# Inequalities for Riemann–Liouville Fractional Integrals of Stronglys,m-Convex Functions

Authors: Fuzhen Zhang; Ghulam Farid; Saira Bano Akbar
Journal: Journal of Mathematics
 (2021)

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## Abstract

The results of this paper provide two Hadamard-type inequalities for stronglys,m-convex functions via Riemann–Liouville fractional integrals and error estimations of well-known fractional Hadamard inequalities. Their special cases are given and connected with the results of some published papers.


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## Body

## 1. Introduction

The most prominent inequality for convex functions is the well-known Hadamard inequality stated in the following.Theorem 1.
(see [1]). Let f:I⊂ℝ⟶ℝ be a convex function on the interval I, where x,y∈I with x<y. Then, the following inequality holds:(1)fx+y2≤1y−x∫xyfudu≤fx+fy2.Convex functions are extended, generalized, and refined in different ways to define new types of convex functions. For instance,s-convex, m-convex, s,m-convex, strongly convex, and strongly s,m-convex functions are extensions of convex functions. The aim of this paper is to establish integral inequalities by using the class of strongly s,m-convex functions. We give definitions of s,m-convex and strongly s,m-convex functions as follows.Definition 1.
(see [2]). A function f:0,∞⟶ℝ is called s,m-convex in the second sense, if the following inequality holds:(2)ftx+m1−ty≤tsfx+m1−tsfy,for every x,y∈0,∞, t∈0,1 and s,m∈0,1×0,1.Definition 2.
(see [3]). A function f:0,∞⟶ℝ is called strongly s,m-convex in the second sense with modulus C≥0, if the following inequality holds:(3)ftx+m1−ty≤tsfx+m1−tsfy−Cmt1−ty−x2,for every x,y∈0,∞, t∈0,1 and s,m∈0,1×0,1.By settings,m=s,1, s,m=1,m, and s,m=1,1 in (2), we get s-convex [4], m-convex [5], and convex functions, respectively, while by setting s,m=s,1, s,m=1,m, and s,m=1,1 in (3), we get strongly s-convex [6], strongly m-convex [7], and strongly convex [6] functions, respectively.Next we give definition of Riemann–Liouville fractional integralsJx+αf and Jy−αf which are utilized to get the desired results of this paper.Definition 3.
(see [8]). Let f∈L1x,y. Then, Riemann–Liouville fractional integral operators of order α>0 are given by(4)Jx+αfu=1Γα∫xuu−tα−1ftdt,u>x,Jy−αfu=1Γα∫uyt−uα−1ftdt,u<y,where Γα=∫0∞e−ttα−1dt is the gamma function and Jx+0fu=Jy−0fu=fu.The following special functions are also involved in the findings of this paper.Definition 4.
The beta function, also referred to as first type of Euler integral, is defined by(5)βα,s=∫01tα−11−ts−1dt,where Reα,Res>0.Close association of the beta function to the gamma function is an important factor of the beta function(6)βα,s=ΓαΓsΓα+s.The beta function is symmetric, i.e.,βα,s=βs,α. A generalization of the beta function, called the incomplete beta function, is defined by(7)βb;α,s=∫0btα−11−ts−1dt,where Reα,Res>0 with 0<b<1. The incomplete beta function βb;α,s weakens to the ordinary βα,s (beta function) by setting b=1.In [8], the Hadamard inequality is studied for Riemann–Liouville fractional integrals which is stated in the following theorem.Theorem 2.
Letf:x,y⟶ℝ be a positive function with 0≤x<y and f∈L1x,y. If f is convex function on x,y, then the following inequality for fractional integrals holds:(8)fx+y2≤Γα+12y−xαJx+αfy+Jy−αfx≤fx+fy2,with α>0.The inequality stated in the aforementioned theorem motivates the researchers to work in this direction by establishing other kinds of inequalities for Riemann–Liouville fractional integrals. In the past decade, several classical inequalities have been extended via different kinds of fractional integral operators. The Hadamard inequality is one of the most studied inequalities for fractional integral operators. For some recent work, we refer the readers to [3, 8–17].This paper is organized as follows. In Section2, two versions of the Hadamard inequality for strongly s,m-convex functions via Riemann–Liouville fractional integrals are given. Their connection with the well-known results is established in the form of corollaries and remarks. In Section 3, the error estimations of Hadamard inequalities for Riemann–Liouville fractional integrals are obtained by using differentiable strongly s,m-convex functions.


## 2. Main Results

Theorem 3.
Letf∈L1x,y be a positive function with 0≤x<y. If f is strongly s,m-convex function on x,my with modulus C≥0, m≠0,0<s≤1, then the following fractional integral inequality holds:(9)2s−1fx+my2+2s−3Cmαα+2x−y2+2my−x/m2αα+1+2x−ymy−x/mα+1≤Γα+12my−xαJx+αfmy+mα+1Jy−αfxm≤αfx+mfy2α+s+mαβα,s+1fy+mfx/m22−Cmαy−x2+my−x/m222α+1α+2,with α>0.Proof.
Sincef is strongly s,m-convex function, for u,v∈x,y, we have(10)fu+mv2≤fu+mfv2s−Cm4u−v2.
By settingu=xt+m1−ty and v=yt+1−tx/m, we have(11)fx+my2≤12sfxt+m1−ty+m2sfyt+1−txm−Cm4tx−y+1−tmy−xm2.
By multiplying inequality (11) with tα−1 on both sides and then integrating over the interval 0,1, we get(12)fx+my2∫01tα−1dt≤12s∫01fxt+m1−tytα−1dt+m2s∫01fyt+1−txmtα−1dt−Cm4∫01tx−y+1−tmy−xm2tα−1dt.
By change of variables, we will get(13)1αfx+my2≤Γα2smy−xα1Γα∫myxmy−uα−1fudu+mα+1Γα∫x/myv−xmα−1fvdv−Cm4x−y2α+2+2my−x/m2αα+1α+2+2x−ymy−x/mα+1α+2.
Further, the above inequality takes the following form:(14)2s−1fx+my2≤Γα+12my−xαJx+αfmy+mα+1Jy−αfxm−2s−1Cmα4x−y2α+2+2my−x/m2αα+1α+2+2x−ymy−x/mα+1α+2.
From the definition of stronglys,m-convex function with modulus C, for t∈0,1, we have the following inequality:(15)ftx+m1−ty+mfyt+1−txm≤tsfx+mfy+m1−tsfy+mfxm2−Cmt1−ty−x2+my−xm22.
By multiplying inequality (15) with tα−1 on both sides and then integrating over the interval 0,1, we get(16)∫01ftx+m1−tytα−1dt+m∫01fyt+1−txmtα−1dt≤fx+mfy∫01ts+α−1dt+mfy+mfxm2∫01tα−11−tsdt−Cmy−x2+my−xm22∫01tα1−tdt.
By change of variables, we will get(17)Γαmy−xα1Γα∫myxmy−uα−1fudu+mα+1Γα∫x/myv−xmα−1fvdv≤fx+mfyα+s+mfy+mfxm2βs+1,α−Cmy−x2+my−x/m22α+1α+2.
Further, the above inequality takes the following form:(18)Γα+12my−xαJx+αfmy+mα+1Jy−αfxm≤αfx+mfy2α+s+mα2fy+mfxm2βs+1,α−Cmαy−x2+my−x/m222α+1α+2.
From inequalities (14) and (18), one can get inequality (9).Remark 1.
(i)
Fors=1 in (9), we have the result for strongly m-convex function [18].(ii)
Form=1 and s=1 in (9), we have the result for strongly convex function.(iii)
Form=1,s=1, and C=0, we get [[16], Theorem 2](iv)
Form=1, s=1, α=1, and C=0, we get the classical Hadamard inequality.(v)
Form=1 and C=0, we get [[17], Theorem 3].Corollary 1.
Form=1, we have the result for Riemann–Liouville fractional integrals of strongly s-convex functions:(19)2s−1fx+y2+2s−1Cαy−x2α2−α+24α+1α+2≤Γα+12y−xαJx+αfy+Jy−αfx≤fx+fy2αα+s+Γα+1Γs+1Γα+s+1−Cαy−x2α+1α+2.Corollary 2.
Forα=1 and m=1, the following inequality holds for strongly s-convex function:(20)2s−1fx+y2+2s−1Cy−x212≤1y−x∫xyfudu≤fx+fys+1−Cy−x26.In the next theorem, we give another version of the Hadamard inequality.Theorem 4.
Under the assumptions of Theorem3, the following fractional integral inequality holds:(21)2s−1fx+my2+Cmα24−sx−y22α+2+my−x/m2α2+5α+82αα+1α+2+x−ymy−x/mα+3α+1α+2≤2α−1Γα+1my−xαJx+my/2+αfym+mα+1Jx+ym/2m−αfxm≤αfx+mfy2s+1α+s+2α−1mαfy+mfxm2β12;s+1,α−Cmαy−x2+my−x/m22α+38α+1α+2,with α>0.Proof.
Lett∈0,1. Using strong s,m-convexity of function f for u=xt/2+m2−t/2y and v=2−t/2x/m+yt/2 in inequality (10), we have(22)fx+my2≤12sfxt2+m2−t2y+m2sf2−t2xm+yt2−Cm4t2x−y+2−t2my−xm2.
By multiplying (22) with tα−1 on both sides and making integration over 0,1, we get(23)fx+my2∫01tα−1dt≤12s∫01fxt2+m2−t2ytα−1dt+m2s∫01f2−t2xm+yt2tα−1dt−Cm4∫01t2x−y+2−t2my−xm2tα−1dt.
By using change of variables and computing the last integral, from (23), we get(24)2sαfx+my2≤2αΓαmy−xα1Γα∫myx+my/2my−uα−1fudu+mα+1Γα∫x/mym+x/2mv−xmα−1fvdv−2sCm4x−y24α+2+my−x/m2α2+5α+84αα+1α+2+x−ymy−x/mα+32α+1α+2.
Further, it takes the following form:(25)2s−1fx+my2≤2α−1Γα+1my−xαJx+my/2+αfym+mα+1Jym+x/2m−αfxm−2s−1Cmα4x−y24α+2+my−x/m2α2+5α+84αα+1α+2+x−ymy−x/mα+32α+1α+2.
The first inequality of (21) can be seen in (25). Now we prove the second inequality of (21). Since f is strongly s,m-convex function and t∈0,1, we have the following inequality:(26)fxt2+m2−t2y+mf2−t2xm+yt2≤t2sfx+mfb+m2−t2sfy+mfxm2−Cmt2−t4y−x2+my−xm22.
By multiplying inequality (26) with tα−1 on both sides and making integration over 0,1, we get(27)∫01fxt2+m2−t2ytα−1dt+m∫01f2−t2xm+yt2tα−1dt≤12sfx+mfy∫01ts+α−1dt+m2sfy+mfxm2∫012−tstα−1dt−Cm4y−x2+my−xm2∫01tα2−tdt.
By using change of variables and computing the last integral, from (27), we get(28)2αΓαmy−xα1Γα∫myx+my/2my−uα−1fudu+mα+1Γα∫x/mmy+x/2mv−xmα−1fvdv≤fx+mfy2sα+s+2αmfy+mfxm2β12:s+1,α−Cmy−x2+my−x/m22α+34α+1α+2.
Further, it takes the following form:(29)2α−1Γα+1my−xαJx+my/2+αfym+mα+1Jx+ym/2m−αfxm≤αfx+mfy2s+1α+s+2α−1mαfy+mfxm2β12:s+1,α−Cmαy−x2+my−x/m22α+38α+1α+2.
From inequalities (25) and (29), we have inequality (21).Remark 2.
(i)
Fors=1 in (21), we get the result for strongly m-convex function [18].(ii)
Form=1, s=1, and C=0, we get [[16], Theorem 2](iii)
Form=1, s=1, α=1, and C=0, we get the classical Hadamard inequality.Corollary 3.
Form=1 and s=1 in (21), we have the result for Riemann–Liouville fractional integrals of strongly convex function:(30)fx+y2+Cy−x22α+1α+2≤2α−1Γα+1y−xαJx+y/2+αfy+Jx+y/2−αfx≤fx+fy2−Cαy−x2α+34α+1α+2.Corollary 4.
Form=1 in (21), we get the result for Riemann–Liouville fractional integrals of strongly s-convex function:(31)2s−1fx+y2+2sCx−y24α+1α+2≤2α−1Γα+1y−xαJx+y/2+αfy+Jx+y/2−αfx≤αfx+fy12s+1α+s+2α−1β12;s+1,α−Cαy−x2α+34α+1α+2.Corollary 5.
Form=1 and C=0 in (21), we get the result for Riemann–Liouville fractional integrals of s-convex function:(32)2s−1fx+y2≤2α−1Γα+1y−xαJx+y/2+αfy+Jx+y/2−αfx≤αfx+fy12s+1α+s+2α−1β12;s+1,α.Corollary 6.
Form=1 and α=1 in (1), we have the Hadamard inequality for strongly s-convex function:(33)2s−1fx+y2+2sCx−y224≤1y−x∫xyfudu≤fx+fys+1−Cy−x26.


## 3. Error Estimations of Riemann–Liouville Fractional Integral Inequalities

The following two lemmas are very useful to obtain the results of this section.Lemma 1.
(see [8]). Let f:x,y⟶ℝ be a differentiable mapping on x,y with x<y. If f′∈Łx,y, then the following fractional integral equality holds:(34)fx+fy2−Γα+12y−xαJx+αfy+Jy−αfx=y−x2∫011−tα−tαf′tx+1−tydt.Lemma 2.
(see [10]). Let f:x,y⟶ℝ be a differentiable mapping on x,y with x<y. If f′∈x,my,m∈0,1, then the following equality for fractional integrals holds:(35)2α−1Γα+1my−xαJx+my/2+αfmy+mα+1Jx+my/2m−αfxm−12fx+my2+mfx+my2m|=mb−a4∫01tαf′xt2+m2−t2ydt+∫01tαf′yt2+2−t2xmdt.Theorem 5.
Letf:x,y⟶ℝ be a differentiable mapping on x,y with x<y. If f′ is a strongly s,m-convex function on x,my with modulus C≥0, m≠0, and 0<s≤1, then the following fractional integral inequality holds:(36)fx+fy2−Γα+12y−xαJx+αfy+Jy−αfx≤y−x2f′xβ12;α+1,s+1−β12;s+1,α+1+1−1/2α+sα+s+1+mf′ymβ12;α+1,s+1−β12;s+1,α+1+1−1/2α+sα+s+1−2Cmy/m−x2α+2α+31−α+42α+2,with α>0.Proof.
Sincef′ is strongly s,m-convex function on x,y, for t∈0,1, we have(37)f′tx+1−ty≤tsf′x+m1−tsf′ym−Cmt1−tym−x2.
By using Lemma1 and (37), we have(38)fx+fy2−Γα+12y−xαJx+αfy+Jy−αfx≤y−x2∫011−tα−tαf′tx+m1−tymdt≤y−x2∫011−tα−tαtsf′x+m1−tsf′ym−Cmt1−tym−x2dt≤y−x2∫01/21−tα−tαtsf′x+m1−tsf′ym−Cmt1−tym−x2dt+∫1/21tα−1−tαtsf′x+m1−tsf′ym−Cmt1−tym−x2dt≤y−x2f′xβ12;α+1,s+1−β12;s+1,α+1+1−1/2α+sα+s+1+mf′ymβ12;α+1,s+1−β12;s+1,α+1+1−1/2α+sα+s+1−2Cmy/m−x2α+2α+31−α+42α+2.
After simplifying the last inequality of (38), we get (36).Remark 3.
(i)
By settingC=0 in inequality (36), one can get result for s,m-convex function.(ii)
By settings=1 in inequality (36), we get [[18], Theorem 8].Corollary 7.
By takingm=1 in (36), we have the result for Riemann–Liouville fractional integrals of strongly s-convex function:(39)fx+fy2−Γα+12y−xαJx+αfy+Jy−αfx≤y−x2f′xβ12;α+1,s+1−β12;s+1,α+1+1−1/2α+sα+s+1+f′yβ12;α+1,s+1−β12;s+1,α+1+1−1/2α+sα+s+1−2Cy−x2α+2α+31−α+42α+2.Corollary 8.
By takingm=1 and s=1 in inequality (36), we have the result for Riemann–Liouville fractional integrals of strongly convex function:(40)fx+fy2−Γα+12y−xαJx+αfy+Jy−αfx≤y−x21−1/2αα+1f′x+f′y−2Cy−x2α+2α+31−α+42α+2.Corollary 9.
By takingm=s=1 and α=1 in inequality (36), we get the following inequality:(41)fx+fy2−1y−x∫xyfudu≤y−x8f′x+f′y−Cy−x332.Inequality (41) provides the refinement of [[19], Theorem 2.2].Theorem 6.
Letf:x,y⟶ℝ be a differentiable mapping on x,y with x<y. If f′q is strongly s,m-convex on x,my with modulus C≥0, s,m∈0,12 for q≥1, then the following inequality for fractional integrals holds:(42)2α−1Γα+1my−xαJx+my/2+αfmy+mα+1Jx+my/2m−αfxm−12fx+my2+mfx+my2m≤my−x4α+11/pf′xq2sα+s+1+2α+1mf′yqβ12;s+1,α+1−Cmy−x2α+44α+2α+31/q+f′yq2sα+s+1+2α+1mf′xm2qβ12;s+1,α+1−Cmx/m2−y2α+44α+2α+31/q.Proof.
By applying Lemma2 and strong s,m-convexity of f′, we have(43)2α−1Γα+1my−xαJx+my/2+αfmy+mα+1Jx+my/2m−αfxm−12fx+my2+mfx+my2m≤my−x4∫01tαf′t2x+m2−t2ydt+∫01tαf′2−t2xm+t2ydt≤my−x4f′x+f′y2s∫01tα+sdt+m2sf′y+f′xm2∫01tα2−tsdt−Cm4y−x2+xm2−y2∫01tα+12−tdt=my−x4f′x+f′y2sα+s+1+2α+1mf′y+f′xm2β12;s+1,α+1−Cmα+44α+2α+3y−x2+xm2−y2.
Now, for strongs,m-convexity of f′q, q>1, using power mean inequality, we get(44)|2α−1Γα+1my−xαJx+my/2+αfmy+mα+1Jx+my/2m−αfxm−12fx+my2+mfx+my2m|≤my−x4∫01tαdt1−1/q∫01tαf′t2x+m2−t2yqdt1/q+∫01tαf′2−t2xm+t2yqdt1/q≤my−x4α+11/pfxq2s∫01tα+sdt+mf′yq2s∫01tα2−tsdt−Cmy−x24∫01tα+12−tdt1/q+f′yq2s∫01tα+sdt+m2sf′xm2q∫01tα2−tsdt−Cmx/m2−y24∫01tα+12−tdt1/q≤my−x4α+11/pf′xq2sα+s+1+2α+1mf′yqβ12;s+1,α+1−Cmy−x2α+44α+2α+31/q+f′yq2sα+s+1+2α+1mf′xm2qβ12;s+1,α+1−Cmx/m2−y2β+44β+2β+31/q.
Hence, we have inequality (42).Remark 4.
(i)
Fors=1 in inequality (42), we have the result for strongly m-convex function [18].(ii)
Fors=1, m=1, and C=0 in inequality (42), we get [[16], Theorem 5].(iii)
Fors=1, m=1, C=0, and α=1 in inequality (42), we get the inequality proved by Kirmaci in [20].Corollary 10.
Fors=1 and m=1 in inequality (42), we have the result for Riemann–Liouville fractional integrals of strongly convex function:(45)2α−1Γα+1y−xαJx+y/2+αfy+Jx+y/2−αfx−fx+y2≤y−x4α+12α+41/qf′xqα+1+f′yqα+3−Cy−x2α+1α+42α+31/q+f′yqα+1+f′xqα+3−Cy−x2α+1α+42α+31/q.Corollary 11.
Form=1 in inequality (42), we have the result for Riemann–Liouville fractional integrals of strongly s-convex function:(46)2α−1Γα+1y−xαJx+y/2+αfy+Jx+y/2−αfx−fx+y2≤y−x4α+11/pf′xq2sα+s+1+2α+1f′yqβ12;s+1,α+1−Cy−x2α+44α+2α+31/q+f″yq2sα+s+1+2α+1f′xqβ12;s+1,α+1−Cy−x2α+44α+2α+31/q.Theorem 7.
Letf:x,y⟶ℝ be a differentiable mapping on x,y with x<y. If f′q is strongly s,m-convex function on x,my with modulus C≥0, s,m∈0,12 for q>1, then the following fractional integral inequality holds:(47)2α−1Γα+1my−xαJx+my/2+αfmy+mα+1Jx+my/2m−αfxm−12fx+my2+mfx+my2m≤my−x2ss+11/p−14αp+11/pf′xq+m2s+1−1f′yq−2sCms+1y−x261/q+m2s+1−1f′xm2q+f′yq−2sCms+1x/m2−y261/q≤my−x2ss+11/p−14αp+11/pf′x+f′y+m2s+1−1f′xm2+f′y−2sCms+16y−x2+xm2−y2,where α>0.Proof.
By applying Lemma2 and then using Hölder inequality and strong s,m-convexity of f′q, we get(48)|2α−1Γα+1my−xαJx+my/2+αfmy+mα+1Jx+my/2m−αfxm−12fx+my2+mfx+my2m|≤my−x4∫01tpαdt1/p∫01f′xt2+m2−t2yqdt1/q+∫01f′yt2+2−t2xmqdt1/q≤my−x41αp+11/pf′xq∫01t2sdt+mf′yq∫012−t2sdt−Cmy−x24∫01t2−tdt1/q+f′yq∫01t2sdt+mf′xm2q∫012−t2sdt−Cmx/m2−y24∫01t2−tdt1/q=my−xs+11/p−122−s1/p−1αp+11/pf′xq+mf′yq2s+1−1−2sCms+1y−x261/q+mf′xm2q2s+1−1+f′yq−2sCms+1x/m2−y261/q≤my−x2ss+11/p−14αp+11/pf′x+f′y+m2s+1−1f′y+f′xm2−2sCms+16y−x2+xm2−y2.
We have usedAq+Bq≤A+Bq, for A≥0,B≥0. This completes the proof.Remark 5.
(i)
Fors=1 in inequality (47), we get [[18], Theorem 10].(ii)
Fors=1 and C=0 in inequality (47), we get [[10], Theorem 2.7].(iii)
Fors=1, m=1, and C=0 in inequality (47), we get [[16], Theorem 6].Corollary 12.
Forα=1 and m=1, we have the result for s-convex function:(49)1y−x∫xyfudu−fx+y2≤y−x2ss+11/p−14p+11/pf′xq+2s+1−1f′yq−2sCs+1y−x261/q+2s+1−1f′xq+f′yq−2sCs+1y−x261/q≤y−x2ss+11/p−14p+11/p2s+1f′x+f′y−2sCs+13y−x2.Corollary 13.
Forα=1 and m=q=1, we have(50)1y−x∫xyfudu−fx+y2≤y−x4s+12f′x+f′y−Cs+13y−x2.Corollary 14.
Forα=1 and m=q=s=1, we have(51)1y−x∫xyfudu−fx+y2≤y−x4|f′x|+f′y−C3y−x2.



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