A GRÜSS RELATED INTEGRAL INEQUALITY AND APPLICATIONS

S.S. DRAGOMIR

Abstract. A Grüss related integral inequality in the general setting of measurable spaces and abstract Lebesgue integrals is proven. Some particular inequalities and applications for Ostrowski and generalized Trapezoid inequality are mentioned as well.

1. Introduction

Let (Ω, A, μ) be a measurable space consisting of a set Ω, a σ – algebra of parts of Ω and a countably additive and positive measure μ on A with values in ℝ∪{∞} .

For a μ−measurable function w : Ω → ℝ, with w (x) ≥ 0 for μ – a.e. x ∈ Ω, assume ∫_{Ω} w (x) dμ (x) > 0. Consider the Lebesgue space L_{w} (Ω, μ) := {f : Ω → ℝ, f is μ−measurable and ∫_{Ω} w (x) |f (x)| dμ (x) < ∞}.

If f, g : Ω → ℝ are μ−measurable functions and f, g, fg ∈ L_{w} (Ω, μ) , then we may consider the Čebyšev functional

(1.1) T_{w} (f, g) := \frac{1}{∫_{Ω} w (x) dμ (x)} ∫_{Ω} w (x) f (x) g (x) dμ (x)

− \frac{1}{∫_{Ω} w (x) dμ (x)} ∫_{Ω} w (x) f (x) dμ (x)

× \frac{1}{∫_{Ω} w (x) dμ (x)} ∫_{Ω} w (x) g (x) dμ (x) .

The following result is known in the literature as the Grüss inequality

(1.2)  |T_{w} (f, g)| ≤ \frac{1}{4} (Γ − γ) (Δ − δ) ,

provided

(1.3)  −∞ < γ ≤ f (x) ≤ Γ < ∞, −∞ < δ ≤ g (x) ≤ Δ < ∞

for μ – a.e. x ∈ Ω.

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

Note that if Ω = {1, …, n} and μ is the discrete measure on Ω, then we obtain the discrete Grüss inequality

(1.4)  |\frac{1}{W_{n}} ∑_{i=1}^{n} w_{i}x_{i}y_{i} − \frac{1}{W_{n}} ∑_{i=1}^{n} w_{i}x_{i} · \frac{1}{W_{n}} ∑_{i=1}^{n} w_{i}y_{i}| ≤ \frac{1}{4} (Γ − γ) (Δ − δ) ,

Date: April 10, 2002.

1991 Mathematics Subject Classification. Primary 26D15; Secondary 26D10.

Key words and phrases. Grüss Inequality, Ostrowski Inequality, Generalized Trapezoid Inequality, Lebesgue Integral.

provided γ ≤ x_{i} ≤ Γ, δ ≤ y_{i} ≤ Δ for each i ∈ {1, …, n} and w_{i} ≥ 0 with W_{n} := ∑_{i=1}^{n} w_{i} > 0.

For some Grüss type inequalities, see the papers [1] – [11] where further references are provided.

2. Some Integral Inequalities

With the above assumption and if f ∈ L_{w} (Ω, μ) then we may define

(2.1)  D_{w} (f) := D_{w,1} (f)

:= \frac{1}{∫_{Ω} w (x) dμ (x)} ∫_{Ω} w (x)

× |f (x) − \frac{1}{∫_{Ω} w (y) dμ (y)} ∫_{Ω} w (y) f (y) dμ (y)| dμ (x) .

The following fundamental result holds [2]. Theorem 1. Let w, f, g : Ω → ℝ be μ−measurable functions with w ≥ 0 μ− a.e. on Ω and ∫_{Ω} w (y) dμ (y) > 0. If f, g, fg ∈ L_{w} (Ω, μ) and there exists the constants δ, Δ such that

(2.2)  −∞ < δ ≤ g (x) ≤ Δ < ∞ for μ − a.e. x ∈ Ω,

then we have the inequality

(2.3)  |T_{w} (f, g)| ≤ \frac{1}{2} (Δ − δ) D_{w} (f) .

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

For f ∈ L_{p,w} (Ω, A, μ) := {f : Ω → ℝ, ∫_{Ω} w (x) |f (x)|^{p} dμ (x) < ∞} , p ≥ 1 we may also define (see [2])

(2.4) D_{w,p} (f) := [\frac{1}{∫_{Ω} w (x) dμ (x)} ∫_{Ω} w (x)

× |f (x) − \frac{1}{∫_{Ω} w (y) dμ (y)} ∫_{Ω} w (y) f (y) dμ (y)|^{p} dμ (x)]^{\frac{1}{p}}

= \begin{matrix}∥f − \frac{1}{∫_{Ω} w(y)dμ(y)} ∫_{Ω} w (y) f (y) dμ (y)∥ & w,p \\ [∫_{Ω} w (x) dμ (x)] & \frac{1}{p} \end{matrix}

where ∥·∥_{w,p} is the usual p−norm on L_{p,w} (Ω, A, μ) , i.e.,

∥h∥_{w,p} := (∫_{Ω} w (x) |h (x)|^{p} dμ (x))^{\frac{1}{p}} , p ≥ 1.

Using Hölder’s inequality we get

(2.5)  D_{w,1} (f) ≤ D_{w,p} (f) for p ≥ 1, f ∈ L_{p,w} (Ω, A, μ) ;

and, in particular for p = 2 (2.6)

D_{w,1} (f) ≤ D_{w,2} (f) = [\frac{∫_{Ω} w (x) f^{2} (x) dμ (x)}{∫_{Ω} w (x) dμ (x)} − (\frac{∫_{Ω} w (x) f (x) dμ (x)}{∫_{Ω} w (x) dμ (x)})^{2}]^{\frac{1}{2}} ,

if f ∈ L_{2,w} (Ω, A, μ) .

For f ∈ L_{∞} (Ω, A, μ) := {f : Ω → ℝ, ∥f∥_{Ω,∞} := ess sup_{α∈Ω} |f (x)| < ∞} we also

have

(2.7)  D_{w,p} (f) ≤ D_{w,∞} (f)

:= ∥f − \frac{1}{∫_{Ω} w (y) dμ (y)} ∫_{Ω} w (y) f (y) dμ (y)∥_{Ω,∞} .

The following corollary may be useful in practice.

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

|T_{w} (f, g)| (2.8)

≤ \frac{1}{2} (Δ − δ) D_{w} (f)

≤ \frac{1}{2} (Δ − δ) D_{w,p} (f) if f ∈ L_{p} (Ω, A, μ) , p > 1;

≤ \frac{1}{2} (Δ − δ) · ∥f − \frac{1}{∫_{Ω} w (y) dμ (y)} ∫_{Ω} w (y) f (y) dμ (y)∥_{Ω,∞} .

Remark 1. The inequalities in (2.8) are in order of increasing coarseness. If we assume that −∞ < γ ≤ f (x) ≤ Γ < ∞ for μ – a.e. x ∈ Ω, then by the Grüss inequality for g = f we obviously deduce for p = 2

(2.9)  [\frac{∫_{Ω} w (x) f^{2} (x) dμ (x)}{∫_{Ω} w (x) dμ (x)} − (\frac{∫_{Ω} w (x) f (x) dμ (x)}{∫_{Ω} w (x) dμ (x)})^{2}]^{\frac{1}{2}} ≤ \frac{1}{2} (Γ − γ)

and then, by (2.8), we deduce the sequence for inequality

|T_{w} (f, g)| (2.10)

≤ \frac{1}{2} (Δ − δ) \frac{1}{∫_{Ω} w (x) dμ (x)} ∫_{Ω} w (x)

× |f (x) − \frac{1}{∫_{Ω} w (y) dμ (y)} ∫_{Ω} w (y) f (y) dμ (y)| dμ (x)

≤ \frac{1}{2} (Δ − δ) [\frac{∫_{Ω} w (x) f^{2} (x) dμ (x)}{∫_{Ω} w (x) dμ (x)} − (\frac{∫_{Ω} w (x) f (x) dμ (x)}{∫_{Ω} w (x) dμ (x)})^{2}]^{\frac{1}{2}}

≤ \frac{1}{4} (Δ − δ) (Γ − γ) ;

for f, g : Ω → ℝ , μ – measurable functions and so that −∞ < γ ≤ f (x) < Γ < ∞, −∞ < δ ≤ g (x) ≤ Δ < ∞ for μ – a.e. x ∈ Ω. Thus, the inequality (2.4) is a refinement of Grüss’ inequality (1.2).

It is well known that if f ∈ L_{2,w} (Ω, A, μ) , then the following Schwartz’s type inequality holds:

(2.11)  \frac{1}{∫_{Ω} w (x) dμ (x)} ∫_{Ω} w (x) f^{2} (x) dμ (x) ≥ (∫_{Ω} w (x) f (x) dμ (x))^{2} .

Using the above results, we may point out the following result.

Proposition 1. Assume that the μ−measurable function f : Ω → ℝ satisfies the assumption:

(2.12)  −∞ < γ ≤ f (x) ≤ Γ < ∞ for a.e. x ∈ Ω.

Then one has the inequality

(2.13)  0 ≤ \frac{1}{∫_{Ω} w (x) dμ (x)} ∫_{Ω} w (x) f^{2} (x) dμ (x)

− (\frac{1}{∫_{Ω} w (x) dμ (x)} ∫_{Ω} w (x) f (x) dμ (x))^{2}

≤ \frac{1}{2} (Γ − γ) \frac{1}{∫_{Ω} w (x) dμ (x)} ∫_{Ω} w (x)

× |f (x) − \frac{1}{∫_{Ω} w (y) dμ (y)} ∫_{Ω} w (y) f (y) dμ (y)| dμ (x)

(≤ \frac{1}{4} (Γ − γ)^{2}) .

The constant \frac{1}{2} is sharp.

The proof follows by the inequality (2.5) for g = f.

In [9], the author obtained the following companion of Grüss inequality.

Theorem 2. Assume that ρ, f, g : Ω → ℝ are Lebesgue μ measurable on Ω with ρ, ρf, ρg, ρfg ∈ L (Ω, μ). If ρ (x) ≥ 0 for μ – a.e. x ∈ Ω with ∫_{Ω} ρ (x) dμ (x) > 0 and there exists the real constants m, M, n, N with the property that

(2.14) −∞ < m ≤ f (x) ≤ M < ∞, −∞ < n ≤ g (x) ≤ N < ∞ for μ – a.e. x ∈ Ω

then we have the inequality

|\frac{1}{∫_{Ω} ρ (x) dμ (x)} ∫_{Ω} ρ (x) f (x) g (x) dμ (x) (2.15)

−\frac{m + M}{2} · \frac{1}{∫_{Ω} ρ (x) dμ (x)} ∫_{Ω} ρ (x) g (x) dμ (x)

−\frac{n + N}{2} · \frac{1}{∫_{Ω} ρ (x) dμ (x)} ∫_{Ω} ρ (x) f (x) dμ (x) + \frac{m + M}{2} · \frac{n + N}{2} |

≤ \frac{1}{4} (M − m) (N − n) .

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

The following corollary is a natural consequence of this theorem.

Corollary 2. Assume that ρ, f : Ω → ℝ are Lebesgue μ – measurable on Ω with ρ, ρf, ρf^{2} ∈ L (Ω, μ). If ρ (x) ≥ 0 for μ – a.e. x ∈ Ω with ∫_{Ω} ρ (x) dμ (x) > 0 and there exists the real constants m, M with the property that

(2.16)  −∞ < m ≤ f (x) ≤ M < ∞, for μ – a.e. x ∈ Ω

then we have the inequality

(2.17) 0 ≤ \frac{1}{∫_{Ω} ρ (x) dμ (x)} ∫_{Ω} ρ (x) f^{2} (x) dμ (x)

− (\frac{1}{∫_{Ω} ρ (x) dμ (x)} ∫_{Ω} ρ (x) f (x) dμ (x))^{2}

≤ \frac{1}{4} (M − m)^{2} − (\frac{1}{∫_{Ω} ρ (x) dμ (x)} ∫_{Ω} ρ (x) f (x) dμ (x) − \frac{m + M}{2} )^{2}

(≤ \frac{1}{4} (M − m)^{2}) .

The constant \frac{1}{4} is best possible in (2.17).

The main aim of this paper is to establish a refinement of the inequality (2.15) in the spirit of the refinement provided for the Grüss inequality incorporated in Theorem 1. Applications for Ostrowski and Generalized Trapezoid inequalities will be provided as well.

3. New Integral Inequalities

We start to the following lemma that is interesting in itself as well. Lemma 1. Assume that ρ, h, l : Ω → ℝ are Lebesgue μ – measurable on Ω with ρ, ρh, ρl, ρhl ∈ L (Ω, μ). If ρ (x) ≥ 0 for μ-a.e. x ∈ Ω with ∫_{Ω} ρ (x) dμ (x) > 0 and there exists the real constants a, A with the property that

(3.1) −∞ < a ≤ h (x) ≤ A < ∞, −∞ < b ≤ l (x) ≤ B < ∞ for μ-a.e. x ∈ Ω

and

(3.2)  ∫_{Ω} ρ (x) l (x) dμ (x) = ∫_{Ω} ρ (x) h (x) dμ (x) = 0,

then we have the inequality

(3.3) |\frac{1}{∫_{Ω} ρ (x) dμ (x)} ∫_{Ω} ρ (x) h (x) l (x) dμ (x) + \frac{a + A}{2} · \frac{b + B}{2} |

≤ \frac{1}{2} (A − a) \frac{1}{∫_{Ω} ρ (x) dμ (x)} ∫_{Ω} ρ (x) |l (x) − \frac{b + B}{2} | dμ (x) .

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

Proof. Firstly, let observe, by the assumption (3.2), that

(3.4) ∫_{Ω} ρ (x) (h (x) − \frac{a + A}{2}) (l (x) − \frac{b + B}{2} ) dμ (x)

= ∫_{Ω} ρ (x) h (x) l (x) dμ (x) + \frac{a + A}{2} · \frac{b + B}{2} ∫_{Ω} ρ (x) dμ (x) .

On the other hand, by (3.1), we have

|h (x) − \frac{a + A}{2}| ≤ \frac{A − a}{2}, for μ − a.e. x ∈ Ω

and thus

|∫_{Ω} ρ (x) (h (x) − \frac{a + A}{2}) (l (x) − \frac{b + B}{2} ) dμ (x)|

≤ ∫_{Ω} ρ (x) |h (x) − \frac{a + A}{2}| |l (x) − \frac{b + B}{2} | dμ (x)
≤ \frac{A − a}{2} · ∫_{Ω} ρ (x) |l (x) − \frac{b + B}{2} | dμ (x) .

Now, using the equality (3.4), the inequality (3.5) and dividing by ∫_{Ω} ρ (x) dμ (x) > 0, we deduce the desired inequality (3.3).

To prove the sharpness of the constant \frac{1}{2}, we assume that (3.3) holds with a constant C > 0, i.e.,

(3.5) |\frac{1}{∫_{Ω} ρ (x) dμ (x)} ∫_{Ω} ρ (x) h (x) l (x) dμ (x) + \frac{a + A}{2} · \frac{b + B}{2} |

≤ C (A − a) \frac{1}{∫_{Ω} ρ (x) dμ (x)} · ∫_{Ω} ρ (x) |l (x) − \frac{b + B}{2} | dμ (x) .

If we consider the functions h, l : [α, β] ⊂ ℝ → ℝ , l = h and

h (x) = \{ \begin{matrix}−1 & if x ∈ [α, \frac{α+β}{2} ] \\ 1 & if x ∈ [\frac{α+β}{2} , β] , \end{matrix}

then for ρ (x) = 1, we have a = b = −1, A = B = 1

∫_{α}^{β} h (x) dx = ∫_{α}^{β} l (x) dx = 0,
∫_{α}^{β} h (x) l (x) dx = 1;

and

\frac{1}{β − α} · ∫_{α}^{β} ρ (x) |l (x) − \frac{b + B}{2} | d (x) = 1

and thus, by (3.3), we deduce C ≥ \frac{1}{2}.

The above lemma gives us the opportunity to state the following companion of Grüss inequality.

Theorem 3. Assume that ρ, f, g : Ω → ℝ are Lebesgue μ measurable on Ω with ρ, ρf, ρg, ρfg ∈ L (Ω, μ). If ρ (x) ≥ 0 for μ-a.e. x ∈ Ω with ∫_{Ω} ρ (x) dμ (x) > 0 and there exists the real constants m, M, n, N with the property that

−∞ < m ≤ f (x) ≤ M < ∞, −∞ < n ≤ g (x) ≤ N < ∞ for μ-a.e. x ∈ Ω

GRÜSS INEQUALITY 7

then we have the inequality

|\frac{1}{∫_{Ω} ρ (x) dμ (x)} ∫_{Ω} ρ (x) f (x) g (x) dμ (x) (3.6)

−\frac{m + M}{2} · \frac{1}{∫_{Ω} ρ (x) dμ (x)} ∫_{Ω} ρ (x) g (x) dμ (x)

−\frac{n + N}{2} · \frac{1}{∫_{Ω} ρ (x) dμ (x)} ∫_{Ω} ρ (x) f (x) dμ (x) + \frac{m + M}{2} · \frac{n + N}{2} |

≤ \frac{1}{2} (M − m) \frac{1}{∫_{Ω} ρ (x) dμ (x)} · ∫_{Ω} ρ (x) |g (x) − \frac{n + N}{2} | dμ (x) .

The constant \frac{1}{2} is best possible. Proof. If we choose in Lemma 1

h (x) = f (x) − \frac{1}{∫_{Ω} ρ (y) dμ (y)} ∫_{Ω} ρ (y) f (y) dμ (y) ,
l (x) = g (x) − \frac{1}{∫_{Ω} ρ (y) dμ (y)} ∫_{Ω} ρ (y) g (y) dμ (y) ;

where x ∈ Ω, and

a = m − \frac{1}{∫_{Ω} ρ (y) dμ (y)} ∫_{Ω} ρ (y) f (y) dμ (y) ,
A = M − \frac{1}{∫_{Ω} ρ (y) dμ (y)} ∫_{Ω} ρ (y) f (y) dμ (y) ,
b = n − \frac{1}{∫_{Ω} ρ (y) dμ (y)} ∫_{Ω} ρ (y) g (y) dμ (y) ;
B = N − \frac{1}{∫_{Ω} ρ (y) dμ (y)} ∫_{Ω} ρ (y) gdμ (y) ;

then a simple calculation will reveal that

\frac{1}{∫_{Ω} ρ (x) dμ (x)} ∫_{Ω} ρ (x) h (x) l (x) dμ (x)

= \frac{1}{∫_{Ω} ρ (x) dμ (x)} ∫_{Ω} ρ (x) f (x) g (x) dμ (x) − \frac{1}{∫_{Ω} ρ (x) dμ (x)} ∫_{Ω} ρ (x) f (x) dμ (x)

· \frac{1}{∫_{Ω} ρ (x) dμ (x)} ∫_{Ω} ρ (x) g (x) dμ (x)

and

\frac{a + A}{2} = \frac{m + M}{2} − \frac{1}{∫_{Ω} ρ (y) dμ (y)} ∫_{Ω} ρ (y) f (y) dμ (y)
\frac{b + B}{2} = \frac{n + N}{2} − \frac{1}{∫_{Ω} ρ (y) dμ (y)} ∫_{Ω} ρ (y) gdμ (y)

and since

A − a = M − m, B − b = N − n

then by (3.3) we deduce the desired inequality (2.15). The sharpness of the constant follows from Lemma 1 and we omit the details.

For f ∈ L_{p,ρ} (Ω, A, μ) := {f : Ω → ℝ, ∫_{Ω} ρ (x) |f (x)|^{p} dμ (x) < ∞} , p ≥ 1 we may also define the quantities

F_{ρ,p} (f; m, M) := [\frac{1}{∫_{Ω} ρ (x) dμ (x)} ∫_{Ω} ρ (x) |f (x) − \frac{m + M}{2} |^{p} dμ (x)]^{\frac{1}{p}}

= \begin{matrix}∥f − \frac{m+M}{2} ∥ & ρ,p \\ [∫_{Ω} ρ (x) dμ (x)] & \frac{1}{p} \end{matrix}

where ∥·∥_{ρ,p} is the usual p−norm on L_{p,ρ} (Ω, A, μ) , i.e.,

∥h∥_{ρ,p} := (∫_{Ω} ρ (x) |h (x)|^{p} dμ (x))^{\frac{1}{p}} , p ≥ 1.

Using Hölder’s inequality we get

F_{ρ,1} (f; m, M) ≤ F_{ρ,p} (f; m, M) for p ≥ 1, f ∈ L_{p,ρ} (Ω, A, μ) ;

For f ∈ L_{∞} (Ω, A, μ) := {f : Ω → ℝ, ∥f∥_{Ω,∞} := ess sup_{α∈Ω} |f (x)| < ∞} we also have

F_{ρ,p} (f; m, M) ≤ F_{ρ,∞} (f; m, M) := ∥f − \frac{m + M}{2} ∥_{ρ,∞} .

The following corollary may be useful in practice.

Corollary 3. With the assumptions of Theorem 3, we have

|\frac{1}{∫_{Ω} ρ (x) dμ (x)} ∫_{Ω} ρ (x) f (x) g (x) dμ (x)

− \frac{m + M}{2} · \frac{1}{∫_{Ω} ρ (x) dμ (x)} ∫_{Ω} ρ (x) g (x) dμ (x)

−\frac{n + N}{2} · \frac{1}{∫_{Ω} ρ (x) dμ (x)} ∫_{Ω} ρ (x) f (x) dμ (x) + \frac{m + M}{2} · \frac{n + N}{2} |

≤ \frac{1}{2} (M − m) F_{ρ} (g; n, N) if f ∈ L_{1,ρ} (Ω, A, μ)

≤ \frac{1}{2} (M − m) F_{ρ,p} (g; n, N) if f ∈ L_{p,ρ} (Ω, A, μ)

≤ \frac{1}{2} (M − m) F_{ρ,∞} (g; n, N) ≤ \frac{1}{4} (M − m) (N − n) if f ∈ L_{p,ρ} (Ω, A, μ) .

The following corollary providing a counterpart for Schwarz’s inequality, is a natural consequence of this theorem

Corollary 4. Assume that ρ, f : Ω → ℝ are Lebesgue μ measurable on Ω with ρ, ρf, ρf^{2} ∈ L (Ω, μ). If ρ (x) ≥ 0 for μ – a.e. x ∈ Ω with ∫_{Ω} ρ (x) dμ (x) > 0 and there exists the real constants m, M with the property that

−∞ < m ≤ f (x) ≤ M < ∞, for μ – a.e. x ∈ Ω

GRÜSS INEQUALITY 9

then we have the inequality

(3.7) 0 ≤ \frac{1}{∫_{Ω} ρ (x) dμ (x)} ∫_{Ω} ρ (x) f^{2} (x) dμ (x)

− (\frac{1}{∫_{Ω} ρ (x) dμ (x)} ∫_{Ω} ρ (x) f (x) dμ (x))^{2}

≤ \frac{1}{2} (M − m) \frac{1}{∫_{Ω} ρ (x) dμ (x)} · ∫_{Ω} ρ (x) |f (x) − \frac{m + M}{2} | dμ (x)

− (\frac{1}{∫_{Ω} ρ (x) dμ (x)} ∫_{Ω} ρ (x) f (x) dμ (x) − \frac{m + M}{2} )^{2} .

The constant \frac{1}{2} is best possible in (3.7).

4. Particular Inequalities

The following particular inequalities are of interest and may be used in applications.

(1) Let f, g : [a, b] → ℝ be Lebesgue measurable functions. If f, g, fg ∈ L [a, b] ,

where L [a, b] is the usual Lebsgue space, and (4.1) − ∞ < m ≤ f (x) ≤ M < ∞, −∞ < n ≤ g (x) ≤ N < ∞

for μ − a.e. x ∈ [a, b] ,

then we have the inequalities

|\frac{1}{b − a} ∫_{a}^{b} f (x) g (x) dx − \frac{n + N}{2} · \frac{1}{b − a} ∫_{a}^{b} f (x) dx

−\frac{m + M}{2} · \frac{1}{b − a} ∫_{a}^{b} g (x) dx + \frac{m + M}{2} · \frac{n + N}{2} |

≤ \frac{1}{2} (M − m) \frac{1}{b − a} ∫_{a}^{b} |g (x) − \frac{n + N}{2} | dx
≤ \frac{1}{2} (M − m) [\frac{1}{b − a} ∫_{a}^{b} |g (x) − \frac{n + N}{2} |^{p} dx]^{\frac{1}{p}} , p > 1
≤ \frac{1}{2} (M − m) ess sup_{x∈[a,b]} |g (x) − \frac{n + N}{2} |
≤ \frac{1}{4} (M − m) (N − n) .

The constant \frac{1}{2} in front of (M − m) is sharp in the inequality (4.2).

The following counterpart of Schwartz’s inequality holds

(4.2)  0 ≤ \frac{1}{b − a} ∫_{a}^{b} f^{2} (x) dx − (\frac{1}{b − a} ∫_{a}^{b} f (x) dx)^{2}

≤ \frac{1}{2} (M − m) \frac{1}{b − a} ∫_{a}^{b} |f (x) − \frac{m + M}{2} | dx

− (\frac{1}{b − a} ∫_{a}^{b} f (x) dx − \frac{m + M}{2} )^{2}

(≤ \frac{1}{4} (M − m)^{2}) ,

provided −∞ < m ≤ f (x) ≤ M < ∞ for a.e. x ∈ [a, b] . The constant \frac{1}{4} is sharp.

2. Let ā = (a_{1}, …, a_{n}) , b̄ = (b_{1}, …, b_{n}) , p̄ = (p_{1}, …, p_{n}) be n−tuples of real numbers with p_{i} ≥ 0 (i ∈ {1, …, n}) and ∑_{i=1}^{n} p_{i} = 1. If

(4.3)  a ≤ a_{i} ≤ A, b ≤ b_{i} ≤ B, i ∈ {1, …, n} ,

then one has the inequality

|∑_{i=1}^{n} p_{i}a_{i}b_{i} − \frac{b + B}{2} · ∑_{i=1}^{n} p_{i}a_{i} − \frac{a + A}{2} · ∑_{i=1}^{n} p_{i}b_{i} + \frac{b + B}{2} · \frac{a + A}{2}| (4.4)

≤ \frac{1}{2} (A − a) ∑_{i=1}^{n} p_{i} |b_{i} − \frac{b + B}{2} |

≤ \frac{1}{2} (A − a) [∑_{i=1}^{n} p_{i} |b_{i} − \frac{b + B}{2} |^{p}]^{\frac{1}{p}} , p > 1

≤ \frac{1}{2} (A − a) max_{i=1,...,n} |b_{i} − \frac{b + B}{2} | ≤ \frac{1}{4} (A − a) (B − b) .

The constant \frac{1}{2} in front of (A − a) is sharp in all the above inequalities. The following counterpart of Schwartz’s inequality also holds

(4.5)  0 ≤ ∑_{i=1}^{n} p_{i}a_{i}^{2} − (∑_{i=1}^{n} p_{i}a_{i})^{2}

≤ \frac{1}{2} (A − a) ∑_{i=1}^{n} p_{i} |b_{i} − \frac{b + B}{2} | − (∑_{i=1}^{n} p_{i}a_{i} − \frac{a + A}{2})^{2}

(≤ \frac{1}{4} (A − a)^{2}) ,

provided a ≤ a_{i} ≤ A for each i ∈ \{1, …, n\} . The constant \frac{1}{2} in front of (A − a) is sharp in the second inequality (4.5).

5. Applications for Ostrowski’s Inequality

If φ : [a, b] → ℝ is an absolutely continuous function on [a, b] such that φ′ ∈ L_{∞} [a, b] , then the following inequality is known in the literature as Ostrowski’s inequality

(5.1) |φ (x) − \frac{1}{b − a} ∫_{a}^{b} φ (t) dt|

≤ [\frac{1}{4} + (\frac{x − \frac{a+b}{2}}{b − a} )^{2}] ∥φ′∥_{∞} (b − a) , x ∈ [a, b] ,

where ∥φ′∥_{∞} := ess sup_{α∈[a,b]} |φ′ (x)| . The constant \frac{1}{4} is best possible.

A simple proof of this fact, may be accomplished by the use of the identity

(5.2)  φ (x) = \frac{1}{b − a} ∫_{a}^{b} φ (t) dt + \frac{1}{b − a} ∫_{a}^{b} K (x, t) φ′ (t) dt,

where the kernel K : [a, b]^{2} → ℝ is defined by

(5.3)  K (x, t) := \{ \begin{matrix}t − a & if a ≤ t ≤ x \\ t − b & if a ≤ x < t ≤ b. \end{matrix}

We will now use the following inequality

|\frac{1}{b − a} ∫_{a}^{b} f (x) g (x) dx − \frac{n + N}{2} · \frac{1}{b − a} ∫_{a}^{b} f (x) dx

−\frac{m + M}{2} · \frac{1}{b − a} ∫_{a}^{b} g (x) dx + \frac{m + M}{2} · \frac{n + N}{2} |

≤ \frac{1}{2} (M − m) \frac{1}{b − a} ∫_{a}^{b} |g (x) − \frac{n + N}{2} | dx

provided −∞ < m ≤ g (x) ≤ M < ∞, −∞ < n ≤ g (x) ≤ N < ∞ for a.e. x ∈ [a, b] and f ∈ L_{∞} [a, b] , to obtain the next result concerning a perturbed version of Ostrowski’s inequality.

Theorem 4. Assume that φ : [a, b] → ℝ is an absolutely continuous function on [a, b] such that φ′ : [a, b] → ℝ satisfies the condition

(5.4)  −∞ < γ ≤ φ′ (x) ≤ Γ < ∞ for a.e. x ∈ [a, b] .

Then we have the inequality

(5.5)  |φ (x) − \frac{1}{b − a} ∫_{a}^{b} φ (t) dt − (x − \frac{a + b}{2}) [φ; a, b]| ≤ \frac{1}{8} (b − a) (Γ − γ)

for any x ∈ [a, b] , where [φ; a, b] = \frac{φ(b)−φ(a)}{b−a} is the divided difference. The constant \frac{1}{8} is best possible in the sense that it cannot be replaced by a smaller one.

Proof. We apply inequality (5.4) for the choices f (t) = K (x, t) defined by (5.3), g (t) = φ′ (t) , t ∈ [a, b] to get

(5.6) |\frac{1}{b − a} ∫_{a}^{b} K (x, t) φ′ (t) dt − \frac{γ + Γ}{2} · \frac{1}{b − a} ∫_{a}^{b} K (x, t) dt

− (x − \frac{a + b}{2}) \frac{1}{b − a} ∫_{a}^{b} φ′ (t) dt + \frac{γ + Γ}{2} (x − \frac{a + b}{2})|

≤ \frac{1}{2} (Γ − γ) · \frac{1}{b − a} ∫_{a}^{b} |K (x, t) − (x − \frac{a + b}{2})| dt

since, obviously,

x − b ≤ K (x, t) ≤ x − a.

We obviously have:

\frac{1}{b − a} ∫_{a}^{b} φ′ (t) dt = \frac{φ (b) − φ (a)}{b − a}.

Also (see [2])

(5.7) I (x) := \frac{1}{b − a} ∫_{a}^{b} |K (x, t) − (x − \frac{a + b}{2})| dt

= \frac{1}{b − a} [∫_{a}^{x} |t − a − x + \frac{a + b}{2}| dt + ∫_{x}^{b} |t − b − x + \frac{a + b}{2}| dt]

= \frac{1}{b − a} [∫_{a}^{x} |t − x + \frac{b − a}{2}| dt + ∫_{x}^{b} |t − x + \frac{b − a}{2}| dt] .

Straight forward substitution of u = t − x + \frac{b−a}{2} and v = t − x − \frac{b−a}{2} gives

I (x) = \frac{1}{b − a} [∫\frac{a+b}{2} −x^{\frac{b−a}{2}} |u| du + ∫_{− \frac{b−a}{2}}^{\frac{a+b}{2} −x} |v| dv]

= \frac{1}{b − a} ∫− \frac{b−a}{2}^{\frac{b−a}{2}} |u| du = \frac{2}{b − a} ∫0^{\frac{b−a}{2}} udu = \frac{b − a}{4}.

Substitution of the above into (5.7) produces (5.5).

Remark 2. The above inequality was firstly proved in a different manner in [3].

6. Application for Generalised Trapezoid Inequality

If φ : [a, b] → ℝ is an absolutely continuous function on [a, b] so that φ′ ∈ L_{∞} [a, b] , then the following inequality is known as the generalised trapezoid inequality

(6.1) |(x − a) φ (a) + (b − x) φ (b) − ∫_{a}^{b} φ (t) dt|

≤ [\frac{1}{4} (b − a)^{2} + (x − \frac{a + b}{2})^{2}] ∥φ′∥_{∞}

for any x ∈ [a, b] . The constant \frac{1}{4} is best possible.

A simple proof of this fact may be done by using the identity

(6.2)  ∫_{a}^{b} φ (t) dt = (x − a) φ (a) + (b − x) φ (b) + ∫_{a}^{b} (x − t) φ′ (t) dt.

Utilising the inequality (5.4) we may point out the following perturbed version of (6.1). Theorem 5. Assume that φ : [a, b] → ℝ is an absolutely continuous function on [a, b] so that φ′ : [a, b] → ℝ satisfies the condition (5.4). Then we have the inequality

(6.3) |\frac{1}{b − a} ∫_{a}^{b} φ (t) dt

− [(\frac{x − a}{b − a}) φ (a) + (\frac{b − x}{b − a}) φ (b)] − (x − \frac{a + b}{2}) [φ; a, b]|

≤ \frac{1}{8} (b − a) (Γ − γ)

for any x ∈ [a, b] . The constant \frac{1}{8} is best possible in the sense that it cannot be replaced by a smaller one.

Proof. We apply inequality (5.4) for the choices f (t) = x−t, g (t) = φ′ (t) , t ∈ [a, b] , to get

(6.4) |\frac{1}{b − a} ∫_{a}^{b} (x − t) φ′ (t) dt − \frac{γ + Γ}{2} \frac{1}{b − a} ∫_{a}^{b} (x − t) dt

− (x − \frac{a + b}{2}) \frac{1}{b − a} ∫_{a}^{b} φ′ (t) dt + \frac{γ + Γ}{2} (x − \frac{a + b}{2})|

≤ \frac{1}{2} (Γ − γ) \frac{1}{b − a} ∫_{a}^{b} |(x − t) − (x − \frac{a + b}{2})| dt.

Since

\frac{1}{b − a} ∫_{a}^{b} (x − t) dt = (x − \frac{a + b}{2}) ,
\frac{1}{b − a} ∫_{a}^{b} φ′ (t) dt = \frac{1}{b − a} ∫_{a}^{b} φ′ (t) dt

and

\frac{1}{b − a} ∫_{a}^{b} |(x − t) − (x − \frac{a + b}{2})| dt = \frac{1}{b − a} ∫_{a}^{b} |x − t − x + \frac{a + b}{2}| dt

= \frac{1}{b − a} ∫_{a}^{b} |t + \frac{a + b}{2}| dt

= \frac{b − a}{4},

from (6.4) we deduce the desired inequality (6.3).

Remark 3. The above inequality (6.3) was firstly proved in [2] by a different argument. In the same paper the authors have shown that the constant \frac{1}{8} is sharp in the sense that it cannot be replaced by a smaller constant.

References

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[2] Cerone, P. and Dragomir, S.S. (2002), A refinement of Grüss’ inequality and applications, (Preprint) RGMIA Res. Rep. Coll., 5(2), Article 14. (ONLINE:

http://rgmia.vu.edu.au/v5n2.html)

[3] Cheng, X.L. and Sun, Jie (2002), A note on the perturbed trapezoid inequality, J. Ineq. Pure & Appl. Math., 3, No. 2, Article 29. [ON LINE: http://jipam.vu.edu.au/v3n2].

[4] Dragomir, S.S. and Booth, G.L. (2000), On a Grüss-Lupaş type inequality and its application for the estimation of p−moments of guessing mappings, Math. Comm., 5, 117-126.

[5] Dragomir, S.S. (2002), Another Grüss type inequality for sequences of vectors in normed linear spaces and applications, J. Comp. Analysis & Appl., 4(2), 157-172.

[6] Dragomir, S.S. (2002), A Grüss type inequality for sequences of vectors in normed linear spaces, (Preprint) RGMIA Res. Rep. Coll., 5(2), Article 9. (ONLINE:

http://rgmia.vu.edu.au/v5n2.html)

[7] Dragomir, S. S. (2001), Integral Grüss inequality for mappings with values in Hilbert spaces and applications. J. Korean Math. Soc. 38, no. 6, 1261–1273.

[8] Dragomir, S. S. (1999), A generalization of Grüss’s inequality in inner product spaces and applications. J. Math. Anal. Appl. 237 , no. 1, 74–82

[9] Dragomir, S.S. (2002), A companion of the Grüss inequality and applications, RGMIA Res. Rep. Coll., 5(Supp), Article 13. (ONLINE: http://rgmia.vu.edu.au/v5(E).html)

[10] Fink, A. M. (1999), A treatise on Grüss’ inequality. Analytic and geometric inequalities and applications, 93–113, Math. Appl., 478, Kluwer Acad. Publ., Dordrecht,

[11] Pečarić, J. (1980), On some inequalities analogous to Grüss inequality. Mat. Vesnik 4(17)(32), no. 2, 197–202.

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