# Some New Generalized Integral Inequalities for GA-s-Convex Functions via Hadamard Fractional Integrals

Authors: İmdat İşcan; Mustafa Aydin
Journal: Chinese Journal of Mathematics
 (2016)

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## Abstract

We prove new generalization of Hadamard, Ostrowski, and Simpson inequalities in the framework of GA-s-convex functions and Hadamard fractional integral.


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## Body

## 1. Introduction

Let a real functionf be defined on a nonempty interval I of real line R. The function f is said to be convex on I if inequality(1)ftx+1-ty≤tfx+1-tfyholds for all x,y∈I and t∈0,1.In [1], Breckner introduced s-convex functions as a generalization of convex functions as follows.Definition 1.
Lets∈(0,1] be a fixed real number. A function f:[0,∞)→[0,∞) is said to be s-convex (in the second sense), or that f belongs to the class Ks2, if (2)ftx+1-ty≤tsfx+1-tsfyfor all x,y∈[0,∞) and t∈[0,1].Of course,s-convexity means just convexity when s=1.The following inequalities are well known in the literature as Hermite-Hadamard inequality, Ostrowski inequality, and Simpson inequality, respectively.Theorem 2.
Letf:I⊆R→R be a convex function defined on the interval I of real numbers and a,b∈I with a<b. The following double inequality holds:(3)fa+b2≤1b-a∫abfxdx≤fa+fb2.Theorem 3.
Letf:I⊆R→R be a mapping differentiable in I∘, the interior of I, and let a,b∈I∘ with a<b. If f′(x)≤M, x∈a,b, then the following inequality holds: (4)fx-1b-a∫abftdt≤Mb-ax-a2+b-x22for all x∈a,b.Theorem 4.
Letf:a,b→R be a four times’ continuously differentiable mapping on a,b and f(4)∞=supx∈a,b⁡f(4)(x)<∞. Then the following inequality holds: (5)13fa+fb2+2fa+b2-1b-a∫abfxdx≤12880f4∞b-a4.We will give definitions of the right and left hand side Hadamard fractional integrals which are used throughout this paper.Definition 5.
Letf∈La,b. The right-sided and left-sided Hadamard fractional integrals Ja+αf and Jb-αf of order α>0 with b>a≥0 are defined by(6)Ja+αfx=1Γα∫axln⁡xtα-1ftdtt,a<x<b,(7)Jb-αfx=1Γα∫xbln⁡txα-1ftdtt,a<x<b,respectively, where Γ(α) is the Gamma function defined by Γ(α)=∫0∞e-ttα-1dt (see [2]).In recent years, many authors have studied errors estimations for Hermite-Hadamard, Ostrowski, and Simpson inequalities; for refinements, counterparts, and generalization see [3–10].Definition 6 (see [11, 12]).
A functionf:I⊆0,∞→R is said to be GA-convex (geometric-arithmetically convex) if (8)fxty1-t≤tfx+1-tfyfor all x,y∈I and t∈0,1.Definition 7 (see [13]).
Fors∈0,1, a function f:I⊆0,∞→R is said to be GA-s-convex (geometric-arithmetically s-convex) if (9)fxty1-t≤tsfx+1-tsfyfor all x,y∈I and t∈0,1.It can be easily seen that ifs=1, GA-s-convexity reduces to GA-convexity.For recent results and generalizations concerning GA-convex and GA-s-convex functions see [13–19].Lemma 8 (see [20]).
Forα>0 and μ>0, one has (10)∫01tα-1μtdt=μ∑k=1∞-1k-1ln⁡μk-1αk<∞,where (11)αk=αα+1α+2⋯α+k-1.Letf:I⊆0,∞→R be a differentiable function on I∘, the interior of I; in sequel of this paper we will take (12)Ifx,λ,α,a,b=1-λlnαxa+lnαbxfx+λfalnαxa+fblnαbx-Γα+1Jx-αfa+Jx+αfb,where a,b∈I with a<b, x∈[a,b], λ∈0,1, α>0, and Γ is Euler Gamma function.In [21], I˙şcan gave Hermite-Hadamard’s inequalities for GA-convex functions in fractional integral forms as follows.Theorem 9.
Letf:I⊆0,∞→R be a function such that f∈L[a,b], where a,b∈I with a<b. If f is a GA-convex function on [a,b], then the following inequalities for fractional integrals hold: (13)fab≤Γα+12ln⁡b/aαJa+αfb+Jb-αfa≤fa+fb2with α>0.In [21], I˙şcan obtained some new inequalities for quasi-geometrically convex functions via fractional integrals by using the following lemma.Lemma 10.
Letf:I⊆0,∞→R be a differentiable function on I∘ such that f′∈L[a,b], where a,b∈I with a<b. Then for all x∈[a,b], λ∈0,1, and α>0 one has (14)Ifx,λ,α,a,b=aln⁡xaα+1∫01tα-λxatf′xta1-tdt-bln⁡bxα+1∫01tα-λxbtf′xtb1-tdt.In this paper, we will use Lemma10 to obtain some new inequalities on generalization of Hadamard, Ostrowski, and Simpson type inequalities for GA-s-convex functions via Hadamard fractional integral.


## 2. Generalized Integral Inequalities for Some GA-s-Convex Functions via Fractional Integrals

Theorem 11.
Letf:I⊂0,∞→R be a differentiable function on I∘ such that f′∈L[a,b], where a,b∈I∘ with a<b. If |f′|q is GA-s-convex on [a,b] in the second sense for some fixed q≥1, x∈[a,b], λ∈0,1, and α>0 then the following inequality for fractional integrals holds:(15)Ifx,λ,α,a,b≤A11-1/qα,λaln⁡xaα+1f′xqA2xaq,α,λ,s+f′aqA3xaq,α,λ,s1/q+bln⁡bxα+1f′xqA2xbq,α,λ,s+f′bqA3xbq,α,λ,s1/q,where (16)A1α,λ=2αλ1+1/α+1α+1-λ,A2xuq,α,λ,s=∫01tα-λxuqttsdt,A3xuq,α,λ,s=∫01tα-λxuqt1-tsdt,u=a,b.Proof.
Using Lemma10, property of the modulus, and the power-mean inequality, we have(17)Ifx,λ,α,a,b≤alnxaα+1∫01tα-λxatf′xta1-tdt+blnbxα+1∫01tα-λxbtf′xtb1-tdt≤alnxaα+1∫01tα-λdt1-1/q∫01tα-λxaqtf′xta1-tqdt1/q+bln⁡bxα+1∫01tα-λdt1-1/q∫01tα-λxbqtf′xtb1-tqdt1/q. Since f′q is GA-s-convex on [a,b], we get(18)∫01tα-λxaqtf′xta1-tqdt≤∫01tα-λxaqttsf′xq+1-tsf′aqdt=f′xqA2xa,α,λ,s,q+f′aqA3xa,α,λ,s,q,(19)∫01tα-λxbqtf′xtb1-tqdt≤∫01tα-λxbqttsf′xq+1-tsf′bqdt=f′xqA2xb,α,λ,s,q+f′bqA3xb,α,λ,s,q,and by a simple computation, we have(20)∫01tα-λdt=∫0λ1/αλ-tαdt+∫λ1/α1tα-λdt=2αλ1+1/α+1α+1-λ.Hence, If we use (18), (19), and (20) in (17), we obtain the desired result. This completes the proof.Corollary 12.
Under the assumptions of Theorem11 with s=1, inequality (15) reduces to the following inequality: (21)Ifx,λ,α,a,b≤A11-1/qα,λaln⁡xaα+1f′xqA2xaq,α,λ,1+f′aqA3xaq,α,λ,11/q+bln⁡bxα+1f′xqA2xbq,α,λ,1+f′bqA3xbq,α,λ,11/q.Corollary 13.
Under the assumptions of Theorem11 with s=1 and α=1, inequality (15) reduces to the following inequality: (22)ln⁡ba-1Ifx,λ,1,a,b=1-λfx+λfaln⁡x/a+fbln⁡b/xln⁡b/a-1ln⁡b/a∫abfuudu≤ln⁡ba-1A11-1/q1,λaln⁡xa2f′xqA2μa,1,λ,1+f′aqA3μa,1,λ,11/q+bln⁡bx2f′xqA2μb,1,λ,1+f′bqA3μb,1,λ,11/q,where (23)A11,λ=2λ2-2λ+12,A2μu,1,λ,1=μu-2λ2μuλln2μu+λμuλln⁡μu-μuλ+1λln⁡μu+λ+4-λ+2μuln⁡μu-μu+1ln⁡μu3,A3μu,1,λ,1=2μuλ+μuln⁡μu-λ1+μuln⁡μu-μu-1ln⁡μu2-A2μu,1,λ,1,μu=xuq,u=a,b.Corollary 14.
Under the assumptions of Theorem11 with q=1, inequality (15) reduces to the following inequality: (24)Ifx,λ,α,a,b≤aln⁡xaα+1f′xA2xa,α,λ,s+f′aA3xa,α,λ,s+bln⁡bxα+1f′xA2xb,α,λ,s+f′bA3xb,α,λ,s.Corollary 15.
Under the assumptions of Theorem11 with x=ab,λ=1/3, from inequality (15), one gets the following Simpson type inequality for fractional integrals: (25)2α-1ln⁡ba-αIfab,13,α,a,b=16fa+4fab+fb-2α-1Γα+1ln⁡b/aαJab-αfa+Jab+αfb≤ln⁡b/a4A11-1/qα,13af′abqA2baq/2,α,13,s+f′aqA3baq/2,α,13,s1/q+bf′abqA2abq/2,α,13,s+f′bqA3abq/2,α,13,s1/q.Corollary 16.
Under the assumptions of Theorem11 with x=ab,λ=0, from inequality (15), one gets (26)2α-1ln⁡ba-αIfab,0,α,a,b=fab-2α-1Γα+1ln⁡b/aαJab-αfa+Jab+αfb≤ln⁡b/a41α+11-1/qaf′abqA2baq/2,α,0,s+f′aqA3baq/2,α,0,s1/q+bf′abqA2abq/2,α,0,s+f′bqA3abq/2,α,0,s1/q.Corollary 17.
Under the assumptions of Theorem11 withx=ab and λ=1, from inequality (15) one gets (27)2α-1ln⁡ba-αIfab,1,α,a,b=fa+fb2-2α-1Γα+1ln⁡b/aαJab-αfa+Jab+αfb≤ln⁡b/a4αα+11-1/qaf′abqA2baq/2,α,1,s+f′aqA3baq/2,α,1,s1/q+bf′abqA2abq/2,α,1,s+f′bqA3abq/2,α,1,s1/q.Corollary 18.
Let the assumptions of Theorem11 hold. If f′(x)≤M for all x∈a,b and λ=0, then from inequality (15), one gets the following Ostrowski type inequality for fractional integrals: (28)ln⁡xaα+ln⁡bxαfx-Γα+1Jx-αfa+Jx+αfb≤M1α+11-1/qaln⁡xaαA2baq/2,α,0,s+A3baq/2,α,0,s1/q+bln⁡bxαA2abq/2,α,0,s+A3abq/2,α,0,s1/qfor all x∈a,b.Theorem 19.
Letf:I⊂0,∞→R be a differentiable function on I∘ such that f′∈L[a,b], where a,b∈I∘ with a<b. If |f′|q is GA-s-convex on [a,b] for some fixed q>1, x∈[a,b], λ∈0,1, and α>0 then the following inequality for fractional integrals holds:(29)Ifx,λ,α,a,b≤C11/pα,λaln⁡xaα+1f′xqC2xaq,s+f′aqC3xaq,s1/q+bln⁡bxα+1f′xqC2xbq,s+f′bqC3xbq,s1/q, where 1/p+1/q=1 and (30)C1α,λ=1αp+1,λ=0λ1+p+1/ααβ1α,p+1+1-λp+1αp+1·F211-1α,1;p+2;1-λ,0<λ≤1,C2xuq,s=xuq∑k=1∞-1k-1ln⁡x/uqk-1s+1k,C3xuq,s=∑k=1∞-1k-1-ln⁡x/uqk-1s+1k,u=a,b.Proof.
Using Lemma10, property of the modulus, the Hölder inequality, and GA-s-convexity of f′q, we have (31)
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,where μa=x/aq, μb=x/bq and(32)∫01tα-λpdt=∫0λ1/αλ-tαpdt+∫λ1/α1tα-λpdt=1αp+1,λ=0λαp+1/ααβ1α,p+1+1-λp+1αp+1·F211-1α,1;p+2;1-λ,0<λ≤1.Using Lemma 8, we have (33)∫01μuttsdt=μu∑k=1∞-1k-1ln⁡μuk-1s+1k,∫01μut1-tsdt=∫01μu1-ttsdt=∑k=1∞-1k-1-ln⁡μuk-1s+1k,u=a,b.Hence, if we use (32)-(33) in (31) and replacing μa=x/aq,μb=x/bq, we obtain the desired result. This completes the proof.Corollary 20.
Under the assumptions of Theorem19 with s=1, inequality (29) reduces to the following inequality: (34)Ifx,λ,α,a,b≤C11/pα,λaln⁡xaα+1f′xqC2xaq,1+f′aqC3xaq,11/q+bln⁡bxα+1f′xqC2xbq,1+f′bqC3xbq,11/q.Corollary 21.
Under the assumptions of Theorem19 with s=1 and α=1, inequality (29) reduces to the following inequality: (35)Ifx,λ,1,a,b=ln⁡ba1-λfx+λfaln⁡xa+fbln⁡bx-∫abfuudu≤λp+1+1-λp+1p+11/paln⁡xa2f′xqC2xaq,1+f′aqC3xaq,11/q+bln⁡bx2f′xqC2xbq,1+f′bqC3xbq,11/q.Corollary 22.
Under the assumptions of Theorem19 with x=ab,λ=1/3, from inequality (29), one gets the following Simpson type inequality for fractional integrals: (36)2α-1ln⁡ba-αIfab,13,α,a,b=16fa+4fab+fb-2α-1Γα+1ln⁡b/aαJab-αfa+Jab+αfb≤ln⁡b/a4C11/pα,13af′abqC2baq/2,s+f′aqC3baq/2,s1/q+bf′abqC2abq/2,s+f′bqC3abq/2,s1/q.Corollary 23.
Under the assumptions of Theorem19 with x=ab,λ=0, from inequality (29), one gets (37)2α-1ln⁡ba-αIfab,0,α,a,b=fab-2α-1Γα+1ln⁡b/aαJab-αfa+Jab+αfb≤ln⁡b/a41αp+11/paf′abqC2baq/2,s+f′aqC3baq/2,s1/q+bf′abqC2abq/2,s+f′bqC3abq/2,s1/q.Corollary 24.
Under the assumptions of Theorem19 withx=ab and λ=1, from inequality (29) one gets (38)2α-1ln⁡ba-αIfab,1,α,a,b=fa+fb2-2α-1Γα+1ln⁡b/aαJab-αfa+Jab+αfb≤ln⁡b/a41αβ1α,p+11/paf′abqC2baq/2,s+f′aqC3baq/2,s1/q+bf′abqC2abq/2,s+f′bqC3abq/2,s1/q.Corollary 25.
Let the assumptions of Theorem19 hold. If f′(x)≤M for all x∈a,b and λ=0, then from inequality (29), one gets the following Ostrowski type inequality for fractional integrals: (39)ln⁡xaα+ln⁡bxαfx-Γα+1Jx-αfa+Jx+αfb≤M1αp+11/paln⁡xaαC2xaq,s+C3xaq,s1/q+bln⁡bxαC2xbq,s+C3xbq,s1/qfor each x∈a,b.



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