# On Right Caputo Fractional Ostrowski Inequalities Involving Three Functions

Authors: Deepak B. Pachpatte
Journal: International Journal of Analysis
 (2013)

---

## Abstract

We establish Ostrowski inequalities involving three functions in right Caputo fractional derivative inLp spaces.


---

## Body

## 1. Introduction

In 1938, Ostrowski proved the following useful inequality.Letf:[a,b]→R be continuous on [a,b] and differentiable on (a,b) whose derivative f′:[a,b]→R is bounded on (a,b), that is, ∥f′∥∞=supt∈(a,b)|f′(t)|<∞. Then
(1)|f(x)-1b-a∫abf(t)dt|≤[14+(x-a+b2)(b-a)2]∥f′∥∞,
for any x∈[a,b]. The constant 1/4 is best possible.In [1, 2] Pachpatte has proved Ostrowski inequality in three independent variables.In past few years many authors have obtained various generalisation and variant of the above type of inequality and other on fractional as well as time scale calculus see [3–6].Here we give some basic definition from fractional calculus used in [7–9].Definition 1.
Letf∈L1[a,b], α>0. The right and left Riemann-Liouville integrals Ia+αf(x) and Ib-αf(x) of order α>0 with a≥0 are defined by
(2)Ia+αf(x)=1Γ(α)∫ax(x-t)(α-1)f(t)dt,x>a,Ib-αf(x)=1Γ(α)∫xb(t-x)(α-1)f(t)dt,x<b,
respectively, where Γ(α)=∫0∞e-ttα-1dt and Ia+0f(x)=Ib-0f(x)=f(x).Definition 2 (see [10, page 2]).
Letf∈ACm([a,b]) (f(m-1) be in AC([a,b)), m∈N, m=[α], α>0 (|·| the ceiling of the number). We define the right Caputo fractional derivative of order α>0 by
(3)Db-αf(x)=(-1)mΓ(m-α)∫xb(t-x)m-α-1f(m)(t)dt,x≤b.
Ifα=m∈N, then
(4)Db-mf(x)=(-1)mf(m)(x),∀x∈[a,b].
Ifx>b, we define Db-mf(x)=0.Definition 3 (see [9, page 74]).
Letf,g∈ACm([a,b]), α>0 with a≥0, and then the fractional integral Ib-αf(x)g(x) is defined by
(5)Ib-αf(x)g(x)=1Γ(α)∫xb(t-x)(α-1)f(t)g(t)dt,x<b.
We give here the theorems proved in [10].Theorem 4.
Letα>0, m=[α], f∈ACm([a,b]). Assume that fk(b)=0, k=1,…,m-1 and Db-af∈L∞([a,b]). Then
(6)|1b-a∫abf(x)dx-f(b)|≤∥Db-αf∥∞,[a,b]Γ(α+2)(b-a)α.Theorem 5.
Letα>0, m=[α], f∈ACm([a,b]). Assume that fk(b)=0, k=1,…,m-1 and Db-af∈L1[a,b]. Then
(7)|1b-a∫abf(x)dx-f(b)|≤∥Db-αf∥L1([a,b])Γ(α+1)(b-a)α-1.Theorem 6.
Letp,q>1; 1/p+1/q=1, α>1-1/p, m=[α], f∈ACm([a,b]). Assume that fk(b)=0,k=1,…,m-1, and Db-αf,h∈Lq[a,b]. Then
(8)|1b-a∫abf(x)dx-f(b)|≤∥Db-αf∥Lq([a,b])Γ(α)(p(α-1)+1)1/p(α+1/p)(b-a)α-1+1/p.


## 2. Main Results

Our main results are given in the following theorems.Theorem 7.
Letα>0, m=[α], f,g,h∈ACm([a,b]). Assume that fk(b)=gk(b)=hk(b)=0, k=1,…,m-1 and Db-af,Db-ag,Db-ah∈L∞([a,b]). Then
(9)|3∫abf(x)g(x)h(x)dx-∫ab[f(b)g(x)h(x)+f(x)g(b)h(x)+f(x)g(x)h(b)]dx{∫abf(x)g(x)h(x)dx}|≤∥Db-αf∥∞Iαα+1|g(b)h(b)|+∥Db-αg∥∞Iαα+1|f(b)h(b)|+∥Db-αh∥∞Iαα+1|f(b)g(b)|.Proof.
Letx∈[a,b] we have
(10)f(x)-f(b)=1γα∫xb(t-x)(α-1)Db-αf(t)dt,(11)g(x)-g(b)=1γα∫xb(t-x)(α-1)Db-αg(t)dt,(12)h(x)-h(b)=1γα∫xb(t-x)(α-1)Db-αh(t)dt.
Multiplying (10), (11), and (12) by g(x)h(x), h(x)f(x), and f(x)g(x), respectively, and adding them, we have
(13)3f(x)g(x)h(x)-f(b)g(x)h(x)-f(x)g(b)h(x)-f(x)g(x)h(b)=g(x)h(x)Γ(γ)∫xb(t-x)α-1Db-αf(t)dt+h(x)f(x)Γ(γ)∫xb(t-x)α-1Db-αg(t)dt+f(x)g(x)Γ(γ)∫xb(t-x)α-1Db-αh(t)dt.
Integrating both sides of (13) with respect to x and rewriting above equation we have
(14)3∫abf(x)g(x)h(x)dx-∫ab[f(b)g(x)h(x)+f(x)g(b)h(x)+f(x)g(x)h(b)]dx=∫abg(x)h(x)Γ(γ)(∫xb(t-x)α-1Db-αf(t)dt)+∫abh(x)f(x)Γ(γ)(∫xb(t-x)α-1Db-αg(t)dt)+∫abf(x)g(x)Γ(γ)(∫xb(t-x)α-1Db-αh(t)dt).
From (14) and using the properties of modulus we have
(15)|3∫abf(x)g(x)h(x)dx-∫ab[f(b)g(x)h(x)+f(x)g(b)h(x)+f(x)g(x)h(b)]dx{∫abf(x)g(x)h(x)dx}|=∫ab|g(x)h(x)|Γ(γ)(∫xb(t-x)α-1|Db-αf(t)dt|)+∫ab|h(x)f(x)|Γ(γ)(∫xb(t-x)α-1|Db-αg(t)dt|)+∫ab|f(x)g(x)|Γ(γ)(∫xb(t-x)α-1|Db-αh(t)dt|).
It is easy to observe that
(16)|3∫abf(x)g(x)h(x)dx-∫ab[f(b)g(x)h(x)+f(x)g(b)h(x)+f(x)g(x)h(b)]dx{∫abf(x)g(x)h(x)dx}|≤∥Db-αf∥∞γα∫ab|g(x)h(x)|(∫xb(t-x)α-1dt)dx+∥Db-αg∥∞γα∫ab|h(x)f(x)|(∫xb(t-x)α-1dt)dx+∥Db-αh∥∞γα∫ab|f(x)g(x)|(∫xb(t-x)α-1dt)dx=∥Db-αf∥∞Γ(α+1)∫ab(b-x)α|g(x)h(x)|dx+∥Db-αg∥∞Γ(α+1)∫ab(b-x)α|h(x)f(x)|dx+∥Db-αh∥∞Γ(α+1)∫ab(b-x)α|f(x)g(x)|dx=∥Db-αf∥∞Ia+α+1|g(b)h(b)|+∥Db-αg∥∞Ia+α+1|f(b)h(b)|+∥Db-αh∥∞Ia+α+1|f(b)g(b)|.
The proof of the theorem is complete.Remark 8.
If we takeh(t)=1, g(t)=1 and hence ∥Db-ah∥∞=0, ∥Db-ag∥∞=0 in Theorem 7, then we get Theorem 4.Theorem 9.
Letα≥0, m=[α], f,g,h∈ACm([a,b]). Assume that fk(b)=gk(b)=hk(b)=0, k=1,…,m-1, and Db-αf,Db-αg,Db-αh∈L1[a,b]. Then
(17)|3∫abf(x)g(x)h(x)dx-∫ab[f(b)g(x)h(x)+f(x)g(b)h(x)+f(x)g(x)h(b)]dx{∫abf(x)g(x)h(x)dx}|≤∥Db-αf∥L1([a,b])Ia+α|g(b)h(b)|+∥Db-αg∥L1([a,b])Ia+α|f(b)h(b)|+∥Db-αh∥L1([a,b])Ia+α|f(b)g(b)|.Proof.
From (15) we have
(18)|3∫abf(x)g(x)h(x)dx-∫ab[f(b)g(x)h(x)+f(x)g(b)h(x)+f(x)g(x)h(b)]dx{∫abf(x)g(x)h(x)dx}|≤∫ab|g(x)h(x)|Γ(α)(∫xb(t-x)α-1|Db-αf(t)dt|)dx+∫ab|h(x)f(x)|Γ(α)(∫xb(t-x)α-1|Db-αg(t)dt|)dx+∫ab|f(x)g(x)|Γ(α)(∫xb(t-x)α-1|Db-αh(t)dt|)dx≤∫ab|g(x)h(x)|Γ(α)(b-x)α-1(∫xb|Db-αf(t)dt|)dx+∫ab|h(x)f(x)|Γ(α)(b-x)α-1(∫xb|Db-αg(t)dt|)dx+∫ab|f(x)g(x)|Γ(α)(b-x)α-1(∫xb|Db-αh(t)dt|)dx=∥Db-αf∥L1([a,b])∫ab|g(x)h(x)|Γ(α)(b-x)α-1dx+∥Db-αg∥L1([a,b])∫ab|h(x)f(x)|Γ(α)(b-x)α-1dx+∥Db-αg∥L1([a,b])∫ab|f(x)g(x)|Γ(α)(b-x)α-1dx≤∥Db-αf∥L1([a,b])Ia+α|g(b)h(b)|+∥Db-αg∥L1([a,b])Ia+α|f(b)h(b)|+∥Db-αh∥L1([a,b])Ia+α|f(b)g(b)|.
This proves the theorem.Remark 10.
If we takeh(t)=1, g(t)=1, and hence ∥Db-ah∥∞=0, ∥Db-ag∥∞=0 in Theorem 9, then we get Theorem 5.Theorem 11.
Letp,q>1; 1/p+1/q=1, α>1-1/p, m=[α], f,g∈ACm([a,b]). Assume that fk(b)=gk(b)=hk(b)=0, k=1,…,m-1, and Db-αf,Db-αg,Db-αh∈Lq[a,b]. Then
(19)|3∫abf(x)g(x)h(x)dx-∫ab[f(b)g(x)h(x)+f(x)g(b)h(x)+f(x)g(x)h(b)]dx{∫abf(x)g(x)h(x)dx}|≤Γ(α+1p)(∥Db-αf∥Lq([a,b])Ia+α|g(b)h(b)|+∥Db-αg∥Lq([a,b])Ia+α|f(b)h(b)|+∥Db-αh∥Lq([a,b])Ia+α|f(b)g(b)|).Proof.
From (15) we have
(20)|3∫abf(x)g(x)h(x)dx-∫ab[f(b)g(x)h(x)+f(x)g(b)h(x)+f(x)g(x)h(b)]dx{∫abf(x)g(x)h(x)dx}|≤∫ab|g(x)h(x)|Γ(α)(∫xb(t-x)α-1|Db-αf(t)dt|)dx+∫ab|h(x)f(x)|Γ(α)(∫xb(t-x)α-1|Db-αg(t)dt|)dx+∫ab|f(x)g(x)|Γ(α)(∫xb(t-x)α-1|Db-αh(t)dt|)dx.
Applying Holder’s inequality to (21), we get
(21)|3∫abf(x)g(x)h(x)dx-∫ab[f(b)g(x)h(x)+f(x)g(b)h(x)+f(x)g(x)h(b)]dx{∫abf(x)g(x)h(x)dx}|≤∫ab|g(x)h(x)|Γ(α)(∫xb(t-x)(α-1)pdt)1/p×(∫xb|Db-αf(t)dt|qdt)1/qdx+∫ab|h(x)f(x)|Γ(α)(∫xb(t-x)(α-1)pdt)1/p×(∫xb|Db-αg(t)dt|qdt)1/qdx+∫ab|f(x)g(x)|Γ(α)(∫xb(t-x)(α-1)pdt)1/p×(∫xb|Db-αh(t)dt|qdt)1/qdx=∥Db-αf∥Γ(α)Lq([a,b])∫ab|g(x)h(x)|×(∫xb(t-x)(α-1)pdt)1/p+∥Db-αg∥Γ(α)Lq([a,b])∫ab|h(x)f(x)|×(∫xb(t-x)(α-1)pdt)1/p+∥Db-αh∥Γ(α)Lq([a,b])∫ab|f(x)g(x)|×(∫xb(t-x)(α-1)pdt)1/p.
We have(22)(∫xb(t-x)(α-1)pdt)1/p=(b-x)α-1+1/p(p(α-1)+1)1/p.
Substituting (22) into (21), we get the required inequality.Remark 12.
If we takeh(t)=1,g(t)=1 and hence ∥Db-ah∥∞=0, ∥Db-ag∥∞=0 in Theorem 11, then we get Theorem 6.



---

*Source: 127061-2013-01-30.xml*