following inequality for vectors in complex inner product spaces (H; (·, ·)).

Theorem 55. Suppose that x, y_{1}, …, y_{n} are vectors in H and c_{1}, …, c_{n} are complex numbers. Then the following inequalities

(3.1)  |∑_{i=1}^{n} c_{i} (x, y_{i})|^{2} ≤ ∥x∥^{2} ∑_{i=1}^{n} |c_{i}|^{2} (∑_{j=1}^{n} |(y_{i}, y_{j})|)

≤ ∥x∥^{2} ∑_{i=1}^{n} |c_{i}|^{2} max_{1≤i≤n} (∑_{j=1}^{n} |(y_{i}, y_{j})|) ,

hold.

He also showed that for c_{i} = \overline{(x, y_{i})}, i ∈ \{1, …, n\} , one gets

(∑_{i=1}^{n} |(x, y_{i})|^{2})^{2} ≤ ∥x∥^{2} ∑_{i=1}^{n} |(x, y_{i})|^{2} (∑_{j=1}^{n} |(y_{i}, y_{j})|)

≤ ∥x∥^{2} ∑_{i=1}^{n} |(x, y_{i})|^{2} max_{1≤i≤n} (∑_{j=1}^{n} |(y_{i}, y_{j})|) ,

which improves Bombieri’s result [3] (see also [11, p. 394])

(3.2)  ∑_{i=1}^{n} |(x, y_{i})|^{2} ≤ ∥x∥^{2} max_{1≤i≤n} (∑_{j=1}^{n} |(y_{i}, y_{j})|) .

Note that (3.2) is in its turn a natural generalisation of Bessel’s inequality

(3.3)  ∑_{i=1}^{n} |(x, e_{i})|^{2} ≤ ∥x∥^{2} , x ∈ H,

which holds for the orthornormal vectors (e_{i})_{1≤i≤n} .

In this section, by following [4], we point out other related results to Pečarić’s inequality (3.1) than the ones stated in the previous sections. Some results of Bombieri type are also mentioned.

3.2. Some Norm Inequalities. We start with the following lemma that is interesting in its own right [4].

Lemma 9. Let z_{1}, …, z_{n} ∈ H and α_{1}, …, α_{n} ∈ 𝕂. Then one has the inequalities:

(3.4)  ∥∑_{i=1}^{n} α_{i}z_{i}∥^{2} ≤ (∑_{i=1}^{n} |α_{i}|^{p} (∑_{j=1}^{n} |(z_{i}, z_{j})|))^{\frac{1}{p}}

× (∑_{i=1}^{n} |α_{i}|^{q} (∑_{j=1}^{n} |(z_{i}, z_{j})|))^{\frac{1}{q}}

≤ \{ \begin{matrix}A \\ B \\ C \end{matrix} ,

where

A := \{ \begin{matrix}max_{1≤i≤n} |α_{i}|^{2} ∑_{i,j=1}^{n} |(z_{i}, z_{j})|; \\ max_{1≤i≤n} & |α_{i}| & (∑_{i=1}^{n} |α_{i}|^{γq}) & \frac{1}{γq} & (∑_{i,j=1}^{n} |(z_{i}, z_{j})|) & \frac{1}{p} \\ × (∑_{i=1}^{n} (∑_{j=1}^{n} |(z_{i}, z_{j})|)^{δ})^{\frac{1}{δq}} , \\ if γ > 1, \frac{1}{γ} + \frac{1}{δ} = 1; \\ max_{1≤i≤n} & |α_{i}| & (∑_{i=1}^{n} |α_{i}|^{q}) & \frac{1}{q} & (∑_{i,j=1}^{n} |(z_{i}, z_{j})|) & \frac{1}{p} \\ × max_{1≤i≤n} (∑_{j=1}^{n} |(z_{i}, z_{j})|)^{\frac{1}{q}} ; \end{matrix}

150 4. GENERALISATIONS OF BESSEL’S INEQUALITY

B := \{ \begin{matrix}max_{1≤i≤n} & |α_{i}| & (∑_{i=1}^{n} |α_{i}|^{αp}) & \frac{1}{αp} \\ × (∑_{i,j=1}^{n} |(z_{i}, z_{j})|)^{\frac{1}{q}} (∑_{i=1}^{n} (∑_{j=1}^{n} |(z_{i}, z_{j})|)^{β})^{\frac{1}{βq}} , \\ if α > 1, \frac{1}{α} + \frac{1}{β} = 1; \\ (∑_{i=1}^{n} |α_{i}|^{αp}) & \frac{1}{αp} & (∑_{i=1}^{n} |α_{i}|^{γq}) & \frac{1}{γq} \\ × (∑_{i=1}^{n} (∑_{j=1}^{n} |(z_{i}, z_{j})|)^{β})^{\frac{1}{pβ}} (∑_{i=1}^{n} (∑_{j=1}^{n} |(z_{i}, z_{j})|)^{δ})^{\frac{1}{δq}} \\ if α > 1, \frac{1}{α} + \frac{1}{β} = 1 and γ > 1, \frac{1}{γ} + \frac{1}{δ} = 1; \\ (∑_{i=1}^{n} |α_{i}|^{q}) & \frac{1}{q} & (∑_{i=1}^{n} |α_{i}|^{αp}) & \frac{1}{αp} & max_{1≤i≤n} & (∑_{j=1}^{n} |(z_{i}, z_{j})|) & \frac{1}{q} \\ × (∑_{i=1}^{n} (∑_{j=1}^{n} |(z_{i}, z_{j})|)^{β})^{\frac{1}{pβ}} , if α > 1, \frac{1}{α} + \frac{1}{β} = 1; \end{matrix}

and

C := \{ \begin{matrix}max_{1≤i≤n} & |α_{i}| & (∑_{i=1}^{n} |α_{i}|^{p}) & \frac{1}{p} & max_{1≤i≤n} & (∑_{j=1}^{n} |(z_{i}, z_{j})|) & \frac{1}{p} \\ × (∑_{i,j=1}^{n} |(z_{i}, z_{j})|)^{\frac{1}{q}} ; \\ (∑_{i=1}^{n} |α_{i}|^{p}) & \frac{1}{p} & (∑_{i=1}^{n} |α_{i}|^{γq}) & \frac{1}{γq} & max_{1≤i≤n} & (∑_{j=1}^{n} |(z_{i}, z_{j})|) & \frac{1}{p} \\ × (∑_{i=1}^{n} (∑_{j=1}^{n} |(z_{i}, z_{j})|)^{δ})^{\frac{1}{δq}} , if γ > 1, \frac{1}{γ} + \frac{1}{δ} = 1; \\ (∑_{i=1}^{n} |α_{i}|^{p})^{\frac{1}{p}} (∑_{i=1}^{n} |α_{i}|^{q})^{\frac{1}{q}} max_{1≤i≤n} (∑_{j=1}^{n} |(z_{i}, z_{j})|) , \end{matrix}

where p > 1, \frac{1}{p} + \frac{1}{q} = 1.

Proof. We observe that

(3.5) ∥∑_{i=1}^{n} α_{i}z_{i}∥^{2} = (∑_{i=1}^{n} α_{i}z_{i}, ∑_{j=1}^{n} α_{j}z_{j})

= ∑_{i=1}^{n} ∑_{j=1}^{n} α_{i}\overline{α_{j}} (z_{i}, z_{j}) = |∑_{i=1}^{n} ∑_{j=1}^{n} α_{i}\overline{α_{j}} (z_{i}, z_{j})|

≤ ∑_{i=1}^{n} ∑_{j=1}^{n} |α_{i}| |α_{j}| |(z_{i}, z_{j})| =: M.

If one uses the Hölder inequality for double sums, i.e., we recall it

∑_{i,j=1}^{n} m_{ij}a_{ij}b_{ij} ≤ (∑_{i,j=1}^{n} m_{ij}a_{ij}^{p})^{\frac{1}{p}} (∑_{i,j=1}^{n} m_{ij}b_{ij}^{q})^{\frac{1}{q}} ,

where m_{ij}, a_{ij}, b_{ij} ≥ 0, \frac{1}{p} + \frac{1}{q} = 1, p > 1; then

(3.6) M≤ (∑_{i,j=1}^{n} |(z_{i}, z_{j})| |α_{i}|^{p})^{\frac{1}{p}} (∑_{i,j=1}^{n} |(z_{i}, z_{j})| |α_{i}|^{q})^{\frac{1}{q}}

=(∑_{i=1}^{n} |α_{i}|^{p} (∑_{j=1}^{n} |(z_{i}, z_{j})|))^{\frac{1}{p}} (∑_{i=1}^{n} |α_{i}|^{q} (∑_{j=1}^{n} |(z_{i}, z_{j})|))^{\frac{1}{q}} ,

and the first inequality in (3.4) is proved.

Observe that

∑_{i=1}^{n} |α_{i}|^{p} (∑_{j=1}^{n} |(z_{i}, z_{j})|)

≤ \{ \begin{matrix}max_{1≤i≤n} |α_{i}|^{p} ∑_{i,j=1}^{n} |(z_{i}, z_{j})|; \\ (∑_{i=1}^{n} |α_{i}|^{αp}) & \frac{1}{α} & (∑_{i=1}^{n} (∑_{j=1}^{n} |(z_{i}, z_{j})|)^{β}) & \frac{1}{β} \\ if α > 1, \frac{1}{α} + \frac{1}{β} = 1; \\ ∑_{i=1}^{n} |α_{i}|^{p} max_{1≤i≤n} (∑_{j=1}^{n} |(z_{i}, z_{j})|); \end{matrix}

giving

(3.7) (∑_{i=1}^{n} |α_{i}|^{p} (∑_{j=1}^{n} |(z_{i}, z_{j})|))^{\frac{1}{p}}

≤ \{ \begin{matrix}max_{1≤i≤n} |α_{i}| (∑_{i,j=1}^{n} |(z_{i}, z_{j})|)^{\frac{1}{p}} ; \\ (∑_{i=1}^{n} |α_{i}|^{αp}) & \frac{1}{αp} & (∑_{i=1}^{n} (∑_{j=1}^{n} |(z_{i}, z_{j})|)^{β}) & \frac{1}{βp} \\ if α > 1, \frac{1}{α} + \frac{1}{β} = 1; \\ (∑_{i=1}^{n} |α_{i}|^{p})^{\frac{1}{p}} max_{1≤i≤n} (∑_{j=1}^{n} |(z_{i}, z_{j})|)^{\frac{1}{p}} . \end{matrix}

Similarly, we have

(3.8) (∑_{i=1}^{n} |α_{i}|^{q} (∑_{j=1}^{n} |(z_{i}, z_{j})|))^{\frac{1}{q}}

≤ \{ \begin{matrix}max_{1≤i≤n} & |α_{i}| & (∑_{i,j=1}^{n} |(z_{i}, z_{j})|) & \frac{1}{q} \\ (∑_{i=1}^{n} |α_{i}|^{γq}) & \frac{1}{γq} & (∑_{i=1}^{n} (∑_{j=1}^{n} |(z_{i}, z_{j})|)^{δ}) & \frac{1}{δq} \\ if γ > 1, \frac{1}{γ} + \frac{1}{δ} = 1; \\ (∑_{i=1}^{n} |α_{i}|^{q})^{\frac{1}{q}} max_{1≤i≤n} (∑_{j=1}^{n} |(z_{i}, z_{j})|)^{\frac{1}{q}} . \end{matrix}

Using (3.5) and (3.7) – (3.8), we deduce the 9 inequalities in the second part of (3.4).

If we choose p = q = 2, then the following result holds [4].

Corollary 47. If z_{1}, …, z_{n} ∈ H and α_{1}, …, α_{n} ∈ 𝕂, then one has

(3.9)  ∥∑_{i=1}^{n} α_{i}z_{i}∥^{2} ≤ ∑_{i=1}^{n} |α_{i}|^{2} (∑_{j=1}^{n} |(z_{i}, z_{j})|)

≤ \{ \begin{matrix}D \\ E \\ F \end{matrix} ,

where

D := \{ \begin{matrix}max_{1≤i≤n} |α_{i}|^{2} ∑_{i,j=1}^{n} |(z_{i}, z_{j})|; \\ max_{1≤i≤n} & |α_{i}| & (∑_{i=1}^{n} |α_{i}|^{2γ}) & \frac{1}{2γ} & (∑_{i,j=1}^{n} |(z_{i}, z_{j})|) & \frac{1}{2} \\ × (∑_{i=1}^{n} (∑_{j=1}^{n} |(z_{i}, z_{j})|)^{δ})^{\frac{1}{2δ}} , \\ if γ > 1, \frac{1}{γ} + \frac{1}{δ} = 1; \\ max_{1≤i≤n} & |α_{i}| & (∑_{i=1}^{n} |α_{i}|^{2}) & \frac{1}{2} & (∑_{i,j=1}^{n} |(z_{i}, z_{j})|) & \frac{1}{2} \\ × max_{1≤i≤n} (∑_{j=1}^{n} |(z_{i}, z_{j})|)^{\frac{1}{2}} ; \end{matrix}

E := \{ \begin{matrix}max_{1≤i≤n} & |α_{i}| & (∑_{i=1}^{n} |α_{i}|^{2α}) & \frac{1}{2α} & (∑_{i,j=1}^{n} |(z_{i}, z_{j})|) & \frac{1}{2} \\ × (∑_{i=1}^{n} (∑_{j=1}^{n} |(z_{i}, z_{j})|)^{β})^{\frac{1}{2β}} , if α > 1, \frac{1}{α} + \frac{1}{β} = 1; \\ (∑_{i=1}^{n} |α_{i}|^{2α}) & \frac{1}{2α} & (∑_{i=1}^{n} |α_{i}|^{2γ}) & \frac{1}{2γ} & (∑_{i=1}^{n} (∑_{j=1}^{n} |(z_{i}, z_{j})|)^{β}) & \frac{1}{2β} \\ × (∑_{i=1}^{n} (∑_{j=1}^{n} |(z_{i}, z_{j})|)^{δ})^{\frac{1}{2δ}} \\ if α > 1, \frac{1}{α} + \frac{1}{β} = 1 and γ > 1, \frac{1}{γ} + \frac{1}{δ} = 1; \\ (∑_{i=1}^{n} |α_{i}|^{2}) & \frac{1}{2} & (∑_{i=1}^{n} |α_{i}|^{2α}) & \frac{1}{2α} & max_{1≤i≤n} & (∑_{j=1}^{n} |(z_{i}, z_{j})|) & \frac{1}{2} \\ × (∑_{i=1}^{n} (∑_{j=1}^{n} |(z_{i}, z_{j})|)^{β})^{\frac{1}{2β}} , if α > 1, \frac{1}{α} + \frac{1}{β} = 1; \end{matrix}

and

F := \{ \begin{matrix}max_{1≤i≤n} & |α_{i}| & (∑_{i=1}^{n} |α_{i}|^{2}) & \frac{1}{2} \\ × max_{1≤i≤n} (∑_{j=1}^{n} |(z_{i}, z_{j})|)^{\frac{1}{2}} (∑_{i,j=1}^{n} |(z_{i}, z_{j})|)^{\frac{1}{2}} ; \\ (∑_{i=1}^{n} |α_{i}|^{2}) & \frac{1}{2} & (∑_{i=1}^{n} |α_{i}|^{2γ}) & \frac{1}{2γ} \\ × max_{1≤i≤n} (∑_{j=1}^{n} |(z_{i}, z_{j})|)^{\frac{1}{2}} (∑_{i=1}^{n} (∑_{j=1}^{n} |(z_{i}, z_{j})|)^{δ})^{\frac{1}{2δ}} , \\ if γ > 1, \frac{1}{γ} + \frac{1}{δ} = 1; \\ ∑_{i=1}^{n} |α_{i}|^{2} max_{1≤i≤n} (∑_{j=1}^{n} |(z_{i}, z_{j})|) . \end{matrix}

3.3. Some Pečarić Type Inequalities. We are now able to point out the following result obtained in [4], which complements and generalises the inequality (3.1) due to J. Pečarić.

Theorem 56. Let x, y_{1}, …, y_{n} be vectors of an inner product space (H; (·, ·)) and c_{1}, …, c_{n} ∈ 𝕂. Then one has the inequalities:

(3.10)  |∑_{i=1}^{n} c_{i} (x, y_{i})|^{2}

≤ ∥x∥^{2} (∑_{i=1}^{n} |c_{i}|^{p} (∑_{j=1}^{n} |(y_{i}, y_{j})|))^{\frac{1}{p}} (∑_{i=1}^{n} |c_{i}|^{q} (∑_{j=1}^{n} |(y_{i}, y_{j})|))^{\frac{1}{q}}

154 4. GENERALISATIONS OF BESSEL’S INEQUALITY

≤ ∥x∥^{2} × \{ \begin{matrix}G \\ H \\ I \end{matrix} ,

where

G := \{ \begin{matrix}max_{1≤i≤n} |c_{i}|^{2} ∑_{i,j=1}^{n} |(y_{i}, y_{j})|; \\ max_{1≤i≤n} & |c_{i}| & (∑_{i=1}^{n} |c_{i}|^{γq}) & \frac{1}{γq} & (∑_{i,j=1}^{n} |(y_{i}, y_{j})|) & \frac{1}{p} \\ × (∑_{i=1}^{n} (∑_{j=1}^{n} |(y_{i}, y_{j})|)^{δ})^{\frac{1}{δq}} , if γ > 1, \frac{1}{γ} + \frac{1}{δ} = 1; \\ max_{1≤i≤n} & |c_{i}| & (∑_{i=1}^{n} |c_{i}|^{q}) & \frac{1}{q} & (∑_{i,j=1}^{n} |(y_{i}, y_{j})|) & \frac{1}{p} \\ × max_{1≤i≤n} (∑_{j=1}^{n} |(y_{i}, y_{j})|)^{\frac{1}{q}} ; \end{matrix}

H := \{ \begin{matrix}max_{1≤i≤n} & |c_{i}| & (∑_{i=1}^{n} |c_{i}|^{αp}) & \frac{1}{αp} & (∑_{i,j=1}^{n} |(y_{i}, y_{j})|) & \frac{1}{q} \\ × (∑_{i=1}^{n} (∑_{j=1}^{n} |(y_{i}, y_{j})|)^{β})^{\frac{1}{pβ}} , if α > 1, \frac{1}{α} + \frac{1}{β} = 1; \\ (∑_{i=1}^{n} |c_{i}|^{αp}) & \frac{1}{αp} & (∑_{i=1}^{n} |c_{i}|^{γq}) & \frac{1}{γq} & (∑_{i=1}^{n} (∑_{j=1}^{n} |(y_{i}, y_{j})|)^{β}) & \frac{1}{pβ} \\ × (∑_{i=1}^{n} (∑_{j=1}^{n} |(y_{i}, y_{j})|)^{δ})^{\frac{1}{δq}} \\ if α > 1, \frac{1}{α} + \frac{1}{β} = 1 and γ > 1, \frac{1}{γ} + \frac{1}{δ} = 1; \\ (∑_{i=1}^{n} |c_{i}|^{q}) & \frac{1}{q} & (∑_{i=1}^{n} |c_{i}|^{αp}) & \frac{1}{αp} & max_{1≤i≤n} & (∑_{j=1}^{n} |(y_{i}, y_{j})|) & \frac{1}{q} \\ × (∑_{i=1}^{n} (∑_{j=1}^{n} |(y_{i}, y_{j})|)^{β})^{\frac{1}{pβ}} , if α > 1, \frac{1}{α} + \frac{1}{β} = 1; \end{matrix}

and

I := \{ \begin{matrix}max_{1≤i≤n} & |c_{i}| & (∑_{i=1}^{n} |c_{i}|^{p}) & \frac{1}{p} & max_{1≤i≤n} & (∑_{j=1}^{n} |(y_{i}, y_{j})|) & \frac{1}{p} \\ × (∑_{i,j=1}^{n} |(y_{i}, y_{j})|)^{\frac{1}{q}} ; \\ (∑_{i=1}^{n} |c_{i}|^{p}) & \frac{1}{p} & (∑_{i=1}^{n} |c_{i}|^{γq}) & \frac{1}{γq} & max_{1≤i≤n} & (∑_{j=1}^{n} |(y_{i}, y_{j})|) & \frac{1}{p} \\ × (∑_{i=1}^{n} (∑_{j=1}^{n} |(y_{i}, y_{j})|)^{δ})^{\frac{1}{δq}} , if γ > 1, \frac{1}{γ} + \frac{1}{δ} = 1; \\ (∑_{i=1}^{n} |c_{i}|^{p})^{\frac{1}{p}} (∑_{i=1}^{n} |c_{i}|^{q})^{\frac{1}{q}} max_{1≤i≤n} (∑_{j=1}^{n} |(y_{i}, y_{j})|); \end{matrix}

for p > 1, \frac{1}{p} + \frac{1}{q} = 1.

Proof. We note that

∑_{i=1}^{n} c_{i} (x, y_{i}) = (x, ∑_{i=1}^{n} \overline{c_{i}}y_{i}) .

Using Schwarz’s inequality in inner product spaces, we have

|∑_{i=1}^{n} c_{i} (x, y_{i})|^{2} ≤ ∥x∥^{2} ∥∑_{i=1}^{n} \overline{c_{i}}y_{i}∥^{2} .

Finally, using Lemma 9 with α_{i} = \overline{c_{i}}, z_{i} = y_{i} (i = 1, …, n) , we deduce the desired inequality (3.10).

Remark 52. If in (3.10) we choose p = q = 2, we obtain amongst others, the result (3.1) due to J. Pečarić.

3.4. More Results of Bombieri Type. The following results of Bombieri type hold [4].

Theorem 57. Let x, y_{1}, …, y_{n} ∈ H. Then one has the inequality:

(3.11) ∑_{i=1}^{n} |(x, y_{i})|^{2} ≤ ∥x∥ [∑_{i=1}^{n} |(x, y_{i})|^{p} (∑_{j=1}^{n} |(y_{i}, y_{j})|)]^{\frac{1}{2p}}

× [∑_{i=1}^{n} |(x, y_{i})|^{q} (∑_{j=1}^{n} |(y_{i}, y_{j})|)]^{\frac{1}{2q}}

≤ ∥x∥ × \{ \begin{matrix}J \\ K \\ L \end{matrix} ,

where

J := \{ \begin{matrix}max_{1≤i≤n} |(x, y_{i})| (∑_{i,j=1}^{n} |(y_{i}, y_{j})|)^{\frac{1}{2}} ; \\ max_{1≤i≤n} & |(x, y_{i})| & \frac{1}{2} & (∑_{i=1}^{n} |(x, y_{i})|^{γq}) & \frac{1}{2γq} & (∑_{i,j=1}^{n} |(y_{i}, y_{j})|) & \frac{1}{2p} \\ × (∑_{i=1}^{n} (∑_{j=1}^{n} |(y_{i}, y_{j})|)^{δ})^{\frac{1}{2δq}} , if γ > 1, \frac{1}{γ} + \frac{1}{δ} = 1; \\ max_{1≤i≤n} & |(x, y_{i})| & \frac{1}{2} & (∑_{i=1}^{n} |(x, y_{i})|^{q}) & \frac{1}{2q} \\ × (∑_{i,j=1}^{n} |(y_{i}, y_{j})|)^{\frac{1}{2p}} max_{1≤i≤n} (∑_{j=1}^{n} |(y_{i}, y_{j})|)^{\frac{1}{2q}} ; \end{matrix}

K := \{ \begin{matrix}max_{1≤i≤n} & |(x, y_{i})| & \frac{1}{2} \\ × (∑_{i=1}^{n} |(x, y_{i})|^{αp})^{\frac{1}{2αβ}} (∑_{i,j=1}^{n} |(y_{i}, y_{j})|)^{\frac{1}{2q}} \\ × (∑_{i=1}^{n} (∑_{j=1}^{n} |(y_{i}, y_{j})|)^{β})^{\frac{1}{pβ}} , if α > 1, \frac{1}{α} + \frac{1}{β} = 1; \\ (∑_{i=1}^{n} |(x, y_{i})|^{αp}) & \frac{1}{2αp} & (∑_{i=1}^{n} |(x, y_{i})|^{γq}) & \frac{1}{2γq} \\ × (∑_{i=1}^{n} (∑_{j=1}^{n} |(y_{i}, y_{j})|)^{β})^{\frac{1}{2pβ}} \\ × (∑_{i=1}^{n} (∑_{j=1}^{n} |(y_{i}, y_{j})|)^{δ})^{\frac{1}{2δq}} \\ if α > 1, \frac{1}{α} + \frac{1}{β} = 1 and γ > 1, \frac{1}{γ} + \frac{1}{δ} = 1; \\ (∑_{i=1}^{n} |(x, y_{i})|^{q}) & \frac{1}{2q} & (∑_{i=1}^{n} |(x, y_{i})|^{αp}) & \frac{1}{2αp} \\ × max_{1≤i≤n} (∑_{j=1}^{n} |(y_{i}, y_{j})|)^{\frac{1}{2p}} \\ × (∑_{i=1}^{n} (∑_{j=1}^{n} |(y_{i}, y_{j})|)^{β})^{\frac{1}{2pβ}} , if α > 1, \frac{1}{α} + \frac{1}{β} = 1; \end{matrix}

and

L := \{ \begin{matrix}max_{1≤i≤n} & |(x, y_{i})| & \frac{1}{2} \\ × (∑_{i=1}^{n} |(x, y_{i})|^{p})^{\frac{1}{2p}} max_{1≤i≤n} (∑_{j=1}^{n} |(y_{i}, y_{j})|)^{\frac{1}{2p}} \\ × (∑_{i,j=1}^{n} |(y_{i}, y_{j})|)^{\frac{1}{2q}} ; \\ (∑_{i=1}^{n} |(x, y_{i})|^{p}) & \frac{1}{2p} & (∑_{i=1}^{n} |(x, y_{i})|^{γq}) & \frac{1}{2γq} \\ × max_{1≤i≤n} (∑_{j=1}^{n} |(y_{i}, y_{j})|)^{\frac{1}{2p}} \\ × (∑_{i=1}^{n} (∑_{j=1}^{n} |(y_{i}, y_{j})|)^{δ})^{\frac{1}{2δq}} , if γ > 1, \frac{1}{γ} + \frac{1}{δ} = 1; \\ (∑_{i=1}^{n} |(x, y_{i})|^{p}) & \frac{1}{2p} & (∑_{i=1}^{n} |(x, y_{i})|^{q}) & \frac{1}{2q} \\ × max_{1≤i≤n} (∑_{j=1}^{n} |(y_{i}, y_{j})|)^{\frac{1}{2}} , \end{matrix}

for p > 1, \frac{1}{p} + \frac{1}{q} = 1.

Proof. The proof follows by Theorem 56 on choosing c_{i} = \overline{(x, y_{i})}, i ∈ {1, …, n} and taking the square root in both sides of the inequalities involved. We omit the details.

Remark 53. We observe, by the last inequality in (3.11), that

\begin{matrix}(∑_{i=1}^{n} |(x, y_{i})|^{2}) & 2 \\ (∑_{i=1}^{n} |(x, y_{i})|^{p}) & \frac{1}{p} & (∑_{i=1}^{n} |(x, y_{i})|^{q}) & \frac{1}{q} \end{matrix} ≤ ∥x∥^{2} max_{1≤i≤n} (∑_{j=1}^{n} |(y_{i}, y_{j})|) ,

where p > 1, \frac{1}{p} + \frac{1}{q} = 1.

If in this inequality we choose p = q = 2, then we recapture Bombieri’s result (3.2) .

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[4] S.S. DRAGOMIR, On Pečarić’s inequality in inner product spaces, RGMIA Res. Rep. Coll., 6(2003), Supplement, Article 17 [ONLINE:

http://rgmia.vu.edu.au/v6(E).html].

[5] S.S. DRAGOMIR, On the Boas-Bellman inequality in inner product spaces, RGMIA Res. Rep. Coll., 6(2003), Supplement, Article 14 [ONLINE:

http://rgmia.vu.edu.au/v6(E).html].

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http://rgmia.vu.edu.au/v6n3.html]

[7] S.S. DRAGOMIR and B. MOND, On the Boas-Bellman generalisation of Bessel’s inequality in inner product spaces, Italian J. of Pure & Appl. Math., 3 (1998), 29–35.

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CHAPTER 5 Some Grüss’ Type Inequalities for n-Tuples of Vectors

1. Introduction

We start by recalling some of the most important Grüss type discrete inequalities for real numbers that are available in the literature.

1. (1950) Biernacki, Pidek, Ryll-Nardzewski [2].

If ā = (a_{1}, …, a_{n}) and b̄ = (b_{1}, …, b_{n}) are n-tuples of real numbers such that there exists the real numbers a, A, b, B with

(1.1)  a ≤ a_{i} ≤ A, b ≤ b_{i} ≤ B, i ∈ {1, …, n} ,

then

|C_{n} (ā, b̄)| ≤ \frac{1}{n} [\frac{n}{2}] (1 − \frac{1}{n} [\frac{n}{2}]) (A − a) (B − b)

= \frac{1}{n^{2}} [\frac{n^{2}}{4} ] (A − a) (B − b)

≤ \frac{1}{4} (A − a) (B − b) .

2. (1988) Andrica-Badea [1].

Let ā, b̄ satisfy (1.1) and p̄ = (p_{1}, …, p_{n}) be an n−tuple of nonnegative numbers with P_{n} > 0. If S is a subset of {1, …, n} that minimises the expression

|∑_{i∈S} q_{i} − \frac{1}{2}Q_{n}| ,

then

C_{n} (p̄, ā, b̄) ≤ \frac{Q_{S}}{Q_{n}} (1 − \frac{Q_{S}}{Q_{n}} ) (A − a) (B − b)

≤ \frac{1}{4} (A − a) (B − b) ,

162 5. SOME GRÜSS’ TYPE INEQUALITIES FOR n-TUPLES OF VECTORS

where Q_{S} := ∑_{i∈S} Q_{i}.

3. (2000) Dragomir-Booth [13].

If ā, b̄ are real n−tuples and p̄ is nonnegative with P_{n} > 0, then

|C_{n} (p̄, ā, b̄)| ≤ max_{1≤j≤n−1} |Δa_{j}| max_{1≤j≤n−1} |Δb_{j}| C_{n} (p̄, ē, ē) ,

where ē = (1, 2, …, n) and Δa_{j} := a_{j+1} − a_{j} is the forward difference, j = 1, …, n − 1. Note that

C_{n} (p̄, ē, ē) = \frac{1}{P_{n}^{2}} ∑_{i=1}^{n} i^{2}p_{i} − (\frac{1}{P_{n}} ∑_{i=1}^{n} ip_{i})^{2} .

In particular, we have

|C_{n} (ā, b̄)| ≤ \frac{1}{12} (n^{2} − 1) max_{1≤j≤n−1} |Δa_{j}| max_{1≤j≤n−1} |Δb_{j}| .

The constant \frac{1}{12} is best possible.

4. (2002) Dragomir [8].

With the assumptions in 3, the following inequality holds

|C_{n} (p̄, ā, b̄)| ≤ \frac{1}{P_{n}^{2}} ∑_{1≤j<i≤n} (i − j) (∑_{k=1}^{n−1} |Δa_{k}|^{p})^{\frac{1}{p}} (∑_{k=1}^{n−1} |Δb_{k}|^{q})^{\frac{1}{q}} ,

where p > 1, \frac{1}{p} + \frac{1}{q} = 1.

In particular, we have

|C_{n} (ā, b̄)| ≤ \frac{1}{6} · \frac{n^{2} − 1}{n} (∑_{k=1}^{n−1} |Δa_{k}|^{p})^{\frac{1}{p}} (∑_{k=1}^{n−1} |Δb_{k}|^{q})^{\frac{1}{q}} .

The constant \frac{1}{6} is best possible.

5. (2002) Dragomir [6].

The following inequality holds, where p̄, ā, b̄ and P_{n} are as in assumption 3,

|C_{n} (p̄, ā, b̄)| ≤ \frac{1}{2} · \frac{1}{P_{n}^{2}} ∑_{i=1}^{n} p_{i} (P_{n} − p_{i}) ∑_{k=1}^{n−1} |Δa_{k}| ∑_{k=1}^{n−1} |Δb_{k}| .

In particular, we have

|C_{n} (ā, b̄)| ≤ \frac{1}{2} (1 − \frac{1}{n}) ∑_{k=1}^{n−1} |Δa_{k}| ∑_{k=1}^{n−1} |Δb_{k}| .

The constant \frac{1}{2} is sharp.

6. (2002) Cerone-Dragomir [3].

If ā, b̄ are real n−tuples and p̄ is a positive n−tuple and there exists m, M ∈ ℝ such that

m ≤ a_{i} ≤ M,

then one has the inequality

|C_{n} (p̄, ā, b̄)| ≤ \frac{1}{2} (M − m) \frac{1}{P_{n}} ∑_{i=1}^{n} p_{i} |b_{i} − \frac{1}{P_{n}} ∑_{j=1}^{n} p_{j}b_{j}| .

The constant \frac{1}{2} is best possible. In particular, we have

|C_{n} (ā, b̄)| ≤ \frac{1}{2} (M − m) · \frac{1}{n} ∑_{i=1}^{n} |b_{i} − \frac{1}{n} ∑_{j=1}^{n} b_{j}| .

The constant \frac{1}{2} is best possible. The main aim of this chapter is to present some extensions of the

above results holding in the general setting of n-tuples of vectors in an inner product space.

2. The Version for Norms

2.1. Preliminary Results. The following lemma is of interest in itself [5].

Lemma 10. Let (H; ⟨·, ·⟩) be an inner product space over the real or complex number field 𝕂, x_{i} ∈ H and p_{i} ≥ 0 (i = 1, …, n) such that ∑_{i=1}^{n} p_{i} = 1 (n ≥ 2) . If x, X ∈ H are such that

(2.1)  Re ⟨X − x_{i}, x_{i} − x⟩ ≥ 0 for all i ∈ {1, …, n} ,

or, equivalently,

∥x_{i} − \frac{x + X}{2} ∥ ≤ \frac{1}{2} ∥X − x∥ for all i ∈ \{1, …, n\} ,

then we have the inequality

(2.2)  0 ≤ ∑_{i=1}^{n} p_{i} ∥x_{i}∥^{2} − ∥∑_{i=1}^{n} p_{i}x_{i}∥^{2} ≤ \frac{1}{4} ∥X − x∥^{2} .

The constant \frac{1}{4} is sharp.

Proof. Define

I_{1} := ⟨X − ∑_{i=1}^{n} p_{i}x_{i}, ∑_{i=1}^{n} p_{i}x_{i} − x⟩

and

I_{2} := ∑_{i=1}^{n} p_{i} ⟨X − x_{i}, x_{i} − x⟩ .

Then

I_{1} = ∑_{i=1}^{n} p_{i} ⟨X, x_{i}⟩ − ⟨X, x⟩ − ∥∑_{i=1}^{n} p_{i}x_{i}∥^{2} + ∑_{i=1}^{n} p_{i} ⟨x_{i}, x⟩

and

I_{2} = ∑_{i=1}^{n} p_{i} ⟨X, x_{i}⟩ − ⟨X, x⟩ − ∑_{i=1}^{n} p_{i} ∥x_{i}∥^{2} + ∑_{i=1}^{n} p_{i} ⟨x_{i}, x⟩ .

Consequently,

(2.3)  I_{1} − I_{2} = ∑_{i=1}^{n} p_{i} ∥x_{i}∥^{2} − ∥∑_{i=1}^{n} p_{i}x_{i}∥^{2} .

Taking the real value in (2.3) , we can state that

(2.4) ∑_{i=1}^{n} p_{i} ∥x_{i}∥^{2} − ∥∑_{i=1}^{n} p_{i}x_{i}∥^{2}

= Re ⟨X − ∑_{i=1}^{n} p_{i}x_{i}, ∑_{i=1}^{n} p_{i}x_{i} − x⟩ − ∑_{i=1}^{n} p_{i} Re ⟨X − x_{i}, x_{i} − x⟩ ,

which is also an identity of interest in itself. Using the assumption (2.1) , we can conclude, by (2.4) , that

(2.5) ∑_{i=1}^{n} p_{i} ∥x_{i}∥^{2} − ∥∑_{i=1}^{n} p_{i}x_{i}∥^{2} ≤ Re ⟨X − ∑_{i=1}^{n} p_{i}x_{i}, ∑_{i=1}^{n} p_{i}x_{i} − x⟩ .

It is known that if y, z ∈ H, then

(2.6)  4 Re ⟨z, y⟩ ≤ ∥z + y∥^{2} ,

with equality iff z = y.

Now, by (2.6) we can state that

Re ⟨X − ∑_{i=1}^{n} p_{i}x_{i}, ∑_{i=1}^{n} p_{i}x_{i} − x⟩ ≤ \frac{1}{4} ∥X − ∑_{i=1}^{n} p_{i}x_{i} + ∑_{i=1}^{n} p_{i}x_{i} − x∥^{2}

= \frac{1}{4} ∥X − x∥^{2} .

Using (2.5) , we obtain (2.2) .

To prove the sharpness of the constant \frac{1}{4}, let us assume that the inequality (2.2) holds with a constant c > 0, i.e.,

(2.7)  0 ≤ ∑_{i=1}^{n} p_{i} ∥x_{i}∥^{2} − ∥∑_{i=1}^{n} p_{i}x_{i}∥^{2} ≤ c ∥X − x∥^{2}

for all p_{i}, x_{i} and n as in the hypothesis of Lemma 10.

Assume that n = 2, p_{1} = p_{2} = \frac{1}{2}, x_{1} = x and x_{2} = X with x, X ∈ H and x ≠ X. Then, obviously,

⟨X − x_{1}, x_{1} − x⟩ = ⟨X − x_{2}, x_{2} − x⟩ = 0,

which shows that the condition (2.1) holds.

If we replace n, p_{1}, p_{2}, x_{1}, x_{2} in (2.7) , we obtain

∑_{i=1}^{2} p_{i} ∥x_{i}∥^{2} − ∥∑_{i=1}^{2} p_{i}x_{i}∥^{2} = \frac{1}{2} (∥x∥^{2} + ∥X∥^{2}) − ∥\frac{x + X}{2} ∥^{2}

= \frac{1}{4} ∥x − X∥^{2}

≤ c ∥x − X∥^{2} ,

from where we deduce that c ≥ \frac{1}{4}, which proves the sharpness of the constant \frac{1}{4}.

Remark 54. The assumption (2.1) can be replaced by the more general condition

(2.8)  ∑_{i=1}^{n} p_{i} Re ⟨X − x_{i}, x_{i} − x⟩ ≥ 0,

and the conclusion (2.2) will still remain valid.

The following corollary is natural.

Corollary 48. Let a_{i} ∈ 𝕂, p_{i} ≥ 0, (i = 1, …, n) (n ≥ 2) with ∑_{i=1}^{n} p_{i} = 1. If a, A ∈ 𝕂 are such that

(2.9)  Re [(A − a_{i}) (ā_{i} − ā)] ≥ 0 for all i ∈ {1, …, n} ,

then we have the inequality

(2.10)  0 ≤ ∑_{i=1}^{n} p_{i} |a_{i}|^{2} − |∑_{i=1}^{n} p_{i}a_{i}|^{2} ≤ \frac{1}{4} |A − a|^{2} .

The constant \frac{1}{4} is sharp.

The proof follows by the above lemma by choosing H = 𝕂, ⟨x, y⟩ := xȳ, x_{i} = a_{i}, x = a and X = A. We omit the details.

Remark 55. The condition (2.9) can be replaced by the more general assumption

∑_{i=1}^{n} p_{i} Re [(A − a_{i}) (ā_{i} − ā)] ≥ 0.

2.2. A Discrete Inequality of Grüss’ Type. The following Grüss type inequality holds [5].

Theorem 58. Let (H; ⟨·, ·⟩) be an inner product space over 𝕂, 𝕂 = ℝ, ℂ, x_{i} ∈ H, a_{i} ∈ 𝕂, p_{i} ≥ 0 (i = 1, …, n) (n ≥ 2) with ∑_{i=1}^{n} p_{i} = 1. If a, A ∈ 𝕂 and x, X ∈ H are such that

(2.11)  Re [(A − a_{i}) (ā_{i} − ā)] ≥ 0, Re ⟨X − x_{i}, x_{i} − x⟩ ≥ 0

for all i ∈ {1, …, n}; then we have the inequality

(2.12) 0 ≤ ∥∑_{i=1}^{n} p_{i}a_{i}x_{i} − ∑_{i=1}^{n} p_{i}a_{i} · ∑_{i=1}^{n} p_{i}x_{i}∥ ≤ \frac{1}{4} |A − a| ∥X − x∥ .

The constant \frac{1}{4} is sharp.

Proof. A simple computation shows that

(2.13) ∑_{i=1}^{n} p_{i}a_{i}x_{i} − ∑_{i=1}^{n} p_{i}a_{i} ∑_{i=1}^{n} p_{i}x_{i} = \frac{1}{2} ∑_{i,j=1}^{n} p_{i}p_{j} (a_{i} − a_{j}) (x_{i} − x_{j}) .

Taking the norm in both parts of (2.13) and using the generalized triangle inequality, we obtain

(2.14) ∥ ∑_{i=1}^{n} p_{i}a_{i}x_{i} − ∑_{i=1}^{n} p_{i}a_{i} ∑_{i=1}^{n} p_{i}x_{i}∥

≤ \frac{1}{2} ∑_{i,j=1}^{n} p_{i}p_{j} |a_{i} − a_{j}| ∥x_{i} − x_{j}∥ .

By the Cauchy-Bunyakovsky-Schwarz discrete inequality for double sums, we obtain

(2.15) (\frac{1}{2} ∑_{i,j=1}^{n} p_{i}p_{j} |a_{i} − a_{j}| ∥x_{i} − x_{j}∥)^{2}

≤ (\frac{1}{2} ∑_{i,j=1}^{n} p_{i}p_{j} |a_{i} − a_{j}|^{2}) (\frac{1}{2} ∑_{i,j=1}^{n} p_{i}p_{j} ∥x_{i} − x_{j}∥^{2}) .

As a simple calculation reveals that

\frac{1}{2} ∑_{i,j=1}^{n} p_{i}p_{j} |a_{i} − a_{j}|^{2} = ∑_{i=1}^{n} p_{i} |a_{i}|^{2} − |∑_{i=1}^{n} p_{i}a_{i}|^{2}

and

\frac{1}{2} ∑_{i,j=1}^{n} p_{i}p_{j} ∥x_{i} − x_{j}∥^{2} = ∑_{i=1}^{n} p_{i} ∥x_{i}∥^{2} − ∥∑_{i=1}^{n} p_{i}x_{i}∥^{2} ,

then, by (2.14) and (2.15) , we conclude that

(2.16) ∥∑_{i=1}^{n} p_{i}a_{i}x_{i} − ∑_{i=1}^{n} p_{i}a_{i} ∑_{i=1}^{n} p_{i}x_{i}∥

≤ (∑_{i=1}^{n} p_{i} |a_{i}|^{2} − |∑_{i=1}^{n} p_{i}a_{i}|^{2})^{\frac{1}{2}} (∑_{i=1}^{n} p_{i} ∥x_{i}∥^{2} − ∥∑_{i=1}^{n} p_{i}x_{i}∥^{2})^{\frac{1}{2}} .

However, from Lemma 10 and Corollary 48, we know that

(2.17)  (∑_{i=1}^{n} p_{i} ∥x_{i}∥^{2} − ∥∑_{i=1}^{n} p_{i}x_{i}∥^{2})^{\frac{1}{2}} ≤ \frac{1}{2} ∥X − x∥

and

(2.18)  (∑_{i=1}^{n} p_{i} |a_{i}|^{2} − |∑_{i=1}^{n} p_{i}a_{i}|^{2})^{\frac{1}{2}} ≤ \frac{1}{2} |A − a| .

Consequently, by using (2.16) − (2.18) , we deduce the desired estimate (2.12) .

To prove the sharpness of the constant \frac{1}{4}, assume that (2.12) holds with a constant c > 0, i.e.,

(2.19)  ∥∑_{i=1}^{n} p_{i}a_{i}x_{i} − ∑_{i=1}^{n} p_{i}a_{i} ∑_{i=1}^{n} p_{i}x_{i}∥ ≤ c |A − a| ∥X − x∥

for all p_{i}, a_{i}, x_{i}, a, A, x, X and n as in the hypothesis of Theorem 58.

If we choose n = 2, a_{1} = a, a_{2} = A, x_{1} = x, x_{2} = X (a ≠ A, x ≠ X) and p_{1} = p_{2} = \frac{1}{2}, then

∑_{i=1}^{2} p_{i}a_{i}x_{i} − ∑_{i=1}^{2} p_{i}a_{i} ∑_{i=1}^{2} p_{i}x_{i} = \frac{1}{2} ∑_{i,j=1}^{2} p_{i}p_{j} (a_{i} − a_{j}) (x_{i} − x_{j})

= \frac{1}{4} (a − A) (x − X) .

Consequently, from (2.19) , we deduce

\frac{1}{4} |a − A| ∥X − x∥ ≤ c |A − a| ∥X − x∥ ,

which implies that c ≥ \frac{1}{4}, and the theorem is completely proved.

Remark 56. The condition (2.11) can be replaced by the more general assumption

∑_{i=1}^{n} p_{i} Re [(A − a_{i}) (ā_{i} − ā)] ≥ 0, ∑_{i=1}^{n} p_{i} Re ⟨X − x_{i}, x_{i} − x⟩ ≥ 0

and the conclusion (2.12) will still be valid.

The following corollary for real or complex numbers holds.

Corollary 49. Let a_{i}, b_{i} ∈ 𝕂 (𝕂 = ℂ, ℝ) , p_{i} ≥ 0 (i = 1, …, n) with ∑_{i=1}^{n} p_{i} = 1. If a, A, b, B ∈ 𝕂 are such that

(2.20)  Re [(A − a_{i}) (ā_{i} − ā)] ≥ 0, Re [(B − b_{i}) (b̄_{i} − b̄)] ≥ 0,

then we have the inequality

(2.21)  0 ≤ |∑_{i=1}^{n} p_{i}a_{i}b_{i} − ∑_{i=1}^{n} p_{i}a_{i} ∑_{i=1}^{n} p_{i}b_{i}| ≤ \frac{1}{4} |A − a| |B − b| ,

where the constant \frac{1}{4} is sharp.

Remark 57. If we assume that a_{i}, b_{i}, a, A, b, B are real numbers, then (2.20) is equivalent to

a ≤ a_{i} ≤ A, b ≤ b_{i} ≤ B for all i ∈ {1, …, n} ,

and (2.21) becomes

0 ≤ |∑_{i=1}^{n} p_{i}a_{i}b_{i} − ∑_{i=1}^{n} p_{i}a_{i} ∑_{i=1}^{n} p_{i}b_{i}| ≤ \frac{1}{4} (A − a) (B − b) ,

which is the classical Grüss inequality for sequences of real numbers.

2.3. Applications for Discrete Fourier Transforms. Let (H; ⟨·, ·⟩) be an inner product space over 𝕂 and x̄ = (x_{1}, …, x_{n}) be a sequence of vectors in H.

For a given w ∈ ℝ, define the discrete Fourier transform as

(2.22)  F_{w} \binom{}{x̄} (m) := ∑_{k=1}^{n} exp (2wimk) × x_{k}, m = 1, …, n.

The following approximation result for the Fourier transform (2.22) holds [5].

Theorem 59. Let (H; ⟨·, ·⟩) and x̄ ∈ H^{n} be as above. If there exists the vectors x, X ∈ H such that

(2.23)  Re ⟨X − x_{k}, x_{k} − x⟩ ≥ 0 for all k ∈ {1, …, n} ,

then we have the inequality

(2.24) ∥F_{w} \binom{}{x̄} (m) − \begin{matrix}sin & (wmn) \\ sin & (wm) \end{matrix} exp [w (n + 1) im] × \frac{1}{n} ∑_{k=1}^{n} x_{k}∥

≤ \frac{1}{2} ∥X − x∥ [n^{2} − \begin{matrix}sin & 2 & (wmn) \\ sin & 2 & (wm) \end{matrix}]^{\frac{1}{2}} ,

for all m ∈ \{1, …, n\} and w ∈ ℝ, w ≠ \frac{l}{m}π, l ∈ ℤ.

Proof. From the inequality (2.16) in Theorem 58, we can state that

(2.25) ∥\frac{1}{n} ∑_{k=1}^{n} a_{k}x_{k} − \frac{1}{n} ∑_{k=1}^{n} a_{k} · \frac{1}{n} ∑_{k=1}^{n} x_{k}∥

≤ (\frac{1}{n} ∑_{k=1}^{n} |a_{k}|^{2} − |\frac{1}{n} ∑_{k=1}^{n} a_{k}|^{2})^{\frac{1}{2}} (\frac{1}{n} ∑_{k=1}^{n} ∥x_{k}∥^{2} − ∥\frac{1}{n} ∑_{k=1}^{n} x_{k}∥^{2})^{\frac{1}{2}}

for all a_{k} ∈ 𝕂, x_{k} ∈ H (k = 1, …, n) .

However, the x_{k} (k = 1, …, n) satisfy (2.23) , and therefore, by Lemma 10, we have

(2.26)  0 ≤ \frac{1}{n} ∑_{k=1}^{n} ∥x_{k}∥^{2} − ∥\frac{1}{n} ∑_{k=1}^{n} x_{k}∥^{2} ≤ \frac{1}{4} ∥X − x∥^{2} .

Consequently, by (2.25) and (2.26) , we conclude that

(2.27) ∥∑_{k=1}^{n} a_{k}x_{k} − ∑_{k=1}^{n} a_{k} · \frac{1}{n} ∑_{k=1}^{n} x_{k}∥

≤ \frac{1}{2} ∥X − x∥ (n ∑_{k=1}^{n} |a_{k}|^{2} − |∑_{k=1}^{n} a_{k}|^{2})^{\frac{1}{2}}

for all a_{k} ∈ 𝕂 (k = 1, …, n) .

We now choose in (2.27) , a_{k} = exp (2wimk) to obtain

(2.28) ∥F_{w} \binom{}{x̄} (m) − ∑_{k=1}^{n} exp (2wimk) × \frac{1}{n} ∑_{k=1}^{n} x_{k}∥

≤ \frac{1}{2} ∥X − x∥ (n ∑_{k=1}^{n} |exp (2wimk)|^{2} − |∑_{k=1}^{n} exp (2wimk)|^{2})^{\frac{1}{2}}

for all m ∈ {1, …, n} .

As a simple calculation reveals that

∑_{k=1}^{n} exp (2wimk) = exp (2wim) × [\frac{exp (2wimn) − 1}{exp (2wim) − 1}]

= exp (2wim) × [\frac{cos (2wmn) + i sin (2wmn) − 1}{cos (2wm) + i sin (2wm) − 1}]

= exp (2wim) × \begin{matrix}sin & (wmn) \\ sin & (wm) \end{matrix} [\frac{cos (wmn) + i sin (wmn)}{cos (wm) + i sin (wm)}]

= \begin{matrix}sin & (wmn) \\ sin & (wm) \end{matrix} × exp (2wim) [\begin{matrix}exp & (iwmn) \\ exp & (iwm) \end{matrix}]

= \begin{matrix}sin & (wmn) \\ sin & (wm) \end{matrix} × exp [w (n + 1) im] ,

∑_{k=1}^{n} |exp (2wimk)|^{2} = n

and

|∑_{k=1}^{n} exp (2wimk)|^{2} = \begin{matrix}sin & 2 & (wmn) \\ sin & 2 & (wm) \end{matrix}, for w ≠ \frac{l}{m}π, l ∈ ℤ,

thus, from (2.28) , we deduce the desired inequality (2.24) .

Remark 58. The assumption (2.23) can be replaced by the more general condition

∑_{i=1}^{n} Re ⟨X − x_{i}, x_{i} − x⟩ ≥ 0,

and the conclusion (2.24) will still remain valid.

The following corollary is an obvious consequence of (2.24) .

Corollary 50. Let a_{i} ∈ 𝕂 (i = 1, …, n) . If a, A ∈ 𝕂 are such that

(2.29)  Re [(A − a_{i}) (ā_{i} − ā)] ≥ 0 for all i ∈ {1, …, n} ,

then we have an approximation of the Fourier transform for the vector \overline{a} = (a_{1}, …, a_{n}) ∈ 𝕂^{n} :

(2.30) ∥F_{w} (\overline{a}) (m) − \begin{matrix}sin & (wmn) \\ sin & (wm) \end{matrix} exp [w (n + 1) im] × \frac{1}{n} ∑_{k=1}^{n} a_{k}∥

≤ \frac{1}{2} |A − a| [n^{2} − \begin{matrix}sin & 2 & (wmn) \\ sin & 2 & (wm) \end{matrix}]^{\frac{1}{2}} ,

for all m ∈ \{1, …, n\} and w ∈ ℝ so that w ≠ \frac{l}{m}π, l ∈ ℤ.

Remark 59. If we assume that 𝕂 = ℝ, then (2.29) is equivalent to

(2.31)  a ≤ a_{i} ≤ A for all i ∈ {1, …, n} .

Consequently, with the assumption (2.31) , we obtain the following approximation of the Fourier transform

∥F_{w} (\overline{a}) (m) − \begin{matrix}sin & (wmn) \\ sin & (wm) \end{matrix} exp [w (n + 1) im] × \frac{1}{n} ∑_{k=1}^{n} a_{k}∥

≤ \frac{1}{2} (A − a) [n^{2} − \begin{matrix}sin & 2 & (wmn) \\ sin & 2 & (wm) \end{matrix}]^{\frac{1}{2}} ,

for all m ∈ \{1, …, n\} and w ≠ \frac{l}{m}π, l ∈ ℤ.

2.4. Applications for the Discrete Mellin Transform. Let (H; ⟨·, ·⟩) be an inner product over ℝ and x̄ = (x_{1}, …, x_{n}) be a sequence of vectors in H.

Define the Mellin transform:

M \binom{}{x̄} (m) := ∑_{k=1}^{n} k^{m−1}x_{k}, m = 1, …, n;

of the sequence x̄ ∈ H^{n}.

The following approximation result holds [5].

Theorem 60. Let H and x̄ be as above. If there exist the vectors x, X ∈ H such that

(2.32)  Re ⟨X − x_{k}, x_{k} − x⟩ ≥ 0 for all k = 1, …, n;

then we have the inequality

(2.33) ∥M \binom{}{x̄} (m) − S_{m−1} (n) · \frac{1}{n} ∑_{k=1}^{n} x_{k}∥

≤ \frac{1}{2} ∥X − x∥ [nS_{2m−2} (n) − S_{m−1}^{2} (n)]^{\frac{1}{2}} , m ∈ \{1, …, n\} ,

where S_{p} (n) , p ∈ ℝ, n ∈ ℕ is the p−powered sum of the first n natural numbers, i.e.,

S_{p} (n) := ∑_{k=1}^{n} k^{p}.

Proof. We apply the inequality (2.27) to obtain

∥∑_{k=1}^{n} k^{m−1}x_{k} − ∑_{k=1}^{n} k^{m−1} · \frac{1}{n} ∑_{k=1}^{n} x_{k}∥

≤ \frac{1}{2} ∥X − x∥ [n ∑_{k=1}^{n} k^{2(m−1)} − (∑_{k=1}^{n} k^{m−1})^{2}]^{\frac{1}{2}}

= \frac{1}{2} ∥X − x∥ [nS_{2m−2} (n) − S_{m−1}^{2} (n)]^{\frac{1}{2}} ,

and the inequality (2.33) is proved.

Consider the following particular values of Mellin Transform

μ_{1} \binom{}{x̄} := ∑_{k=1}^{n} kx_{k}

and

μ_{2} \binom{}{x̄} := ∑_{k=1}^{n} k^{2}x_{k}.

The following corollary holds.

Corollary 51. Let H and x̄ be as in Theorem 60. Then we have the inequalities:

(2.34)  ∥μ_{1} \binom{}{x̄} − \frac{n + 1}{2} · ∑_{k=1}^{n} x_{k}∥ ≤ \frac{1}{2} ∥X − x∥ n [\frac{n (n + 1)}{2}]^{\frac{1}{2}}

and

(2.35) ∥μ_{2} \binom{}{x̄} − \frac{(n + 1) (2n + 1)}{6} · ∑_{k=1}^{n} x_{k}∥

≤ \frac{1}{12\sqrt{5}} ∥X − x∥ n\sqrt{(n − 1) (n + 1) (2n + 1) (8n + 1)}.

Remark 60. If we assume that \overline{p} = (p_{1}, …, p_{n}) is a probability distribution, i.e., p_{k} ≥ 0 (k = 1, …, n) and ∑_{k=1}^{n} p_{k} = 1 and p ≤ p_{k} ≤ P (k = 1, …, n) , then by (2.34) and (2.35) , we get the inequalities

|∑_{k=1}^{n} kp_{k} − \frac{n + 1}{2}| ≤ \frac{1}{2} (P − p) n [\frac{n (n + 1)}{2}]^{\frac{1}{2}}

and

|∑_{k=1}^{n} k^{2}p_{k} − \frac{(n + 1) (2n + 1)}{6}|

≤ \frac{1}{12\sqrt{5}} (P − p) n\sqrt{(n − 1) (n + 1) (2n + 1) (8n + 1)}.

2.5. Applications for Polynomials. Let (H; ⟨·, ·⟩) be an inner product space over 𝕂 and c̄ = (c_{0}, …, c_{n}) be a sequence of vectors in H.

Define the polynomial P : ℂ →H with the coefficients c̄ = (c_{0}, …, c_{n}) by

P (z) = c_{0} + zc_{1} + ⋯ + z^{n}c_{n}, z ∈ ℂ, c_{n} ≠ 0.

The following approximation result for the polynomial P holds [5].

Theorem 61. Let H, c̄ and P be as above. If there exist the vectors c, C ∈ H such that

(2.36)  Re ⟨C − c_{k}, c_{k} − c⟩ ≥ 0 for all k ∈ {0, …, n} ,

then we have the inequality

(2.37) ∥P (z) − \frac{z^{n+1} − 1}{z − 1} × \frac{c_{0} + c_{1} + ⋯ + c_{n}}{n + 1} ∥

≤ \frac{1}{2} ∥C − c∥ [(n + 1) \frac{|z|^{2n+2} − 1}{|z|^{2} − 1} − \frac{|z|^{2n+2} − 2 Re (z^{n+1}) + 1}{|z|^{2} − 2 Re (z) + 1}]^{\frac{1}{2}}

for all z ∈ ℂ, |z| ≠ 1.

Proof. Using the inequality (2.27) , we can state that

∥∑_{k=0}^{n} z^{k}c_{k} − ∑_{k=0}^{n} z^{k} · \frac{1}{n + 1} ∑_{k=0}^{n} c_{k}∥ (2.38)

≤ \frac{1}{2} ∥C − c∥ ((n + 1) ∑_{k=0}^{n} |z|^{2k} − |∑_{k=0}^{n} z^{k}|^{2})^{\frac{1}{2}}

= \frac{1}{2} ∥C − c∥ [(n + 1) \frac{|z|^{2n+2} − 1}{|z|^{2} − 1} − |\frac{z^{n+1} − 1}{z − 1}|^{2}]^{\frac{1}{2}}

= \frac{1}{2} ∥C − c∥[(n + 1) \frac{|z|^{2n+2} − 1}{|z|^{2} − 1} − \frac{|z|^{2n+2} − 2 Re (z^{n+1}) + 1}{|z|^{2} − 2 Re (z) + 1}]^{\frac{1}{2}}

and the inequality (2.37) is proved.

The following result for the complex roots of the unity also holds [5].

Theorem 62. Let z_{k} := cos (\frac{kπ}{n+1}) + i sin (\frac{kπ}{n+1}) , k ∈ \{0, …, n\} be the complex (n + 1) −roots of the unity. Then we have the inequality

(2.39)  ∥P (z_{k})∥ ≤ \frac{1}{2} (n + 1) ∥C − c∥ , k ∈ \{1, …, n\};

where the coefficients c̄ = (c_{0}, …, c_{n}) ∈ H^{n+1} satisfy the assumption (2.36) .

Proof. From the inequality (2.38) , we can state that

∥P (z_{k}) − \frac{z^{n+1} − 1}{z − 1} · \frac{1}{n + 1} ∑_{k=0}^{n} c_{k}∥

≤ \frac{1}{2} ∥C − c∥ [(n + 1) ∑_{k=0}^{n} |z|^{2k} − |\frac{z^{n+1} − 1}{z − 1}|^{2}]^{\frac{1}{2}}

for all z ∈ ℂ, z ≠ 1.

Putting z = z_{k}, k ∈ {1, …, n} and taking into account that z_{k}^{n+1} = 1, |z_{k}| = 1, we get the desired result (2.39) .

The following corollary is a natural consequence of Theorem 62.

Corollary 52. Let P (z) := ∑_{k=0}^{n} a_{k}z^{k} be a polynomial with real coefficients and z_{k} the (n + 1)-roots of the unity as defined above. If a ≤ a_{k} ≤ A, k = 0, …, n, then we have the inequality:

|P (z_{k})| ≤ \frac{1}{2} (n + 1) (A − a) .

2.6. Applications for Lipschitzian Mappings. Let (H; ⟨·, ·⟩) be as above and F : H → B a mapping defined on the inner product space H with values in the normed linear space B which satisfy the Lipschitzian condition:

(2.40)  |F (x) − F (y)| ≤ L ∥x − y∥ , for all x, y ∈ H,

where | · | denotes the norm on B and ∥ · ∥ is the Euclidean norm on H.

The following theorem holds [5].

Theorem 63. Let F : H → B be as above and x_{i} ∈ H, p_{i} ≥ 0 (i = 1, …, n) with P_{n} := ∑_{i=1}^{n} p_{i} > 0. If there exists two vectors x, X ∈ H such that

(2.41)  Re ⟨X − x_{i}, x_{i} − x⟩ ≥ 0 for all i ∈ {1, …, n} ,

then we have the inequality

(2.42)  |\frac{1}{P_{n}} ∑_{i=1}^{n} p_{i}F (x_{i}) − F (\frac{1}{P_{n}} ∑_{i=1}^{n} p_{i}x_{i})| ≤ \frac{1}{2} · L ∥X − x∥ .

Proof. As F is Lipschitzian, we have (2.40) for all x, y ∈ H. Choose x = \frac{1}{P_{n}} ∑_{i=1}^{n} p_{i}x_{i} and y = x_{j} (j = 1, …, n) , to get

(2.43)  |F (\frac{1}{P_{n}} ∑_{i=1}^{n} p_{i}x_{i}) − F (x_{j})| ≤ L ∥\frac{1}{P_{n}} ∑_{i=1}^{n} p_{i}x_{i} − x_{j}∥ ,

for all j ∈ {1, …, n}.

If we multiply (2.43) by p_{j} ≥ 0 and sum over j from 1 to n, we obtain

(2.44)  ∑_{j=1}^{n} p_{j} |F (\frac{1}{P_{n}} ∑_{i=1}^{n} p_{i}x_{i}) − F (x_{j})|

≤ L ∑_{j=1}^{n} p_{j} ∥\frac{1}{P_{n}} ∑_{i=1}^{n} p_{i}x_{i} − x_{j}∥ .

176 5. SOME GRÜSS’ TYPE INEQUALITIES FOR n-TUPLES OF VECTORS

Using the generalized triangle inequality, we have

(2.45) ∑_{j=1}^{n} p_{j} |F (\frac{1}{P_{n}} ∑_{i=1}^{n} p_{i}x_{i}) − F (x_{j})|

≥ |P_{n}F (\frac{1}{P_{n}} ∑_{i=1}^{n} p_{i}x_{i}) − ∑_{j=1}^{n} p_{j}F (x_{j})| .

By the Cauchy-Bunyakovsky-Schwarz inequality, we also have

∑_{j=1}^{n} p_{j} ∥\frac{1}{P_{n}} ∑_{i=1}^{n} p_{i}x_{i} − x_{j}∥ (2.46)

≤ [∑_{j=1}^{n} p_{j} ∥\frac{1}{P_{n}} ∑_{i=1}^{n} p_{i}x_{i} − x_{j}∥^{2}]^{\frac{1}{2}} Pn^{\frac{1}{2}}

= Pn^{\frac{1}{2}} [∑_{j=1}^{n} p_{j} [∥\frac{1}{P_{n}} ∑_{i=1}^{n} p_{i}x_{i}∥^{2}

−2 Re ⟨\frac{1}{P_{n}} ∑_{i=1}^{n} p_{i}x_{i}, x_{j}⟩ + ∥x_{j}∥^{2}]]^{\frac{1}{2}}

= Pn^{\frac{1}{2}} [P_{n} ∥\frac{1}{P_{n}} ∑_{i=1}^{n} p_{i}x_{i}∥^{2}

−2 Re ⟨\frac{1}{P_{n}} ∑_{i=1}^{n} p_{i}x_{i}, ∑_{j=1}^{n} p_{j}x_{j}⟩ + ∑_{j=1}^{n} p_{j} ∥x_{j}∥^{2}]^{\frac{1}{2}}

= P_{n} [\frac{1}{P_{n}} ∑_{i=1}^{n} p_{i} ∥x_{i}∥^{2} − ∥\frac{1}{P_{n}} ∑_{i=1}^{n} p_{i}x_{i}∥^{2}]^{\frac{1}{2}} .

Combining the above inequalities (2.44)−(2.46) we deduce, by dividing with P_{n} > 0, that

(2.47) |F (\frac{1}{P_{n}} ∑_{i=1}^{n} p_{i}x_{i}) − \frac{1}{P_{n}} ∑_{i=1}^{n} p_{i}F (x_{i})|

≤ L · [\frac{1}{P_{n}} ∑_{i=1}^{n} p_{i} ∥x_{i}∥^{2} − ∥\frac{1}{P_{n}} ∑_{i=1}^{n} p_{i}x_{i}∥^{2}]^{\frac{1}{2}} .

Finally, using Lemma 10, we obtain the desired result.

Remark 61. The condition (2.41) can be substituted by the more general condition

∑_{i=1}^{n} p_{i} Re ⟨X − x_{i}, x_{i} − x⟩ ≥ 0,

and the conclusion (2.42) will still remain valid.

The following corollary is a natural consequence of the above findings.

Corollary 53. Let x_{i} ∈ H (i = 1, …, n) and x, X ∈ H be such that the condition (2.41) holds. Then we have the inequality

0 ≤ \frac{1}{P_{n}} ∑_{i=1}^{n} p_{i} ∥x_{i}∥ − ∥\frac{1}{P_{n}} ∑_{i=1}^{n} p_{i}x_{i}∥ ≤ \frac{1}{2} ∥X − x∥ .

The proof follows by Theorem 63 by choosing F : H → ℝ, F (x) = ∥x∥ which is Lipschitzian with the constant L = 1, as

|F (x) − F (y)| = |∥x∥ − ∥y∥| ≤ ∥x − y∥ ,

for all x, y ∈ H.

3. The Version for Inner-Products

3.1. A Discrete Inequality of Grüss Type. The following Grüss type inequality holds [7].

Theorem 64. Let (H; ⟨·, ·⟩) be an inner product space over 𝕂; 𝕂 = ℂ, ℝ, x_{i}, y_{i} ∈ H, p_{i} ≥ 0 (i = 0, …, n) (n ≥ 2) with ∑_{i=1}^{n} p_{i} = 1. If x, X, y, Y ∈ H are such that

(3.1)  Re ⟨X − x_{i}, x_{i} − x⟩ ≥ 0 and Re ⟨Y − y_{i}, y_{i} − y⟩ ≥ 0

for all i ∈ {1, …, n} , then we have the inequality

(3.2) |∑_{i=1}^{n} p_{i} ⟨x_{i}, y_{i}⟩ − ⟨∑_{i=1}^{n} p_{i}x_{i}, ∑_{i=1}^{n} p_{i}y_{i}⟩| ≤ \frac{1}{4} ∥X − x∥ ∥Y − y∥ .

The constant \frac{1}{4} is sharp.

Proof. A simple calculation shows that

(3.3)  ∑_{i=1}^{n} p_{i} ⟨x_{i}, y_{i}⟩ − ⟨∑_{i=1}^{n} p_{i}x_{i}, ∑_{i=1}^{n} p_{i}y_{i}⟩

= \frac{1}{2} ∑_{i,j=1}^{n} p_{i}p_{j} ⟨x_{i} − x_{j}, y_{i} − y_{j}⟩ .

Taking the modulus in both parts of (3.3) and using the generalized triangle inequality, we obtain

|∑_{i=1}^{n} p_{i} ⟨x_{i}, y_{i}⟩ − ⟨∑_{i=1}^{n} p_{i}x_{i}, ∑_{i=1}^{n} p_{i}y_{i}⟩|

≤ \frac{1}{2} ∑_{i,j=1}^{n} p_{i}p_{j} |⟨x_{i} − x_{j}, y_{i} − y_{j}⟩| .

By Schwarz’s inequality in inner product spaces we have

|⟨x_{i} − x_{j}, y_{i} − y_{j}⟩| ≤ ∥x_{i} − x_{j}∥ ∥y_{i} − y_{j}∥ ,

for all i, j ∈ {1, …, n} , and therefore

|∑_{i=1}^{n} p_{i} ⟨x_{i}, y_{i}⟩ − ⟨∑_{i=1}^{n} p_{i}x_{i}, ∑_{i=1}^{n} p_{i}y_{i}⟩|

≤ \frac{1}{2} ∑_{i,j=1}^{n} p_{i}p_{j} ∥x_{i} − x_{j}∥ ∥y_{i} − y_{j}∥ .

Using the Cauchy-Bunyakovsky-Schwarz inequality for double sums, we can state that

\frac{1}{2} ∑_{i,j=1}^{n} p_{i}p_{j} ∥x_{i} − x_{j}∥ ∥y_{i} − y_{j}∥

≤ (\frac{1}{2} ∑_{i,j=1}^{n} p_{i}p_{j} ∥x_{i} − x_{j}∥^{2})^{\frac{1}{2}} (\frac{1}{2} ∑_{i,j=1}^{n} p_{i}p_{j} ∥y_{i} − y_{j}∥^{2})^{\frac{1}{2}}

and, as a simple calculation shows that,

\frac{1}{2} ∑_{i,j=1}^{n} p_{i}p_{j} ∥x_{i} − x_{j}∥^{2} = ∑_{i=1}^{n} p_{i} ∥x_{i}∥^{2} − ∥∑_{i=1}^{n} p_{i}x_{i}∥^{2}

and

\frac{1}{2} ∑_{i,j=1}^{n} p_{i}p_{j} ∥y_{i} − y_{j}∥^{2} = ∑_{i=1}^{n} p_{i} ∥y_{i}∥^{2} − ∥∑_{i=1}^{n} p_{i}y_{i}∥^{2} ,

we obtain

(3.4) |∑_{i=1}^{n} p_{i} ⟨x_{i}, y_{i}⟩ − ⟨∑_{i=1}^{n} p_{i}x_{i}, ∑_{i=1}^{n} p_{i}y_{i}⟩|

≤ (∑_{i=1}^{n} p_{i} ∥x_{i}∥^{2} − ∥∑_{i=1}^{n} p_{i}x_{i}∥^{2})^{\frac{1}{2}} (∑_{i=1}^{n} p_{i} ∥y_{i}∥^{2} − ∥∑_{i=1}^{n} p_{i}y_{i}∥^{2})^{\frac{1}{2}} .

Using Lemma 10, we know that

(∑_{i=1}^{n} p_{i} ∥x_{i}∥^{2} − ∥∑_{i=1}^{n} p_{i}x_{i}∥^{2})^{\frac{1}{2}} ≤ \frac{1}{2} ∥X − x∥

and

(∑_{i=1}^{n} p_{i} ∥y_{i}∥^{2} − ∥∑_{i=1}^{n} p_{i}y_{i}∥^{2})^{\frac{1}{2}} ≤ \frac{1}{2} ∥Y − y∥ ,

and then, by (3.4), we deduce the desired inequality (3.3).

To prove the sharpness of the constant \frac{1}{4}, let us assume that (3.2) holds with a constant c > 0, i.e.,

(3.5) |∑_{i=1}^{n} p_{i} ⟨x_{i}, y_{i}⟩ − ⟨∑_{i=1}^{n} p_{i}x_{i}, ∑_{i=1}^{n} p_{i}y_{i}⟩| ≤ c ∥X − x∥ ∥Y − y∥

under the above assumptions for p_{i}, x_{i}, y_{i}, x, X, y, Y and n ≥ 2.

If we choose n = 2, x_{1} = x, x_{2} = X, y_{1} = y, y_{2} = Y (x ≠ X, y ≠ Y) and p_{1} = p_{2} = \frac{1}{2}, then

∑_{i=1}^{2} p_{i} ⟨x_{i}, y_{i}⟩ − ⟨∑_{i=1}^{2} p_{i}x_{i}, ∑_{i=1}^{2} p_{i}y_{i}⟩ = \frac{1}{2} ∑_{i,j=1}^{2} p_{i}p_{j} ⟨x_{i} − x_{j}, y_{i} − y_{j}⟩

= ∑_{1≤i<j≤2} p_{i}p_{j} ⟨x_{i} − x_{j}, y_{i} − y_{j}⟩

= \frac{1}{4} ⟨x − X, y − Y ⟩

and then

|∑_{i=1}^{2} p_{i} ⟨x_{i}, y_{i}⟩ − ⟨∑_{i=1}^{2} p_{i}x_{i}, ∑_{i=1}^{2} p_{i}y_{i}⟩| = \frac{1}{4} |⟨x − X, y − Y ⟩| .

Choose X − x = z, Y − y = z, z ≠ 0. Then using (3.5), we derive

\frac{1}{4} ∥z∥^{2} ≤ c ∥z∥^{2} , z ≠ 0

which implies that c ≥ \frac{1}{4}, and the theorem is proved.

Remark 62. The condition (3.1) can be replaced by the more general assumption

∑_{i=1}^{n} p_{i} Re ⟨X − x_{i}, x_{i} − x⟩ ≥ 0, ∑_{i=1}^{n} p_{i} Re ⟨Y − y_{i}, y_{i} − y⟩ ≥ 0

and the conclusion (3.2) still remains valid.

The following corollary for real or complex numbers holds.

Corollary 54. Let a_{i}, b_{i} ∈ 𝕂 (𝕂 = ℂ, ℝ) , p_{i} ≥ 0 (i = 1, …, n) with ∑_{i=1}^{n} p_{i} = 1. If a, A, b, B ∈ 𝕂 are such that

(3.6)  Re [(A − a_{i}) (ā_{i} − ā)] ≥ 0, Re [(B − b_{i}) (b̄_{i} − b̄)] ≥ 0,

then we have the inequality

(3.7)  |∑_{i=1}^{n} p_{i}a_{i}b̄_{i} − ∑_{i=1}^{n} p_{i}a_{i} ∑_{i=1}^{n} p_{i}b̄_{i}| ≤ \frac{1}{4} |A − a| |B − b|

and the constant \frac{1}{4} is sharp.

The proof is obvious by Theorem 64 applied for the inner product space (ℂ, ⟨·, ·⟩) , where ⟨x, y⟩ = x · ȳ. We omit the details.

Remark 63. The condition (3.6) can be replaced by the more general condition

∑_{i=1}^{n} p_{i} Re [(A − a_{i}) (ā_{i} − ā)] ≥ 0, ∑_{i=1}^{n} p_{i} Re [(B − b_{i}) (b̄_{i} − b̄)] ≥ 0

and the conclusion of the above corollary will still remain valid.

Remark 64. If we assume that a_{i}, b_{i}, a, b, A, B are real numbers, then (3.6) is equivalent to

a ≤ a_{i} ≤ A, b ≤ b_{i} ≤ B for all i ∈ {1, …, n}

and (3.7) becomes

0 ≤ |∑_{i=1}^{n} p_{i}a_{i}b_{i} − ∑_{i=1}^{n} p_{i}a_{i} ∑_{i=1}^{n} p_{i}b_{i}| ≤ \frac{1}{4} (A − a) (B − b) ,

which is the classical Grüss inequality for sequences of real numbers.

3.2. Applications for Convex Functions. Let (H; ⟨·, ·⟩) be a real inner product space and F : H → ℝ a Fréchet differentiable convex mapping on H. Then we have the “gradient inequality”

(3.8)  F (x) − F (y) ≥ ⟨▽F (y) , x − y⟩

for all x, y ∈ H, where ∇F : H → H is the gradient operator associated to the differentiable convex function F.

The following theorem holds [7].

Theorem 65. Let F : H → ℝ be as above and x_{i} ∈ H (i = 1, …, n). Suppose that there exists the vectors x, X ∈ H such that ⟨x_{i} − x, X − x_{i}⟩ ≥ 0 for all i ∈ {1, …, m} and y, Y ∈ H such that

⟨▽F (x_{i}) − y, Y − ▽F (x_{i})⟩ ≥ 0

for all i ∈ {1, …, m}. Then for all p_{i} ≥ 0 (i = 1, …, m) with P_{m} := ∑_{i=1}^{m} p_{i} > 0, we have the inequality

(3.9) 0 ≤ \frac{1}{P_{m}} ∑_{i=1}^{m} p_{i}F (x_{i}) − F (\frac{1}{P_{m}} ∑_{i=1}^{m} p_{i}x_{i}) ≤ \frac{1}{4} ∥X − x∥ ∥Y − y∥ .

Proof. Choose in (3.8), x = \frac{1}{P_{M}} ∑_{i=1}^{m} p_{i}x_{i} and y = x_{j} to obtain

(3.10) F (\frac{1}{P_{m}} ∑_{i=1}^{m} p_{i}x_{i}) − F (x_{j}) ≥ ⟨▽F (x_{j}) , \frac{1}{P_{m}} ∑_{i=1}^{m} p_{i}x_{i} − x_{j}⟩

for all j ∈ {1, …, n}.

If we multiply (3.10) by p_{j} ≥ 0 and sum over j from 1 to m, we have

P_{m}F (\frac{1}{P_{m}} ∑_{i=1}^{m} p_{i}x_{i}) − ∑_{j=1}^{m} p_{j}F (x_{j})

≥ \frac{1}{P_{m}} ⟨∑_{j=1}^{m} ▽F (x_{j}) , ∑_{i=1}^{m} p_{i}x_{i}⟩ − ∑_{i=1}^{m} ⟨▽F (x_{j}) , x_{j}⟩ .

Dividing by P_{m} > 0, we obtain the inequality

(3.11)  0 ≤ \frac{1}{P_{m}} ∑_{i=1}^{m} p_{i}F (x_{i}) − F (\frac{1}{P_{m}} ∑_{i=1}^{m} p_{i}x_{i})

≤ \frac{1}{P_{m}} ∑_{i=1}^{m} p_{i} ⟨▽F (x_{i}) , x_{i}⟩

− ⟨\frac{1}{P_{m}} ∑_{i=1}^{m} p_{i}▽F (x_{i}) , \frac{1}{P_{m}} ∑_{i=1}^{m} p_{i}x_{i}⟩ ,

which is a generalisation for inner product spaces of the result by Dragomir-Goh established in 1996 for the case of differentiable mappings defined on ℝ^{n} [14].

Applying Theorem 64 for real inner product spaces, and y_{i} = ∇F (x_{i}), we easily deduce

(3.12) \frac{1}{P_{m}} ∑_{i=1}^{m} p_{i} ⟨x_{i}, ▽F (x_{i})⟩ − ⟨\frac{1}{P_{m}} ∑_{i=1}^{m} p_{i}x_{i}, \frac{1}{P_{m}} ∑_{i=1}^{m} p_{i}▽F (x_{i})⟩

≤ \frac{1}{4} ∥X − x∥ ∥Y − y∥

and then, by (3.11) and (3.12) we can conclude that the desired inequality (3.9) holds.

3.3. Applications for Some Discrete Transforms. Let (H; ⟨·, ·⟩) be an inner product space over 𝕂, 𝕂 = ℂ, ℝ and x̄ = (x_{1}, …, x_{n}) be a sequence of vectors in H.

For a given m ∈ 𝕂, define the discrete Fourier Transform

F_{w} \binom{}{x̄} (m) = ∑_{k=1}^{n} exp (2wimk) × x_{k}, m = 1, …, n.

The complex number ∑_{k=1}^{n} exp (2wimk) ⟨x_{k}, y_{k}⟩ is actually the usual Fourier transform of the vector (⟨x_{1}, y_{1}⟩ , …, ⟨x_{n}, y_{n}⟩) ∈ 𝕂^{n} and will be denoted by

F_{w} (x̄ · ȳ) (m) = ∑_{k=1}^{n} exp (2wimk) ⟨x_{k}, y_{k}⟩ , m = 1, …, n.

The following result holds [7].

Theorem 66. Let x̄, ȳ ∈ H^{n} be sequences of vectors such that there exists the vectors c, C, y, Y ∈ H with the properties

Re ⟨C − exp (2wimk) x_{k}, exp (2wimk) x_{k} − c⟩ ≥ 0, k, m = 1, …, n

and

(3.13)  Re ⟨Y − y_{k}, y_{k} − y⟩ ≥ 0, k = 1, …, n.

Then we have the inequality

|F_{w} (x̄ · ȳ) (m) − ⟨F_{w} \binom{}{x̄} (m) , \frac{1}{n} ∑_{k=1}^{n} y_{k}⟩| ≤ \frac{n}{4} ∥C − c∥ ∥Y − y∥ ,

for all m ∈ {1, …, n}.

The proof follows by Theorem 64 applied for p_{k} = \frac{1}{n} and for the sequences x_{k} → c_{k} = exp (2wimk) x_{k} and y_{k} (k = 1, …, n). We omit the details.

We can also consider the Mellin transform

M \binom{}{x̄} (m) := ∑_{k=1}^{n} k^{m−1}x_{k}, m = 1, …, n,

of the sequence x̄ = (x_{1}, …, x_{n}) ∈ H^{n}.

We remark that the complex number ∑_{k=1}^{n} k^{m−1} ⟨x_{k}, y_{k}⟩ is actually the Mellin transform of the vector (⟨x_{1}, y_{1}⟩ , …, ⟨x_{n}, y_{n}⟩) ∈ 𝕂^{n} and will be denoted by

M (x̄ · ȳ) (m) := ∑_{k=1}^{n} k^{m−1} ⟨x_{k}, y_{k}⟩ .

The following theorem holds [7].

Theorem 67. Let x̄, ȳ ∈ H^{n} be sequences of vectors such that there exist the vectors d, D, y, Y ∈ H with the properties

Re ⟨D − k^{m−1}x_{k}, k^{m−1}x_{k} − d⟩ ≥ 0

for all k, m ∈ {1, …, n} , and (3.13) is fulfilled.

Then we have the inequality

|M (x̄ · ȳ) (m) − ⟨M \binom{}{x̄} (m) , \frac{1}{n} ∑_{k=1}^{n} y_{k}⟩| ≤ \frac{n}{4} ∥D − d∥ ∥Y − y∥

for all m ∈ {1, …, n}.

The proof follows by Theorem 64 applied for p_{k} = \frac{1}{n} and for the sequences x_{k} → d_{k} = kx_{k} and y_{k} (k = 1, …, n). We omit the details.

Another result which connects the Fourier transforms for different parameters w also holds [7].

Theorem 68. Let x̄, ȳ ∈ H^{n} and w, z ∈ 𝕂. If there exist the vectors e, E, f, F ∈ H such that

Re ⟨E − exp (2wimk) x_{k}, exp (2wimk) x_{k} − e⟩ ≥ 0, k, m = 1, …, n

and

Re ⟨F − exp (2zimk) y_{k}, exp (2zimk) y_{k} − f⟩ ≥ 0, k, m = 1, …, n then we have the inequality:

|\frac{1}{n}F_{w+z} (x̄ · ȳ) (m) − ⟨\frac{1}{n}F_{w} \binom{}{x̄} (m) , \frac{1}{n}F_{z} \binom{}{ȳ} (m)⟩|

≤ \frac{1}{4} ∥E − e∥ ∥F − f∥ ,

for all m ∈ {1, …, n}.

The proof follows by Theorem 64 for the sequences exp (2wimk) x_{k}, exp (2zimk) y_{k} (k = 1, …, n). We omit the details.

4. More Grüss’ Type Inequalities

4.1. Introduction. In the recent paper [11], the author has obtained the following Grüss type inequality for forward difference.

Theorem 69. Let \overline{x} = (x_{1}, …, x_{n}) , \overline{y} = (y_{1}, …, y_{n}) ∈ H^{n} and \overline{p} ∈ ℝ_{+}^{n} be a probability sequence. Then one has the inequalities

(4.1) |∑_{i=1}^{n} p_{i} ⟨x_{i}, y_{i}⟩ − ⟨∑_{i=1}^{n} p_{i}x_{i}, ∑_{i=1}^{n} p_{i}y_{i}⟩|

≤ \{ \begin{matrix}[∑_{i=1}^{n} i^{2}p_{i} − (∑_{i=1}^{n} ip_{i})^{2}] \\ × max_{k=1,...,n−1} ∥Δx_{k}∥ max_{k=1,...,n−1} ∥Δy_{k}∥; \\ ∑_{1≤j<i≤n} p_{i}p_{j} (i − j) (∑_{k=1}^{n−1} ∥Δx_{k}∥^{p})^{\frac{1}{p}} (∑_{k=1}^{n−1} ∥Δy_{k}∥^{q})^{\frac{1}{q}} \\ if p > 1, \frac{1}{p} + \frac{1}{q} = 1; \\ \frac{1}{2} [∑_{i=1}^{n} p_{i} (1 − p_{i})] ∑_{k=1}^{n−1} ∥Δx_{k}∥ ∑_{k=1}^{n−1} ∥Δy_{k}∥ . \end{matrix}

The constants 1, 1 and \frac{1}{2} in the right hand side of the inequality (4.1) are best in the sense that they cannot be replaced by smaller constants.

If one chooses p_{i} = \frac{1}{n} (i = 1, …, n) in (4.1), then the following unweighted inequalities hold:

(4.2) |\frac{1}{n} ∑_{i=1}^{n} ⟨x_{i}, y_{i}⟩ − ⟨\frac{1}{n} ∑_{i=1}^{n} x_{i}, \frac{1}{n} ∑_{i=1}^{n} y_{i}⟩|

≤ \{ \begin{matrix}\frac{n^{2}−1}{12} max_{k=1,...,n−1} ∥Δx_{k}∥ max_{k=1,...,n−1} ∥Δy_{k}∥; \\ \frac{n^{2}−1}{6n} & (∑_{k=1}^{n−1} ∥Δx_{k}∥^{p}) & \frac{1}{p} & (∑_{k=1}^{n−1} ∥Δy_{k}∥^{q}) & \frac{1}{q} \\ if p > 1, \frac{1}{p} + \frac{1}{q} = 1; \\ \frac{n−1}{2n} ∑_{k=1}^{n−1} ∥Δx_{k}∥ ∑_{k=1}^{n−1} ∥Δy_{k}∥ . \end{matrix}

Here, the constants \frac{1}{12}, \frac{1}{6} and \frac{1}{2} are also best possible in the above sense.

The following reverse of the Cauchy-Bunyakovsky-Schwarz inequality for sequences of vectors in inner product spaces holds.

Corollary 55. With the assumptions in Theorem 69 for \overline{x} and \overline{p} one has the inequalities

(4.3) 0 ≤ ∑_{i=1}^{n} p_{i} ∥x_{i}∥^{2} − ∥∑_{i=1}^{n} p_{i}x_{i}∥^{2}

≤ \{ \begin{matrix}[∑_{i=1}^{n} i^{2}p_{i} − (∑_{i=1}^{n} ip_{i})^{2}] max_{k=\overline{1,n−1}} ∥Δx_{k}∥^{2} ; \\ ∑_{1≤j<i≤n} p_{i}p_{j} (i − j) ( ∑_{k=1}^{n−1} ∥Δx_{k}∥^{p})^{\frac{1}{p}} ( ∑_{k=1}^{n−1} ∥Δx_{k}∥^{q})^{\frac{1}{q}} \\ if p > 1, \frac{1}{p} + \frac{1}{q} = 1; \\ \frac{1}{2} [∑_{i=1}^{n} p_{i} (1 − p_{i})] ( ∑_{k=1}^{n−1} ∥Δx_{k}∥)^{2} . \end{matrix}

The constants 1, 1 and \frac{1}{2} are best possible in the above sense.

The following particular inequalities that may be deduced from (4.3) on choosing the equal weights p_{i} = \frac{1}{n}, i = 1, …, n are also of interest

(4.4)  0 ≤ \frac{1}{n} ∑_{i=1}^{n} ∥x_{i}∥^{2} − ∥\frac{1}{n} ∑_{i=1}^{n} x_{i}∥^{2}

≤ \{ \begin{matrix}\frac{n^{2} − 1}{12} max_{k=\overline{1,n−1}} ∥Δx_{k}∥^{2} ; \\ \frac{n^{2} − 1}{6n} & ( ∑_{k=1}^{n−1} ∥Δx_{k}∥^{p}) & \frac{1}{p} & ( ∑_{k=1}^{n−1} ∥Δx_{k}∥^{q}) & \frac{1}{q} \\ if p > 1, \frac{1}{p} + \frac{1}{q} = 1; \\ \frac{n − 1}{2n} ( ∑_{k=1}^{n−1} ∥Δx_{k}∥)^{2} . \end{matrix}

Here the constants \frac{1}{12}, \frac{1}{6} and \frac{1}{2} are also best possible.

The main aim of this section is to present, by following [10], a different class of Grüss type inequalities for sequences of vectors in inner product spaces and to apply them for obtaining a reverse of Jensen’s inequality for convex functions defined on such spaces.

4.2. More Grüss Type Inequalities. The following lemma holds (see also [12]).

Lemma 11. Let a, x, A be vectors in the inner product space (H; ⟨·, ·⟩) over the real or complex number field 𝕂 (𝕂 = ℝ, ℂ) with a ≠ A. The following statements are equivalent:

(i) Re ⟨A − x, x − a⟩ ≥ 0; (ii) ∥x − \frac{a+A}{2} ∥ ≤ \frac{1}{2} ∥A − a∥ .

The following inequality of Grüss type for sequences of vectors in inner product spaces holds [10].

Theorem 70. Let (H; ⟨·, ·⟩) be an inner product over 𝕂 (𝕂 = ℂ, ℝ), and \overline{x} = (x_{1}, …, x_{n}) , \overline{y} = (y_{1}, …, y_{n}) ∈ H^{n}, \overline{p} ∈ ℝ_{+}^{n} with ∑_{i=1}^{n} p_{i} = 1. If x, X ∈ H are such that

(4.5)  Re ⟨X − x_{i}, x_{i} − x⟩ ≥ 0 for each i ∈ {1, …, n} ,

or, equivalently,

(4.6)  ∥x_{i} − \frac{x + X}{2} ∥ ≤ \frac{1}{2} ∥X − x∥ for each i ∈ \{1, …, n\} ,

then one has the inequality

|∑_{i=1}^{n} p_{i} ⟨x_{i}, y_{i}⟩ − ⟨∑_{i=1}^{n} p_{i}x_{i}, ∑_{i=1}^{n} p_{i}y_{i}⟩| (4.7)

≤ \frac{1}{2} ∥X − x∥ ∑_{i=1}^{n} p_{i} ∥y_{i} − ∑_{j=1}^{n} p_{j}y_{j}∥

≤ \frac{1}{2} ∥X − x∥ [∑_{i=1}^{n} p_{i} ∥y_{i}∥^{2} − ∥∑_{i=1}^{n} p_{i}y_{i}∥^{2}]^{\frac{1}{2}} .

The constant \frac{1}{2} is best possible in the first and second inequality in the sense that it cannot be replaced by a smaller constant.

Proof. It is easy to see that the following identity holds true

(4.8) ∑_{i=1}^{n} p_{i} ⟨x_{i}, y_{i}⟩ − ⟨∑_{i=1}^{n} p_{i}x_{i}, ∑_{i=1}^{n} p_{i}y_{i}⟩

= ∑_{i=1}^{n} p_{i} ⟨x_{i} − \frac{x + X}{2} , y_{i} − ∑_{j=1}^{n} p_{j}y_{j}⟩ .

Taking the modulus in (4.8) and using the Schwarz inequality in the inner product space (H; ⟨·, ·⟩) , we have

|∑_{i=1}^{n} p_{i} ⟨x_{i}, y_{i}⟩ − ⟨∑_{i=1}^{n} p_{i}x_{i}, ∑_{i=1}^{n} p_{i}y_{i}⟩|

≤ ∑_{i=1}^{n} p_{i} |⟨x_{i} − \frac{x + X}{2} , y_{i} − ∑_{j=1}^{n} p_{j}y_{j}⟩|

≤ ∑_{i=1}^{n} p_{i} ∥x_{i} − \frac{x + X}{2} ∥ ∥y_{i} − ∑_{j=1}^{n} p_{j}y_{j}∥

≤ \frac{1}{2} ∥X − x∥ ∑_{i=1}^{n} p_{i} ∥y_{i} − ∑_{j=1}^{n} p_{j}y_{j}∥ ,

and the first inequality in (4.7) is proved.

Using the Cauchy-Bunyakovsky-Schwarz inequality for positive sequences and the calculation rules in inner product spaces, we have

∑_{i=1}^{n} p_{i} ∥y_{i} − ∑_{j=1}^{n} p_{j}y_{j}∥ ≤ [∑_{i=1}^{n} p_{i} ∥y_{i} − ∑_{j=1}^{n} p_{j}y_{j}∥^{2}]^{\frac{1}{2}}

and

∑_{i=1}^{n} p_{i} ∥y_{i} − ∑_{j=1}^{n} p_{j}y_{j}∥^{2} = ∑_{i=1}^{n} p_{i} ∥y_{i}∥^{2} − ∥∑_{i=1}^{n} p_{i}y_{i}∥^{2}

giving the second part of (4.7).

To prove the sharpness of the constant \frac{1}{2} in the first inequality in (4.7), let us assume that, under the assumptions of the theorem, the inequality holds with a constant C > 0, i.e.,

(4.9) |∑_{i=1}^{n} p_{i} ⟨x_{i}, y_{i}⟩ − ⟨∑_{i=1}^{n} p_{i}x_{i}, ∑_{i=1}^{n} p_{i}y_{i}⟩|

≤ C ∥X − x∥ ∑_{i=1}^{n} p_{i} ∥y_{i} − ∑_{j=1}^{n} p_{j}y_{j}∥ .

Consider n = 2 and observe that

∑_{i=1}^{2} p_{i} ⟨x_{i}, y_{i}⟩ − ⟨∑_{i=1}^{2} p_{i}x_{i}, ∑_{i=1}^{2} p_{i}y_{i}⟩ = p_{2}p_{1} ⟨x_{2} − x_{1}, y_{2} − y_{1}⟩ ,

∑_{i=1}^{2} p_{i} ∥y_{i} − ∑_{j=1}^{2} p_{j}y_{j}∥ = 2p_{2}p_{1} ∥y_{2} − y_{1}∥

and then, by (4.9), we deduce

(4.10)  p_{2}p_{1} |⟨x_{2} − x_{1}, y_{2} − y_{1}⟩| ≤ 2C ∥X − x∥ p_{2}p_{1} ∥y_{2} − y_{1}∥ .

If we choose p_{1}, p_{2} > 0, y_{2} = x_{2}, y_{1} = x_{1} and x_{2} = X, x_{1} = x with x ≠ X, then (4.6) holds and from (4.10) we deduce C ≥ \frac{1}{2}.

The fact that \frac{1}{2} is best possible in the second inequality may be proven in a similar manner and we omit the details.

Remark 65. If \overline{x} and \overline{y} satisfy the assumptions of Theorem 70, or, equivalently,

(4.11) ∥x_{i} − \frac{x + X}{2} ∥ ≤ \frac{1}{2} ∥X − x∥ , ∥y_{i} − \frac{y + Y}{2} ∥ ≤ \frac{1}{2} ∥Y − y∥ ,

for each i ∈ {1, …, n} , then by Theorem 70 we may state the following sequence of inequalities improving the Grüss inequality (4.8)

(4.12)  0 ≤ |∑_{i=1}^{n} p_{i} ⟨x_{i}, y_{i}⟩ − ⟨∑_{i=1}^{n} p_{i}x_{i}, ∑_{i=1}^{n} p_{i}y_{i}⟩|

≤ \frac{1}{2} ∥X − x∥ ∑_{i=1}^{n} p_{i} ∥y_{i} − ∑_{j=1}^{n} p_{j}y_{j}∥

≤ \frac{1}{2} ∥X − x∥ (∑_{i=1}^{n} p_{i} ∥y_{i}∥^{2} − ∥∑_{i=1}^{n} p_{i}y_{i}∥^{2})^{\frac{1}{2}}

≤ \frac{1}{4} ∥X − x∥ ∥Y − y∥ .

In particular, for x_{i} = y_{i} (i = 1, …, n) , one has

(4.13)  0 ≤ ∑_{i=1}^{n} p_{i} ∥x_{i}∥^{2} − ∥∑_{i=1}^{n} p_{i}x_{i}∥^{2}

≤ \frac{1}{2} ∥X − x∥ ∑_{i=1}^{n} p_{i} ∥x_{i} − ∑_{j=1}^{n} p_{j}x_{j}∥

and the constant \frac{1}{2} is best possible.

The following result also holds [10].

Theorem 71. Let (H; ⟨·, ·⟩) and 𝕂 be as above and \overline{x} = (x_{1}, …, x_{n}) ∈ H^{n}, \overline{α} = (α_{1}, …, α_{n}) ∈ 𝕂^{n} and \overline{p} a probability vector. If x, X ∈ H are such that (4.5) or, equivalently, (4.6) holds, then we have the inequality

(4.14)  0 ≤ ∥∑_{i=1}^{n} p_{i}α_{i}x_{i} − ∑_{i=1}^{n} p_{i}α_{i} · ∑_{i=1}^{n} p_{i}x_{i}∥

≤ \frac{1}{2} ∥X − x∥ ∑_{i=1}^{n} p_{i} |α_{i} − ∑_{j=1}^{n} p_{j}α_{j}|

≤ \frac{1}{2} ∥X − x∥ [∑_{i=1}^{n} p_{i} |α_{i}|^{2} − |∑_{i=1}^{n} p_{i}α_{i}|^{2}]^{\frac{1}{2}} .

The constant \frac{1}{2} in the first and second inequalities is best possible in the sense that it cannot be replaced by a smaller constant.

Proof. We start with the following equality that may be easily verified by direct calculation

(4.15) ∑_{i=1}^{n} p_{i}α_{i}x_{i} − ∑_{i=1}^{n} p_{i}α_{i} · ∑_{i=1}^{n} p_{i}x_{i}

= ∑_{i=1}^{n} p_{i} (α_{i} − ∑_{j=1}^{n} p_{j}α_{j}) (x_{i} − \frac{x + X}{2} ) .

If we take the norm in (4.15), we deduce

∥∑_{i=1}^{n} p_{i}α_{i}x_{i} − ∑_{i=1}^{n} p_{i}α_{i} · ∑_{i=1}^{n} p_{i}x_{i}∥

≤ ∑_{i=1}^{n} p_{i} |α_{i} − ∑_{j=1}^{n} p_{j}α_{j}| ∥x_{i} − \frac{x + X}{2} ∥

≤ \frac{1}{2} ∥X − x∥ ∑_{i=1}^{n} p_{i} |α_{i} − ∑_{j=1}^{n} p_{j}α_{j}|

≤ \frac{1}{2} ∥X − x∥ (∑_{i=1}^{n} p_{i} (α_{i} − ∑_{j=1}^{n} p_{j}α_{j})^{2})^{\frac{1}{2}}

= \frac{1}{2} ∥X − x∥ (∑_{i=1}^{n} p_{i} |α_{i}|^{2} − |∑_{i=1}^{n} p_{i}α_{i}|^{2})^{\frac{1}{2}} ,

proving the inequality (4.14).

The fact that the constant \frac{1}{2} is sharp may be proven in a similar manner to the one embodied in the proof of Theorem 70. We omit the details.

Remark 66. If \overline{x} and \overline{α} satisfy the assumption

∥α_{i} − \frac{a + A}{2}∥ ≤ \frac{1}{2} |A − a| , ∥x_{i} − \frac{x + X}{2} ∥ ≤ \frac{1}{2} ∥X − x∥ ,

for each i ∈ {1, …, n} , then by Theorem 70 we may state the following sequence of inequalities improving the Grüss inequality

0 ≤ ∥∑_{i=1}^{n} p_{i}α_{i}x_{i} − ∑_{i=1}^{n} p_{i}α_{i} · ∑_{i=1}^{n} p_{i}x_{i}∥

≤ \frac{1}{2} ∥X − x∥ ∑_{i=1}^{n} p_{i} |α_{i} − ∑_{j=1}^{n} p_{j}α_{j}|

≤ \frac{1}{2} ∥X − x∥ (∑_{i=1}^{n} p_{i} |α_{i}|^{2} − |∑_{i=1}^{n} p_{i}α_{i}|^{2})^{\frac{1}{2}}

≤ \frac{1}{4} |A − a| ∥X − x∥ .

Remark 67. If in (4.14) we choose x_{i} = α_{i} ∈ ℂ and assume that |α_{i} − \frac{a+A}{2} | ≤ \frac{1}{2} |A − a| , where a, A ∈ ℂ, then we get the following interesting inequality for complex numbers

0 ≤ |∑_{i=1}^{n} p_{i}α_{i}^{2} − (∑_{i=1}^{n} p_{i}α_{i})^{2}|

≤ \frac{1}{2} |A − a| ∑_{i=1}^{n} p_{i} |α_{i} − ∑_{j=1}^{n} p_{j}α_{j}|

≤ \frac{1}{2} |A − a| [∑_{i=1}^{n} p_{i} |α_{i}|^{2} − |∑_{i=1}^{n} p_{i}α_{i}|^{2}]^{\frac{1}{2}} .

4.3. Applications for Convex Functions. Let (H; ⟨·, ·⟩) be a real inner product space and F : H → ℝ a Fréchet differentiable convex function on H. If ▽F : H → H denotes the gradient operator associated to F, then we have the inequality

(4.16)  F (x) − F (y) ≥ ⟨▽F (y) , x − y⟩

for each x, y ∈ H.

The following result holds [10].

Theorem 72. Let F : H → ℝ be as above and z_{i} ∈ H, i ∈ {1, …, n} . Suppose that there exist the vectors m, M ∈ H such that either

⟨▽F (z_{i}) − m, M − ▽F (z_{i})⟩ ≥ 0 for each i ∈ {1, …, n};

or, equivalently,

∥▽F (z_{i}) − \frac{m + M}{2} ∥ ≤ \frac{1}{2} ∥M − m∥ for each i ∈ \{1, …, n\} ,

holds.

If q_{i} ≥ 0 (i ∈ {1, …, n}) with Q_{n} := ∑_{i=1}^{n} q_{i} > 0, then we have the following reverse of Jensen’s inequality

(4.17)  0 ≤ \frac{1}{Q_{n}} ∑_{i=1}^{n} q_{i}F (z_{i}) − F (\frac{1}{Q_{n}} ∑_{i=1}^{n} q_{i}z_{i})

≤ \frac{1}{2} ∥M − m∥ \frac{1}{Q_{n}} ∑_{i=1}^{n} q_{i} ∥z_{i} − \frac{1}{Q_{n}} ∑_{j=1}^{n} q_{j}z_{j}∥

≤ \frac{1}{2} ∥M − m∥ [\frac{1}{Q_{n}} ∑_{i=1}^{n} q_{i} ∥z_{i}∥^{2} − ∥\frac{1}{Q_{n}} ∑_{i=1}^{n} q_{i}z_{i}∥^{2}]^{\frac{1}{2}} .

Proof. We know, see for example [7, Eq. (4.4)], that the following reverse of Jensen’s inequality for Fréchet differentiable convex functions

(4.18) 0 ≤ \frac{1}{Q_{n}} ∑_{i=1}^{n} q_{i}F (z_{i}) − F (\frac{1}{Q_{n}} ∑_{i=1}^{n} q_{i}z_{i})

≤ \frac{1}{Q_{n}} ∑_{i=1}^{n} q_{i} ⟨▽F (z_{i}) , z_{i}⟩ − ⟨\frac{1}{Q_{n}} ∑_{i=1}^{n} q_{i}▽F (z_{i}) , \frac{1}{Q_{n}} ∑_{i=1}^{n} q_{i}z_{i}⟩

holds.

Now, if we use Theorem 70 for the choices x_{i} = ∇F (z_{i}) , y_{i} = z_{i} and p_{i} = \frac{1}{Q_{n}} q_{i}, i ∈ \{1, …, n\} , we can state the inequality

\frac{1}{Q_{n}} ∑_{i=1}^{n} q_{i} ⟨▽F (z_{i}) , z_{i}⟩ − ⟨\frac{1}{Q_{n}} ∑_{i=1}^{n} q_{i}▽F (z_{i}) , \frac{1}{Q_{n}} ∑_{i=1}^{n} q_{i}z_{i}⟩ (4.19)

≤ \frac{1}{2} ∥M − m∥ \frac{1}{Q_{n}} ∑_{i=1}^{n} q_{i} ∥z_{i} − \frac{1}{Q_{n}} ∑_{j=1}^{n} q_{j}z_{j}∥

≤ \frac{1}{2} ∥M − m∥ [\frac{1}{Q_{n}} ∑_{i=1}^{n} q_{i} ∥z_{i}∥^{2} − ∥\frac{1}{Q_{n}} ∑_{i=1}^{n} q_{i}z_{i}∥^{2}]^{\frac{1}{2}} .

Utilizing (4.18) and (4.19), we deduce the desired result (4.17).

If more information is available about the vector sequence \overline{z} = (z_{1}, …, z_{n}) ∈ H^{n}, then we may state the following corollary.

Corollary 56. With the assumptions in Theorem 72 and if there exist the vectors z, Z ∈ H such that either

(4.20)  ⟨z_{i} − z, Z − z_{i}⟩ ≥ 0 for each i ∈ {1, …, n};

or, equivalently,

(4.21)  ∥z_{i} − \frac{z + Z}{2} ∥ ≤ \frac{1}{2} ∥Z − z∥ for each i ∈ \{1, …, n\} ,

holds, then we have the inequality

(4.22)  0 ≤ \frac{1}{Q_{n}} ∑_{i=1}^{n} q_{i}F (z_{i}) − F (\frac{1}{Q_{n}} ∑_{i=1}^{n} q_{i}z_{i})

≤ \frac{1}{2} ∥M − m∥ \frac{1}{Q_{n}} ∑_{i=1}^{n} q_{i} ∥z_{i} − \frac{1}{Q_{n}} ∑_{j=1}^{n} q_{j}z_{j}∥

≤ \frac{1}{2} ∥M − m∥ [\frac{1}{Q_{n}} ∑_{i=1}^{n} q_{i} ∥z_{i}∥^{2} − ∥\frac{1}{Q_{n}} ∑_{i=1}^{n} q_{i}z_{i}∥^{2}]^{\frac{1}{2}}

≤ \frac{1}{4} ∥M − m∥ ∥Z − z∥ .

Remark 68. Note that the inequality between the first term and the last term in (4.22) was first proved in [7, Theorem 4.1]. Consequently, the above corollary provides an improvement of the reverse of Jensen’s inequality established in [7].

5. Some Inequalities for Forward Difference

5.1. Introduction. In [4], we have proved the following generalisation of the Grüss inequality.

Theorem 73. Let (H, ⟨·, ·⟩) be an inner product space over 𝕂, 𝕂 = ℂ, ℝ and e ∈ H, ∥e∥ = 1. If ϕ, Φ, γ, Γ ∈ 𝕂 and x, y ∈ H are such that

Re ⟨Φe − x, x − ϕe⟩ ≥ 0 and Re ⟨Γe − y, y − γe⟩ ≥ 0

hold, then we have the inequality

|⟨x, y⟩ − ⟨x, e⟩ ⟨e, y⟩| ≤ \frac{1}{4} |Φ − ϕ| |Γ − γ| .

The constant \frac{1}{4} is the best possible.

A Grüss type inequality for sequences of vectors in inner product spaces was pointed out in [5].

Theorem 74. Let H and 𝕂 be as in Theorem 73 and x_{i} ∈ H, a_{i} ∈ 𝕂, p_{i} ≥ 0 (i = 1, …, n) (n ≥ 2) with ∑_{i=1}^{n} p_{i} = 1. If a, A ∈ 𝕂 and x, X ∈ H are such that:

Re [(A − a_{i}) (\overline{a_{i}} − \overline{a})] ≥ 0, Re ⟨X − x_{i}, x_{i} − x⟩ ≥ 0

for any i ∈ {1, …, n} , then we have the inequality

0 ≤ ∥∑_{i=1}^{n} p_{i}a_{i}x_{i} − ∑_{i=1}^{n} p_{i}a_{i} · ∑_{i=1}^{n} p_{i}x_{i}∥

≤ \frac{1}{4} |A − a| ∥X − x∥ .

The constant \frac{1}{4} is best possible.

A complementary result for two sequences of vectors in inner product spaces is the following result that has been obtained in [7].

Theorem 75. Let H and 𝕂 be as above, x_{i}, y_{i} ∈ H, p_{i} ≥ 0 (i = 1, …, n) (n ≥ 2) with ∑_{i=1}^{n} p_{i} = 1. If x, X, y, Y ∈ H are such that:

Re ⟨X − x_{i}, x_{i} − x⟩ ≥ 0 and Re ⟨Y − y_{i}, y_{i} − y⟩ ≥ 0

for all i ∈ {1, …, n} , then we have the inequality

0 ≤ |∑_{i=1}^{n} p_{i} ⟨x_{i}, y_{i}⟩ − ⟨∑_{i=1}^{n} p_{i}x_{i}, ∑_{i=1}^{n} p_{i}y_{i}⟩| ≤ \frac{1}{4} ∥X − x∥ ∥Y − y∥ .

In the general case of normed linear spaces, the following Grüss type inequality in terms of the forward difference is known, see [13].

Theorem 76. Let (E, ∥·∥) be a normed linear space over 𝕂 = ℂ, ℝ, x_{i} ∈ E, α_{i} ∈ 𝕂 and p_{i} ≥ 0 (i = 1, …, n) such that ∑_{i=1}^{n} p_{i} = 1. Then we have the inequality

(5.1)0 ≤ ∥∑_{i=1}^{n} p_{i}α_{i}x_{i} − ∑_{i=1}^{n} p_{i}α_{i} · ∑_{i=1}^{n} p_{i}x_{i}∥

≤ max_{1≤j≤n−1} |Δα_{j}| max_{1≤j≤n−1} ∥Δx_{j}∥ [∑_{i=1}^{n} i^{2}p_{i} − (∑_{i=1}^{n} ip_{i})^{2}] ,

where Δα_{j} = α_{j+1} − α_{j} and Δx_{j} = x_{j+1} − x_{j} (j = 1, …, n − 1) are the forward differences of the vectors having the components α_{j} and x_{j} (j = 1, …, n − 1) , respectively.

The inequality (5.1) is sharp in the sense that the multiplicative constant C = 1 in the right hand side cannot be replaced by a smaller one.

An important particular case is the one where all the weights are equal, giving the following corollary [13].

Corollary 57. Under the above assumptions for α_{i}, x_{i} (i = 1, …, n) we have the inequality

(5.2)  0 ≤ ∥\frac{1}{n} ∑_{i=1}^{n} α_{i}x_{i} − \frac{1}{n} ∑_{i=1}^{n} α_{i} · \frac{1}{n} ∑_{i=1}^{n} x_{i}∥

≤ \frac{n^{2} − 1}{12} max_{1≤j≤n−1} |Δα_{j}| max_{1≤j≤n−1} ∥Δx_{j}∥ .

The constant \frac{1}{12} is best possible.

Another result of this type was proved in [6].

Theorem 77. With the assumptions of Theorem 76, one has the inequality

(5.3)  0 ≤ ∥∑_{i=1}^{n} p_{i}α_{i}x_{i} − ∑_{i=1}^{n} p_{i}α_{i} · ∑_{i=1}^{n} p_{i}x_{i}∥

≤ \frac{1}{2} ∑_{j=1}^{n−1} |Δα_{j}| ∑_{j=1}^{n−1} ∥Δx_{j}∥ ∑_{i=1}^{n} p_{i} (1 − p_{i}) .

As a useful particular case, we have the following corollary [6]. Corollary 58. If α_{i}, x_{i} (i = 1, …, n) are as in Theorem 76, then

0 ≤ ∥\frac{1}{n} ∑_{i=1}^{n} α_{i}x_{i} − \frac{1}{n} ∑_{i=1}^{n} α_{i} · \frac{1}{n} ∑_{i=1}^{n} x_{i}∥

≤ \frac{1}{2} (1 − \frac{1}{n}) ∑_{i=1}^{n−1} |Δα_{i}| ∑_{i=1}^{n−1} ∥Δx_{i}∥ .

The constant \frac{1}{2} is the best possible.

Finally, the following result is also known [8]. Theorem 78. With the assumptions in Theorem 76, we have the

inequality:

(5.4)  0 ≤ ∥∑_{i=1}^{n} p_{i}α_{i}x_{i} − ∑_{i=1}^{n} p_{i}α_{i} · ∑_{i=1}^{n} p_{i}x_{i}∥

≤ (∑_{j=1}^{n−1} |Δα_{j}|^{p})^{\frac{1}{p}} (∑_{j=1}^{n−1} ∥Δx_{j}∥^{q})^{\frac{1}{q}} ∑_{1≤i<j≤n} (j − i) p_{i}p_{j},

where p > 1, \frac{1}{p} + \frac{1}{q} = 1.

The constant c = 1 in the right hand side of (5.4) is sharp. The case of equal weights is embodied in the following corollary [8]. Corollary 59. With the above assumptions for α_{i}, x_{i} (i = 1, …, n)

one has

0 ≤ ∥\frac{1}{n} ∑_{i=1}^{n} α_{i}x_{i} − \frac{1}{n} ∑_{i=1}^{n} α_{i} · \frac{1}{n} ∑_{i=1}^{n} x_{i}∥

≤ \frac{n^{2} − 1}{6n} (∑_{j=1}^{n−1} |Δα_{j}|^{p})^{\frac{1}{p}} (∑_{j=1}^{n−1} ∥Δx_{j}∥^{q})^{\frac{1}{q}} ,

where p > 1, \frac{1}{p} + \frac{1}{q} = 1.

The constant \frac{1}{6} is the best possible. The main aim of this section is to establish some similar bounds for

the absolute value of the difference

∑_{i=1}^{n} p_{i} ⟨x_{i}, y_{i}⟩ − ⟨∑_{i=1}^{n} p_{i}x_{i}, ∑_{i=1}^{n} p_{i}y_{i}⟩

provided that x_{i}, y_{i} (i = 1, …, n) are vectors in an inner product space H, and p_{i} ≥ 0 (i = 1, …, n) with ∑_{i=1}^{n} p_{i} = 1.

5.2. The Main Results. We assume that (H, ⟨·, ·⟩) is an inner product space over 𝕂, 𝕂 = ℂ or 𝕂 = ℝ. The following discrete inequality of Grüss’ type holds.

Theorem 79. If x_{i}, y_{i} ∈ H, p_{i} ≥ 0 (i = 1, …, n) with ∑_{i=1}^{n} p_{i} = 1, then one has the inequalities:

|∑_{i=1}^{n} p_{i} ⟨x_{i}, y_{i}⟩ − ⟨∑_{i=1}^{n} p_{i}x_{i}, ∑_{i=1}^{n} p_{i}y_{i}⟩| (5.5)

≤ \begin{cases}[∑_{i=1}^{n} i^{2}p_{i} − (∑_{i=1}^{n} ip_{i})^{2}] max_{k=1,...,n−1} ∥Δx_{k}∥ max_{k=1,...,n−1} ∥Δy_{k}∥; \\ [∑_{1≤j<i≤n} p_{i}p_{j} (i − j)] & (∑_{k=1}^{n−1} ∥Δx_{k}∥^{p}) & \frac{1}{p} & (∑_{k=1}^{n−1} ∥Δy_{k}∥^{q}) & \frac{1}{q} \\ if p > 1, \frac{1}{p} + \frac{1}{q} = 1; \\ \frac{1}{2} [∑_{i=1}^{n} p_{i} (1 − p_{i})] ∑_{k=1}^{n−1} ∥Δx_{k}∥ ∑_{k=1}^{n−1} ∥Δy_{k}∥ . \end{cases}

All the inequalities in (5.5) are sharp.

The following particular case for equal vectors holds.

Corollary 60. With the assumptions of Theorem 79, one has the inequalities

0 ≤ ∑_{i=1}^{n} p_{i} ∥x_{i}∥^{2} − ∥∑_{i=1}^{n} p_{i}x_{i}∥^{2}

≤ \{ \begin{matrix}[∑_{i=1}^{n} i^{2}p_{i} − (∑_{i=1}^{n} ip_{i})^{2}] max_{k=1,...,n−1} ∥Δx_{k}∥^{2} ; \\ ∑_{1≤j<i≤n} p_{i}p_{j} (i − j) (∑_{k=1}^{n−1} ∥Δx_{k}∥^{p})^{\frac{1}{p}} (∑_{k=1}^{n−1} ∥Δx_{k}∥^{q})^{\frac{1}{q}} \\ if p > 1, \frac{1}{p} + \frac{1}{q} = 1; \\ \frac{1}{2} ∑_{i=1}^{n} p_{i} (1 − p_{i}) (∑_{k=1}^{n−1} ∥Δx_{k}∥)^{2} . \end{matrix}

The following particular case for equal weights may be useful in practice.

Corollary 61. If x_{i}, y_{i} ∈ H (i = 1, …, n), then one has the inequalities:

|\frac{1}{n} ∑_{i=1}^{n} ⟨x_{i}, y_{i}⟩ − ⟨\frac{1}{n} ∑_{i=1}^{n} x_{i}, \frac{1}{n} ∑_{i=1}^{n} y_{i}⟩|

≤ \{ \begin{matrix}\frac{n^{2}−1}{12} max_{k=1,...,n−1} ∥Δx_{k}∥ max_{k=1,...,n−1} ∥Δy_{k}∥; \\ \frac{n^{2}−1}{6n} & (∑_{k=1}^{n−1} ∥Δx_{k}∥^{p}) & \frac{1}{p} & (∑_{k=1}^{n−1} ∥Δy_{k}∥^{q}) & \frac{1}{q} \\ if p > 1, \frac{1}{p} + \frac{1}{q} = 1; \\ \frac{n−1}{2n} ∑_{k=1}^{n−1} ∥Δx_{k}∥ ∑_{k=1}^{n−1} ∥Δy_{k}∥ . \end{matrix}

The constants \frac{1}{12}, \frac{1}{6} and \frac{1}{2} are best possible.

In particular, the following corollary holds.

Corollary 62. If x_{i} ∈ H (i = 1, …, n) , then one has the inequality

0 ≤ \frac{1}{n} ∑_{i=1}^{n} ∥x_{i}∥^{2} − ∥\frac{1}{n} ∑_{i=1}^{n} x_{i}∥^{2}

≤ \{ \begin{matrix}\frac{n^{2}−1}{12} max_{k=\overline{1,n}} ∥Δx_{k}∥^{2} ; \\ \frac{n^{2}−1}{6n} & (∑_{k=1}^{n−1} ∥Δx_{k}∥^{p}) & \frac{1}{p} & (∑_{k=1}^{n−1} ∥Δx_{k}∥^{q}) & \frac{1}{q} \\ if p > 1, \frac{1}{p} + \frac{1}{q} = 1; \\ \frac{n−1}{2n} (∑_{k=1}^{n−1} ∥Δx_{k}∥)^{2} . \end{matrix}

The constants \frac{1}{12}, \frac{1}{6} and \frac{1}{2} are best possible.

5.3. Proof of the Main Result. It is well known that, the following identity holds in inner product spaces:

∑_{i=1}^{n} p_{i} ⟨x_{i}, y_{i}⟩ − ⟨∑_{i=1}^{n} p_{i}x_{i}, ∑_{i=1}^{n} p_{i}y_{i}⟩ (5.6)

= \frac{1}{2} ∑_{i,j=1}^{n} p_{i}p_{j} ⟨x_{i} − x_{j}, y_{i} − y_{j}⟩

= ∑_{1≤j<i≤n} p_{i}p_{j} ⟨x_{i} − x_{j}, y_{i} − y_{j}⟩ .

We observe, for i > j, we can write that

(5.7)  x_{i} − x_{j} = ∑_{k=j}^{i−1} Δx_{k},  y_{i} − y_{j} = ∑_{k=j}^{i−1} Δy_{k}.

Taking the modulus in (5.6) and by the use of (5.7) and Schwarz’s inequality in inner product spaces, i.e., we recall that |⟨z, u⟩| ≤ ∥z∥ ∥u∥ ,

198 5. SOME GRÜSS’ TYPE INEQUALITIES FOR n-TUPLES OF VECTORS

z, u ∈ H, we have:

|∑_{i=1}^{n} p_{i} ⟨x_{i}, y_{i}⟩ − ⟨∑_{i=1}^{n} p_{i}x_{i}, ∑_{i=1}^{n} p_{i}y_{i}⟩|

≤ ∑_{1≤j<i≤n} p_{i}p_{j} |⟨x_{i} − x_{j}, y_{i} − y_{j}⟩|

≤ ∑_{1≤j<i≤n} p_{i}p_{j} ∥x_{i} − x_{j}∥ ∥y_{i} − y_{j}∥

= ∑_{1≤j<i≤n} p_{i}p_{j} ∥∑_{k=j}^{i−1} Δx_{k}∥ ∥∑_{l=j}^{i−1} Δy_{l}∥

≤ ∑_{1≤j<i≤n} p_{i}p_{j} ∑_{k=j}^{i−1} ∥Δx_{k}∥ ∑_{l=j}^{i−1} ∥Δy_{l}∥

:= M.

It is obvious that

∑_{k=j}^{i−1} ∥Δx_{k}∥ ≤ (i − j) max_{k=j,...,i−1} ∥Δx_{k}∥ ≤ (i − j) max_{k=1,...,n} ∥Δx_{k}∥

and

∑_{k=j}^{i−1} ∥Δy_{k}∥ ≤ (i − j) max_{k=j,...,i−1} ∥Δy_{k}∥ ≤ (i − j) max_{k=1,...,n} ∥Δy_{k}∥ ,

giving that

M ≤ ∑_{1≤j<i≤n} p_{i}p_{j} (i − j)^{2} · max_{k=1,...,n} ∥Δx_{k}∥ max_{k=1,...,n} ∥Δy_{k}∥ ,

and since

∑_{1≤j<i≤n} p_{i}p_{j} (i − j)^{2} = \frac{1}{2} ∑_{i,j=1}^{n} p_{i}p_{j} (i − j)^{2} = ∑_{i=1}^{n} p_{i}i^{2} − (∑_{i=1}^{n} ip_{i})^{2} ,

the first inequality in (5.5) is proved.

Using Hölder’s discrete inequality, we can state that

∑_{k=j}^{i−1} ∥Δx_{k}∥ ≤ (i − j)^{\frac{1}{q}} (∑_{k=j}^{i−1} ∥Δx_{k}∥^{p})^{\frac{1}{p}} ≤ (i − j)^{\frac{1}{q}} (∑_{k=1}^{n−1} ∥Δx_{k}∥^{p})^{\frac{1}{p}}

and

∑_{k=j}^{i−1} ∥Δy_{k}∥ ≤ (i − j)^{\frac{1}{p}} (∑_{k=j}^{i−1} ∥Δy_{k}∥^{q})^{\frac{1}{q}} ≤ (i − j)^{\frac{1}{p}} (∑_{k=1}^{n−1} ∥Δy_{k}∥^{q})^{\frac{1}{q}} ,

for p > 1, \frac{1}{p} + \frac{1}{q} = 1, giving that:

M ≤ ∑_{1≤j<i≤n} p_{i}p_{j} (i − j) · (∑_{k=1}^{n−1} ∥Δx_{k}∥^{p})^{\frac{1}{p}} (∑_{k=1}^{n−1} ∥Δy_{k}∥^{q})^{\frac{1}{q}}

and the second inequality in (5.5) is proved.

Also, observe that

∑_{k=j}^{i−1} ∥Δx_{k}∥ ≤ ∑_{k=1}^{n−1} ∥Δx_{k}∥ and ∑_{k=j}^{i−1} ∥Δy_{k}∥ ≤ ∑_{k=1}^{n−1} ∥Δy_{k}∥

and thus

M ≤ ∑_{1≤j<i≤n} p_{i}p_{j} (i − j) ∑_{k=1}^{n−1} ∥Δx_{k}∥ ∑_{k=1}^{n−1} ∥Δy_{k}∥ .

Since

∑_{1≤j<i≤n} p_{i}p_{j} = \frac{1}{2} [∑_{i,j=1}^{n} p_{i}p_{j} − ∑_{k=1}^{n} p_{k}^{2}]

= \frac{1}{2} (1 − ∑_{k=1}^{n} p_{k}^{2})

= \frac{1}{2} ∑_{i=1}^{n} p_{i} (1 − p_{i}) ,

the last part of (5.5) is also proved.

Now, assume that the first inequality in (5.5) holds with a constant c > 0, i.e.,

∑_{i=1}^{n} p_{i} ⟨x_{i}, y_{i}⟩ − ⟨∑_{i=1}^{n} p_{i}x_{i}, ∑_{i=1}^{n} p_{i}y_{i}⟩

≤ c [∑_{i=1}^{n} i^{2}p_{i} − (∑_{i=1}^{n} ip_{i})^{2}] max_{k=1,...,n−1} ∥Δx_{k}∥ max_{k=1,...,n−1} ∥Δy_{k}∥

and choose n = 2 to get

(5.8)  p_{1}p_{2} |⟨x_{2} − x_{1}, y_{2} − y_{1}⟩| ≤ cp_{1}p_{2} ∥x_{2} − x_{1}∥ ∥y_{2} − y_{1}∥

for any p_{1}, p_{2} > 0 and x_{1}, x_{2}, y_{1}, y_{2} ∈ H.

If in (5.8) we choose y_{2} = x_{2}, y_{1} = x_{1} and x_{2} ≠ x_{1}, then we deduce c ≥ 1, which proves the sharpness of the constant in the first inequality in (5.5).

In a similar way one may show that the other two inequalities are sharp, and the theorem is completely proved.

5.4. A Reverse for Jensen’s Inequality. Let (H; ⟨·, ·⟩) be a real inner product space and F : H → ℝ a Fréchet differentiable convex function on H. If ▽F : H → H denotes the gradient operator associated to F, then we have the inequality

F (x) − F (y) ≥ ⟨▽F (y) , x − y⟩

for each x, y ∈ H.

The following result holds.

Theorem 80. Let F : H → ℝ be as above and z_{i} ∈ H, i ∈ {1, …, n} . If q_{i} ≥ 0 (i ∈ {1, …, n}) with ∑_{i=1}^{n} q_{i} = 1, then we have the following reverse of Jensen’s inequality

(5.9) 0 ≤ ∑_{i=1}^{n} q_{i}F (z_{i}) − F (∑_{i=1}^{n} q_{i}z_{i})

≤ \{ \begin{matrix}[∑_{i=1}^{n} i^{2}q_{i} − (∑_{i=1}^{n} iq_{i})^{2}] \\ × max_{k=1,...,n−1} ∥Δ (▽F (z_{i}))∥ max_{k=1,...,n−1} ∥Δz_{i}∥; \\ [∑_{1≤j<i≤n} q_{i}q_{j} (i − j)] \\ × (∑_{i=1}^{n−1} ∥Δ (▽F (z_{i}))∥^{p})^{\frac{1}{p}} (∑_{i=1}^{n−1} ∥Δz_{i}∥^{q})^{\frac{1}{q}} \\ if p > 1, \frac{1}{p} + \frac{1}{q} = 1; \\ \frac{1}{2} [∑_{i=1}^{n} q_{i} (1 − q_{i})] ∑_{i=1}^{n−1} ∥Δ (▽F (z_{i}))∥ ∑_{i=1}^{n−1} ∥Δz_{i}∥ . \end{matrix}

Proof. We know, see for example [7, Eq. (4.4)], that the following reverse of Jensen’s inequality for Fréchet differentiable convex functions

(5.10)  0 ≤ ∑_{i=1}^{n} q_{i}F (z_{i}) − F (∑_{i=1}^{n} q_{i}z_{i})

≤ ∑_{i=1}^{n} q_{i} ⟨▽F (z_{i}) , z_{i}⟩ − ⟨∑_{i=1}^{n} q_{i}▽F (z_{i}) , ∑_{i=1}^{n} q_{i}z_{i}⟩

holds.

Now, if we apply Theorem 79 for the choices x_{i} = ▽F (z_{i}) , y_{i} = z_{i} and p_{i} = q_{i} (i = 1, ..., n) , then we may state

|∑_{i=1}^{n} q_{i} ⟨▽F (z_{i}) , z_{i}⟩ − ⟨∑_{i=1}^{n} q_{i}▽F (z_{i}) , ∑_{i=1}^{n} q_{i}z_{i}⟩| (5.11)

≤ \{ \begin{matrix}[∑_{i=1}^{n} i^{2}q_{i} − (∑_{i=1}^{n} iq_{i})^{2}] \\ × max_{k=1,...,n−1} ∥Δ (▽F (z_{k}))∥ max_{k=1,...,n−1} ∥Δz_{k}∥; \\ [∑_{1≤j<i≤n} q_{i}q_{j} (i − j)] \\ × (∑_{k=1}^{n−1} ∥Δ (▽F (z_{k}))∥^{p})^{\frac{1}{p}} (∑_{k=1}^{n−1} ∥Δz_{k}∥^{q})^{\frac{1}{q}} \\ if p > 1, \frac{1}{p} + \frac{1}{q} = 1; \\ \frac{1}{2} [∑_{i=1}^{n} q_{i} (1 − p_{i})] ∑_{k=1}^{n−1} ∥Δ (▽F (z_{k}))∥ ∑_{k=1}^{n−1} ∥Δz_{k}∥ . \end{matrix}

Finally, on making use of the inequalities (5.10) and (5.11), we deduce the desired result (5.9).

The unweighted case may useful in application and is incorporated in the following corollary.

Corollary 63. Let F : H → ℝ be as above and z_{i} ∈ H, i ∈ {1, …, n} . Then we have the inequalities

0 ≤ \frac{1}{n} ∑_{i=1}^{n} F (z_{i}) − F (\frac{1}{n} ∑_{i=1}^{n} z_{i})

≤ \{ \begin{matrix}\frac{n^{2}−1}{12} max_{k=1,...,n−1} ∥Δ (▽F (z_{k}))∥ max_{k=1,...,n−1} ∥Δz_{k}∥; \\ \frac{n^{2}−1}{6n} & ( ∑_{k=1}^{n−1} ∥Δ (▽F (z_{k}))∥^{p}) & \frac{1}{p} & (∑_{k=1}^{n−1} ∥Δz_{k}∥^{q}) & \frac{1}{q} \\ if p > 1, \frac{1}{p} + \frac{1}{q} = 1; \\ \frac{n−1}{2n} ∑_{k=1}^{n−1} ∥Δ (▽F (z_{k}))∥ ∑_{k=1}^{n−1} ∥Δz_{k}∥ . \end{matrix}

6. Bounds for a Pair of n-Tuples of Vectors

6.1. Introduction. Let (H; ⟨·, ·⟩) be an inner product over the real or complex number field 𝕂. For p = (p_{1}, …, p_{n}) ∈ ℝ^{n} and x = (x_{1}, …, x_{n}) , y = (y_{1}, …, y_{n}) ∈ H^{n}, define the Čebyšev functional

(6.1)  T_{n} (p̄; x̄, ȳ) := P_{n} ∑_{i=1}^{n} p_{i} ⟨x_{i}, y_{i}⟩ − ⟨∑_{i=1}^{n} p_{i}x_{i}, ∑_{i=1}^{n} p_{i}y_{i}⟩ ,

where P_{n} := ∑_{i=1}^{n} p_{i}.

The following Grüss type inequality has been obtained in [7].

Theorem 81. Let H, x, y be as above and p_{i} ≥ 0 (i ∈ {1, …, n}) with ∑_{i=1}^{n} p_{i} = 1, i.e., p is a probability sequence. If x, X, y, Y ∈ H are such that

(6.2)  Re ⟨X − x_{i}, x_{i} − x⟩ ≥ 0, Re ⟨Y − y_{i}, y_{i} − y⟩ ≥ 0

for each i ∈ {1, …, n} , or, equivalently, (see [10])

(6.3) ∥x_{i} − \frac{x + X}{2} ∥ ≤ \frac{1}{2} ∥X − x∥ ,  ∥y_{i} − \frac{y + Y}{2} ∥ ≤ \frac{1}{2} ∥Y − y∥

for each i ∈ {1, …, n} , then we have the inequality

(6.4)  |T_{n} (p̄; x̄, ȳ)| ≤ \frac{1}{4} ∥X − x∥ ∥Y − y∥ .

The constant \frac{1}{4} is best possible in the sense that it cannot be replaced by a smaller constant.

In [11], the following Grüss type inequality for the forward difference of vectors was established.

Theorem 82. Let x = (x_{1}, …, x_{n}) , y = (y_{1}, …, y_{n}) ∈ H^{n} and p ∈ ℝ_{+}^{n} be a probability sequence. Then one has the inequality:

|T_{n} (p̄; x̄, ȳ)| (6.5)

≤ \{ \begin{matrix}[∑_{i=1}^{n} i^{2}p_{i} − (∑_{i=1}^{n} ip_{i})^{2}] & max_{1≤k≤n−1} & ∥Δx_{k}∥ \\ × max_{1≤k≤n−1} ∥Δy_{k}∥; \\ ∑_{1≤j<i≤n} p_{i}p_{j} (i − j) \\ × (∑_{k=1}^{n−1} ∥Δx_{k}∥^{p})^{\frac{1}{p}} (∑_{k=1}^{n−1} ∥Δy_{k}∥^{q})^{\frac{1}{q}} \\ if p > 1, \frac{1}{p} + \frac{1}{q} + 1 \\ \frac{1}{2} [∑_{i=1}^{n} p_{i} (1 − p_{i})] ∑_{k=1}^{n−1} ∥Δx_{k}∥ ∑_{k=1}^{n−1} ∥Δy_{k}∥ . \end{matrix}

The constants 1, 1 and \frac{1}{2} in the right hand side of inequality (6.5) are best in the sense that they cannot be replaced by smaller constants.

Another result is incorporated in the following theorem (see [10]).

Theorem 83. Let x, y and p be as in Theorem 82. If there exist x, X ∈ H such that

(6.6)  Re ⟨X − x_{i}, x_{i} − x⟩ ≥ 0 for each i ∈ {1, …, n} ,

or, equivalently,

(6.7)  ∥x_{i} − \frac{x + X}{2} ∥ ≤ \frac{1}{2} ∥X − x∥ for each i ∈ \{1, …, n\} ,

then one has the inequality

(6.8) |T_{n} (p̄; x̄, ȳ)| ≤ \frac{1}{2} ∥X − x∥ ∑_{i=1}^{n} p_{i} ∥y_{i} − ∑_{j=1}^{n} p_{j}y_{j}∥

≤ \frac{1}{2} ∥X − x∥ [∑_{i=1}^{n} p_{i} ∥y_{i}∥^{2} − ∥∑_{i=1}^{n} p_{i}y_{i}∥^{2}]^{\frac{1}{2}} .

The constant \frac{1}{2} is best possible in the first and second inequalities in the sense that it cannot be replaced by a smaller constant.

Remark 69. If x and y satisfy the assumptions of Theorem 81, then we have the following sequence of inequalities improving the Grüss inequality (6.4):

(6.9) |T_{n} (p̄; x̄, ȳ)| ≤ \frac{1}{2} ∥X − x∥ ∑_{i=1}^{n} p_{i} ∥y_{i} − ∑_{j=1}^{n} p_{j}y_{j}∥

≤ \frac{1}{2} ∥X − x∥ (∑_{i=1}^{n} p_{i} ∥y_{i}∥^{2} − ∥∑_{i=1}^{n} p_{i}y_{i}∥^{2})^{\frac{1}{2}}

≤ \frac{1}{4} ∥X − x∥ ∥Y − y∥ .

Now, if we consider the Čebyšev functional for the uniform probability distribution u = (\frac{1}{n}, …, \frac{1}{n}) ,

T_{n} (x̄, ȳ) := \frac{1}{n} ∑_{i=1}^{n} ⟨x_{i}, y_{i}⟩ − ⟨\frac{1}{n} ∑_{i=1}^{n} x_{i}, \frac{1}{n} ∑_{i=1}^{n} y_{i}⟩ ,

then, with the assumptions of Theorem 81, we have

(6.10)  |T_{n} (x̄, ȳ)| ≤ \frac{1}{4} ∥X − x∥ ∥Y − y∥ .

Theorem 82 will provide the following inequalities

(6.11) |T_{n} (x̄, ȳ)| ≤ \begin{cases}\frac{1}{12} (n^{2} − 1) max_{1≤k≤n−1} ∥Δx_{k}∥ max_{1≤k≤n−1} ∥Δy_{k}∥; \\ \frac{1}{6} & (n − \frac{1}{n}) & (∑_{k=1}^{n−1} ∥Δx_{k}∥^{p}) & \frac{1}{p} & (∑_{k=1}^{n−1} ∥Δy_{k}∥^{q}) & \frac{1}{q} \\ if p > 1, \frac{1}{p} + \frac{1}{q} + 1; \\ \frac{1}{2} (1 − \frac{1}{n}) ∑_{k=1}^{n−1} ∥Δx_{k}∥ ∑_{k=1}^{n−1} ∥Δy_{k}∥ . \end{cases}

Here the constants \frac{1}{12}, \frac{1}{6} and \frac{1}{2} are best possible in the above sense.

Finally, from (6.9), we have

(6.12)  |T_{n} (x̄, ȳ)| ≤ \frac{1}{2n} ∥X − x∥ ∑_{i=1}^{n} ∥y_{i} − \frac{1}{n} ∑_{j=1}^{n} y_{j}∥

≤ \frac{1}{2} ∥X − x∥ (\frac{1}{n} ∑_{i=1}^{n} ∥y_{i}∥^{2} − ∥\frac{1}{n} ∑_{i=1}^{n} y_{i}∥^{2})^{\frac{1}{2}}

≤ \frac{1}{4} ∥X − x∥ ∥Y − y∥ .

It is the main aim of this section to point out other bounds for the Čebyšev functionals T_{n} (p, x, y) and T_{n} (x, y) .

6.2. Identities for Inner Products. For p = (p_{1}, …, p_{n}) ∈ ℝ^{n} and a = (a_{1}, …, a_{n}) ∈ H^{n} we define

P_{i} := ∑_{k=1}^{i} p_{k},  P̄_{i} = P_{n} − P_{i}, i ∈ {1, …, n − 1}

and the vectors

A_{i} (p) = ∑_{k=1}^{i} p_{k}a_{k},  Ā_{i} (p) = A_{n} (p) − A_{i} (p)

for i ∈ {1, …, n − 1} .

The following result holds [9].

Theorem 84. Let (H; ⟨·, ·⟩) be an inner product space over 𝕂, p = (p_{1}, …, p_{n}) ∈ ℝ^{n} and a = (a_{1}, …, a_{n}) , b = (b_{1}, …, b_{n}) ∈ H^{n}. Then we have the identities

(6.13)  T_{n} (p; a, b) = ∑_{i=1}^{n−1} ⟨P_{i}A_{n} (p) − P_{n}A_{i} (p) , Δb_{i}⟩

= P_{n} ∑_{i=1}^{n−1} P_{i} ⟨\frac{1}{P_{n}} A_{n} (p) − \frac{1}{P_{i}} A_{i} (p) , Δb_{i}⟩

(if P_{i} ≠ 0, i ∈ {1, …, n})

= ∑_{i=1}^{n−1} P_{i}P̄_{i} ⟨\frac{1}{P̄_{i}} Ā_{i} (p) − \frac{1}{P_{i}} A_{i} (p) , Δb_{i}⟩

(if P_{i}, P̄_{i} ≠ 0, i ∈ {1, …, n − 1}),

where Δx_{i} = x_{i+1} − x_{i} (i ∈ {1, …, n − 1}) is the forward difference.

Proof. We use the following summation by parts formula for vectors in inner product spaces

(6.14)  ∑_{l=p}^{q−1} ⟨d_{l}, Δv_{l}⟩ = ⟨d_{l}, v_{l}⟩ |_{p}^{q} − ∑_{l=p}^{q−1} ⟨v_{l+1}, Δd_{l}⟩ ,

where d_{l}, v_{l} are vectors in H, l = p, …, q (q > p; p, q are natural numbers).

If we choose in (6.14), p = 1, q = n, d_{i} = P_{i}A_{n} (p) − P_{n}A_{i} (p) and v_{i} = b_{i} (i ∈ {1, …, n − 1}) , then we get

∑_{i=1}^{n−1} ⟨P_{i}A_{n} (p) − P_{n}A_{i} (p) , Δb_{i}⟩

= ⟨P_{i}A_{n} (p) − P_{n}A_{i} (p) , b_{i}⟩ |_{1}^{n} − ∑_{i=1}^{n−1} ⟨Δ (P_{i}A_{n} (p) − P_{n}A_{i} (p)) , b_{i+1}⟩

= ⟨P_{n}A_{n} (p) − P_{n}A_{n} (p) , b_{n}⟩ − ⟨P_{1}A_{n} (p) − P_{n}A_{1} (p) , b_{1}⟩

− ∑_{i=1}^{n−1} ⟨P_{i+1}A_{n} (p) − P_{n}A_{i+1} (p) − P_{i}A_{n} (p) + P_{n}A_{i} (p) , b_{i+1}⟩

= P_{n}p_{1} ⟨a_{1}, x_{1}⟩ − p_{1} ⟨A_{n} (p) , b_{1}⟩ − ⟨A_{n} (p) , ∑_{i=1}^{n−1} p_{i+1}b_{i+1}⟩

+ P_{n} ∑_{i=1}^{n−1} p_{i+1} ⟨a_{i+1}, b_{i+1}⟩

= P_{n} ∑_{i=1}^{n} p_{i} ⟨a_{i}, b_{i}⟩ − ⟨∑_{i=1}^{n} p_{i}a_{i}, ∑_{i=1}^{n} p_{i}b_{i}⟩

= T_{n} (p; a, b) ,

proving the first identity in (6.13).

The second and third identities are obvious and we omit the details.

The following lemma is of interest in itself [9].

Lemma 12. Let p = (p_{1}, …, p_{n}) ∈ ℝ^{n} and a = (a_{1}, …, a_{n}) ∈ H. Then we have the equality

(6.15)  P_{i}A_{n} (p) − P_{n}A_{i} (p) = ∑_{j=1}^{n−1} P_{min{i,j}}P̄_{max{i,j}}Δa_{j}

for each i ∈ {1, …, n − 1} .

206 5. SOME GRÜSS’ TYPE INEQUALITIES FOR n-TUPLES OF VECTORS

Proof. Define, for i ∈ {1, …, n − 1} , the vector

K (i) := ∑_{j=1}^{n−1} P_{min{i,j}}P̄_{max{i,j}} · Δa_{j}.

We have

(6.16)  K (i) = ∑_{j=1}^{i} P_{min{i,j}}P̄_{max{i,j}} · Δa_{j}

+ ∑_{j=i+1}^{n−1} P_{min{i,j}}P̄_{max{i,j}} · Δa_{j}

= ∑_{j=1}^{i} P_{j}P̄_{i} · Δa_{j} + ∑_{j=i+1}^{n−1} P_{i}P̄_{j} · Δa_{j}

= P̄_{i} ∑_{j=1}^{i} P_{j} · Δa_{j} + P_{i} ∑_{j=i+1}^{n−1} P̄_{j} · Δa_{j}.

Using the summation by parts formula, we have

(6.17)  ∑_{j=1}^{i} P_{j} · Δa_{j} = P_{j}a_{j}|_{1}^{i+1} − ∑_{j=1}^{i} (P_{j+1} − P_{j}) a_{j+1}

= P_{i+1}a_{i+1} − p_{1}a_{1} − ∑_{j=1}^{i} p_{j+1}a_{j+1}

= P_{i+1}a_{i+1} − ∑_{j=1}^{i+1} p_{j}a_{j}

and

(6.18) ∑_{j=i+1}^{n−1}P̄_{j} · Δa_{j}

= P̄_{j}a_{j}|_{i+1}^{n} − ∑_{j=i+1}^{n−1} (P̄_{j+1} − P̄_{j}) a_{j+1}

= P̄_{n}a_{n} − P̄_{i+1}a_{i+1} − ∑_{j=i+1}^{n−1} (P_{n} − P_{j+1} − P_{n} + P_{j}) a_{j+1}

= −P̄_{i+1}a_{i+1} + ∑_{j=i+1}^{n−1} p_{j+1}a_{j+1}.

Using (6.17) and (6.18), we have

K (i) = P̄_{i} (P_{i+1}a_{i+1} − ∑_{j=1}^{i+1} p_{j}a_{j}) + P_{i} ( ∑_{j=i+1}^{n−1} p_{j+1}a_{j+1} − P̄_{i+1}a_{i+1})

= P̄_{i}P_{i+1}a_{i+1} − P̄_{i}P̄_{i+1}a_{i+1} − P̄_{i} ∑_{j=1}^{i+1} p_{j}a_{j} + P_{i} ∑_{j=i+1}^{n−1} p_{j+1}a_{j+1}

= [(P_{n} − P_{i}) P_{i+1} − P_{i} (P_{n} − P_{i+1})] a_{i+1}

+ P_{i} ∑_{j=i+1}^{n−1} p_{j+1}a_{j+1} − P̄_{i} ∑_{j=1}^{i+1} p_{j}a_{j}

= P_{n}p_{i+1}a_{i+1} + P_{i} ∑_{j=i+1}^{n−1} p_{j+1}a_{j+1} − P̄_{i} ∑_{j=1}^{i+1} p_{j}a_{j}

= (P_{i} + P̄_{i}) p_{i+1}a_{i+1} + P_{i} ∑_{j=i+1}^{n−1} p_{j+1}a_{j+1} − P̄_{i} ∑_{j=1}^{i+1} p_{j}a_{j}

= P_{i} ∑_{j=i+1}^{n−1} p_{j}a_{j} − P̄_{i} ∑_{j=1}^{i} p_{j}a_{j}

= P_{i}Ā_{i} (p) − P̄_{i}A_{i} (p) = P_{i}A_{n} (p) − P_{n}A_{i} (p) ,

and the identity is proved.

We are able now to state and prove the second identity for the Čebyšev functional [9].

Theorem 85. With the assumptions of Theorem 84, we have the identity

(6.19)  T_{n} (p; a, b) = ∑_{i=1}^{n−1} ∑_{j=1}^{n−1} P_{min{i,j}}P̄_{max{i,j}} · ⟨Δa_{j}, Δb_{i}⟩ .

Proof. Follows by Theorem 84 and Lemma 12 and we omit the details.

6.3. New Inequalities. The following result holds [9].

Theorem 86. Let (H; ⟨·, ·⟩) be an inner product space over the real or complex number field 𝕂; p = (p_{1}, …, p_{n}) ∈ ℝ^{n} and a = (a_{1}, …, a_{n}) , b = (b_{1}, …, b_{n}) ∈ H^{n}. Then we have the inequalities

(6.20) |T_{n} (p; a, b)|

≤ \{ \begin{matrix}max_{1≤i≤n−1} ∥P_{i}A_{n} (p) − P_{n}A_{i} (p)∥ ∑_{j=1}^{n−1} ∥Δb_{j}∥; \\ (∑_{i=1}^{n−1} ∥P_{i}A_{n} (p) − P_{n}A_{i} (p)∥^{q}) & \frac{1}{q} & (∑_{j=1}^{n−1} ∥Δb_{j}∥^{p}) & \frac{1}{p} \\ for p > 1, \frac{1}{p} + \frac{1}{q} = 1; \\ ∑_{i=1}^{n−1} ∥P_{i}A_{n} (p) − P_{n}A_{i} (p)∥ · max_{1≤j≤n−1} ∥Δb_{j}∥ . \end{matrix}

All the inequalities in (6.20) are sharp in the sense that the constants 1 cannot be replaced by smaller constants.

Proof. Using the first identity in (6.13) and Schwarz’s inequality in H, i.e., |⟨u, v⟩| ≤ ∥u∥ ∥v∥ , u, v ∈ H, we have successively:

|T_{n} (p; a, b)| ≤ ∑_{i=1}^{n−1} |⟨P_{i}A_{n} (p) − P_{n}A_{i} (p) , Δb_{i}⟩|

≤ ∑_{i=1}^{n−1} ∥P_{i}A_{n} (p) − P_{n}A_{i} (p)∥ ∥Δb_{i}∥ .

Using Hölder’s inequality, we deduce the desired result (6.20).

Let us prove, for instance, that the constant 1 in the second inequality is best possible.

Assume, for c > 0, we have that

(6.21) |T_{n} (p; a, b)|

≤ c (∑_{i=1}^{n−1} ∥P_{i}A_{n} (p) − P_{n}A_{i} (p)∥^{q})^{\frac{1}{q}} (∑_{j=1}^{n−1} ∥Δb_{j}∥^{p})^{\frac{1}{p}}

for p > 1, \frac{1}{p} + \frac{1}{q} = 1, n ≥ 2.

If we choose n = 2, then we get

T_{2} (p; a, b) = p_{1}p_{2} ⟨a_{2} − a_{1}, b_{2} − b_{1}⟩ .

Also, for n = 2,

(∑_{i=1}^{n−1} ∥P_{i}A_{n} (p) − P_{n}A_{i} (p)∥^{q})^{\frac{1}{q}} = |p_{1}p_{2}| ∥a_{2} − a_{1}∥

and

(∑_{j=1}^{n−1} ∥Δb_{j}∥^{p})^{\frac{1}{p}} = ∥b_{2} − b_{1}∥ ,

and then, from (6.21), for n = 2, we deduce

(6.22)  |p_{1}p_{2}| |⟨a_{2} − a_{1}, b_{2} − b_{1}⟩| ≤ c |p_{1}p_{2}| ∥a_{2} − a_{1}∥ ∥b_{2} − b_{1}∥ .

If in (6.22) we choose a_{2} = b_{2}, a_{2} = b_{1} and b_{2} ≠ b_{1}, p_{1}, p_{2} ≠ 0, we deduce c ≥ 1, proving that 1 is the best possible constant in that inequality.

The following corollary for the uniform distribution of the probability p holds.

Corollary 64. With the assumptions of Theorem 86 for a and b, we have the inequalities

(6.23) 0 ≤ |T_{n} (a, b)|

≤ \frac{1}{n^{2}} \{ \begin{matrix}max_{1≤i≤n−1} ∥i ∑_{k=1}^{n} a_{k} − n ∑_{k=1}^{i} a_{k}∥ ∑_{j=1}^{n−1} ∥Δb_{j}∥; \\ (∑_{i=1}^{n−1} ∥i ∑_{k=1}^{n} a_{k} − n ∑_{k=1}^{i} a_{k}∥^{q}) & \frac{1}{q} \\ × (∑_{j=1}^{n−1} ∥Δb_{j}∥^{p})^{\frac{1}{p}} for p > 1, \frac{1}{p} + \frac{1}{q} = 1; \\ ∑_{i=1}^{n−1} ∥i ∑_{k=1}^{n} a_{k} − n ∑_{k=1}^{i} a_{k}∥ · max_{1≤j≤n−1} ∥Δb_{j}∥ . \end{matrix}

The following result may be stated as well [9].

Theorem 87. With the assumptions of Theorem 86 and if P_{i} ≠ 0 (i = 1, …, n) , then we have the inequalities

(6.24) |T_{n} (p; a, b)|

≤ |P_{n}| × \{ \begin{matrix}max_{1≤i≤n−1} ∥\frac{1}{P_{n}} A_{n} (p) − \frac{1}{P_{i}} A_{i} (p)∥ ∑_{i=1}^{n−1} |P_{i}| ∥Δb_{i}∥; \\ (∑_{i=1}^{n−1} |P_{i}| ∥\frac{1}{P_{n}} A_{n} (p) − \frac{1}{P_{i}} A_{i} (p)∥^{q}) & \frac{1}{q} \\ × (∑_{i=1}^{n} |P_{i}| ∥Δb_{i}∥^{p})^{\frac{1}{p}} for p > 1, \frac{1}{p} + \frac{1}{q} = 1; \\ ∑_{i=1}^{n−1} |P_{i}| ∥\frac{1}{P_{n}} A_{n} (p) − \frac{1}{P_{i}} A_{i} (p)∥ · max_{1≤i≤n−1} ∥Δb_{i}∥ . \end{matrix}

All the inequalities in (6.24) are sharp in the sense that the constant 1 cannot be replaced by a smaller constant.

Proof. Using the second equality in (6.13) and Schwarz’s inequality, we have

|T_{n} (p; a, b)| ≤ |P_{n}| ∑_{i=1}^{n−1} |P_{i}| |⟨\frac{1}{P_{n}} A_{n} (p) − \frac{1}{P_{i}} A_{i} (p) , Δb_{i}⟩|

≤ |P_{n}| ∑_{i=1}^{n−1} |P_{i}| ∥\frac{1}{P_{n}} A_{n} (p) − \frac{1}{P_{i}} A_{i} (p)∥ ∥Δb_{i}∥ .

Using Hölder’s weighted inequality, we deduce (6.24).

The sharpness of the constant may be proven in a similar manner to the one in Theorem 86. We omit the details.

The following corollary containing the unweighted inequalities holds.

Corollary 65. With the above assumptions for a and b, one has

(6.25) |T_{n} (a, b)|

≤ \frac{1}{n} \{ \begin{matrix}max_{1≤i≤n−1} ∥\frac{1}{n} ∑_{k=1}^{n} a_{k} − \frac{1}{i} ∑_{k=1}^{i} a_{k}∥ ∑_{i=1}^{n−1} i ∥Δb_{i}∥; \\ (∑_{i=1}^{n−1} i ∥\frac{1}{n} ∑_{k=1}^{n} a_{k} − \frac{1}{i} ∑_{k=1}^{i} a_{k}∥^{q}) & \frac{1}{q} \\ × (∑_{i=1}^{n−1} i ∥Δb_{i}∥^{p})^{\frac{1}{p}} for p > 1, \frac{1}{p} + \frac{1}{q} = 1; \\ ∑_{i=1}^{n−1} i ∥\frac{1}{n} ∑_{k=1}^{n} a_{k} − \frac{1}{i} ∑_{k=1}^{i} a_{k}∥ · max_{1≤i≤n−1} ∥Δb_{i}∥ . \end{matrix}

The inequalities (6.25) are sharp in the sense mentioned above.

Another type of inequality may be stated if ones used the third identity in (6.13) and Hölder’s weighted inequality with the weights: |P_{i}| |P̄_{i}| , i ∈ {1, …, n − 1} [9].

Theorem 88. With the assumptions in Theorem 86 and if P_{i}, P̄_{i} ≠ 0, i ∈ {1, …, n − 1} , then we have the inequalities

(6.26) |T_{n} (p; a, b)|

≤ |P_{n}| × \{ \begin{matrix}max_{1≤i≤n−1} ∥\frac{1}{P̄_{i}} Ā_{i} (p) − \frac{1}{P_{i}} A_{i} (p)∥ ∑_{i=1}^{n−1} |P_{i}| |P̄_{i}| ∥Δb_{i}∥; \\ (∑_{i=1}^{n−1} |P_{i}| |P̄_{i}| ∥\frac{1}{P̄_{i}} Ā_{i} (p) − \frac{1}{P_{i}} A_{i} (p)∥^{q}) & \frac{1}{q} \\ × (∑_{i=1}^{n−1} |P_{i}| |P̄_{i}| ∥Δb_{i}∥^{p})^{\frac{1}{p}} for p > 1, \frac{1}{p} + \frac{1}{q} = 1; \\ ∑_{i=1}^{n−1} |P_{i}| |P̄_{i}| ∥\frac{1}{P̄_{i}} Ā_{i} (p) − \frac{1}{P_{i}} A_{i} (p)∥ · max_{1≤i≤n−1} ∥Δb_{i}∥ . \end{matrix}

In particular, if p_{i} = \frac{1}{n}, i ∈ \{1, …, n\} , then we have

(6.27) |T_{n} (a, b)|

≤ \frac{1}{n^{2}} \{ \begin{matrix}max_{1≤i≤n−1} ∥\frac{1}{n−i} ∑_{k=i+1}^{n} a_{k} − \frac{1}{i} ∑_{k=1}^{i} a_{k}∥ ∑_{i=1}^{n−1} i (n − i) ∥Δb_{i}∥; \\ (∑_{i=1}^{n−1} i (n − i) ∥\frac{1}{n−i} ∑_{k=i+1}^{n} a_{k} − \frac{1}{i} ∑_{k=1}^{i} a_{k}∥^{q}) & \frac{1}{q} \\ × (∑_{i=1}^{n−1} i (n − i) ∥Δb_{i}∥^{p})^{\frac{1}{p}} for p > 1, \frac{1}{p} + \frac{1}{q} = 1; \\ ∑_{i=1}^{n−1} i (n − i) ∥\frac{1}{n−i} ∑_{k=i+1}^{n} a_{k} − \frac{1}{i} ∑_{k=1}^{i} a_{k}∥ · max_{1≤i≤n−1} ∥Δb_{i}∥ . \end{matrix}

The inequalities in (6.26) and (6.27) are sharp in the above mentioned sense.

A different approach may be considered if one uses the representation in terms of double sums for the Čebyšev functional provided by Theorem 85.

The following result holds [9].

Theorem 89. With the above assumptions of Theorem 86, we have the inequalities

(6.28) |T_{n} (p; a, b)|

≤ |P_{n}| × \{ \begin{matrix}max_{1≤i,j≤n−1} & \{|P_{min\{i,j\}}| , |P̄_{max\{i,j\}}|\} \\ × ∑_{i=1}^{n−1} ∥Δa_{i}∥ ∑_{i=1}^{n−1} ∥Δb_{i}∥; \\ (∑_{i=1}^{n−1} ∑_{j=1}^{n−1} |P_{min\{i,j\}}|^{q} |P̄_{max\{i,j\}}|^{q}) & \frac{1}{q} \\ × (∑_{i=1}^{n−1} ∥Δa_{i}∥^{p})^{\frac{1}{p}} (∑_{i=1}^{n−1} ∥Δb_{i}∥^{p})^{\frac{1}{p}} \\ for p > 1, \frac{1}{p} + \frac{1}{q} = 1; \\ ∑ & {n−1 \atop i=1} & ∑ & {n−1 \atop j=1} & |P_{min\{i,j\}}| & |P̄_{max\{i,j\}}| \\ × max_{1≤i≤n−1} ∥Δa_{i}∥ max_{1≤i≤n−1} ∥Δb_{i}∥ . \end{matrix}

The inequalities are sharp in the sense mentioned above.

The proof follows by the identity (6.19) on using Hölder’s inequality for double sums and we omit the details.

Now, define

k_{∞} := max_{1≤i,j≤n−1} \{\frac{min \{i, j\}}{n} (1 − \frac{max \{i, j\}}{n})\} , n ≥ 2.

Using the elementary inequality

ab ≤ \frac{1}{4} (a + b)^{2} , a, b ∈ ℝ;

we deduce

min \{i, j\} (n − max \{i, j\}) ≤ \frac{1}{4} (n − |i − j|)^{2}

for 1 ≤ i, j ≤ n − 2. Consequently, we deduce

k_{∞} ≤ \frac{1}{4n^{2}} max_{1≤i,j≤n−1} \{(n − |i − j|)^{2}\} = \frac{1}{4}.

We may now state the following corollary of Theorem 89 [9].

Corollary 66. With the assumptions of Theorem 86 for a and b, we have the inequality

(6.29)  |T_{n} (a, b)| ≤ k_{∞} ∑_{i=1}^{n−1} ∥Δa_{i}∥ ∑_{i=1}^{n−1} ∥Δb_{i}∥

≤ \frac{1}{4} ∑_{i=1}^{n−1} ∥Δa_{i}∥ ∑_{i=1}^{n−1} ∥Δb_{i}∥ .

The constant \frac{1}{4} cannot be replaced in general by a smaller constant.

Remark 70. The inequality (6.29) is better than the third inequality in (6.11).

Consider now, for q > 1, the number

k_{q} := \frac{1}{n^{2}} (∑_{i=1}^{n−1} ∑_{j=1}^{n−1} [min \{i, j\} (n − max \{i, j\})]^{q})^{\frac{1}{q}} .

We observe, by symmetry of the terms under the summation symbol, that we have

k_{q} = \frac{1}{n^{2}} (2 ∑_{1≤i<j≤n−1} i^{q} (n − j)^{q} + ∑_{i=1}^{n−1} i^{q} (n − i)^{q})^{\frac{1}{q}} .

Note that the quantity k_{q} may be computed exactly if q = 2 or another natural number.

Since, as above,

[min \{i, j\} (n − max \{i, j\})]^{q} ≤ \frac{1}{4^{q}} (n − |i − j|)^{2q} ,

6. BOUNDS FOR A PAIR OF N-TUPLES OF VECTORS 213

we deduce

k_{q} ≤ \frac{1}{4n^{2}} (∑_{i=1}^{n−1} ∑_{j=1}^{n−1} (n − |i − j|)^{2q})^{\frac{1}{q}}

≤ \frac{1}{4n^{2}} [(n − 1)^{2} n^{2q}]^{\frac{1}{q}}

= \frac{1}{4} (n − 1)^{\frac{2}{q}} .

Consequently, we may state the following corollary as well [9].

Corollary 67. With the assumptions of Theorem 86 for a and b, we have the inequalities

(6.30) |T_{n} (a, b)| ≤ k_{q} (∑_{i=1}^{n−1} ∥Δa_{i}∥^{p})^{\frac{1}{p}} (∑_{i=1}^{n−1} ∥Δb_{i}∥^{p})^{\frac{1}{p}}

≤ \frac{1}{4} (n − 1)^{\frac{2}{q}} (∑_{i=1}^{n−1} ∥Δa_{i}∥^{p})^{\frac{1}{p}} (∑_{i=1}^{n−1} ∥Δb_{i}∥^{p})^{\frac{1}{p}} ,

provided p > 1 , \frac{1}{p} + \frac{1}{q} = 1. The constant \frac{1}{4} cannot be replaced in general by a smaller constant.

Finally, if we denote

k_{1} := \frac{1}{n^{2}} ∑_{i=1}^{n−1} ∑_{j=1}^{n−1} [min \{i, j\} (n − max \{i, j\})] ,

then we observe, for u = (\frac{1}{n}, …, \frac{1}{n}) , e = (1, 2, …, n) , that

k_{1} = |T_{n} (u; e, e)| = \frac{1}{n} ∑_{i=1}^{n} i^{2} − (\frac{1}{n} ∑_{i=1}^{n} i)^{2} = \frac{1}{12} (n^{2} − 1) ,

and by Theorem 89, we deduce the inequality

|T_{n} (a, b)| ≤ \frac{1}{12} (n^{2} − 1) max_{1≤j≤n−1} ∥Δa_{j}∥ max_{1≤j≤n−1} ∥Δb_{j}∥ .

Note that, the above inequality has been discovered using a different method in [10]. The constant \frac{1}{12} is best possible.

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[15] D.S. MITRINOVIĆ, J.E. PEČARIĆ and A.M. FINK, Classical and New Inequalities in Analysis, Kluwer Academic Publishers, 1993.

[16] J.E. PEČARIĆ, F. PROSCHAN and Y.L. TONG, Convex Functions, Partial Orderings and Statistical Applications, Academic Press, San Diego, 1992.

CHAPTER 6 Other Inequalities in Inner Product Spaces

1. The Ostrowski Inequality

1.1. Introduction. In 1951, A.M. Ostrowski [16, p. 289] proved the following result (see also [15, p. 92]):

Theorem 90. Suppose that a, b and x are real n-tuples such that a ≠ 0 and

∑_{i=1}^{n} a_{i}x_{i} = 0 and ∑_{i=1}^{n} b_{i}x_{i} = 1.

Then

∑_{i=1}^{n} x_{i}^{2} ≥ \frac{∑_{i=1}^{n} a_{i}^{2}}{∑_{i=1}^{n} a_{i}^{2} ∑_{i=1}^{n} b_{i}^{2} − (∑_{i=1}^{n} a_{i}b_{i})^{2}}

with equality if and only if

x_{k} = \frac{b_{k} ∑_{i=1}^{n} a_{i}^{2} − a_{k} ∑_{i=1}^{n} a_{i}b_{i}}{∑_{i=1}^{n} a_{i}^{2} ∑_{i=1}^{n} b_{i}^{2} − (∑_{i=1}^{n} a_{i}b_{i})^{2}} ,

for k ∈ {1, …, n} .

An integral version of this inequality was obtained by Pearce, Pečarić and Varošanec in 1998, [17].

H. Šikić and T. Šikić in 2001, [18], by the use of an argument based on orthogonal projections in inner product spaces have observed that Ostrowski’s inequality may be naturally stated in an abstract setting as follows:

Theorem 91. Let (H; ⟨·, ·⟩) be a real or complex inner product space and a, b ∈ H two linearly independent vectors. If x ∈ H is such that

⟨x, a⟩ = 0 and ⟨x, b⟩ = 1,

then one has the inequality

(1.1)  ∥x∥^{2} ≥ \frac{∥a∥^{2}}{∥a∥^{2} ∥b∥^{2} − |⟨a, b⟩|^{2}} ,

with equality if and only if

x = \frac{∥a∥^{2} b − \overline{⟨a, b⟩} · a}{∥a∥^{2} ∥b∥^{2} − |⟨a, b⟩|^{2}} .

In the present section, by the use of elementary arguments and Schwarz’s inequality in inner product spaces, we show that Ostrowski’s inequality (1.1) holds true for a larger class of elements x ∈ H. The case of equality is analyzed. Applications for complex sequences and integrals are also provided.

1.2. The General Inequality. The following theorem holds [6].

Theorem 92. Let (H; ⟨·, ·⟩) be a real or complex inner product space and a, b ∈ H two linearly independent vectors. If x ∈ H is such that

(1.2)  ⟨x, a⟩ = 0, and |⟨x, b⟩| = 1;

then one has the inequality

(1.3)  ∥x∥^{2} ≥ \frac{∥a∥^{2}}{∥a∥^{2} ∥b∥^{2} − |⟨a, b⟩|^{2}} .

The equality holds in (1.3) if and only if

x = μ (b − \frac{\overline{⟨a, b⟩}}{∥a∥^{2}} · a)

where μ ∈ 𝕂 (𝕂 = ℝ,ℂ) is such that

(1.4)  |μ| = \frac{∥a∥^{2}}{∥a∥^{2} ∥b∥^{2} − |⟨a, b⟩|^{2}} .

Proof. We use Schwarz’s inequality in the inner product space (H; ⟨·, ·⟩) , i.e.,

(1.5)  ∥u∥^{2} ∥v∥^{2} ≥ |⟨u, v⟩|^{2} ; u, v ∈ H

with equality iff there exists a scalar α ∈ 𝕂 such that u = αv. If we apply (1.5) for

u = z − \frac{⟨z, c⟩}{∥c∥^{2}} · c, v = d − \frac{⟨d, c⟩}{∥c∥^{2}} · c,

where c ≠ 0 and c, d, z ∈ H, then we have

(1.6) ∥z − \frac{⟨z, c⟩}{∥c∥^{2}} · c∥^{2} ∥d − \frac{⟨d, c⟩}{∥c∥^{2}} · c∥^{2}

≥ |⟨z − \frac{⟨z, c⟩}{∥c∥^{2}} · c, d − \frac{⟨d, c⟩}{∥c∥^{2}} · c⟩|^{2}

with equality iff there is a scalar β ∈ 𝕂 such that

(1.7)  z = \frac{⟨z, c⟩}{∥c∥^{2}} · c + β (d − \frac{⟨d, c⟩}{∥c∥^{2}} · c) .

Since simple calculations show that

∥z − \frac{⟨z, c⟩}{∥c∥^{2}} · c∥^{2} = \frac{∥z∥^{2} ∥c∥^{2} − |⟨z, c⟩|^{2}}{∥c∥^{2}} ,

∥d − \frac{⟨d, c⟩}{∥c∥^{2}} · c∥^{2} = \frac{∥d∥^{2} ∥c∥^{2} − |⟨d, c⟩|^{2}}{∥c∥^{2}} ,

and

⟨z − \frac{⟨z, c⟩}{∥c∥^{2}} · c, d − \frac{⟨d, c⟩}{∥c∥^{2}} · c⟩ = \frac{⟨z, d⟩ ∥c∥^{2} − ⟨z, c⟩ ⟨c, d⟩}{∥c∥^{2}},

then, by (1.6), we deduce

(1.8) [∥z∥^{2} ∥c∥^{2} − |⟨z, c⟩|^{2}] [∥d∥^{2} ∥c∥^{2} − |⟨d, c⟩|^{2}]
≥ |⟨z, d⟩ ∥c∥^{2} − ⟨z, c⟩ ⟨c, d⟩|^{2} ,

with equality if and only if there is a β ∈ 𝕂 such that (1.7) holds.

If a, x, b satisfy (1.2) then by (1.8) and (1.7) for the choices z = x, c = a and d = b we deduce the inequality (1.3) with equality iff there exists a μ ∈ 𝕂 such that

x = μ (b − \frac{\overline{⟨a, b⟩}}{∥a∥^{2}} · a)

and, by the second condition in (1.2) ,

(1.9)  |μ ⟨b − \frac{\overline{⟨a, b⟩}}{∥a∥^{2}} · a, b⟩| = 1.

Since (1.9) is clearly equivalent to (1.4), the theorem is completely proved.

1.3. Applications for Sequences and Integrals. The following particular cases hold.

1. If a, b, x ∈ ℓ^{2} (𝕂) , where ℓ^{2} (𝕂) := {x = (x_{i})_{i∈ℕ} , ∑_{i=1}^{∞} |x_{i}|^{2} < ∞} , with a, b linearly independent and

∑_{i=1}^{∞} x_{i}\overline{a_{i}} = 0 and |∑_{i=1}^{∞} x_{i}\overline{b_{i}}| = 1,

then one has the inequality

∑_{i=1}^{∞} |x_{i}|^{2} ≥ \frac{∑_{i=1}^{∞} |a_{i}|^{2}}{∑_{i=1}^{∞} |a_{i}|^{2} ∑_{i=1}^{∞} |b_{i}|^{2} − |∑_{i=1}^{∞} a_{i}\overline{b_{i}}|^{2}} ,

with equality iff

x_{i} = μ [b_{i} − \frac{∑_{k=1}^{∞} \overline{a_{k}}b_{k}}{∑_{k=1}^{∞} |a_{k}|^{2}} · a_{i}] , i ∈ ℕ

and μ ∈ 𝕂 with the property

|μ| = \frac{∑_{i=1}^{∞} |a_{i}|^{2}}{∑_{i=1}^{∞} |a_{i}|^{2} ∑_{i=1}^{∞} |b_{i}|^{2} − |∑_{i=1}^{∞} a_{i}\overline{b_{i}}|^{2}} .

2. If f, g, h ∈ L^{2} (Ω, m), where Ω is a measurable space and L^{2} (Ω, m) := {f : Ω → 𝕂, ∫_{Ω} |f (x)|^{2} dm (x) < ∞} , with f, g linearly independent and

∫_{Ω} h (x) \overline{f (x)}dm (x) = 0, |∫_{Ω} h (x) \overline{g (x)}dm (x)| = 1,

then one has the inequality

∫_{Ω} |h (x)|^{2} dm (x)

≥ \frac{∫_{Ω} |f (x)|^{2} dm (x)}{∫_{Ω} |f (x)|^{2} dm (x) ∫_{Ω} |g (x)|^{2} dm (x) − |∫_{Ω} f (x) \overline{g (x)}dm (x)|^{2}}

with equality iff

h (x) = ν [g (x) − \frac{∫_{Ω} f\overline{(x)}g (x) dm (x)}{∫_{Ω} |f (x)|^{2} dm (x)} · f (x)]

for m− a.e. x ∈ Ω, and ν ∈ 𝕂 with

|ν| = \frac{∫_{Ω} |f (x)|^{2} dm (x)}{∫_{Ω} |f (x)|^{2} dm (x) ∫_{Ω} |g (x)|^{2} dm (x) − |∫_{Ω} f (x) \overline{g (x)}dm (x)|^{2}} .

2. Another Ostrowski Type Inequality

2.1. Introduction. Another result due to Ostrowski which is far less known than the one incorporated in Theorem 90 and obtained in the same work [16, p. 130] (see also [15, p. 94]), is the following one.

Theorem 93. Let a, b and x be n−tuples of real numbers with a ≠ 0 and

∑_{i=1}^{n} a_{i}x_{i} = 0 and ∑_{i=1}^{n} x_{i}^{2} = 1.

Then

(2.1)  \frac{∑_{i=1}^{n} a_{i}^{2} ∑_{i=1}^{n} b_{i}^{2} − (∑_{i=1}^{n} a_{i}b_{i})^{2}}{∑_{i=1}^{n} a_{i}^{2}} ≥ (∑_{i=1}^{n} b_{i}x_{i})^{2} .

If a and b are not proportional, then the equality holds in (2.1) iff

x_{k} = q · \frac{b_{k} ∑_{i=1}^{n} a_{i}^{2} − a_{k} ∑_{i=1}^{n} a_{i}b_{i}}{(∑_{k=1}^{n} a_{k}^{2})^{\frac{1}{2}} [∑_{i=1}^{n} a_{i}^{2} ∑_{i=1}^{n} b_{i}^{2} − (∑_{i=1}^{n} a_{i}b_{i})^{2}]^{\frac{1}{2}}} ,

k ∈ {1, …, n} , with q ∈ {−1, 1, } .

The case of equality which was neither mentioned in [16] nor in [15] is considered in Remark 71.

In the present section, by the use of an elementary argument based on Schwarz’s inequality, a natural generalisation in inner product spaces of (2.1) is given. The case of equality is analyzed. Applications for sequences and integrals are also provided.

2.2. The General Result. The following theorem holds [9].

Theorem 94. Let (H, ⟨·, ·⟩) be a real or complex inner product space and a, b ∈ H two linearly independent vectors. If x ∈ H is such that

(i)  ⟨x, a⟩ = 0 and ∥x∥ = 1,

then

(2.2)  \frac{∥a∥^{2} ∥b∥^{2} − |⟨a, b⟩|^{2}}{∥a∥^{2}} ≥ |⟨x, b⟩|^{2} .

The equality holds in (2.2) iff

x = ν (b − \frac{\overline{⟨a, b⟩}}{∥a∥^{2}} · a) ,

where ν ∈ 𝕂 (ℂ, ℝ) is such that

|ν| = \frac{∥a∥}{[∥a∥^{2} ∥b∥^{2} − |⟨a, b⟩|^{2}]^{\frac{1}{2}}} .

Proof. We use Schwarz’s inequality in the inner product space H, i.e.,

(2.3)  ∥u∥^{2} ∥v∥^{2} ≥ |⟨u, v⟩|^{2} , u, v ∈ H

with equality iff there is a scalar α ∈ 𝕂 such that

u = αv.

If we apply (2.3) for u = z − \frac{⟨z,c⟩}{∥c∥^{2}} · c, v = d − \frac{⟨d,c⟩}{∥c∥^{2}} · c, where c ≠ 0 and c, d, z ∈ H, then we deduce the inequality

(2.4) [∥z∥^{2} ∥c∥^{2} − |⟨z, c⟩|^{2}] [∥d∥^{2} ∥c∥^{2} − |⟨d, c⟩|^{2}]

≥ |⟨z, d⟩ ∥c∥^{2} − ⟨z, c⟩ ⟨c, d⟩|^{2}

with equality iff there is a β ∈ 𝕂 such that

z = \frac{⟨z, c⟩}{∥c∥^{2}} · c + β (d − \frac{⟨d, c⟩}{∥c∥^{2}} · c) .

If in (2.4) we choose z = x, c = a and d = b, where a and x statisfy (i), then we deduce

∥a∥^{2} [∥a∥^{2} ∥b∥^{2} − |⟨a, b⟩|^{2}] ≥ [⟨x, b⟩ ∥a∥^{2}]^{2}

which is clearly equivalent to (2.2).

The equality holds in (2.2) iff

x = ν (b − \frac{\overline{⟨a, b⟩}}{∥a∥^{2}} · a) ,

where ν ∈ 𝕂 satisfies the condition

1 = ∥x∥ = |ν| ∥b − \frac{\overline{⟨a, b⟩}}{∥a∥^{2}} · a∥ = |ν| [\frac{∥a∥^{2} ∥b∥^{2} − |⟨a, b⟩|^{2}}{∥a∥^{2}} ]^{\frac{1}{2}} ,

and the theorem is thus proved.

2.3. Applications for Sequences and Integrals. The following particular cases hold.

1. If a, b, x ∈ ℓ^{2} (𝕂) , 𝕂 = ℂ, ℝ, where

ℓ^{2} (𝕂) := {x = (x_{i})_{i∈ℕ} , ∑_{i=1}^{∞} |x_{i}|^{2} < ∞}

with a, b linearly independent and

∑_{i=1}^{∞} x_{i}\overline{a_{i}} = 0, ∑_{i=1}^{∞} |x_{i}|^{2} = 1,

then

(2.5)  \frac{∑_{i=1}^{∞} |a_{i}|^{2} ∑_{i=1}^{∞} |b_{i}|^{2} − |∑_{i=1}^{∞} a_{i}\overline{b_{i}}|^{2}}{∑_{i=1}^{∞} |a_{i}|^{2}} ≥ |∑_{i=1}^{∞} x_{i}\overline{b_{i}}|^{2} .

The equality holds in (2.5) iff

x_{i} = ν [b_{i} − \frac{∑_{k=1}^{∞} a_{k}\overline{b_{k}}}{∑_{k=1}^{∞} |a_{k}|^{2}} · a_{i}] , i ∈ \{1, 2, … \}

with ν ∈ 𝕂 such that

|ν| = \begin{matrix}(∑_{k=1}^{∞} |a_{k}|^{2}) & \frac{1}{2} \\ [∑_{k=1}^{∞} |a_{k}|^{2} ∑_{k=1}^{∞} |b_{k}|^{2} − |∑_{k=1}^{∞} a_{k}\overline{b_{k}}|^{2}] & \frac{1}{2} \end{matrix} .

Remark 71. The case of equality in (2.1) is obviously a particular case of the above. We omit the details.

2. If f, g, h ∈ L^{2} (Ω, m) , where Ω is an m−measurable space and

L^{2} (Ω, m) := {f : Ω → 𝕂, ∫_{Ω} |f (x)|^{2} dm (x) < ∞} ,

with f, g being linearly independent and

∫_{Ω} h (x) \overline{f (x)}dm (x) = 0, ∫_{Ω} |h (x)|^{2} dm (x) = 1,

then

(2.6) \frac{∫_{Ω} |f (x)|^{2} dm (x) ∫_{Ω} |g (x)|^{2} dm (x) − |∫_{Ω} f (x) \overline{g (x)}dm (x)|^{2}}{∫_{Ω} |f (x)|^{2} dm (x)}

≥ |∫_{Ω} h (x) \overline{g (x)}dm (x)|^{2} .

The equality holds in (2.6) iff

h (x) = ν [g (x) − \frac{∫_{Ω} g (x) \overline{f (x)}dm (x)}{∫_{Ω} |f (x)|^{2} dm (x)}f (x)] for a.e. x ∈ Ω

and ν ∈ 𝕂 with

|ν| = \begin{matrix}(∫_{Ω} |f (x)|^{2} dm (x)) & \frac{1}{2} \\ [∫_{Ω} |f (x)|^{2} dm (x) ∫_{Ω} |g (x)|^{2} dm (x) − |∫_{Ω} f (x) \overline{g (x)}dm (x)|^{2}] & \frac{1}{2} \end{matrix} .

3. The Wagner Inequality in Inner Product Spaces

3.1. Introduction. In 1965, S.S. Wagner [19] (see also [14] or [15, p. 85]) pointed out the following generalisation of Cauchy-Bunyakovsky-Schwarz’s inequality for real numbers.

Theorem 95. Let a = (a_{1}, …, a_{n}) and b = (b_{1}, …, b_{n}) be two ntuples of real numbers. Then for any x ∈ [0, 1] , one has the inequality

(∑_{k=1}^{n} a_{k}b_{k} + x · ∑_{1≤i≠j≤n} a_{i}b_{j})^{2}

≤ (∑_{k=1}^{n} a_{k}^{2} + 2x · ∑_{1≤i<j≤n} a_{i}a_{j}) · (∑_{k=1}^{n} b_{k}^{2} + 2x · ∑_{1≤i<j≤n} b_{i}b_{j}) .

For x = 0, we recapture the Cauchy-Bunyakovsky-Schwarz’s inequality, i.e., (see for example [15, p. 84])

(∑_{k=1}^{n} a_{k}b_{k})^{2} ≤ ∑_{k=1}^{n} a_{k}^{2} ∑_{k=1}^{n} b_{k}^{2},

with equality if and only if there exists a real number r such that a_{k} = rb_{k} for each k ∈ {1, …, n}.

In this section we extend the above result for sequences of vectors in real or complex inner product spaces.

3.2. The Results. Let (H; ⟨·, ·⟩) be an inner product space over 𝕂, where 𝕂 = ℝ or 𝕂 = ℂ. The following result holds [3].

Theorem 96. Let x = (x_{1}, …, x_{n}) and y = (y_{1}, …, y_{n}) be two n-tuples of vectors in H. Then for any α ∈ [0, 1] one has the inequality

(3.1) [∑_{k=1}^{n} Re ⟨x_{k}, y_{k}⟩ + α · ∑_{1≤i≠j≤n} Re ⟨x_{i}, y_{j}⟩]^{2}

≤ [∑_{k=1}^{n} ∥x_{k}∥^{2} + 2α · ∑_{1≤i<j≤n} Re ⟨x_{i}, x_{j}⟩]

× [∑_{k=1}^{n} ∥y_{k}∥^{2} + 2α · ∑_{1≤i<j≤n} Re ⟨y_{i}, y_{j}⟩] .

Proof. Following the proof by P. Flor [14], we may consider the function f : ℝ → ℝ, given by

(3.2)  f (t) = (1 − α) · ∑_{k=1}^{n} ∥tx_{k} − y_{k}∥^{2} + α · ∥∑_{k=1}^{n} (tx_{k} − y_{k})∥^{2} .

Then

(3.3) f (t) = [(1 − α) · ∑_{k=1}^{n} ∥x_{k}∥^{2} + α · ∥∑_{k=1}^{n} x_{k}∥^{2}] t^{2}

+ 2 [(1 − α) · ∑_{k=1}^{n} Re ⟨x_{k}, y_{k}⟩ + α · Re ⟨∑_{k=1}^{n} x_{k}, ∑_{k=1}^{n} y_{k}⟩] t

+ [(1 − α) · ∑_{k=1}^{n} ∥y_{k}∥^{2} + α · ∥∑_{k=1}^{n} y_{k}∥^{2}] .

Observe that

(3.4)  ∥∑_{k=1}^{n} x_{k}∥^{2} = ∑_{k=1}^{n} ∥x_{k}∥^{2} + 2 · ∑_{1≤i<j≤n} Re ⟨x_{i}, x_{j}⟩

and

(3.5)  ∥∑_{k=1}^{n} y_{k}∥^{2} = ∑_{k=1}^{n} ∥y_{k}∥^{2} + 2 · ∑_{1≤i<j≤n} Re ⟨y_{i}, y_{j}⟩ .

Also

(3.6) Re ⟨∑_{k=1}^{n} x_{k}, ∑_{k=1}^{n} y_{k}⟩ = ∑_{k=1}^{n} Re ⟨x_{k}, y_{k}⟩ + ∑_{1≤i≠j≤n} Re ⟨x_{i}, y_{j}⟩ .

Using (3.3) − (3.6) , we deduce that

(3.7) f (t) = [∑_{k=1}^{n} ∥x_{k}∥^{2} + 2α · ∑_{1≤i<j≤n} Re ⟨x_{i}, x_{j}⟩] t^{2}

+ 2 [∑_{k=1}^{n} Re ⟨x_{k}, y_{k}⟩ + α · ∑_{1≤i≠j≤n} Re ⟨x_{i}, y_{j}⟩] t

+ [∑_{k=1}^{n} ∥y_{k}∥^{2} + 2α · ∑_{1≤i<j≤n} Re ⟨y_{i}, y_{j}⟩] .

Since, by (3.2) , f (t) ≥ 0 for any t ∈ ℝ, it follows that the discriminant of the quadratic function given by (3.7) is negative, which is clearly equivalent with the desired inequality (3.1) .

One may obtain an interesting inequality if x and y are assumed to incorporate orthogonal vectors.

Corollary 68. Assume that {x_{i}}_{i=1,...,n} are orthogonal, i.e., x_{i} ⊥ x_{j} for any i, j ∈ {1, …, n}, i ≠ j; and {y_{i}}_{i=1,...,n} are also orthogonal in the real inner product space (H; ⟨·, ·⟩) . Then

sup_{α∈[0,1]} [∑_{k=1}^{n} ⟨x_{k}, y_{k}⟩ + α · ∑_{1≤i≠j≤n} ⟨x_{i}, y_{j}⟩]^{2} ≤ ∑_{k=1}^{n} ∥x_{k}∥^{2} ∑_{k=1}^{n} ∥y_{k}∥^{2} .

3.3. Applications.

1. If we assume that H = ℂ, with the inner product ⟨x, y⟩ = x · ȳ, then by (3.1) we may deduce the following Wagner type inequality for complex numbers

[∑_{k=1}^{n} Re (a_{k}b̄_{k}) + α · ∑_{1≤i≠j≤n} Re (a_{i}b̄_{j})]^{2}

≤ [∑_{k=1}^{n} |a_{k}|^{2} + 2α · ∑_{1≤i<j≤n} Re (a_{i}ā_{j})]

× [∑_{k=1}^{n} |b_{k}|^{2} + 2α · ∑_{1≤i<j≤n} Re (b_{i}b̄_{j})] ,

where α ∈ [0, 1] and a = (a_{1}, …, a_{n}), b = (b_{1}, …, b_{n}) ∈ ℂ^{n}.

2. Consider the Hilbert space

L_{2} (Ω, μ) := {f : Ω → ℂ, ∫_{Ω} |f (x)|^{2} dμ (x) < ∞} , where Ω is a μ-measurable space and μ : Ω → [0, ∞] is a positive measure on Ω. Then for H = L_{2} (Ω, μ) and since the inner product generating the norm is given by

⟨f, g⟩ = ∫_{Ω} f (x) ḡ (x) dμ (x) ,

we get the inequality

[∑_{k=1}^{n} ∫_{Ω} Re (f_{k} (x) ḡ_{k} (x)) dμ (x)

+α · ∑_{1≤i≠j≤n} ∫_{Ω} Re (f_{i} (x) ḡ_{j} (x)) dμ (x)]^{2}

≤ [∑_{k=1}^{n} ∫_{Ω} |f_{k} (x)|^{2} dμ (x) + 2α · ∑_{1≤i<j≤n} ∫_{Ω} Re (f_{i} (x) f̄_{j} (x)) dμ (x)]

× [∑_{k=1}^{n} ∫_{Ω} |g_{k} (x)|^{2} dμ (x) + 2α · ∑_{1≤i<j≤n} ∫_{Ω} Re (g_{i} (x) ḡ_{j} (x)) dμ (x)] ,

where f_{i}, g_{i} ∈ L_{2} (Ω, μ) , i ∈ {1, …, n} and α ∈ [0, 1] .

4. A Monotoniciy Property of Bessel’s Inequality

Let X be a linear space over the real or complex number field 𝕂. A mapping (·, ·) : X × X → 𝕂 is said to be a positive hermitian form if the following conditions are satisfied:

(i) (αx + βy, z) = α (x, z) + β (y, z) for all x, y, z ∈ X and α, β ∈

𝕂;

(ii) (y, x) = \overline{(x, y)} for all x, y ∈ X; (iii) (x, x) ≥ 0 for all x ∈ X.

If ∥x∥ := (x, x)^{\frac{1}{2}} denotes the semi-norm associated to this form and (e_{i})_{i∈I} is an orthonormal family of vectors in X, that is, (e_{i}, e_{j}) = δ_{ij} (i, j ∈ I), then one has [20]:

(4.1)  ∥x∥^{2} ≥ ∑_{i∈I} |(x, e_{i})|^{2} for all x ∈ X,

which is well known in the literature as Bessel’s inequality.

The main aim of the section is to point out an improvement for this result as follows [4].

Theorem 97. Let X be a linear space and (·, ·)_{2} , (·, ·)_{1} two hermitian forms on X such that ∥·∥_{2} is greater than or equal to ∥·∥_{1}, that is, ∥x∥_{2} ≥ ∥x∥_{1} for all x ∈ X. Assume that (e_{i})_{i∈I} is an orthonormal family in (X; (·, ·)_{2}) and (f_{j})_{j∈J} is an orthonormal family in (X; (·, ·)_{1}) such that for any i ∈ I there exists a finite K ⊂ J such that

(4.2)  e_{i} = ∑_{j∈K} α_{j}f_{j}, α_{j} ∈ 𝕂, (j ∈ K) ,

then one has the inequality:

(4.3) ∥x∥_{2}^{2} − ∑_{i∈I} |(x, e_{i})_{2}|^{2} ≥ ∥x∥_{1}^{2} − ∑_{j∈J} |(x, f_{j})_{1}|^{2} ≥ 0, for all x ∈ X.

In order to prove this, we require the following lemma.

Lemma 13. Let X be a linear space endowed with a positive hermitian form (·, ·) and (g_{k}) , k ∈ {1, …, n} be an orthonormal family in (X; (·, ·)) . Then

∥x − ∑_{k=1}^{n} λ_{k}g_{k}∥^{2} ≥ ∥x∥^{2} − ∑_{k=1}^{n} |(x, g_{k})|^{2} ≥ 0,

for all λ_{k} ∈ 𝕂, k ∈ {1 …, n} and x ∈ X.

The proof follows by mathematical induction.

Proof of Theorem 97. Let H be a finite subset of I. Since ∥·∥_{2} is greater than ∥·∥_{1}, we have:

∥x∥_{2}^{2} − ∑_{i∈H} |(x, e_{i})_{2}|^{2} = ∥x − ∑_{i∈H} (x, e_{i})_{2} e_{i}∥_{2}^{2}

≥ ∥x − ∑_{i∈H} (x, e_{i})_{2} e_{i}∥_{1}^{2} , x ∈ X.

Since, by (4.2), we may state that for any i ∈ H there exists a finite K ⊂ J with

e_{i} = ∑_{j∈K} (e_{i}, f_{j})_{1} f_{j},

we have, for all x ∈ X

∥x − ∑_{i∈H} (x, e_{i})_{2} e_{i}∥_{1}^{2} = ∥x − ∑_{i∈H} (x, e_{i})_{2} ∑_{j∈K} (e_{i}, f_{j})_{1} f_{j}∥_{1}^{2}

= ∥x − ∑_{j∈K} (∑_{i∈H} (x, e_{i})_{2} e_{i}, f_{j})_{1} f_{j}∥_{1}^{2} .

Applying Lemma 13, we can conclude that

∥x − ∑_{j∈K} λ_{j}f_{j}∥_{1}^{2} ≥ ∥x∥_{1}^{2} − ∑_{j∈K} |(x, f_{j})_{1}|^{2} , x ∈ X,

where

λ_{j} = (∑_{i∈H} (x, e_{i})_{2} e_{i}, f_{j})_{1} ∈ 𝕂, (j ∈ K) .

Consequently, we have

∥x∥_{2}^{2} − ∑_{i∈H} |(x, e_{i})_{2}|^{2} ≥ ∥x∥_{1}^{2} − ∑_{j∈K} |(x, f_{j})_{1}|^{2}

≥ ∥x∥_{1}^{2} − ∑_{j∈J} |(x, f_{j})_{1}|^{2}

for all x ∈ X and H a finite subset of I, from which (4.3) results.

Corollary 69. Let ∥·∥_{1} , ∥·∥_{2} : X → ℝ_{+} be as above. Then for all x, y ∈ X, we have the inequality:

(4.4)  ∥x∥_{2}^{2} ∥y∥_{2}^{2} − |(x, y)_{2}|^{2} ≥ ∥x∥_{1}^{2} ∥y∥_{1}^{2} − |(x, y)_{1}|^{2} ≥ 0,

which is an improvement of the well known Cauchy-Schwarz inequality.

Remark 72. For a different proof of (4.4), see also [10] or [11].

Now, we will give some natural applications of the above theorem.

(1) Let (X; (·, ·)) be an inner product space and (e_{i})_{i∈I} an orthonormal family in X. Assume that A : X → X is a linear operator such that ∥Ax∥ ≤ ∥x∥ for all x ∈ X and (Ae_{i}, Ae_{j}) = δ_{ij} for all i, j ∈ I. Then one has the inequality

∥x∥^{2} − ∑_{i∈I} |(x, e_{i})|^{2} ≥ ∥Ax∥^{2} − ∑_{i∈I} |(Ax, Ae_{i})|^{2}
≥ 0, for all x ∈ X.

(2) If A : X → X is such that ∥Ax∥ ≥ ∥x∥ for all x ∈ X, then, with the previous assumptions, we also have

0 ≤ ∥x∥^{2} − ∑_{i∈I} |(x, e_{i})|^{2}

≤ ∥Ax∥^{2} − ∑_{i∈I} |(Ax, Ae_{i})|^{2} , for all x ∈ X.

(3) Suppose that A : X → X is a symmetric positive definite operator with (Ax, x) ≥ ∥x∥^{2} for all x ∈ X. If (e_{i})_{i∈I} is an orthonormal family in X such that (Ae_{i}, Ae_{j}) = δ_{ij} for all i, j ∈ I, then one has the inequality

0 ≤ ∥x∥^{2} − ∑_{i∈I} |(x, e_{i})|^{2} ≤ (Ax, x) − ∑_{i∈I} |(Ax, e_{i})|^{2} ,

for any x ∈ X.

5. Other Bombieri Type Inequalities

5.1. Introduction. In 1971, E. Bombieri [1] gave the following generalisation of Bessel’s inequality:

(5.1)  ∑_{i=1}^{n} |(x, y_{i})|^{2} ≤ ∥x∥^{2} max_{1≤i≤n} {∑_{j=1}^{n} |(y_{i}, y_{j})|} ,

where x, y_{1}, …, y_{n} are vectors in the inner product space (H; (·, ·)) .

It is obvious that if (y_{i})_{1≤i≤n} = (e_{i})_{1≤i≤n} , where (e_{i})_{1≤i≤n} are orthornormal vectors in H, i.e., (e_{i}, e_{j}) = δ_{ij} (i, j = 1, …, n) , where δ_{ij} is the Kronecker delta, then (5.1) provides Bessel’s inequality

∑_{i=1}^{n} |(x, e_{i})|^{2} ≤ ∥x∥^{2} , x ∈ H.

In this section we point out some Bombieri type inequalities that complement the results obtained in Chapter 4.

5.2. The Results. The following lemma, which is of interest in itself, holds [7].

Lemma 14. Let z_{1}, …, z_{n} ∈ H and α_{1}, …, α_{n} ∈ 𝕂. Then one has the inequalities:

∥∑_{i=1}^{n} α_{i}z_{i}∥^{2} ≤ \{ \begin{matrix}max_{1≤i≤n} |α_{i}|^{2} ∑_{i,j=1}^{n} |(z_{i}, z_{j})|; \\ (∑_{i=1}^{n} |α_{i}|^{p})^{\frac{2}{p}} (∑_{i,j=1}^{n} |(z_{i}, z_{j})|^{q})^{\frac{1}{q}} , \\ where p > 1, \frac{1}{p} + \frac{1}{q} = 1; \\ (∑_{i=1}^{n} |α_{i}|)^{2} max_{1≤i,j≤n} |(z_{i}, z_{j})|; \end{matrix}

≤ \{ \begin{matrix}max_{1≤i≤n} |α_{i}|^{2} (∑_{i=1}^{n} ∥z_{i}∥)^{2} ; \\ (∑_{i=1}^{n} |α_{i}|^{p})^{\frac{2}{p}} (∑_{i=1}^{n} ∥z_{i}∥^{q})^{\frac{2}{q}} , \\ where p > 1, \frac{1}{p} + \frac{1}{q} = 1; \\ (∑_{i=1}^{n} |α_{i}|)^{2} max_{1≤i≤n} ∥z_{i}∥^{2} . \end{matrix}

Proof. We observe that

∥∑_{i=1}^{n} α_{i}z_{i}∥^{2} = (∑_{i=1}^{n} α_{i}z_{i}, ∑_{j=1}^{n} α_{j}z_{j}) = ∑_{i=1}^{n} ∑_{j=1}^{n} α_{i}\overline{α_{j}} (z_{i}, z_{j})

= |∑_{i=1}^{n} ∑_{j=1}^{n} α_{i}\overline{α_{j}} |(z_{i}, z_{j})||

≤ ∑_{i=1}^{n} ∑_{j=1}^{n} |α_{i}| |α_{j}| |(z_{i}, z_{j})| =: M.

Firstly, we have

M ≤ max_{1≤i,j≤n} {|α_{i}| |α_{j}|} ∑_{i,j=1}^{n} |(z_{i}, z_{j})|

= max_{1≤i≤n} |α_{i}|^{2} ∑_{i,j=1}^{n} |(z_{i}, z_{j})| .

Secondly, by the Hölder inequality for double sums, we obtain

M ≤ [∑_{i,j=1}^{n} (|α_{i}| |α_{j}|)^{p}]^{\frac{1}{p}} (∑_{i,j=1}^{n} |(z_{i}, z_{j})|^{q})^{\frac{1}{q}}

= (∑_{i=1}^{n} |α_{i}|^{p} ∑_{j=1}^{n} |α_{j}|^{p})^{\frac{1}{p}} (∑_{i,j=1}^{n} |(z_{i}, z_{j})|^{q})^{\frac{1}{q}}

= (∑_{i=1}^{n} |α_{i}|^{p})^{\frac{2}{p}} (∑_{i,j=1}^{n} |(z_{i}, z_{j})|^{q})^{\frac{1}{q}} ,

where p > 1, \frac{1}{p} + \frac{1}{q} = 1.

Finally, we have

M ≤ max_{1≤i,j≤n} |(z_{i}, z_{j})| ∑_{i,j=1}^{n} |α_{i}| |α_{j}| = (∑_{i=1}^{n} |α_{i}|)^{2} max_{1≤i,j≤n} |(z_{i}, z_{j})|

and the first part of the lemma is proved.

The second part is obvious on taking into account, by Schwarz’s inequality in H, that we have

|(z_{i}, z_{j})| ≤ ∥z_{i}∥ ∥z_{j}∥ ,

for any i, j ∈ {1, …, n} . We omit the details.

Corollary 70. With the assumptions in Lemma 14, one has

(5.2)  ∥∑_{i=1}^{n} α_{i}z_{i}∥^{2} ≤ ∑_{i=1}^{n} |α_{i}|^{2} (∑_{i,j=1}^{n} |(z_{i}, z_{j})|^{2})^{\frac{1}{2}}

≤ ∑_{i=1}^{n} |α_{i}|^{2} ∑_{i=1}^{n} ∥z_{i}∥^{2} .

The proof follows by Lemma 14 on choosing p = q = 2. Note also that (5.2) provides a refinement of the well known Cauchy-

Bunyakovsky-Schwarz inequality for sequences of vectors in inner product spaces, namely

∥∑_{i=1}^{n} α_{i}z_{i}∥^{2} ≤ ∑_{i=1}^{n} |α_{i}|^{2} ∑_{i=1}^{n} ∥z_{i}∥^{2} .

The following lemma also holds [7].

Lemma 15. Let x, y_{1}, …, y_{n} ∈ H and c_{1}, …, c_{n} ∈ 𝕂. Then one has the inequalities:

|∑_{i=1}^{n} c_{i} (x, y_{i})|^{2} (5.3)

≤ ∥x∥^{2} × \{ \begin{matrix}max_{1≤i≤n} |c_{i}|^{2} ∑_{i,j=1}^{n} |(y_{i}, y_{j})|; \\ (∑_{i=1}^{n} |c_{i}|^{p})^{\frac{2}{p}} (∑_{i,j=1}^{n} |(y_{i}, y_{j})|^{q})^{\frac{1}{q}} , \\ where p > 1, \frac{1}{p} + \frac{1}{q} = 1; \\ (∑_{i=1}^{n} |c_{i}|)^{2} max_{1≤i,j≤n} |(y_{i}, y_{j})|; \end{matrix}

≤ ∥x∥^{2} × \{ \begin{matrix}max_{1≤i≤n} |c_{i}|^{2} (∑_{i=1}^{n} ∥y_{i}∥)^{2} ; \\ (∑_{i=1}^{n} |c_{i}|^{p})^{\frac{2}{p}} (∑_{i=1}^{n} ∥y_{i}∥^{q})^{\frac{2}{q}} , \\ where p > 1, \frac{1}{p} + \frac{1}{q} = 1; \\ (∑_{i=1}^{n} |c_{i}|)^{2} max_{1≤i≤n} ∥y_{i}∥^{2} . \end{matrix}

Proof. We have, by Schwarz’s inequality in the inner product (H; (·, ·)) , that

|∑_{i=1}^{n} c_{i} (x, y_{i})|^{2} = |(x, ∑_{i=1}^{n} \overline{c_{i}}y_{i})|^{2} ≤ ∥x∥^{2} ∥∑_{i=1}^{n} \overline{c_{i}}y_{i}∥^{2} .

5. OTHER BOMBIERI TYPE INEQUALITIES 233

Now, applying Lemma 14 for α_{i} = \overline{c_{i}}, z_{i} = y_{i} (i = 1, …, n) , the inequality (5.3) is proved.

Corollary 71. With the assumptions in Lemma 15, one has

(5.4)  |∑_{i=1}^{n} c_{i} (x, y_{i})|^{2} ≤ ∥x∥^{2} ∑_{i=1}^{n} |c_{i}|^{2} (∑_{i,j=1}^{n} |(y_{i}, y_{j})|^{2})^{\frac{1}{2}}

≤ ∥x∥^{2} ∑_{i=1}^{n} |c_{i}|^{2} ∑_{i=1}^{n} ∥y_{i}∥^{2} .

The proof follows by Lemma 15, on choosing p = q = 2. Remark 73. The inequality (5.4) was firstly obtained in [12, In-

equality (7)].

The following theorem incorporating three Bombieri type inequalities holds [7].

Theorem 98. Let x, y_{1}, …, y_{n} ∈ H. Then one has the inequalities:

(5.5) ∑_{i=1}^{n} |(x, y_{i})|^{2}

≤ ∥x∥ × \{ \begin{matrix}max_{1≤i≤n} |(x, y_{i})| (∑_{i,j=1}^{n} |(y_{i}, y_{j})|)^{\frac{1}{2}} ; \\ (∑_{i=1}^{n} |(x, y_{i})|^{p})^{\frac{1}{p}} (∑_{i,j=1}^{n} |(y_{i}, y_{j})|^{q})^{\frac{1}{2q}} , \\ where p > 1, \frac{1}{p} + \frac{1}{q} = 1; \\ ∑_{i=1}^{n} |(x, y_{i})| max_{1≤i,j≤n} |(y_{i}, y_{j})|^{\frac{1}{2}} . \end{matrix}

Proof. Choosing c_{i} = \overline{(x, y_{i})} (i = 1, …, n) in (5.3) we deduce

(5.6) (∑_{i=1}^{n} |(x, y_{i})|^{2})^{2}

≤ ∥x∥^{2} × \{ \begin{matrix}max_{1≤i≤n} |(x, y_{i})|^{2} (∑_{i,j=1}^{n} |(y_{i}, y_{j})|); \\ (∑_{i=1}^{n} |(x, y_{i})|^{p})^{\frac{2}{p}} (∑_{i,j=1}^{n} |(y_{i}, y_{j})|^{q})^{\frac{1}{q}} , \\ where p > 1, \frac{1}{p} + \frac{1}{q} = 1; \\ (∑_{i=1}^{n} |(x, y_{i})|)^{2} max_{1≤i,j≤n} |(y_{i}, y_{j})|; \end{matrix}

which, by taking the square root, is clearly equivalent to (5.5).

Remark 74. If (y_{i})_{1≤i≤n} = (e_{i})_{1≤i≤n} , where (e_{i})_{1≤i≤n} are orthornormal vectors in H, then by (5.5) we deduce

(5.7) ∑_{i=1}^{n} |(x, e_{i})|^{2} ≤ ∥x∥ \{ \begin{matrix}\sqrt{n} max_{1≤i≤n} |(x, e_{i})|; \\ n^{\frac{1}{2q}} (∑_{i=1}^{n} |(x, e_{i})|^{p})^{\frac{1}{p}} , \\ where p > 1, \frac{1}{p} + \frac{1}{q} = 1; \\ ∑_{i=1}^{n} |(x, e_{i})| . \end{matrix}

If in (5.6) we take p = q = 2, then we obtain the following inequality which was formulated in [12, p. 81].

Corollary 72. With the assumptions in Theorem 98, we have:

(5.8)  ∑_{i=1}^{n} |(x, y_{i})|^{2} ≤ ∥x∥^{2} (∑_{i,j=1}^{n} |(y_{i}, y_{j})|^{2})^{\frac{1}{2}} .

Remark 75. Observe, that by the monotonicity of power means, we may write

(\frac{∑_{i=1}^{n} |(x, y_{i})|^{p}}{n} )^{\frac{1}{p}} ≤ (\frac{∑_{i=1}^{n} |(x, y_{i})|^{2}}{n} )^{\frac{1}{2}} , 1 < p ≤ 2.

Taking the square in both sides, one has

(\frac{∑_{i=1}^{n} |(x, y_{i})|^{p}}{n} )^{\frac{2}{p}} ≤ \frac{∑_{i=1}^{n} |(x, y_{i})|^{2}}{n} ,

giving

(5.9)  (∑_{i=1}^{n} |(x, y_{i})|^{p})^{\frac{2}{p}} ≤ n^{\frac{2}{p} −1} ∑_{i=1}^{n} |(x, y_{i})|^{2} .

Using (5.9) and the second inequality in (5.6) we may deduce the following result

(5.10)  ∑_{i=1}^{n} |(x, y_{i})|^{2} ≤ n^{\frac{2}{p} −1} ∥x∥^{2} (∑_{i,j=1}^{n} |(y_{i}, y_{j})|^{q})^{\frac{1}{q}} ,

for 1 < p ≤ 2, \frac{1}{p} + \frac{1}{q} = 1.

Note that for p = 2 (q = 2) we recapture (5.8).

Remark 76. Let us compare Bombieri’s result

(5.11)  ∑_{i=1}^{n} |(x, y_{i})|^{2} ≤ ∥x∥^{2} max_{1≤i≤n} {∑_{j=1}^{n} |(y_{i}, y_{j})|}

with our general result (5.10).

To do that, denote

M_{1} := max_{1≤i≤n} {∑_{j=1}^{n} |(y_{i}, y_{j})|}

and

M_{2} := n^{\frac{2}{p} −1} (∑_{i,j=1}^{n} |(y_{i}, y_{j})|^{q})^{\frac{1}{q}} , 1 < p ≤ 2, \frac{1}{p} + \frac{1}{q} = 1.

Consider the inner product space H = ℝ, (x, y) = x · y, n = 2 and y_{1} = a > 0, y_{2} = b > 0. Then

M_{1} = max {a^{2} + ab, ab + b^{2}} = (a + b) max {a, b} ,
M_{2} = 2^{\frac{2}{p} −1} (a^{q} + b^{q})^{\frac{2}{q}} = 2^{\frac{2}{p} −1} (a^{\frac{p}{p−1}} + b^{\frac{p}{p−1}} )^{\frac{2(p−1)}{p}} , 1 < p ≤ 2.

Assume that a = 1, b ∈ [0, 1] , p ∈ (1, 2]. Utilizing Maple 6, one may easily see by plotting the function

f (b, p) := M_{2} − M_{1} = 2^{\frac{2}{p} −1} (1 + b^{\frac{p}{p−1}} )^{\frac{2(p−1)}{p}} − 1 − b

that it has positive and negative values in the box [0, 1] × [1, 2], showing that the inequalities (5.10) and (5.11) cannot be compared. This means that one is not always better than the other.

6. Some Pre-Grüss Inequalities

6.1. Introduction. Let f, g be two functions defined and integrable on [a,b]. Assume that

φ ≤ f (x) ≤ Φ and γ ≤ g (x) ≤ Γ

for each x ∈ [a, b], where φ, Φ, γ, Γ are given real constants. Then we have the following inequality which is well known in the literature as the Grüss inequality ([15, pp. 296])

|\frac{1}{b − a} ∫_{a}^{b} f (x) g (x) dx − \frac{1}{b − a}∫_{a}^{b}f (x) dx·\frac{1}{b − a}∫_{a}^{b} g (x) dx|

≤ \frac{1}{4} |Φ − φ| · |Γ − γ| .

In this inequality, G. Grüss has proven that the constant \frac{1}{4} is the best possible in the sense that it cannot be replaced by a smaller one, and this is achieved when

f (x) = g (x) = sgn (x − \frac{a + b}{2}) .

Recently, S.S. Dragomir proved the following Grüss’ type inequality in real or complex inner product spaces [2].

Theorem 99. Let (H, ⟨·, ·⟩) be an inner product space over 𝕂 (𝕂 = ℝ, ℂ) and e ∈ H, ∥e∥ = 1. If φ, γ, Φ, Γ are real or complex numbers and x, y are vectors in H such that the conditions

Re ⟨Φe − x, x − φe⟩ ≥ 0 and Re ⟨Γe − x, x − γe⟩ ≥ 0

hold, then we have the inequality

(6.1)  |⟨x, y⟩ − ⟨x, e⟩ ⟨e, y⟩| ≤ \frac{1}{4} |Φ − φ| · |Γ − γ| .

The constant \frac{1}{4} is best possible in sense that it cannot be replaced by a smaller constant.

In [8], by using the following lemmas

Lemma 16. Let x, e ∈ H with ∥e∥ = 1 and δ, Δ ∈ 𝕂 with δ ≠ Δ. Then

Re ⟨Δe − x, x − δe⟩ ≥ 0

if and only if

∥x − \frac{δ + Δ}{2}e∥ ≤ \frac{1}{2} |Δ − δ| .

and

Lemma 17. Let x, e ∈ H with ∥e∥ = 1. Then one has the following representation

0 ≤ ∥x∥^{2} − |⟨x, e⟩|^{2} = inf_{λ∈K} ∥x − λe∥^{2} .

the author gave an alternative proof for (6.1) and also obtained the following refinement of it, namely

Theorem 100. Let (H, ⟨·, ·⟩) be an inner product space over 𝕂 (𝕂 = ℝ,ℂ) and e ∈ H, ∥e∥ = 1. If φ, γ, Φ, Γ are real or complex numbers and x, y are vectors in H such that either the conditions

Re ⟨Φe − x, x − φe⟩ ≥ 0, Re ⟨Γe − y, y − γe⟩ ≥ 0,

or, equivalently,

∥x − \frac{φ + Φ}{2} · e∥ ≤ \frac{1}{2} |Φ − φ| , ∥y − \frac{γ + Γ}{2} · e∥ ≤ \frac{1}{2} |Γ − γ| ,

hold, then we have the inequality

|⟨x, y⟩ − ⟨x, e⟩ ⟨e, y⟩|

≤ \frac{1}{4} |Φ − φ| · |Γ − γ| − [Re ⟨Φe − x, x − φe⟩]^{\frac{1}{2}} [Re ⟨Γe − y, y − γe⟩]^{\frac{1}{2}}

≤ (\frac{1}{4} |Φ − φ| · |Γ − γ|) .

The constant \frac{1}{4} is best possible.

Further, as a generalization for orthonormal families of vectors in inner product spaces, S.S. Dragomir proved, in [5], the following reverse of Bessel’s inequality:

Theorem 101. Let {e_{i}} , i ∈ I be a family of orthonormal vectors in H, F a finite part of I, φ_{i}, Φ_{i} ∈ 𝕂, i ∈ F and x a vector in H such that either the condition

Re ⟨∑_{i∈F} Φ_{i}e_{i} − x, x − ∑_{i∈F} φ_{i}e_{i}⟩ ≥ 0,

or, equivalently,

∥x − ∑_{i∈F} \frac{Φ_{i} + φ_{i}}{2} e_{i}∥ ≤ \frac{1}{2} (∑_{i∈F} |Φ_{i} − φ_{i}|^{2})^{\frac{1}{2}} ,

holds, then we have the following reverse of Bessel’s inequality

(6.2) ∥x∥^{2} − ∑_{i∈F} |⟨x, e_{i}⟩|^{2}

≤ \frac{1}{4} ∑_{i∈F} |Φ_{i} − φ_{i}|^{2} − ∑_{i∈F} |\frac{φ_{i} + Φ_{i}}{2} − ⟨x, e_{i}⟩|^{2} .

The constant \frac{1}{4} is best possible.

The corresponding Grüss type inequality is embodied in the following theorem:

Theorem 102. Let {e_{i}}_{i∈I} be a family of orthornormal vectors in H, F a finite part of I, ϕ_{i}, γ_{i}, Φ_{i}, Γ_{i} ∈ ℝ (i ∈ F), and x, y ∈ H. If either

Re ⟨∑_{i=1}^{n} Φ_{i}e_{i} − x, x − ∑_{i=1}^{n} ϕ_{i}e_{i}⟩ ≥ 0,

Re ⟨∑_{i=1}^{n} Γ_{i}e_{i} − y, y − ∑_{i=1}^{n} γ_{i}e_{i}⟩ ≥ 0,

or, equivalently,

∥x − ∑_{i∈F} \frac{Φ_{i} + ϕ_{i}}{2} e_{i}∥ ≤ \frac{1}{2} (∑_{i∈F} |Φ_{i} − ϕ_{i}|^{2})^{\frac{1}{2}} ,

∥y − ∑_{i∈F} \frac{Γ_{i} + γ_{i}}{2} e_{i}∥ ≤ \frac{1}{2} (∑_{i∈F} |Γ_{i} − γ_{i}|^{2})^{\frac{1}{2}} ,

hold true, then

0 ≤ |⟨x, y⟩ − ∑_{i=1}^{n} ⟨x, e_{i}⟩ ⟨e_{i}, y⟩|

≤ \frac{1}{4} (∑_{i=1}^{n} |Φ_{i} − ϕ_{i}|^{2})^{\frac{1}{2}} · (∑_{i=1}^{n} |Γ_{i} − γ_{i}|^{2})^{\frac{1}{2}}

− ∑_{i∈F} |\frac{Φ_{i} + ϕ_{i}}{2} − ⟨x, e_{i}⟩| |\frac{Γ_{i} + γ_{i}}{2} − ⟨y, e_{i}⟩|

(≤ \frac{1}{4} (∑_{i=1}^{n} |Φ_{i} − ϕ_{i}|^{2})^{\frac{1}{2}} · (∑_{i=1}^{n} |Γ_{i} − γ_{i}|^{2})^{\frac{1}{2}} ) .

The constant \frac{1}{4} is best possible in the sense that it cannot be replaced by a smaller constant.

The main aim here is to provide some similar inequalities which, giving refinements of the usual Grüss’ inequality, are known in the literature as pre-Grüss type inequalities. Applications for Lebesgue integrals in general measure spaces are also given.

6.2. Pre-Grüss Inequalities in Inner Product Spaces. We start with the following result [13]:

Theorem 103. Let (H, ⟨·, ·⟩) be an inner product space over 𝕂, (𝕂 = ℝ, ℂ) and e ∈ H, ∥e∥ = 1. If φ, Φ are real or complex numbers and x, y are vectors in H such that either the condition

Re ⟨Φe − x, x − φe⟩ ≥ 0,

or, equivalently,

(6.3)  ∥x − \frac{φ + Φ}{2}e∥ ≤ \frac{1}{2} |Φ − φ| ,

holds true, then we have the inequalities

(6.4)  |⟨x, y⟩ − ⟨x, e⟩ ⟨e, y⟩| ≤ \frac{1}{2} |Φ − φ| · \sqrt{∥y∥^{2} − |⟨y, e⟩|^{2}}

and

(6.5) |⟨x, y⟩ − ⟨x, e⟩ ⟨e, y⟩|

≤ \frac{1}{2} |Φ − φ| · ∥y∥ − (Re ⟨Φe − x, x − φe⟩)^{\frac{1}{2}} · |⟨y, e⟩| .

Proof. It is obvious that:

⟨x, y⟩ − ⟨x, e⟩ ⟨e, y⟩ = ⟨x − ⟨x, e⟩ e, y − ⟨y, e⟩ e⟩ .

Using Schwarz’s inequality in inner product spaces |⟨u, v⟩| ≤ ∥u∥ · ∥v∥ for the vectors x − ⟨x, e⟩ e and y − ⟨y, e⟩ e, we deduce:

(6.6) |⟨x, y⟩ − ⟨x, e⟩ ⟨e, y⟩|^{2} ≤ (∥x∥^{2} − |⟨x, e⟩|^{2}) · (∥y∥^{2} − |⟨y, e⟩|^{2}) .

Now, the inequality (6.4) is a simple consequence of (6.1) for x = y, or of Lemma 17 and (6.3).

Since (see for instance [2]),

(6.7) ∥x∥^{2} − |⟨x, e⟩|^{2} = Re ((Φ − ⟨x, e⟩) · (⟨e, x⟩ − φ̄)) − Re ⟨Φe − x, x − φe⟩ ,

then making use of the elementary inequality 4 Re (ab̄) ≤ |a + b|^{2} with a, b ∈ 𝕂 (𝕂 = ℝ,ℂ) , we can state that

(6.8)  Re ((Φ − ⟨x, e⟩) · (⟨e, x⟩ − φ̄)) ≤ \frac{1}{4} |Φ − φ|^{2} .

Using (6.7) and (6.8) we have

(6.9) ∥x∥^{2} − |⟨x, e⟩|^{2} ≤ (\frac{1}{2} |Φ − φ|)^{2} − ((Re ⟨Φe − x, x − φe⟩)^{\frac{1}{2}} )^{2} .

Taking into account the inequalities (6.6) and (6.9), we get that

|⟨x − ⟨x, e⟩ e, y − ⟨y, e⟩ e⟩|^{2}

≤ ((\frac{1}{2} |Φ − φ|)^{2} − ((Re ⟨Φe − x, x − φe⟩)^{\frac{1}{2}} )^{2})·(∥y∥^{2} − |⟨y, e⟩|^{2}) .

Finally, using the elementary inequality for positive real numbers:

(6.10)  (m^{2} − n^{2}) · (p^{2} − q^{2}) ≤ (mp − nq)^{2} ,

we have:

((\frac{1}{2} |Φ − φ|)^{2} − ((Re ⟨Φe − x, x − φe⟩)^{\frac{1}{2}} )^{2}) · (∥y∥^{2} − |⟨y, e⟩|^{2})

≤ (\frac{1}{2} |Φ − φ| · ∥y∥ − (Re ⟨Φe − x, x − φe⟩)^{\frac{1}{2}} · |⟨y, e⟩|)^{2} ,

giving the desired inequality (6.5).

A similar version for Bessel’s inequality is incorporated in the following theorem [13]:

Theorem 104. Let {e_{i}}_{i∈I} , be a family of orthonormal vectors in H, F a finite part of I, φ_{i}, Φ_{i} ∈ 𝕂, i ∈ F and x, y vectors in H such that either the condition

Re ⟨∑_{i∈F} Φ_{i}e_{i} − x, x − ∑_{i∈F} φ_{i}e_{i}⟩ ≥ 0,

or, equivalently,

∥x − ∑_{i∈F} \frac{Φ_{i} + φ_{i}}{2} e_{i}∥ ≤ \frac{1}{2} (∑_{i∈F} |Φ_{i} − φ_{i}|^{2})^{\frac{1}{2}}

holds. Then we have inequalities

(6.11) |⟨x, y⟩ − ∑_{i∈F} ⟨x, e_{i}⟩ ⟨e_{i}, y⟩|

≤ \frac{1}{2} (∑_{i∈F} |Φ_{i} − φ_{i}|^{2})^{\frac{1}{2}} \sqrt{∥y∥^{2} − ∑_{i∈F} |⟨y, e_{i}⟩|^{2}}

and

(6.12) |⟨x, y⟩ − ∑_{i∈F} ⟨x, e_{i}⟩ ⟨e_{i}, y⟩| ≤ \frac{1}{2} (∑_{i∈F} |Φ_{i} − φ_{i}|^{2})^{\frac{1}{2}} · ∥y∥

− (∑_{i∈F} |\frac{Φ_{i} + φ_{i}}{2} − ⟨x, e_{i}⟩|^{2})^{\frac{1}{2}} · (∑_{i∈F} |⟨y, e_{i}⟩|^{2})^{\frac{1}{2}} .

Proof. It is obvious (see for example [5]) that:

⟨x, y⟩ − ∑_{i∈F} ⟨x, e_{i}⟩ ⟨e_{i}, y⟩ = ⟨x − ∑_{i∈F} ⟨x, e_{i}⟩ e_{i}, y − ∑_{i∈F} ⟨y, e_{i}⟩ e_{i}⟩ .

Using Schwarz’s inequality in inner product spaces, we have:

|⟨x − ∑_{i∈F} ⟨x, e_{i}⟩ e_{i}, y − ∑_{i∈F} ⟨y, e_{i}⟩ e_{i}⟩|^{2} (6.13)

≤ ∥x − ∑_{i∈F} ⟨x, e_{i}⟩ e_{i}∥^{2} · ∥x − ∑_{i∈F} ⟨y, e_{i}⟩ e_{i}∥^{2}

= (∥x∥^{2} − ∑_{i∈F} |⟨x, e_{i}⟩|^{2}) · (∥y∥^{2} − ∑_{i∈F} |⟨y, e_{i}⟩|^{2}) .

In a similar manner to the one in the proof of Theorem 103 we may conclude that (6.11) holds true.

Now, using (6.2) and (6.13) we also have:

|⟨x − ∑_{i∈F} ⟨x, e_{i}⟩ e_{i}, y − ∑_{i∈F} ⟨y, e_{i}⟩ e_{i}⟩|^{2}

≤ (\frac{1}{2} ((∑_{i∈F} |Φ_{i} − φ_{i}|^{2})^{\frac{1}{2}} )^{2} − ((∑_{i∈F} |\frac{φ_{i} + Φ_{i}}{2} − ⟨x, e_{i}⟩|^{2})^{\frac{1}{2}} )^{2})

× (∥y∥^{2} − ∑_{i∈F} |⟨y, e_{i}⟩|^{2}) .

Finally, utilizing the elementary inequality (6.10), we have

(6.14) (\frac{1}{2}((∑_{i∈F} |Φ_{i} − φ_{i}|^{2})^{\frac{1}{2}})^{2}−((∑_{i∈F} \frac{φ_{i} + Φ_{i}}{2} − ⟨x, e_{i}⟩^{2})^{\frac{1}{2}})^{2})

× (∥y∥^{2} − ∑_{i∈F} |⟨y, e_{i}⟩|^{2}) ≤ (\frac{1}{2} (∑_{i∈F} |Φ_{i} − φ_{i}|^{2})^{\frac{1}{2}} · ∥y∥^{2}

− (∑_{i∈F} |\frac{φ_{i} + Φ_{i}}{2} − ⟨x, e_{i}⟩|^{2})^{\frac{1}{2}} · ∑_{i∈F} |⟨y, e_{i}⟩|^{2})^{2} ,

which gives the desired result (6.12).

6.3. Applications for Integrals. Let (Ω, Σ, μ) be a measure space consisting of a set Ω, Σ a σ−algebra of parts and μ a countably additive and positive measure on Σ with values in ℝ ∪ {∞}. Denote by L^{2} (Ω,𝕂) the Hilbert space of all real or complex valued functions f defined on Ω and 2–integrable on Ω, i. e.

∫_{Ω} |f (s)|^{2} dμ (s) < ∞.

The following proposition holds [13].

Proposition 33. If f, g, h ∈ L^{2} (Ω,𝕂) and φ, Φ ∈ 𝕂 are such that ∫_{Ω} |h (s)|^{2} dμ (s) = 1 and, either

(6.15)  ∫_{Ω} Re ((Φh (s) − f (s)) (f̄(s) −\overline{φ}h̄ (s))) dμ (s) ≥ 0,

or, equivalently,

(∫_{Ω} |f (s) − \frac{Φ + φ}{2}h (s)|^{2} dμ (s))^{\frac{1}{2}} ≤ \frac{1}{2} |Φ − φ|

holds, then we have the inequalities

|∫_{Ω} f (s) ḡ (s) dμ (s) − ∫_{Ω} f (s) h̄ (s) dμ (s) ∫_{Ω} h (s) ḡ (s) dμ (s)|

≤ \frac{1}{2} |Φ − φ| · \sqrt{(_{∫Ω} |g (s)|^{2} dμ (s) − |∫_{Ω} h (s) ḡ (s) dμ (s)|^{2})}

and

|∫_{Ω} f (s) ḡ (s) dμ (s) − ∫_{Ω} f (s) h̄ (s) dμ (s) ∫_{Ω} h (s) ḡ (s) dμ (s)|

≤ \frac{1}{2} |Φ − φ| · (_{∫Ω} |g (s)|^{2} dμ (s))^{\frac{1}{2}}

− (∫_{Ω} Re ((Φh (s) − f (s)) (h (s) f̄(s) − φh (s))) dμ (s))^{\frac{1}{2}}

× |∫_{Ω} h (s) ḡ (s) dμ (s)| .

Proof. The proof follows by Theorem 103 on choosing H = L^{2} (Ω, K) with the inner product

⟨f, g⟩ = ∫_{Ω} f (s) ḡ (s) dμ (s) .

Remark 77. We observe that a sufficient condition for (6.15) to hold is:

(6.16)  Re (Φh (s) − f (s)) (f̄(s) −\overline{φ}h̄ (s)) ≥ 0,

for μ−a.e. s ∈ Ω.

If the functions are real-valued, then, for Φ and φ real numbers, a sufficient condition for (6.16) to hold is

Φh (s) ≥ f (s) ≥ φh (s)

for μ−a.e. s ∈ Ω.

In this way we can see the close connection that exists between the classical Grüss inequality and the results obtained above.

Now, consider the family {f_{i}}_{i∈I} of functions in L^{2} (Ω, 𝕂) with the properties that

∫_{Ω} f_{i} (s) \overline{f_{j}} (s) dμ (s) = δ_{ij}, i, j ∈ I,

where δ_{ij} is 0 if i ≠ j and δ_{ij} = 1 if i = j. {f_{i}}_{i∈I} is an orthornormal family in L^{2} (Ω, 𝕂) .

The following proposition holds [13].

Proposition 34. Let {f_{i}}_{i∈I} be an orthornormal family of functions in L^{2} (Ω, 𝕂) , F a finite subset of I, ϕ_{i}, Φ_{i} ∈ 𝕂 (i ∈ F) and f ∈ L^{2} (Ω, 𝕂) , such that either

(6.17) ∫_{Ω}Re[(∑_{i∈F} Φ_{i}f_{i} (s) − f (s))(\overline{f} (s) − ∑_{i∈F} \overline{ϕ_{i}} \overline{f_{i}} (s))] dμ (s) ≥ 0

or, equivalently,

∫_{Ω} |f (s) − ∑_{i∈F} \frac{Φ_{i} + ϕ_{i}}{2} f_{i} (s)|^{2} dμ (s) ≤ \frac{1}{4} ∑_{i∈F} |Φ_{i} − ϕ_{i}|^{2} .

holds. Then we have the inequalities

|∫_{Ω} f (s) \overline{g (s)}dμ (s) − ∑_{i∈F} ∫_{Ω} f (s) \overline{f_{i}} (s) dμ (s) ∫_{Ω} f_{i} (s) \overline{g (s)}dμ (s)|

≤ \frac{1}{2} (∑_{i∈F} |Φ_{i} − ϕ_{i}|^{2})^{\frac{1}{2}}

× (∫_{Ω} |g (s)|^{2} dμ (s) − ∑_{i∈F} |∫_{Ω} g (s) \overline{f_{i} (s)}dμ (s)|^{2})^{\frac{1}{2}}

244 6. OTHER INEQUALITIES IN INNER PRODUCT SPACES

and

|∫_{Ω} f (s) \overline{g (s)}dμ (s) − ∑_{i∈F} ∫_{Ω} f (s) \overline{f_{i}} (s) dμ (s) ∫_{Ω} f_{i} (s) \overline{g (s)}dμ (s)|

≤ \frac{1}{2} (∑_{i∈F} |Φ_{i} − ϕ_{i}|^{2})^{\frac{1}{2}} (∫_{Ω} |g (s)|^{2} dμ (s))^{\frac{1}{2}}

− (∑_{i∈F} |\frac{Φ_{i} + ϕ_{i}}{2} − ∫_{Ω} f (s) \overline{f_{i}} (s) dμ (s)|^{2})^{\frac{1}{2}}

× (∑_{i∈F} |∫_{Ω} f (s) \overline{f_{i}} (s) dμ (s)|^{2})^{\frac{1}{2}} .

The proof is obvious by Theorem 103 and we omit the details. Remark 78. In the real case, we observe that a sufficient condition

for (6.17) to hold is

∑_{i∈F} Φ_{i}f_{i} (s) ≥ f (s) ≥ ∑_{i∈F} φ_{i}f_{i} (s)

for μ−a.e. s ∈ Ω.

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ticle 3 [ON LINE: http://rgmia.vu.edu.au/v6(E).html].

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Index

Čebyšev inequality, 37, 38

Andrica-Badea inequality, 159

Bessel inequality, 71, 73, 74, 77, 79, 82, 85, 89, 92, 95, 109, 116, 123, 131, 132, 144, 146, 147, 225, 228, 235, 238

reverse, 84, 96, 99, 102, 105, 107,

109, 111, 114 reverses, 72

Biernacki, Pidek, Ryll-Nardzewski in-

equality, 159 Boas-Bellman inequality, 123, 124, 126,

129–132 Bombieri inequality, 132, 133, 141, 145–

147, 153, 228, 231, 232 bounds, 1, 2, 39, 71, 86, 87, 127, 129,

136, 140, 193, 202

Cassels inequality, 9, 10 Cauchy-Bunyakovsky-Schwarz, 2, 7–

9, 14, 16, 90, 98, 111, 115, 126, 164, 176, 183, 185, 222, 230

Cerone-Dragomir inequality, 161 companion inequality

Grüss, 46, 52, 62, 78, 90, 104 complex roots, 172 convex function, 179, 183, 189, 198

Diaz-Metcalf inequality, 2, 3 discrete Fourier transform, 166 discrete inequality, 41

CBS, 164

Grüss, 159, 194

Hölder, 196 discrete measure, 52, 61 double sums, 125, 149, 164, 176, 209,

229 Dragomir inequality, 160 Dragomir-Booth inequality, 160 Dragomir-Goh inequality, 180

Euclidean norm, 173

forward difference, 160, 182, 192, 200,

202 Fourier coefficients, 129, 145 Fourier transform, 166, 169, 180, 181 Fréchet differentiable, 179, 189, 198

Grüss inequality, 37, 38, 41–44, 46, 47, 49, 52, 53, 56, 58, 61, 62, 72, 73, 75, 78, 80–82, 87, 90, 93–95, 101, 104, 107, 108, 112, 113, 117, 159, 164, 166, 175, 178, 182, 184, 186, 188, 191, 192, 194, 200, 201, 233–236

gradient inequality, 179

Greub-Reinboldt inequality, 3, 9

Hölder inequality, 98, 125, 135, 136, 139, 149, 196, 206, 209, 229

weighted, 208 Heilbronn inequality, 132, 133 Hilbert space, 7, 8, 16, 25, 71, 80–82, 94, 95, 108, 109, 112–114, 116, 117, 224, 239

inequality Čebyšev, 37, 38 Andrica-Badea, 159 Bessel, 71, 73, 74, 77, 79, 82, 85, 89, 92, 95, 109, 116, 123, 131, 132, 144, 146, 147, 225, 228, 235, 238 reverse, 72, 84, 96, 99, 102, 105,

107, 109, 111, 114 Biernacki, Pidek, Ryll-Nardzewski,

159 Boas-Bellman, 123, 124, 126, 129–

131 Bombieri, 132, 133, 141, 145–147,

153, 228, 231, 232 Cassels, 9, 10 Cauchy-Bunyakovsky-Schwarz, 2, 7–

9, 14, 16, 90, 98, 111, 115, 126, 164, 176, 183, 185, 222, 230

Cerone-Dragomir, 161 companion

Grüss, 46, 52, 62, 78, 90, 104 Diaz-Metcalf, 2, 3 discrete, 41

CBS, 164

Grüss, 159, 194

Hölder, 196 Dragomir, 160 Dragomir-Booth, 160 Dragomir-Goh, 180 Grüss, 37, 38, 41–44, 46, 47, 49, 52,

53, 56, 58, 61, 62, 72, 73, 75, 78, 80–82, 87, 90, 93–95, 101, 104, 107, 108, 112, 113, 117, 159, 164, 166, 175, 178, 182, 184, 186, 188, 191, 192, 194, 200, 201, 233–236

gradient, 179 Greub-Reinboldt, 3, 9 Hölder, 98, 125, 135, 136, 139, 149,

196, 206, 208, 209, 229 Heilbronn, 132, 133 Jensen, 183, 189, 190, 198 Klamkin-McLenaghan, 3 Lupaş, 38 Mitrinović-Pečarić-Fink, 128–130 Niculescu, 10, 12 Ostrowski, 38, 215, 216, 219 Ozeki, 2 Pólya-Szegö, 2, 10 Pečarić, 132, 139, 141, 146, 147,

151, 153 pre-Grüss, 233, 236 reverse, 23

additive, 3 Bessel, 71–74, 77, 79, 82, 84, 85, 89, 92, 95, 96, 98, 99, 102, 107, 109, 111, 114, 235 CBS, 2, 7–9, 14, 16, 183 Jensen, 183, 190, 198 Schwarz, 3, 6, 18, 27, 100 triangle, 24, 30, 33

Schwarz, 1–3, 6, 10, 18, 19, 21, 23, 27, 43, 100, 113, 129, 140, 153, 176, 185, 195, 206, 208, 216, 219, 220, 229, 230, 237, 238

Selberg, 132, 133 Shisha-Mond, 2 triangle, 1, 23, 24, 30, 33, 164, 174,

176 Wagner, 222, 224

inner product, 1, 7, 8, 14, 15, 17, 18, 25, 49, 58, 86, 129, 169, 184, 202, 224, 225, 230, 240

inner product space, 3, 10, 20–25, 27– 30, 41–44, 46–48, 52, 53, 55, 61– 63, 65, 71–73, 76, 80, 88, 93, 100, 103, 112, 123, 128–132, 139, 140, 144, 146, 151, 153, 161, 164, 166, 171, 173, 175, 176, 178–180, 183– 185, 189, 191, 193–195, 198, 202, 203, 205, 215, 216, 219, 220, 222, 224, 227, 228, 230, 233–238 isotonic linear functional, 3, 5, 6, 10,

13

Jensen inequality, 183, 189, 190, 198 reverse, 198

Klamkin-McLenaghan inequality, 3

Lebesgue integral, 41, 236 linearly dependent, 1 linearly independent, 215, 216, 218,

219, 221 Lipschitzian function, 39 Lupaş inequality, 38

measurable function, 7, 80 measurable space, 218, 221, 225 measure, 80, 239 measure space, 7, 41, 73, 80, 236, 239 Mellin transform, 169, 170, 181 Mitrinović-Pečarić-Fink inequality, 128–

130 n-tuples, 2, 159–161, 215, 219, 222,

223 Niculescu inequality, 10, 12 norm, 1, 25, 39, 129, 164, 173, 187,

225 normalised, 5 normed linear space, 173, 192

orthonormal family, ii, 71, 75, 87, 101,

112, 117, 130, 131, 225–228, 235 orthonormal vector, 71, 123, 132, 144,

146, 235, 238 Ostrowski inequality, 38, 215, 216, 219 Ozeki inequality, 2

p-powered sum, 170 Pólya-Szegö inequality, 2, 10 Pečarić inequality, 132, 139, 141, 146,

147, 151, 153 polynomial, 171, 172 positive linear functional, 6 positive measure, 225 power means, 137, 232 pre-Grüss inequality, 233, 236

reverse inequality, 23 additive, 3 Bessel, 71–74, 77, 79, 82, 84, 85, 89,

92, 95, 96, 98, 99, 102, 107, 109, 111, 114, 235

CBS, 2, 7–9, 14, 16, 183

Jensen, 183, 190, 198

Schwarz, 3, 6, 18, 19, 21, 27, 100 triangle, 23, 24, 30, 33

Schwarz inequality, 1, 2, 43, 113, 129, 140, 153, 176, 185, 195, 206, 208, 216, 219, 220, 229, 230, 237, 238

quadratic reverse, 18

reverse, 3, 6, 10, 19, 21, 23, 27, 100 Selberg inequality, 132, 133 semi-norm, 225 Shisha-Mond inequality, 2 Stieltjes integral, 38, 39

triangle inequality, 1, 164, 174, 176 reverse, 23, 24, 30, 33

Wagner inequality, 222, 224