Ostrowski and Ostrowski-Gruss type inequalities for vector-valued functions with k points via a parameter

Adv. Inequal. Appl. 2018, 2018:10 https://doi.org/10.28919/aia/3621 ISSN: 2050-7461

OSTROWSKI AND OSTROWSKI–GR ¨USS TYPE INEQUALITIES FOR

VECTOR-VALUED FUNCTIONS WITH k POINTS VIA A PARAMETER

SETH KERMAUSUOR

Department of Mathematics and Computer Science, Alabama State University, Montgomery, AL 36101, USA

Copyright c2018 Seth Kermausuor. This is an open access article distributed under the Creative Commons Attribution License, which permits

unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited.

Abstract. In this paper, we provide an extension of the Ostrowski’s inequality for vector-valued functions fork

points via a parameter. We also provide a sharp Ostrowski-Gr¨uss type inequality for vector valued functions fork

points. Our results generalize some of the results for the real-valued case in the literature.

Keywords:Montgomery’s identity; Ostrowski’s inequality; Gr¨uss inequality; Bochner integral; Banach spaces.

2010 AMS Subject Classification:26D15, 26D10, 46B25.

1. Introduction

Throughout this paper,(X,k · k)denotes a Banach space over the real numbers.

Definition 1.1.X is said to be a Radon-Nikodym space if every absolutely continuousX-valued

function is almost everywhere differentiable.

It is well-known that every reflexive space is a Radon-Nikodym space. For example every

Hilbert space is a Radon-Nikodym space. However, the spaceL₁([0,1])of all integrable

func-tions defined on the interval[0,1], endowed with the norm

kgk1:= Z 1

0

|g(t)|dt

E-mail address: [email protected] Received December 19, 2017

is not a Radon-Nikodym space.

Definition 1.2. Let a,b∈_R,a<band f :[a,b]→X be anX-valued function. f is said to be

Bochner integrable if and only if the real-valued functionkfk:[a,b]→_Rdefined bykfk(t):=

kf(t)k, is Lebesgue integrable on[a,b].

We will denote the Lebesgue integral of a real-valued function gby Rb

ag(t)dt and denote the

Bochner integral of anX-valued function f by(B)Rb

a f(t)dt.

The integration by parts formula holds for the Bochner integral under the following conditions.

Theorem 1.1. Leta,b∈_R,a<b,g:[a,b]→_Rand f :[a,b]→X. Suppose f andgare both

differentiable, and the functionsg f0andg0f are Bochner integrable on[a,b], then

(B)

Z b

a

g(t)f0(t)dt=g(b)f(b)−g(a)f(b)−(B)

Z b

a

g0(t)f(t)dt

Theorem 1.2.If f :[a,b]→X is Bochner integrable then

(B)

Z b

a

f(t)dt

≤ Z b

a

kf(t)kdt.

For more information on the Bochner integral we refer the reader to [4] and [8].

In 1938, the Ukranian mathematician Alexander Ostrowski [11] obtained an inequality known

in the literature as Ostrowski’s inequality to provide a bound for the difference between a

real-valued function and its integral mean. This result is stated in the following theorem;

Theorem 1.3. Let f :[a,b]→ _R be continuous on [a,b] and differentiable in (a,b) and its derivative f0:(a,b)→_Ris bounded in(a,b). IfM:=sup_t∈(a,b)|f0(t)|<∞, then we have

f(x)− 1

b−a Z b

a

f(t)dt

≤ 1

4+

x−a+b

2 2

(b−a)2 !

(b−a)M (1.1)

for allx∈[a,b].The inequality is sharp in the sense that the constant 1₄ cannot be replaced by

a smaller one.

This inequality has been studied and extended in several different ways by many authors over

the past years. For more information about the Ostrowski inequality and its associates, we refer

the reader to [1, 2, 5, 6, 7, 9, 10, 12]. In 2002, Barnett et al [2] extended Theorem 1.3 for

Theorem 1.4. Let (X,k · k) be a Banach space with the Radon-Nikodym property and f :

[a,b]→X an absolutely continuous function on[a,b]with the property that f0∈L_∞([a,b],X),

i.e f0:[a,b]→X and

k|f0|k_[a,b],∞:=ess sup

t∈[a,b]

kf0(t)k<∞

Then we have the inequality

f(s)− 1 b−a(B)

Z b

a

f(t)dt

≤ 1

4+

s−a+₂b2

(b−a)2 !

(b−a)k|f0|k_[a,b],∞ (1.2)

for alls∈[a,b].

The purpose of this paper is to extend Theorem 1.4 for kpoints via a parameter which will be

done in Section 2. In Section 3, we provide a sharp Ostrowski-Gr¨uss type inequality for

vector-valued functions by using a sharp Gr¨uss inequality obtained by Barnett et al in [3]. Our results

extends some of the results in the literature for real-valued functions to the case of vector-valued

functions. In addition, we consider some particular cases as examples.

2. Ostrowski inequality for vector-valued functions for

k

points

To prove our main results, we obtained the following generalized Montgomery identity for

vector-valued functions. In what follows, we assume that (X,k · k) has the Radon-Nikodym

property. We first define the following kernel function which was first provided by Xu and Fang

[12];

Definition 2.1. Let

(1) a,b∈_R, λ ∈[0,1], Ik:a=x0<x1<· · ·<xk−1<xk=bis a partition of the interval

[a,b],

(2) αi∈R (i=0,1,· · ·,k+1) isk+2 points so that α0=a, αi∈[xi−1,xi] (i=1,· · ·,k)

we define the kernel functionK(,I_k):[a,b]→_Ras follows;

K(t,I_k) =

                    

                   

t−α1−λα1₂−a

, t∈[a,α1),

t−α1+λα2−₂α1

, t∈[α1,x1),

t−α2−λα2−₂α1

, t∈[x₁,α2),

.. .

t−αk−1+λαk−₂αk−1

, t∈[αk−1,xk−1),

t−αk−λαk−₂αk−1

, t∈[x_k−1,αk),

t−αk+λαk+1₂−αk

, t∈[αk,b],

(2.1)

for allt ∈[a,b].

Lemma 2.1. [Montgomery Identity for vector-valued functions with k points.] Suppose that

f :[a,b]→X is an absolutely continuous function on [a,b]. Then for any λ ∈[0,1], we have

the identity

(B)

Z b

a

K(t,I_k)f0(t)dt= (1−λ)

k

∑

(αi+1−αi)f(xi)

+λ

2

k

∑

(αi+1−αi)

f(αi) + f(αi+1)

−(B)

Z b

a

f(t)dt

(2.2)

whereK(·,I_k)is given in Definition 2.1.

Proof.First, we observe that

(B)

Z b

a

K(t,I_k)f0(t)dt =

k−1

∑

h

(B)

Z αi+1

xi

(t−(αi+1−λ

αi+1−αi

2 ))f

0_(t)dt

+ (B)

Z x_i+1

αi+1

(t−(αi+1+λ

αi+2−αi+1

2 ))f

0_(t)dti_.

(B)

Z b

a

K(t,I_k)f0(t)dt =

k−1

∑

(αi+1−(αi+1−λ

αi+1−αi

2 ))f(αi+1)

−(x_i−(αi+1−λ

αi+1−αi

2 ))f(xi)−(B)

Z αi+1

xi

f(t)dt

+ (xi+1−(αi+1+λαi+2−αi+1

2 ))f(xi+1)

−(αi+1−(αi+1+λ

αi+2−αi+1

2 ))f(αi+1)−(B)

Z x_i+1

αi+1

f(t)dt

=

k−1

∑

λαi+1−αi

2 f(αi+1)−(xi−αi+1)f(xi)−λ

αi+1−αi

2 f(xi) + (xi+1−αi+1)f(xi+1)

−λαi+2−αi+1

2 f(xi+1) +λ

αi+2−αi+1

2 f(αi+1)−(B)

Z x_i+1

xi

f(t)dt

.

It follows that,

(B)

Z b

a

K(t,I_k)f0(t)dt =

k−1

∑

λαi+2−αi

2 f(αi+1)−(B)

Z x_i+1

xi

f(t)dt

+

k−1

∑

−(x_i−αi+1)f(xi) + (xi+1−αi+1)f(xi+1)

+

k−1

∑

−λαi+1−αi

2 f(xi)−λ

αi+2−αi+1

2 f(xi+1)

=

k−1

∑

λαi+2−αi

2 f(αi+1)−(B)

Z b

a

f(t)dt

−x₀f(x₀) +x_kf(x_k) +

k−1

∑

αi+1(f(xi)−f(xi+1))

+

k−1

∑

−λ

2

(αi+1−αi)f(xi) + (αi+2−αi+1)f(xi+1)

.

That is,

(B)

Z b

a

K(t,Ik)f0(t)dt = k−1

∑

λαi+2−αi

2 f(αi+1)−(B)

Z b

a

f(t)dt

+ (α1−a)f(a) + (b−αk)f(b) + k−1

∑

(αi+1−αi)f(xi)

−λ

2

(α1−a)f(a) + (b−αk)f(b) +2 k−1

∑

=

k−1

∑

λαi+2−αi

2 f(αi+1)−(B)

Z b

a

f(t)dt

+ (1−λ)

k−1

∑

(αi+1−αi)f(xi) + (1−

λ

2)

(α1−a)f(a) + (b−αk)f(b)

.

(2.3)

Now, consider the following

k−1

∑

(αi+2−αi)f(αi+1) = k−1

∑

(αi+2−αi+1)f(αi+1) + k−1

∑

(αi+1−αi)f(αi+1)

=

k

∑

(αi+1−αi)f(αi) + k−1

∑

(αi+1−αi)f(αi+1)

=

k

∑

(αi+1−αi)f(αi)−(α1−α0)f(α0) + k

∑

(αi+1−αi)f(αi+1)−(αk+1−αk)f(αk+1)

=

k

∑

(αi+1−αi)(f(αi) + f(αi+1))−

(α1−α0)f(α0) + (αk+1−αk)f(αk+1)

.

So,

k−1

∑

λ

2(αi+2−αi)f(αi+1) =

λ

2

k

∑

(αi+1−αi)

f(αi) + f(αi+1)

−λ

2

(α1−a)f(a) + (b−αk)f(b)

.

(2.4)

Substituting (2.4) in (2.3) gives the identity

(B)

Z b

a

K(t,I_k)f0(t)dt= (1−λ)

k

∑

(αi+1−αi)f(xi)

+λ

2

k

∑

(αi+1−αi)

f(αi) + f(αi+1)

−(B)

Z b

a

f(t)dt.

This completes the proof.

Corollary 2.1.If we takeλ =0 in Lemma 2.1, we have the identity

(B)

Z b

a

K(t,I_k)f0(t)dt =

k

∑

(αi+1−αi)f(xi)−(B) Z b

a

where

K(t,I_k) =

            

           

t−α1, t∈[a,x1),

t−α2, t∈[x1,x2),

.. .

t−αk−1, t∈[xk−2,xk−1),

t−α_k, t∈[x_k−1,b].

(2.6)

Corollary 2.2. Let a,b∈_R,a<band let f :[a,b]→X be an absolutely continuous function

on[a,b]. Then for anyx∈[a,b]we have the identity

1

b−a(B) Z b

a

K(t,x)f0(t)dt= (1−λ)f(x) +λ

2

f(a) +f(b)− 1 b−a(B)

Z b

a

f(t)dt (2.7)

where

K(t,x) =

  

 

t−a+λb−₂a

, t∈[a,x),

t−b−λb−₂a

, t∈[x,b].

(2.8)

Proof. The proof follows directly from Lemma 2.1 by takingk=2,x₀=α0=α1=a,x1=x

andx₂=α2=α3=b.

Theorem 2.1. Suppose f satisfies the conditions of Lemma 2.1, and f0∈L_∞([a,b],X). Then we have the inequality

k(1−λ)

k

∑

(αi+1−αi)f(xi) +

λ

2

k

∑

(αi+1−αi)

f(αi) + f(αi+1)

−(B)

Z b

a

f(t)dtk

≤M k−1

∑

h1

4

x_i+1−xi 2

+αi+1−

xi+1+xi

2

2i

+λ

2

4

h

(b−αk)2−(α1−a)2

+2

k−1

∑

αi+1−αi 2i

+λ

2

k−1

∑

h

(αi+1−xi+1)(αi+2−αi+1)−(αi+1−xi)(αi+1−αi)

i

(2.9)

where

M=ess sup

t∈[a,b]

kf0(t)k<∞.

Z b

a

|K(t,Ik)|dt= k−1

∑

i=0 hZ αi+1

xi

t−(αi+1−λ

αi+1−αi

2 ) dt+

Z x_i+1

αi+1

t−(αi+1+λ

αi+2−αi+1 2 ) dt i =

k−1

∑

i=0

hZ αi+1−λαi+1₂−αi

xi

(αi+1−λαi+1−αi

2 )−t

dt+

Z αi+1

αi+1−λαi+1₂−αi

t−(αi+1−λαi+1−αi

2 )

dt

+

Z αi+1+λαi+2−₂αi+1

αi+1

(αi+1+λαi+2−αi+1

2 )−t

dt+

Z x_i+1

αi+1+λαi+2−₂αi+1

t−(αi+1+λαi+2−αi+1

2 ) dt i =1 2

k−1

∑

i=0 h

(αi+1−λαi+1−αi

2 )−xi 2

+λ 2

4

αi+1−αi

2

+(αi+1+λ

αi+2−αi+1

2 )−xi+1 2

+λ 2

4

αi+2−αi+1 2i

=1 2

k−1

∑

i=0 h

(αi+1−xi

2

−λ(αi+1−xi)(αi+1−αi) + λ2

2

αi+1−αi

2

+

(αi+1−xi+1 2

+λ(αi+1−xi+1)(αi+2−αi+1) + λ2

2

αi+2−αi+1 2i

=1 2

k−1

∑

i=0 h

αi+1−xi

2

+αi+1−xi+1 2

+λ 2

2

αi+1−αi

2 +λ

2

2

αi+2−αi+1 2

−λ(αi+1−xi)(αi+1−αi) +λ(αi+1−xi+1)(αi+2−αi+1) i

.

That is,

Z b

a

|K(t,Ik)|dt=

1 2

k−1

∑

i=0 h

αi+1−xi

2 +

αi+1−xi+1 2i

+λ 2

4

k−1

∑

i=0 h

αi+1−αi

2

+αi+2−αi+1 2i

+λ 2

k−1

∑

i=0 h

(αi+1−xi+1)(αi+2−αi+1)−(αi+1−xi)(αi+1−αi)

i

.

By applying the parallelogram law for numbers on the terms in the first sum and rearranging the terms

in the second sum, we have

Z b

a

|K(t,Ik)|dt= k−1

∑

i=0 h1

4

xi+1−xi

2

+αi+1−

xi+1+xi

2 2i +λ 2 4 h

(b−αk)2−(α1−a)2

+2

k−1

∑

i=0

αi+1−αi

2i

+λ 2

k−1

∑

i=0 h

(αi+1−xi+1)(αi+2−αi+1)−(αi+1−xi)(αi+1−αi)

i

.

Now from Lemma 2.1 we have

(1−λ) k

∑

i=0

(αi+1−αi)f(xi) + λ

2

k

∑

i=0

(αi+1−αi)

f(αi) +f(αi+1)

−(B)

Z b

a

f(t)dt

= (B)

Z b

a

K(t,Ik)f0(t)dt.

(2.11)

By taking the norm on both sides of (2.11) and applying Theorem 1.2, we obtain

k(1−λ) k

∑

i=0

(αi+1−αi)f(xi) + λ

2

k

∑

i=0

(αi+1−αi)

f(αi) +f(αi+1)

−(B)

Z b

a

f(t)dtk

≤M Z b

a

|K(t,Ik)|dt

(2.12)

where

M=ess sup

t∈[a,b]

kf0(t)k<∞

The desired inequality follows by substituting (2.10) in (2.12).

Corollary 2.3.If we takeλ =0 in Theorem 2.1, then we have the inequality

k

∑

i=0

(αi+1−αi)f(xi)−(B)

Z b

a

f(t)dt

≤M

k−1

∑

i=0 h1

4

xi+1−xi

2

+αi+1−

xi+1+xi

2 2i

(2.13)

where

M=ess sup

t∈[a,b]

kf0(t)k<∞.

Remark 2.1.Ifk=1 in Corollary 2.3, then we obtain Theorem 1.4.

Corollary 2.4. Leta,b∈R,a<band let f :[a,b]→X be an absolutely continuous function on [a,b]

such that f0∈L_∞([a,b],X). Then for anyx∈[a,b]we have the inequality

(1−λ)f(x) +λ 2

f(a) +f(b)− 1

b−a(B) Z b

a

f(t)dt

≤M

1

4(b−a)((1−λ) 2₊

λ2) +

x−a+b

2 2

(2.14)

where

M=ess sup

t∈[a,b]

kf0(t)k<∞.

Proof. The proof follows directly from Theorem 2.1 by taking k=2,x0=α0=α1 =a,x1=x and x₂=α2=α3=b.

The following lemma is a particular case of Theorem 2 in [3].

Lemma 3.1.Leta,b∈Rwitha<b. Ifg:[a,b]→Xis Bochner integrable on[a,b]and there existv∈X

andr>0 such thatg(x)∈B(v,r):={y∈X:ky−vk ≤r}for a.ex∈[a,b]andα:[a,b]→Ra Lebesgue

integrable function withαgBochner integrable on[a,b], then we have the sharp inequality

(B)

Z b

a

α(t)g(t)dt− 1

b−a Z b

a

α(t)dt·(B) Z b

a

g(t)dt ≤r Z b a

α(t)− 1

b−a Z b

a

α(s)ds

dt.

(3.1)

Theorem 3.1.Leta,b∈_Rwitha<b, If f:[a,b]→X be absolutely continuous on[a,b]such that f0is Bochner integrable on[a,b]. If there existv∈Xandr>0 such thatf0(x)∈B(v,r):={y∈X:ky−vk ≤r}

for a.ex∈[a,b], then we have the sharp inequality

(1−λ) k

∑

i=0

(αi+1−αi)f(xi) + λ

2

k

∑

i=0

(αi+1−αi)

f(αi) +f(αi+1)

−(B)

Z b

a

f(t)dt

− 1

8(b−a)

k−1

∑

i=0

λ2

αi+1−αi

2

−2xi−λ αi+ (λ−2)αi+1 2

+2xi+1−λ αi+2+ (λ−2)αi+1 2

−λ2

αi+2−αi+1 2

f(b)−f(a) ≤r Z b a

K(t,Ik)−

1 8(b−a)

k−1

∑

i=0

λ2

αi+1−αi

2

−2xi−λ αi+ (λ−2)αi+1 2

+2xi+1−λ αi+2+ (λ−2)αi+1 2

−λ2

αi+2−αi+1 2

dt.

(3.2)

Proof.By applying Lemma 3.1 to the functionsα(t):=K(t,Ik)andg(t):= f0(t)we have

(B)

Z b

a

K(t,Ik)f0(t)dt−

1 b−a

Z b

a

K(t,Ik)dt·(B)

Z b

a

f0(t)dt ≤r Z b a

K(t,Ik)−

1 b−a

Z b

a

K(s,Ik)ds

dt.

(3.3)

But we observe that

(B)

Z b

a

f0(t)dt= f(b)−f(a)(3.4)

and

Z b

a

K(t,Ik)dt=

1 8

k−1

∑

i=0

λ2

αi+1−αi

2

−2xi−λ αi+ (λ−2)αi+1 2

+

2xi+1−λ αi+2+ (λ−2)αi+1 2

−λ2

αi+2−αi+1 2

.

From Lemma 2.1, we have

(B)

Z b

a

K(t,Ik)f0(t)d= (1−λ) k

∑

i=0

(αi+1−αi)f(xi)t

+λ 2

k

∑

i=0

(αi+1−αi)

f(αi) +f(αi+1)

−(B)

Z b

a

f(t)dt

(3.6)

The desired inequality is obtained from (3.3) by using (3.4)–(3.6).

Corollary 3.1.Leta,b∈_R,a<b,f :[a,b]→Xbe an absolutely continuous function on[a,b]and f0is Bochner integrable. If there exist vectorsv,V ∈X such that

f0(x)−1

2(v+V)

≤1

2kV−vka.e x∈[a,b]

then we have the inequality

(1−λ) k

∑

i=0

(αi+1−αi)f(xi) + λ

2

k

∑

i=0

(αi+1−αi)

f(αi) +f(αi+1)

−(B)

Z b

a

f(t)dt

− 1

8(b−a)

k−1

∑

i=0

λ2

αi+1−αi

2

−2xi−λ αi+ (λ−2)αi+1 2

+

2xi+1−λ αi+2+ (λ−2)αi+1 2

−λ2

αi+2−αi+1 2

f(b)−f(a)

≤1

2kV−vk Z b a

K(t,Ik)−

1 8(b−a)

k−1

∑

i=0

λ2

αi+1−αi

2

−2xi−λ αi+ (λ−2)αi+1 2

+2xi+1−λ αi+2+ (λ−2)αi+1 2

−λ2

αi+2−αi+1 2

dt.

(3.7)

Corollary 3.2.If we takeλ =0 in Theorem 3.1, then we have the inequality

k

∑

i=0

(αi+1−αi)f(xi)−(B)

Z b

a

f(t)dt

− 1

2(b−a)

b2−a2− k−1

∑

i=0

αi+1

xi+1−xi

2

f(b)−f(a) ≤r Z b a

K(t,Ik)−

1 2(b−a)

b2−a2− k−1

∑

i=0

αi+1

xi+1−xi

2

dt,

(3.8)

whereK(·,Ik)is given by (2.6).

X:ky−vk ≤r}for a.ex∈[a,b], then for anyx∈[a,b], we have the sharp inequality

(1−λ)f(x) +λ

2

f(a) +f(b)

− 1

b−a(B) Z b

a

f(t)dt

− 1

8(b−a)

2x−λa+ (λ−2)b

2

−2x−λb+ (λ−2)a

2

f(b)−f(a)

≤r Z b

a

K(t,x)− 1

8(b−a)

2x−λa+ (λ−2)b 2

−2x−λb+ (λ−2)a 2

dt.

(3.8)

whereK(·,x)is given by (2.8).

Proof. The proof follows directly from Theorem 3.1 by taking k=2,x0=α0=α1 =a,x1=x and x₂=α2=α3=b.

4. Conclusion

Two main results have been established. The first result extends a result of Barnet et al in [2] fork

points via a parameterλ ∈[0,1]. In the second part, a sharp Ostrowski–G¨uss type inequality for vector valued functions has been provided. Some particular cases have been considered as examples. These

results generalizes the results in the literature for vector-valued functions.

Conflict of Interests

The author declares that there is no conflict of interests.

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