FURTHER REVERSES OF THE SCHWARZ INEQUALITY IN INNER PRODUCT SPACES

S.S. DRAGOMIR

Abstract. Some new refinements of the Schwarz inequality in inner product spaces and applications for vector-valued sequnces and integrals are given.

1. Introduction

Let (H; ⟨·, ·⟩) be an inner product space over the real or complex number field 𝕂. One of the most important inequalities in inner product spaces with numerous applications, is the Schwarz inequality

(1.1)  |⟨x, y⟩|^{2} ≤ ∥x∥^{2} ∥y∥^{2} , x, y ∈ H,

with equality if and only if there exists a scalar α ∈ 𝕂 with x = αy.

If B̄ (y, r) := {z ∈ H| ∥z − y∥ ≤ r} is the closed ball of radius r > 0 centered in y and x ∈ B̄ (y, r) , then the following reverse of the Schwarz inequality holds [1]

(1.2)  (0 ≤) ∥x∥ ∥y∥ − |⟨x, y⟩| ≤ ∥x∥ ∥y∥ − |Re ⟨x, y⟩|

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

The constant \frac{1}{2} is best possible in (1.2).

If we assume that ∥x − y∥ ≤ r < ∥y∥ , then the following quadratic reverse of the Schwarz inequality also holds [2]

(1.3)  (0 ≤) ∥x∥^{2} ∥y∥^{2} − |⟨x, y⟩|^{2} ≤ ∥x∥^{2} ∥y∥^{2} − [Re ⟨x, y⟩]^{2}
≤ r^{2} ∥x∥^{2} .

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

Date: 23 July, 2004.

2000 Mathematics Subject Classification. Primary 46C05; Secondary 26D15.

Key words and phrases. Schwarz inequality, Triangle inequality, Bessel inequality.

If ∥x − y∥ ≤ r < ∥y∥ , then the following multiplicative version of the Schwarz inequality is valid (see [2])

(1.4)  (1 ≤) \begin{matrix}∥x∥ & ∥y∥ \\ Re & ⟨x, y⟩ \end{matrix} ≤ \frac{∥y∥}{\sqrt{∥y∥^{2} − r^{2}}} .

After simple calculation, (1.4) is clearly equivalent to:

(0 ≤) ∥x∥ ∥y∥ − Re ⟨x, y⟩ (1.5)

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

Other types of reverse results of the Schwarz inequality are presented in the following.

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

(1.6)  Re ⟨Ay − x, x − ay⟩ ≥ 0,

or, equivalently,

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

holds, then one has the inequality [3]

(1.8)  (0 ≤) ∥x∥^{2} ∥y∥^{2} − |⟨x, y⟩|^{2} ≤ \frac{1}{4} |A − a|^{2} ∥y∥^{4} .

The constant \frac{1}{4} is sharp in (1.8).

This inequality has been refined in [4] as follows (1.9)

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

provided (1.6) or (1.7) holds. A different refinement of (1.8) has been obtained in [5]: (1.10)

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

It has been shown in [4] that the bounds provided by (1.9) and (1.10) for the Schwarz difference ∥x∥^{2} ∥y∥^{2} − |⟨x, y⟩|^{2} cannot be compared in general, meaning that, sometimes one bound is better than the other.

If A, a ∈ 𝕂 satisfy either (1.6) or (1.7) and Re (Aā) > 0, then the following inequality also holds (see [6] and [7])

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

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

In the particular case where A = M ≥ m = a > 0, the inequality (1.11) becomes

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

provided

(1.13)  Re ⟨My − x, x − my⟩ ≥ 0

or, equivalently,

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

The constant \frac{1}{2} is sharp in (1.12).

An equivalent additive version of (1.12) is the following result

(1.15) (0 ≤) ∥x∥ ∥y∥ − Re ⟨x, y⟩ ≤ \frac{1}{2} · \begin{matrix}(\sqrt{M} − \sqrt{m}) & 2 \\ √ & \overline{Mm} \end{matrix} Re ⟨x, y⟩ .

In this inequality \frac{1}{2} is also sharp.

As pointed out in [2], the quadratic version in (1.11), i.e.,

(1.16)  ∥x∥^{2} ∥y∥^{2} ≤ \frac{1}{4} · \begin{matrix}|A − a| & 2 \\ Re & (Aā) \end{matrix} |⟨x, y⟩|^{2}

has the following equivalent additive version [2]

(1.17)  (0 ≤) ∥x∥^{2} ∥y∥^{2} − |⟨x, y⟩|^{2} ≤ \frac{1}{4} · \begin{matrix}|A − a| & 2 \\ Re & (Aā) \end{matrix} |⟨x, y⟩|^{2} ,

in which \frac{1}{4} is the best possible multiplicative constant in (1.17).

Finally, if A, a ∈ 𝕂 and x, y ∈ H satisfy either (1.6) or (1.7) and A ≠ −a, then we have the inequality [1]

(0 ≤) ∥x∥ ∥y∥ − |⟨x, y⟩| (1.18)

≤ ∥x∥ ∥y∥ − |Re [\frac{Ā + ā}{|A + a|} ⟨x, y⟩]|

≤ ∥x∥ ∥y∥ − Re [\frac{Ā + ā}{|A + a|} ⟨x, y⟩]

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

The constant \frac{1}{4} is best possible in (1.18).

The main aim of this paper is to provide new reverse inequalities for the Schwarz result. Some applications in relation to the triangle and Bessel inequalities are also provided.

2. Some New Reverses

The following result holds.

Theorem 1. Let (H; ⟨·, ·⟩) be an inner product space over the real or complex number field 𝕂 and x, y ∈ H, ρ_{1}, ρ_{2} > 0. If

(2.1)  ρ_{1} ≤ |∥x∥ − ∥y∥| (≤) ∥x − y∥ ≤ ρ_{2},

then we have the inequalities

(2.2)  (0 ≤) ∥x∥ ∥y∥ − |⟨x, y⟩| ≤ ∥x∥ ∥y∥ − |Re ⟨x, y⟩|

≤ ∥x∥ ∥y∥ − Re ⟨x, y⟩

≤ \frac{1}{2} (ρ_{2}^{2} − ρ_{1}^{2}) .

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

Proof. We need only to prove that the last inequality holds and the sharpness of the constant occurs.

Taking the square in the second inequality in (2.1), we have

∥x∥^{2} − 2 Re ⟨x, y⟩ + ∥y∥^{2} ≤ ρ_{2}^{2}.

This is equivalent to

(2.3)  2 (∥x∥ ∥y∥ − Re ⟨x, y⟩) + (∥x∥ − ∥y∥)^{2} ≤ ρ_{2}^{2}.

Utilising the first inequality in (2.1), we have

(2.4)  ρ_{1}^{2} ≤ (∥x∥ − ∥y∥)^{2} .

Now, making use of (2.3) and (2.4), we deduce (2.2).

To prove the sharpness of the constant \frac{1}{2}, let us assume that there exists a constant C > 0 such that

(2.5)  ∥x∥ ∥y∥ − Re ⟨x, y⟩ ≤ C (ρ_{2}^{2} − ρ_{1}^{2})

provided the condition (2.1) is satisfied.

Let e ∈ H with ∥e∥ = 1 and for ρ_{2} > ρ_{1} > 0, define

(2.6)  x = \frac{ρ_{2} + ρ_{1}}{2} · e,  y = \frac{ρ_{1} − ρ_{2}}{2} · e.

Then

∥x − y∥ = ρ_{2} and |∥x∥ − ∥y∥| = ρ_{1},

showing that the condition (2.1) is satisfied with equality on both sides.

If we replace in (2.5) the vectors x and y as defined by (2.6), we deduce

\frac{ρ_{2}^{2} − ρ_{1}^{2}}{2} ≥ C (ρ_{2}^{2} − ρ_{1}^{2}) ,

which implies C ≥ \frac{1}{2}, and the theorem is completely proved.

Remark 1. If ρ_{1} = 0, then (2.2) will produce the known inequality (1.2).

The following corollary holds.

Corollary 1. With the assumptions of Theorem 1, we have the inequality:

(2.7)  (0 ≤) ∥x∥ + ∥y∥ − ∥x + y∥ ≤ \sqrt{ρ_{2}^{2} − ρ_{1}^{2}}.

Proof. We have, by (2.2), that

(∥x∥ + ∥y∥)^{2} − ∥x + y∥^{2} = 2 (∥x∥ ∥y∥ − Re ⟨x, y⟩) ≤ ρ_{2}^{2} − ρ_{1}^{2}

giving

(2.8)  (∥x∥ + ∥y∥)^{2} ≤ ∥x + y∥^{2} + ρ_{2}^{2} − ρ_{1}^{2}.

Taking the square root in (2.8) and taking into account that

\sqrt{α + β} ≤ \sqrt{α} + \sqrt{β} for α, β ≥ 0,

we deduce the desired inequality (2.7).

If (H; ⟨·, ·⟩) is a Hilbert space and {e_{i}}_{i∈I} is an orthornormal family in H, i.e., we recall that ⟨e_{i}, e_{j}⟩ = δ_{ij} for any i, j ∈ I, where δ_{ij} is Kronecker’s delta, then we have the following inequality which is known widely in the literature as Bessel’s inequality

(2.9)  ∑_{i∈I} |⟨x, e_{i}⟩|^{2} ≤ ∥x∥^{2} for each x ∈ H.

Here, 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} .

The following result provides a reverse of Bessel’s inequality.

Theorem 2. Let (H; ⟨·, ·⟩) be a Hilbert space and {e_{i}}_{i∈I} an orthornormal family in H. If x ∈ H\ {0} and for ρ_{2}, ρ_{1} > 0 we have

(2.10) ρ_{1} ≤ ∥x∥ − (∑_{i∈I} |⟨x, e_{i}⟩|^{2})^{\frac{1}{2}} (≤) ∥x − ∑_{i∈I} ⟨x, e_{i}⟩ e_{i}∥ ≤ ρ_{2},

then we have the inequality:

(2.11)  (0 ≤) ∥x∥ − (∑_{i∈I} |⟨x, e_{i}⟩|^{2})^{\frac{1}{2}} ≤ \frac{1}{2} · \frac{ρ_{2}^{2} − ρ_{1}^{2}}{(∑_{i∈I} |⟨x, e_{i}⟩|^{2})^{\frac{1}{2}}} .

The constant \frac{1}{2} is best possible in (2.11).

Proof. Consider y := ∑_{i∈I} ⟨x, e_{i}⟩ e_{i}. Obviously, since H is a Hilbert space, y ∈ H. We also note that

∥y∥ = ∥∑_{i∈I} ⟨x, e_{i}⟩ e_{i}∥ = \sqrt{∥∑_{i∈I} ⟨x, e_{i}⟩ e_{i}∥^{2}} = \sqrt{∑_{i∈I} |⟨x, e_{i}⟩|^{2}},

and thus (2.10) is in fact (2.1) of Theorem 1.

Since

∥x∥ ∥y∥ − Re ⟨x, y⟩ = ∥x∥ (∑_{i∈I} |⟨x, e_{i}⟩|^{2})^{\frac{1}{2}} − Re ⟨x, ∑_{i∈I} ⟨x, e_{i}⟩ e_{i}⟩

= (∑_{i∈I} |⟨x, e_{i}⟩|^{2})^{\frac{1}{2}} [∥x∥ − (∑_{i∈I} |⟨x, e_{i}⟩|^{2})^{\frac{1}{2}} ] ,

then, by (2.2), we deduce the desired result (2.11).

We will prove the sharpness on the constant for the case of one element, i.e., I = {1} , e_{1} = e ∈ H, ∥e∥ = 1. For this, assume that there exists a constant D > 0 such that

(2.12)  ∥x∥ − |⟨x, e⟩| ≤ D · \frac{ρ_{2}^{2} − ρ_{1}^{2}}{|⟨x, e⟩|} ,

provided x ∈ H\ {0} satisfies the condition

ρ_{1} ≤ ∥x∥ − |⟨x, e⟩| (≤) ∥x − ⟨x, e⟩ e∥ ≤ ρ_{2}.

SCHWARZ INEQUALITY IN INNER PRODUCT SPACES 7

Assume that x = λe + μf with e, f ∈ H, ∥e∥ = ∥f∥ = 1 and e ⊥ f.

We look for positive numbers λ, μ such that

(2.13)  ∥x − ⟨x, e⟩ e∥ = ρ_{2} > ρ_{1} = ∥x∥ − |⟨x, e⟩| .

Since (for λ, μ > 0)

∥x − ⟨x, e⟩ e∥ = μ

and

∥x∥ − |⟨x, e⟩| = \sqrt{λ^{2} + μ^{2}} − λ

hence, by (2.13), we get μ = ρ_{2} and

\sqrt{λ^{2} + ρ_{2}^{2}} − λ = ρ_{1}

giving

λ^{2} + ρ_{2}^{2} = (λ + ρ_{1})^{2} = λ^{2} + 2λρ_{1} + ρ_{1}^{2}

from where we deduce

λ = \frac{ρ_{2}^{2} − ρ_{1}^{2}}{2ρ_{1}} > 0.

With these values for λ and μ, we have

∥x∥ − |⟨x, e⟩| = ρ_{1}, |⟨x, e⟩| = \frac{ρ_{2}^{2} − ρ_{1}^{2}}{2ρ_{1}} ,

and thus, from (2.12), we deduce

ρ_{1} ≤ D · \frac{ρ_{2}^{2} − ρ_{1}^{2}}{\frac{ρ_{2}^{2}−ρ_{1}^{2}}{2ρ_{1}}} ,

giving D ≥ \frac{1}{2}, and the theorem is completely proved.

The following corollary follows immediately. Corollary 2. Let x, y ∈ H with ⟨x, y⟩ ≠ 0 and ρ_{2} ≥ ρ_{1} > 0 such that

(2.14) ∥y∥ ρ_{1} ≤ ∥x∥ ∥y∥ − |⟨x, y⟩| ≤ ∥∥y∥ x − \frac{⟨x, y⟩}{∥y∥} · y∥ ≤ ρ_{2} ∥y∥ .

Then the following reverse of Schwarz’s inequality is valid:

(2.15)  (0 ≤) ∥x∥ ∥y∥ − |⟨x, y⟩| ≤ \frac{1}{2} (ρ_{2}^{2} − ρ_{1}^{2}) \begin{matrix}∥y∥ & 2 \\ | & ⟨x, y⟩ & | \end{matrix}.

The constant \frac{1}{2} is best possible.

3. Other Reverses Related to the Triangle Inequality

The following result holds.

Theorem 3. Let (H; ⟨·, ·⟩) be an inner product space and R ≥ 1. For x, y ∈ H, the following statements are equivalent:

(i) The ensuing reverse of the triangle inequality holds

(3.1)  ∥x∥ + ∥y∥ ≤ R ∥x + y∥;

(ii) The subsequent reverse of Schwarz’s inequality holds

(3.2)  (0 ≤) ∥x∥ ∥y∥ − Re ⟨x, y⟩ ≤ \frac{1}{2} (R^{2} − 1) ∥x + y∥^{2} .

The constant \frac{1}{2} is best possible in (3.2).

Proof. Taking the square in (3.1), we have, after some calculation,

(3.3) 2 ∥x∥ ∥y∥ ≤ (R^{2} − 1) ∥x∥^{2} + 2R^{2} Re ⟨x, y⟩ + (R^{2} − 1) ∥y∥^{2} .

Subtracting from both sides of (3.3) the quantity 2 Re ⟨x, y⟩ , we get

2 (∥x∥ ∥y∥ − Re ⟨x, y⟩) ≤ (R^{2} − 1) [∥x∥^{2} + 2 Re ⟨x, y⟩ + ∥y∥^{2}]
= (R^{2} − 1) ∥x + y∥^{2}

which is equivalent to (3.2).

To prove the best constant, let us assume that there exists a positive number E > 0 such that

(3.4)  0 ≤ ∥x∥ ∥y∥ − Re ⟨x, y⟩ ≤ E (R^{2} − 1) ,

provided x, y, R satisfy (3.1).

Assume R > 1 and choose x = \frac{1−R}{2}e, y = \frac{1+R}{2}e where e ∈ H with ∥e∥ = 1. Then

x + y = e, \frac{∥x∥ + ∥y∥}{R} = 1

and thus (3.1) holds with equality.

From (3.4), for the above choices for x and y, we have \frac{1}{2} (R^{2} − 1) ≤ D (R^{2} − 1) , which shows that D ≥ \frac{1}{2}.

The following corollary holds.

Corollary 3. Let x, y, e ∈ H with ∥e∥ = 1 and R ≥ 1 such that

(3.5)  ∥x∥ ≤ R Re ⟨x, e⟩ and ∥y∥ ≤ R Re ⟨y, e⟩ .

Then we have the inequality:

(3.6)  (0 ≤) ∥x∥ ∥y∥ − Re ⟨x, y⟩ ≤ \frac{1}{2} (R^{2} − 1) ∥x + y∥^{2} .

Proof. Adding the two inequalities in (3.5), we get

∥x∥ + ∥y∥ ≤ R Re ⟨x + y, e⟩ ≤ R |Re ⟨x + y, e⟩|
≤ R |⟨x + y, e⟩| ≤ R |⟨x + y, e⟩|
≤ R ∥x + y∥ ∥e∥ = R ∥x + y∥ .

Utilising the implication “(i) ⟹ (ii)” of Theorem 3, we deduce the desired inequality (3.6).

Remark 2. It is an open problem whether \frac{1}{2} is the best possible constant in (3.6). Preliminary investigations show that \frac{1}{4} may be the candidate for the best constant, but we do not have any analytic proof of this fact.

The following result also holds.

Theorem 4. Let (H; ⟨·, ·⟩) be an inner product space and r ∈ (0, 1], x, y ∈ H. The following statements are equivalent

(i) We have the reverse of the continuity of the norm inequality

(3.7)  |∥x∥ − ∥y∥| ≥ r ∥x − y∥;

(ii) We have the subsequent reverse of Schwarz’s inequality

(3.8)  0 ≤ ∥x∥ ∥y∥ − Re ⟨x, y⟩ ≤ \frac{1}{2} (1 − r^{2}) ∥x − y∥^{2} .

The constant \frac{1}{2} is best possible in (3.8).

Proof. Taking the square in (3.7), we have

∥x∥^{2} − 2 ∥x∥ ∥y∥ + ∥y∥^{2} ≥ r^{2} [∥x∥^{2} − 2 Re ⟨x, y⟩ + ∥y∥^{2}]

which is clearly equivalent to

(1 − r^{2}) [∥x∥^{2} − 2 Re ⟨x, y⟩ + ∥y∥^{2}] ≥ 2 (∥x∥ ∥y∥ − Re ⟨x, y⟩) ,

or with (3.8).

To prove the sharpness of the constant, assume that there exists a positive number F such that

(3.9)  0 ≤ ∥x∥ ∥y∥ − Re ⟨x, y⟩ ≤ F (1 − r^{2}) ∥x − y∥^{2} ,

provided x, y and r satisfy (3.7).

For r ∈ (0, 1) , consider the vectors

x = \frac{1 + r}{2}e, y = \frac{1 − r}{2}e, where e ∈ H, ∥e∥ = 1.

Then

|∥x∥ − ∥y∥| = r, ∥x − y∥ = 1,

and by (3.9), we get

\frac{1 − r^{2}}{4} − \frac{r^{2} − 1}{4} ≤ F (1 − r^{2}) ,

from where we deduce F ≥ \frac{1}{2}.

4. Discrete Inequalities

Assume that (K; (·, ·)) is a Hilbert space over the real or complex number field. Assume also that p_{i} ≥ 0, i ∈ ℕ with ∑_{i=1}^{∞} p_{i} = 1 and define

ℓ_{p}^{2} (K) := {x := (x_{i})_{i∈ℕ}| x_{i} ∈ 𝕂, i ∈ ℕ and ∑_{i=1}^{∞} p_{i} ∥x_{i}∥^{2} < ∞} .

It is well known that ℓ_{p}^{2} (K) endowed with the inner product ⟨·, ·⟩_{p} defined by

⟨x, y⟩_{p} := ∑_{i=1}^{∞} p_{i} (x_{i}, y_{i})

and generating the norm

∥x∥_{p} := (∑_{i=1}^{∞} p_{i} ∥x_{i}∥^{2})^{\frac{1}{2}}

is a Hilbert space over 𝕂.

We may state the following discrete inequality providing a reverse of the Cauchy-Bunyakovsky-Schwarz result.

Proposition 1. Assume that x, y ∈ ℓ_{p}^{2} (K) and ρ_{2} > ρ_{1} ≥ 0 such that

(4.1)  ∥x_{i} − y_{i}∥ ≤ ρ_{2} for each i ∈ ℕ

and

(4.2)  |(∑_{i=1}^{∞} p_{i} ∥x_{i}∥^{2})^{\frac{1}{2}} − (∑_{i=1}^{∞} p_{i} ∥y_{i}∥^{2})^{\frac{1}{2}} | ≥ ρ_{1}.

SCHWARZ INEQUALITY IN INNER PRODUCT SPACES 11

Then we have the inequalities

(0 ≤) (∑_{i=1}^{∞} p_{i} ∥x_{i}∥^{2})^{\frac{1}{2}} (∑_{i=1}^{∞} p_{i} ∥y_{i}∥^{2})^{\frac{1}{2}} − |∑_{i=1}^{∞} p_{i} (x_{i}, y_{i})| (4.3)

≤ (∑_{i=1}^{∞} p_{i} ∥x_{i}∥^{2})^{\frac{1}{2}} (∑_{i=1}^{∞} p_{i} ∥y_{i}∥^{2})^{\frac{1}{2}} − |∑_{i=1}^{∞} p_{i} Re (x_{i}, y_{i})|

≤ (∑_{i=1}^{∞} p_{i} ∥x_{i}∥^{2})^{\frac{1}{2}} (∑_{i=1}^{∞} p_{i} ∥y_{i}∥^{2})^{\frac{1}{2}} − ∑_{i=1}^{∞} p_{i} Re (x_{i}, y_{i})

≤ \frac{1}{2} (ρ_{2}^{2} − ρ_{1}^{2}) .

The constant \frac{1}{2} is best possible in (5.3). Proof. From (4.1) we have

∥x − y∥_{p} = (∑_{i=1}^{∞} p_{i} ∥x_{i} − y_{i}∥^{2})^{\frac{1}{2}} ≤ (∑_{i=1}^{∞} p_{i}ρ_{2}^{2})^{\frac{1}{2}} = ρ_{2}.

Since (4.2) is in fact the following inequality

|∥x∥ − ∥y∥| ≥ ρ_{1}

then, on applying Theorem 1 for ℓ_{p}^{2} (K) and ⟨·, ·⟩_{p} , we deduce the desired inequality (4.3). Corollary 4. Under the assumptions of Proposition 1, we have

(0 ≤) (∑_{i=1}^{∞} p_{i} ∥x_{i}∥^{2})^{\frac{1}{2}} + (∑_{i=1}^{∞} p_{i} ∥y_{i}∥^{2})^{\frac{1}{2}} − (∑_{i=1}^{∞} p_{i} ∥x_{i} + y_{i}∥^{2})^{\frac{1}{2}}

≤ \sqrt{ρ_{2}^{2} − ρ_{1}^{2}}. Remark 3. Similar results may be stated if one uses the other inequalities obtained in Sections 2 and 3, but we omit the details.

5. Integral Inequalities

Assume that (K; (·, ·)) is a Hilbert space over the real or complex number field 𝕂. If ρ : [a, b] ⊂ ℝ → [0, ∞) is a Lebesgue integrable function with ∫_{a}^{b} ρ (t) dt = 1, then we may consider the space L_{ρ}^{2} ([a, b] ; K) of all functions f : [a, b] → K, that are Bochner measurable and ∫_{a}^{b} ρ (t) ∥f (t)∥^{2} dt < ∞. It is well known that L_{ρ}^{2} ([a, b] ; K) endowed with the inner product ⟨·, ·⟩_{ρ} defined by

⟨f, g⟩_{ρ} := ∫_{a}^{b} ρ (t) (f (t) , g (t)) dt

and generating the norm

∥f∥_{ρ} := (∫_{a}^{b} ρ (t) ∥f (t)∥^{2} dt)^{\frac{1}{2}}

is a Hilbert space over 𝕂.

The following reverse of the vector-valued Cauchy-Bunyakovsky-Schwarz integral inequalities holds.

Proposition 2. Assume that f, g ∈ L_{ρ}^{2} ([a, b] ; K) and ρ_{2} > ρ_{1} ≥ 0 such that

(5.1)  ∥f (t) − g (t)∥ ≤ ρ_{2} for a.e. t ∈ [a, b]

and

(5.2)  |(∫_{a}^{b} ρ (t) ∥f (t)∥^{2} dt)^{\frac{1}{2}} − (∫_{a}^{b} ρ (t) ∥g (t)∥^{2} dt)^{\frac{1}{2}} | ≥ ρ_{1}.

Then we have the inequality

(5.3)  0 ≤ (∫_{a}^{b} ρ (t) ∥f (t)∥^{2} dt)^{\frac{1}{2}} (∫_{a}^{b} ρ (t) ∥g (t)∥^{2} dt)^{\frac{1}{2}}
− |∫_{a}^{b} ρ (t) (f (t) , g (t)) dt|

≤ (∫_{a}^{b} ρ (t) ∥f (t)∥^{2} dt)^{\frac{1}{2}} (∫_{a}^{b} ρ (t) ∥g (t)∥^{2} dt)^{\frac{1}{2}}

− |∫_{a}^{b} ρ (t) Re (f (t) , g (t)) dt|

≤ (∫_{a}^{b} ρ (t) ∥f (t)∥^{2} dt)^{\frac{1}{2}} (∫_{a}^{b} ρ (t) ∥g (t)∥^{2} dt)^{\frac{1}{2}}

− ∫_{a}^{b} ρ (t) Re (f (t) , g (t)) dt

≤ \frac{1}{2} (ρ_{2}^{2} − ρ_{1}^{2}) .

The constant \frac{1}{2} is best possible.

Proof. From (5.1), we have

∥f − g∥_{ρ} = (∫_{a}^{b} ρ (t) (∥f (t) − g (t)∥)^{2} dt)^{\frac{1}{2}} ≤ (∫_{a}^{b} ρ (t) ρ_{2}^{2}dt)^{\frac{1}{2}} = ρ_{2}.

Since (5.2) is in fact the inequality

|∥f∥_{ρ} − ∥g∥_{ρ}| ≥ ρ_{1},

then, on applying Theorem 1 for ℓ_{p}^{2} (K) on ⟨·, ·⟩_{ρ} , we deduce the desired inequality (5.3).

Corollary 5. Under the assumptions of Proposition 2, we have:

(0 ≤) (∫_{a}^{b} ρ (t) ∥f (t)∥^{2} dt)^{\frac{1}{2}} + (∫_{a}^{b} ρ (t) ∥g (t)∥^{2} dt)^{\frac{1}{2}}

− (∫_{a}^{b} ρ (t) ∥f (t) + g (t)∥^{2} dt)^{\frac{1}{2}}

≤ \sqrt{ρ_{2}^{2} − ρ_{1}^{2}}.

Remark 4. Similar results may be stated if one uses the other inequalities obtained in Sections 2 and 3, but we omit the details.

References

[1] S.S. DRAGOMIR, New reverses of Schwarz, triangle and Bessel inequalities in inner product spaces, RGMIA Res. Rep. Coll., 6(Supp.) (2003), Art. 20.

[ONLINE http://rgmia.vu.edu.au/v6(E).html]

[2] S.S. DRAGOMIR, Reverses of Schwarz, triangle and Bessel inequalities in inner product spaces, RGMIA Res. Rep. Coll., 6(Supp.) (2003), Art. 19. [ONLINE

http://rgmia.vu.edu.au/v6(E).html]

[3] S.S. DRAGOMIR, A counterpart of Schwarz’s inequality in inner product spaces, RGMIA Res. Rep. Coll., 6(Supp.) (2003), Art. 18. [ONLINE

http://rgmia.vu.edu.au/v6(E).html]

[4] S.S. DRAGOMIR, On Bessel and Grüss inequalities for orthornormal families in inner product spaces, RGMIA Res. Rep. Coll., 6(Supp.) (2003), Art. 12.

[ONLINE http://rgmia.vu.edu.au/v6(E).html]

[5] S.S. DRAGOMIR, Some Grüss type inequalities in inner product spaces, J. Inequal. in Pure and Appl. Math., 4(2) (2003), Art. 42. [ONLINE

http://jipam.vu.edu.au/article.php?sid=280]

[6] S.S. DRAGOMIR, A generalisation of the Cassels and Greub-Reinboldt inequalities in inner product spaces, Preprint Mathematics Archive, math. CA10307130.

[7] S.S. DRAGOMIR, Some companions of the Grüss inequality in inner product spaces, RGMIA Res. Rep. Coll., 6(Supp.) (2003), Art. 8. [ONLINE

http://rgmia.vu.edu.au/v6(E).html]

School of Computer Science and Mathematics, Victoria University of Technology, PO Box 14428, Melbourne VIC 8001, Australia.

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