SOME GRÜSS’ TYPE INEQUALITIES IN INNER PRODUCT SPACES

S.S. DRAGOMIR

Abstract. Some new Grüss type inequalities in inner product spaces and applications for integrals are given.

1. Introduction

In [1], the author has proved the following Grüss’ type inequality in real or complex inner product spaces. Theorem 1. 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

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

hold, then we have the inequality

(1.2)  |⟨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 can not be replaced by a smaller constant.

Some particular cases of interest for integrable functions with real or complex values and the corresponding discrete versions are listed bellow. Corollary 1. Let f, g : [a, b] → 𝕂 (𝕂 = ℝ,ℂ) be Lebesgue integrable and so that

(1.3)  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

(1.4) |\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 2. Let x, y ∈𝕂^{n} and φ, γ, Φ, Γ are real or complex numbers so that

Date: February 21, 2003.

1991 Mathematics Subject Classification. Primary 26D15, 46D05 Secondary 46C99.

Key words and phrases. Grüss’ Inequality, Inner products, Integral inequalities, Discrete Inequalities.

(1.5)  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

(1.6)  |\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 1, see the recent paper [2].

In the present paper we show that the condition (1.1) may be replaced by an equivalent but simpler assumption and a new proof of Theorem 1 is produced. A refinement of the Grüss’ type inequality (1.2) , some companions and applications for integrals are pointed out as well.

2. An Equivalent Assumption

The following lemma holds. Lemma 1. 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 3. 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 1. 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 also holds.

Lemma 2. Let x, e ∈ H with ∥e∥ = 1. Then one has the following representation

(2.1)  0 ≤ ∥x∥^{2} − |⟨x, e⟩|^{2} = inf_{λ∈𝕂} ∥x − λe∥^{2} .

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

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

Taking the infimum in (2.2) 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 (2.1) is proved.

We are able now to provide a different proof for the Grüss’ type inequality in inner product spaced mentioned in Introduction, than the one from paper [1]. Theorem 2. 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 (1.1) hold, or, equivalently, the following assumptions

(2.3)  ∥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

(2.4)  |⟨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 [1]) that

(2.5)  |⟨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 2 and the conditions (2.3) 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 (2.5) the desired inequality (2.4) is obtained.

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

3. A Refinement of Grüss Inequality

The following result improving (1.1) holds

Theorem 3. 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 (1.1), or, equivalently, (2.3) hold, then we have the inequality

(3.1) |⟨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}} .

Proof. As in [1], we have

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

(3.3) ∥x∥^{2} − |⟨x, e⟩|^{2} = Re [(Φ − ⟨x, e⟩) (\overline{⟨x, e⟩} − \overline{φ})] − Re ⟨Φe − x, x − φe⟩

and

(3.4) ∥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

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

and

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

Consequently, by (3.2) − (3.6) we may state that

(3.7) |⟨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 (3.1) .

4. Some Companion Inequalities

The following companion of Grüss inequality in inner product spaces holds. Theorem 4. Let (H, ⟨., .⟩) be an inner product space over 𝕂 (𝕂 = ℝ,ℂ) and e ∈ H, ∥e∥ = 1. If γ, Γ ∈ 𝕂 and x, y ∈ H are so that

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

or, equivalently,

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

then we have the inequality

(4.3)  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 well known inequality

(4.4)  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 (4.4) we may write

(4.5)  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

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

Making use of (4.5) and (4.6) we deduce (4.3) .

The fact that \frac{1}{4} is the best possible constant in (4.3) follows by the fact that if in (4.1) we choose x = y, then it becomes Re ⟨Γe − x, x − γe⟩ ≥ 0, implying 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 4. Let (H, ⟨., .⟩) be an inner product space over 𝕂 (𝕂 = ℝ,ℂ) and e ∈ H, ∥e∥ = 1. If γ, Γ ∈ 𝕂 and x, y ∈ H are so that

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

or, equivalently,

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

then we have the inequality

(4.9)  |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)

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

or, equivalently,

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

implies

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

In both inequalities (4.9) and (4.12) , 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 4, we get

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

showing that

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

Making use of (4.3) and (4.13) we deduce the desired result (4.9) .

Finally, we may state and proof the following dual result as well Proposition 1. Let (H, ⟨., .⟩) be an inner product space over 𝕂 (𝕂 = ℝ,ℂ) and e ∈ H, ∥e∥ = 1. If φ, Φ ∈ 𝕂 and x, y ∈ H are so that

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

then we have the inequalities

(4.15)  ∥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 (3.3))

(4.16) ∥x∥^{2} − |⟨x, e⟩|^{2} = Re [(Φ − ⟨x, e⟩) (\overline{⟨x, e⟩} − \overline{φ})] + Re ⟨x − Φe, x − φe⟩ .

Using the assumption (4.14) and the fact that

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

by (4.16) we deduce the first inequality in (4.15) .

The second inequality in (4.15) 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.

5. Integral Inequalities

Let (Ω, Σ, μ) be a measure space consisting of a set Ω, a σ−algebra of parts Σ and a countably additive and positive measure μ 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

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

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

or, equivalently

(5.2)  (∫_{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 refinement of Grüss integral inequality

(5.3) |∫_{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)} − φ\overline{h (s)})] dμ (s)

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

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

The proof follows by Theorem 3 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 2. It is obvious that a sufficient condition for (5.1) to hold is

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

and

Re [(Γh (s) − g (s)) (\overline{g (s)} − 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 5. If z, Z, t, T ∈ 𝕂, ρ ∈ L (Ω, ℝ) , μ (Ω) < ∞ and f, g ∈ L^{2} (Ω, 𝕂) are such that:

(5.4)  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

(5.5)  |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

GRÜSS’ TYPE INEQUALITIES 9

then we have the inequality

(5.6) |\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 4 we may state the following result as well. Proposition 3. If f, g, h ∈ L^{2} (Ω, 𝕂) and γ, Γ ∈ 𝕂 are such that ∫_{Ω} |h (s)|^{2} dμ (s) = 1 and

(5.7) ∫_{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,

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

then we have the inequality

(5.9)  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 (5.7) and (5.8) hold with “ ± ” instead of “ + ”, then

(5.10)  |I| ≤ \frac{1}{4} |Γ − γ|^{2} .

Remark 3. It is obvious that a sufficient condition for (5.7) to hold is

(5.11)  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

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

(5.13)  Re [(Z − \frac{f (s) + g (s)}{2}) (\frac{\overline{f (s)} + \overline{g (s)}}{2} − z)] ≥ 0 for a.e. s ∈ Ω

or, equivalently

(5.14)  |\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 (5.13) and (5.14) hold with “ ± ” instead of “ + ”, then

(5.15)  |J| ≤ \frac{1}{4} |Z − z|^{2} .

Remark 4. 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.

References

[1] S.S. Dragomir, A generalization of Grüss’ inequality in inner product spaces and applications, J. Math. Anal. Appl., 237(1999), 74-82.

[2] S.S. Dragomir and I. Gomm, Some integral and discrete versions of the Grüss inequality for real and complex functions and sequences, RGMIA Res. Rep. Coll., 5(2003), No. 3, Article 9

   [ON LINE http://rgmia.vu.edu.au/v5n3.html]

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

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