# Generalized Hadamard Fractional Integral Inequalities for Stronglys,m-Convex Functions

Authors: Chao Miao; Ghulam Farid; Hafsa Yasmeen; Yanhua Bian
Journal: Journal of Mathematics
 (2021)

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## Abstract

This article deals with Hadamard inequalities for stronglys,m-convex functions using generalized Riemann–Liouville fractional integrals. Several generalized fractional versions of the Hadamard inequality are presented; we also provide refinements of many known results which have been published in recent years.


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## Body

## 1. Introduction

Fractional calculus is related to the integrals and derivatives of any arbitrary real or complex order. Its history starts from the end of the seventeenth century, but now it has many applications in almost every field of mathematics, science, and engineering such as electromagnetic, viscoelasticity, fluid mechanics, and signal processing. Fractional integral and derivative operators are of great importance in fractional calculus. The Riemann–Liouville fractional integrals are playing key role in its development. Sarikaya et al. [1, 2] studied Hadamard inequality through Riemann–Liouville fractional integrals of convex functions. This study has encouraged a number of researchers to work further in the field of mathematical inequalities by using fractional integral operators. As a consequence, Hadamard’s inequality is generalized and extended by fractional integral operators in many ways (see [3–9] and the references therein). The following inequality is the well-known Hadamard inequality for convex functions which is stated in [10].Letf:I⟶ℝ be a convex function defined on an interval I⊂ℝ and x,y∈I where x<y. Then, the following inequality holds:(1)fx+y2≤1y−x∫xyfvdv≤fx+fy2.For the history of this inequality, we refer the readers to [11, 12]. Use of convex functions in the fields of statistics [13], economics [14], and optimization [15] is of prime importance because they play an important role in development of new concepts and notions. Various scholars extended the research on integral inequalities to fractal sets [16]. In this paper, the Hadamard inequality is studied for generalized Riemann–Liouville fractional integrals of strongly s,m-convex functions; also, by using two integral identities, some error bounds of already established fractional inequalities are studied. Bracamonte et al. [17] defined the strongly s,m-convex function as follows.Definition 1.
A functionf:0,+∞⟶ℝ is said to be strongly s,m-convex function with modulus c≥0 in second sense, where s,m∈0,12, if(2)fxt+m1−ty≤tsfx+m1−tsfy−cmt1−ty−x2,holds for all x,y∈0,+∞ and t∈0,1.The well-known definition of Riemann–Liouville fractional integral is given as follows.Definition 2.
(see [18]) (see also [19]). Let f∈La,b. Then, left-sided and right-sided Riemann–Liouville fractional integrals of a function f of order μ where ℜμ>0 are given by(3)Ia+μfx=1Γμ∫axx−tμ−1ftdt,x>a,(4)Ib−μfx=1Γμ∫xbt−xμ−1ftdt,x<b,where ℜμ is real part of μ and Γμ=∫0∞e−zzμ−1dz. The following theorems are the fractional versions of Hadamard inequality by Riemann–Liouville fractional integrals.Theorem 1.
(see [1]). Let f:a,b⟶ℝ be a positive function with 0≤a<b and f∈La,b. If f is a convex function on a,b, then the following fractional integral inequality holds:(5)fa+b2≤Γμ+12b−aμIa+μfb+Ib−μfa≤fa+fb2,with μ>0.Theorem 2.
(see [2]). Under the assumptions of Theorem 1, the following fractional integral inequality holds:(6)fa+b2≤2μ−1Γμ+1b−aμIa+b/2+μfb+Ia+b/2−μfa≤fa+fb2,with μ>0.By establishing an integral identity, the following error estimation of inequality (6) is proved.Theorem 3.
(see [1]). Let f:a,b⟶ℝ be a differentiable mapping on a,b with a<b. If f′ is convex on a,b, then the following fractional integral inequality holds:(7)fa+fb2−Γμ+12b−aμIa+μfb+Ib−μfa≤b−a2μ+11−12μf′a+f′b.Ak-analogue of Riemann–Liouville integral is defined as follows.Definition 3.
(see [20]). Let f∈La,b. Then, k-fractional Riemann–Liouville integrals of order μ where ℜμ>0, k>0, are defined as(8)Ia+μkfx=1kΓkμ∫axx−tμ/k−1ftdt,x>a,(9)Ib−μkfx=1kΓkμ∫xbt−xμ/k−1ftdt,x<b,where Γk. is defined by [21](10)Γkμ=∫0∞tμ−1e−tk/kdt,ℜμ>0.Ifk=1, (8) and (9) coincide with (3) and (4).Twok-fractional versions of Hadamard inequality for k-fractional Riemann–Liouville integrals are given in the next two theorems.Theorem 4.
(see [22]). Let f:a,b⟶ℝ be a positive function with 0≤a<b. If f is a convex function on a,b, then the following inequality for k-fractional integrals holds:(11)fa+b2≤Γkμ+k2b−aμ/kIa+μkfb+Ib−μkfa≤fa+fb2.Theorem 5.
(see [23]). Under the assumption of Theorem 4, the following inequality for k-fractional integrals holds:(12)fa+b2≤2μ/k−1Γkμ+kb−aμ/kIa+b/2+μkfb+Ia+b/2−μkfa≤fa+fb2.By establishing an integral identity, in the following theorem, the error estimation of Theorem4 is proved.Theorem 6.
(see [22]). Let f:a,b⟶ℝ be a differentiable mapping on a,b with 0≤a<b. If f′ is convex on a,b, then the following inequality for k-fractional integrals holds:(13)fa+fb2−Γkμ+k2b−aμ/kIa+μkfb+Ib−μkfa≤b−a2μ/k+11−12μ/kf′a+f′b.In the following, we recall the definition of generalized Riemann–Liouville fractional integrals by a monotonically increasing function.Definition 4.
(see [24]). Let f:a,b⟶ℝ be an integrable function. Also, let ψ be an increasing and positive function on a,b, having a continuous derivative ψ′ on a,b. The left-sided and right-sided fractional integrals of a function f with respect to another function ψ on a,b of order μ where ℜμ>0 are given by(14)Ia+μ,ψfx=1Γμ∫axψ′tψx−ψtμ−1ftdt,x>a,(15)Ib−μ,ψfx=1Γμ∫xbψ′tψt−ψxμ−1ftdt,x<b.Ifψ is identity function, then (14) and (15) coincide with (3) and (4).Thek-analogue of generalized Riemann–Liouville fractional integral is defined as follows.Definition 5.
(see [25]). Let f:a,b⟶ℝ be an integrable function. Also, let ψ be an increasing and positive function on a,b, having a continuous derivative ψ′ on a,b. The left-sided and right-sided fractional integrals of a function f with respect to another function ψ on a,b of order μ where ℜμ>0, k>0, are defined by(16)Ia+μ,ψkfx=1kΓkμ∫axψ′tψx−ψtμ/k−1ftdt,x>a,(17)Ib−μ,ψkfx=1kΓkμ∫xbψ′tψt−ψxμ/k−1ftdt,x<b.For further study of fractional integrals, see [26, 27]. We will utilize the following well-known hypergeometric, beta, and incomplete beta functions in our results [28].(18)F12a,b;c;z=1Bb,c−b∫01tb−11−tc−b−11−zt−adt,c>b>0,z<1,Bx,y=∫01tx−11−ty−1dt=ΓxΓyΓx+y,Bx,y;z=∫0ztx−11−ty−1dt.The rest of the paper is organized as follows. InSection 2, we obtain Hadamard inequalities for generalized Riemann-Liouville fractional integrals of strongly s,m-convex functions. Many specific cases are given as outcomes of these inequalities; they are related to the results which have been published in different papers. In Section 3, by using two integral identities for generalized fractional integrals, the error bounds of fractional Hadamard inequalities are established for differentiable strongly s,m-convex functions. This paper reproduces the results which are explicitly given in [1, 2, 22, 23, 29–37].


## 2. Main Results

This section is dedicated to the Hadamard inequality for stronglys,m-convex functions via generalized Riemann–Liouville fractional integrals. We will give two versions of this inequality. First one is stated and proved in the following theorem.Theorem 7.
Letf:a,b⟶ℝ, a,b⊂0,+∞ be a positive function and f∈L1a,b. Also, let f be strongly s,m-convex function on a,b with modulus c, such that a/m,a/m2,mb∈a,b. Then, for k>0 and s,m∈0,12, the following k-fractional integral inequalities hold for operators given in (16) and (17):(19)fa+mb2+cm4μ+kμ+2kμμ+kb−a2+2k2am−mb2+2kμb−aam−mb≤Γkμ+k2smb−aμ/kIψ−1a+μ,ψkf∘ψψ−1mb+mμ/k+1Iψ−1b−μ,ψkf°ψψ−1am≤μfa+mfb2sμ+sk+mμBμ/k,s+1fb+mfa/m2k2s−cmkμb−a2+mb−a/m222sμ+kμ+2k,with μ>0.Proof.
The following inequality holds for stronglys,m-convex functions.(20)fx+my2≤fx+mfy2s−cmy−x24.
By settingx=at+m1−tb, y=a/m1−t+tb, t∈0,1, in (20), multiplying resulting inequality with tμ/k−1, and then integrating with respect to t, we get(21)kμfa+mb2≤12s∫01fat+m1−tbtμ/k−1dt+m∫01fam1−t+bttμ/k−1dt−cm4kb−a2μ+2k+2k3a/m−mb2μμ+kμ+2k+2k2b−aa/m−mbμ+kμ+2k.
By settingψu=at+m1−tb and ψv=a/m1−t+bt in (21) and by applying Definition 5, we get the following inequality:(22)fa+mb2≤Γkμ+k2smb−aμ/kIψ−1a+μ,ψkf∘ψψ−1mb+mμ/k+1Iψ−1b−μ,ψkf∘ψψ−1am−cmμ4b−a2μ+2k+2k2a/m−mb2μμ+kμ+2k+2kb−aa/m−mbμ+kμ+2k.
The above inequality leads to the first inequality of (19). On the other hand, f is strongly s,m-convex function with modulus c; for t∈0,1, we have the following inequality:(23)fta+m1−tb+mfam1−t+tb≤tsfa+mfb+m1−tsfb+mfam2−cmt1−tb−a2+mb−am22.
By integrating (23) over 0,1 after multiplying with tμ/k−1, the following inequality holds:(24)∫01tμ/k−1fta+m1−tbdt+m∫01tμ/k−1fam1−t+tbdt≤fa+mfbkμ+sk+mfb+mfam2Bμk,s+1−cmk2b−a2+mb−a/m22μ+kμ+2k.
Again using substitutions as considered in (21), we get(25)kΓkμmb−aμ/kIψ−1a+μ,ψkf°ψψ−1mb+mμ/k+1Iψ−1b−μ,ψkf°ψψ−1am≤fa+mfbkμ+sk+mfb+mfam2Bμk,s+1−cmk2b−a2+mb−a/m22μ+kμ+2k.
This leads to the second inequality of (19).Remark 1.
Under the assumption of Theorem7, the following outcomes are noted.(i)
Ifs=1, m=1, then the inequality stated in [[32], Theorem 9] is obtained.(ii)
Ifc=0, s=1, m=1, and ψ is the identity function in (19), then Theorem 4 is obtained.(iii)
Ifc=0, s=1, m=1, k=1, and ψ is the identity function in (19), then Theorem 1 is obtained.(iv)
Ifk=1, s=1, m=1, and ψ is the identity function in (19), then refinement of Theorem 1 is obtained.(v)
Ifμ=1, k=1, s=1, m=1, c=0, and ψ is the identity function in (19), then Hadamard inequality is obtained.(vi)
Ifm=1, s=1, and c=0 in (19), then the inequality [[34], Theorem 1] is obtained.(vii)
Ifc=0, k=1, m=1, and s=1 in (19), then the inequality stated in [[33], Theorem 2.1] is obtained.(viii)
Ifs=1,k=1, and ψ is the identity function in (19), then the inequality stated in [[31], Theorem 6] is obtained.(ix)
Ifk=1, m=1, s=1, μ=1, and ψ is the identity function in (19), then the inequality stated in [[35], Theorem 6] is obtained.(x)
Ifα=1, k=1, c=0, and ψ is the identity function in (19), then the inequality stated in [[29], Theorem 2.1] is obtained.Corollary 1.
Under the assumption of Theorem7 with c=0 in (19), the following inequality holds:(26)fa+mb2≤Γkμ+k2smb−aμ/kIψ−1a+μ,ψkf°ψψ−1mb+mμ/k+1Iψ−1b−μ,ψkf°ψψ−1am≤μfa+mfb2sμ+sk+mμBμ/k,s+1fb+mfa/m2k2s.Theorem 8.
Under the assumption of Theorem7, the following k-fractional integral inequality holds:(27)fa+mb2+cm16μ+kμ+2kμb−a2+am−mb2μ2+5kμ+8k2+2μb−aam−mbμ+3k≤2μ/k−sΓkμ+kmb−aμ/kIψ−1a+mb/2+μ,ψkf°ψψ−1mb+mμ/k+1Iψ−1a+mb/2m−μ,ψkf°ψψ−1am≤μfa+mfb22sμ+sk+m2F1−s,μ/k,μ+k/k;1/2fb+mfa/m22s−cmμ+3kμb−a2+mb−a/m22s+2μ+kμ+2k,with μ>0.Proof.
By settingx=at/2+m2−t/2b, y=a/m2−t/2+bt/2, t∈0,1 in (20), multiplying resulting inequality with tμ/k−1, and then integrating with respect to t, we get(28)kμfa+mb2≤12s∫01fat2+m2−t2btμ/k−1dt+m∫01fam2−t2+bt2tμ/k−1dt−cm4b−a2k4μ+2k+ka/m−mb2μ2+5kμ+8k24μμ+kμ+2k+b−aa/m−mbμ+3kk2μ+kμ+2k.
By settingψu=at/2+bm2−t/2 and ψv=a/m2−t/2+bt/2 and by applying Definition 5, we get the following inequality:(29)kμfa+mb2≤2μ/kkΓkμ2smb−aμ/kIψ−1a+mb/2+μ,ψkf°ψψ−1mb+mμ/k+1Iψ−1a+mb/2m−μ,ψkf°ψψ−1am−cm4kb−a24μ+2k+ka/m−mb2μ2+5kμ+8k24μμ+kμ+2k+kb−aa/m−mbμ+3k2μ+kμ+2k.
The above inequality leads to the first inequality of (27). On the other hand, f is strongly s,m-convex function with modulus c; for t∈0,1, we have the following inequality:(30)fat2+m2−t2b+mfam2−t2+bt2≤t2sfa+mfb+m2−t2sfb+mfam2−cmt2−tb−a2+mb−a/m224.
By integrating (30) over 0,1 after multiplying with tμ/k−1, the following inequality holds:(31)∫01fat2+m2−t2btμ/k−1dt+m∫01fam2−t2+bt2tμ/k−1dt≤kfa+mfb2ssk+μ+mkfb+fa/m22F1−s,1+μ/k,2+μ+k/k;1/2μ−cmkμ+3kb−a2+mb−a/m224μ+kμ+2k.
Again using substitutions as considered in (28), we get(32)2μ/kkΓkμmb−aμ/kIψ−1a+mb/2+μ,ψkf°ψψ−1mb+mμ/k+1Iψ−1a+mb/2m−μ,ψkf°ψψ−1am≤kfa+mfb2ssk+μ+mkfb+mfa/m22F1−s,μ/k,μ+k/k;1/2μ−cmkμ+3kb−a2+mb−a/m224μ+kμ+2k.
This leads to the second inequality of (27).Remark 2.
Under the assumption of Theorem8, the following outcomes are noted.(i)
Ifs=1 and m=1 in (27), then the inequality stated in [[32], Theorem 10] is obtained.(ii)
Ifs=1, m=1, k=1, c=0, and ψ is the identity function in (27), then Theorem 2 is obtained.(iii)
Ifs=1, m=1, k=1, and ψ is the identity function in (27), then refinement of Theorem 2 is obtained.(iv)
Ifs=1, m=1, k=1, μ=1,c=0, and ψ is the identity function in (27), then Hadamard inequality is obtained.(v)
Ifs=1, m=1, c=0, and ψ is the identity function in inequality (27), then Theorem 5 is obtained.(vi)
Ifs=1, m=1, and c=0 in (27), then the inequality stated in [[32], Corollary 5] is obtained.(vii)
Ifs=1, k=1, and ψ is the identity function in inequality (27), then the inequality stated in [[31], Theorem 7] is obtained.Corollary 3.
Under the assumption of Theorem8 with c=0 in (27), the following inequality holds:(33)fa+mb2≤2μ/k−sΓkμ+kmb−aμ/kIψ−1a+mb/2+μ,ψkf°ψψ−1mb+mμ/k+1Iψ−1a+mb/2m−μ,ψkf°ψψ−1am≤μfa+mfb22sμ+sk+mfb+mfa/m22F1−s,μ/k,μ+k/k;1/22s.


## 3. Error Estimations of Hadamard Inequalities via Stronglys,m-Convex Functions

In this section, we will study error estimations of Hadamard inequalities for generalized Riemann–Liouville fractional integrals of stronglys,m-convex functions. The estimations obtained here provide refinements of many well-known results. The Mathematica program is used for integration. We recall the well-known Hölder’s integral inequality.Theorem 9.
(see [38]). Let p>1 and 1/p+1/q=1. If f and g are real functions defined on a,b and if fp and gq are integrable functions on a,b, then(34)∫abfxgxdx≤∫abfxpdx1/p∫abgxqdx1/q,with equality holding iff Afxp=Bgxq almost everywhere, where A and B are constants.In order to prove the next result, the following lemma is useful.Lemma 1.
(see [34]). Let a<b and f:a,b⟶ℝ be a differentiable mapping on a,b. Also, suppose that f′∈La,b, ψx is an increasing and positive monotone function on a,b, having a continuous derivative ψ′x on a,b, and α∈0,1. Then, for k>0, the following identity holds:(35)fa+fb2−Γkμ+k2b−aμ/kIψ−1a+μ,ψkf°ψψ−1b+Iψ−1b−μ,ψkf°ψψ−1a=b−a2∫011−tμ/k−tμ/kf′ta+1−tbdt.Theorem 10.
Letf:a,b⟶ℝ, a,b⊂0,+∞ be a differentiable mapping on a,b such that f′∈L1a,b. Also, suppose that f′ is strongly s,m-convex on a,b with modulus c. Then, for k>0 and s,m∈0,12, the following k-fractional integral inequality holds for operators given in (16) and (17):(36)fa+fb2−Γkμ+k2b−aμ/kIψ−1a+μ,ψkf°ψψ−1b+Iψ−1b−μ,ψkf°ψψ−1a≤b−a2f′a2B12;s+1,μk+1+1−1/2s+μ/ks+μ/k+1−B1+s,1+μk+mf′bm1−1/2s+μ/k+1s+μ/k+1−F12−s,1+μ/k,2+μ/k;1/221+μ/kμ/k+1+B1+s,1+μ/k−kk+μ+2sk+ks+μ2F1−s,1+μ/k,2+μ/k;1/22s+μ/k+1sk+μ+kμ+k−cb/m−a2μ/k+2μ/k+31−μ/k+42μ/k+2,with μ>0 and F12−s,1+μ/k,2+μ/k;1/2 being the hypergeometric function.Proof.
By Lemma1, it follows that(37)fa+fb2−Γkμ+k2b−aμ/kIψ−1a+μ,ψkf°ψψ−1b+Iψ−1b−μ,ψkf°ψψ−1b≤b−a2∫011−tμ/k−tμ/kf′ta+1−tbdt.
Sincef′ is strongly s,m-convex function on a,b, for t∈0,1, we have(38)f′ta+1−tb≤tsf′a+m1−tsf′bm−cmt1−tbm−a2.
Now using (38) in (37), we have(39)fa+fb2−Γkμ+k2b−aμ/kIψ−1a+μ,ψkf°ψψ−1b+Iψ−1b−μ,ψkf°ψψ−1b≤b−a2∫011−tμ/k−tμ/ktsf′a+m1−tsf′bm−cmt1−tbm−a2dt.fa+fb2−Γkμ+k2b−aμ/kIψ−1a+μ,ψkf°ψψ−1b+Iψ−1b−μ,ψkf°ψψ−1b≤b−a2∫01/21−tμ/k−tμ/ktsf′a+m1−tsf′bm−cmt1−tbm−a2+∫1/21tμ/k−1−tμ/ktsf′a+m1−tsf′bm−cmt1−tbm−a2dt≤b−a2f′a∫01/2ts1−tμ/k−tμ/kdt+∫1/21tstμ/k−1−tμ/kdt+mf′bm∫01/21−ts1−tμ/k−tμ/kdt+∫1211−tstμ/k−1−tμ/kdt−cmbm−a2∫01/2t1−t1−tμ/k−tμ/kdt+∫1/21t1−ttμ/k−1−tμ/kdt.
We now evaluate integrals that appear on the right side of the above inequality:(40)∫01/2ts1−tμ/k−tμ/kdt+∫1/21tstμ/k−1−tμ/kdt=2B12;s+1,μk+1+1−1/2s+μ/ks+μ/k+1−B1+s,1+μk,(41)∫01/21−ts1−tμ/k−tμ/kdt+∫1/211−tstμ/k−1−tμ/kdt=1−1/2s+μ/k+1s+μ/k+1−1/21+μ/kF12−s,1+μ/k,2+μ/k;1/2μ/k+1+B1+s,1+μ/k−1/2s+μ/k+1kk+μ+2sk+ks+μ2F1−s,1+μ/k,2+μ/k;1/2sk+μ+kμ+k,(42)∫01/2t1−t1−tμ/k−tμ/kdt+∫1/21t1−ttμ/k−1−tμ/kdt=1−μ/k+4/2μ/k+2μ/k+2μ/k+3.
Using (40)–(42) in (39), we get (36).Remark 3.
Under the assumption of Theorem10, the following outcomes are noted.(i)
Ifs=1 and m=1 in (45), then the inequality stated in [[32], Theorem 11] is obtained.(ii)
Ifs=1, m=1, and c=0 in (45), then the inequality stated in [[32], Corollary 10] is obtained.(iii)
Ifs=1, m=1, k=1, and ψ is the identity function in (45), then a refined error estimation of the fractional Hadamard inequality is obtained.(iv)
Ifm=1 and c=0 in (45), then the inequality stated in [[34], Theorem 2] is obtained.(v)
Ifm=1, s=1, c=0, and ψ is the identity function in (45), then Theorem 6 is obtained.(vi)
Ifk=1, m=1, s=1, c=0, and ψ is the identity function in (45), then Theorem 3 is obtained.(vii)
Ifk=1, s=1, and ψ is the identity function in (45), then the inequality stated in [[31], Theorem 8] is obtained.Corollary 5.
Under the assumption of Theorem10 with c=0 in (3.10), the following inequality holds:(43)fa+fb2−Γkμ+k2b−aμ/kIψ−1a+μ,ψkf°ψψ−1b+Iψ−1b−μ,ψkf°ψψ−1a≤b−a2f′a2B12;s+1,μk+1+1−1/2s+μ/ks+μ/k+1−B1+s,1+μk+mf′bm1−1/2s+μ/k+1s+μ/k+1−F12−s,1+μ/k,2+μ/k;1/221+μ/kμ/k+1+B1+s,1+μk−kk+μ+2sk+ks+μ2F1−s,1+μ/k,2+μ/k;1/22s+μ/k+1sk+μ+kμ+k.Corollary 6.
Under the assumption of Theorem10 with k=1, μ=1, s=1, m=1, and ψ as the identity function in (3.10), the following inequality holds:(44)fa+fb2−1b−a−∫abfνdν≤b−a8f′a+f′b−cb−a332.Lemma 2.
Letf:a,b⟶ℝ be a differentiable mapping on a,b such that f′∈La,b. Then, for k>0 and m∈0,1, the following identity holds for operators given in (16) and (17):(45)2μ/k−1Γkμ+kmb−aμ/kIψ−1a+mb/2+μ,ψkf°ψψ−1mb+mμ/k+1Iψ−1a+mb/2m−μ,ψkf°ψψ−1am−12fa+mb2+mfa+mb2m=mb−a4∫01tμ/kf′at2+m2−t2bdt−∫01tμ/kf′am2−t2+bt2dt.Proof.
Let(46)I1=2μ/k−1Γkμ+kmb−aμ/kIψ−1a+mb/2+μ,ψkf°ψψ−1mb,I2=mμ/k+12μ/k−1Γkμ+kmb−aμ/kIψ−1a+mb/2m−μ,ψkf°ψψ−1am.
First, we evaluateI1:(47)I1=2μ/k−1μkmb−aμ/k∫ψ−1a+mb/2ψ−1mbψ′umb−ψuμ/k−1f°ψudu=−2μ/k−1mb−aμ/k∫ψ−1a+mb/2ψ−1mbdmb−ψuμ/kfψu.
Now integrating by parts, we have(48)I1=12fa+mb2+12∫ψ−1a+mb/2ψ−1mb2mb−ψumb−aμ/kψ′uf′ψudu.
Substitutingt=2mb−ψu/mb−a, so that ψu=at/2+m2−t/2b in (48), we get the following inequality:(49)I1=12fa+mb2+mb−a4∫01tμ/kf′at2+m2−t2bdt.
Now, we evaluateI2:(50)I2=mμ/k+12μ/k−1Γkμ+kmb−aμ/kIψ−1a+mb/2m−μ,ψkf°ψψ−1am=mμ/k+1μ2μ/k−1kmb−aμ/k∫ψ−1a/mψ−1a+mb/2mψ′vψv−amμ/k−1f°ψvdv=mμ/k+12μ/k−1mb−aμ/k∫ψ−1a/mψ−1a+mb/2mfψvdψv−amμ/k.
Integrating by parts, we get(51)I2=m2fa+mb2m−m2∫ψ−1a/mψ−1a+mb/2m2mψv−a/mmb−aμ/kψ′vf′ψvdv.
Substitutings=2mψv−a/m/mb−a, so that ψv=a/m2−t/2+bt/2 in (51), we get the following inequality:(52)I2=m2fa+mb2m−mb−a4∫01sμ/kf′am2−s2+bs2ds.
Adding (49) and (52), (45) is obtained.Remark 4.
Under the assumption of Lemma2, the following outcomes are noted.(i)
Ifk=1 and ψ is the identity function in (45), then the identity stated in [[30], Lemma 2.3] is obtained.(ii)
Ifm=1 in (45), then the identity stated in [[32], Lemma 2] is obtained.(iii)
Ifm=1, k=1, and ψ is the identity function in (45), then the identity stated in [[2], Lemma 3] is obtained.(iv)
Ifm=1, k=1, μ=1, and ψ is the identity function in (45), then the identity stated in [[2], Corollary 1] is obtained.(v)
Ifm=1 and ψ is the identity function in (45), then the identity stated in [[23], Lemma 3.1] is obtained.Theorem 11.
Letf:a,b⟶ℝ, a,b⊂0,+∞ be a differentiable mapping on a,b such that f′∈Ła,b. Also, suppose that f′q is strongly s,m-convex function on a,b for q≥1. Then, for k>0 and s,m∈0,12, the following k-fractional integral inequality holds for operators given in (16) and (17):(53)2μ/k−1Γkμ+kmb−aμ/kIψ−1a+mb/2+μ,ψkf°ψψ−1mb+mμ/k+1Iψ−1a+mb/2m−μ,ψkf°ψψ−1am−12fa+mb2+mfa+mb2m≤mb−a4μ/k+112μ/k+21/q2μ/k+1μ/k+2f′aq2sμ/k+s+1+2mf′bqμk+2×F12−s,1+μk,2+μk;12−cmb−a2μ/k+1μ/k+42μ/k+31/q+2mμk+2f′am2qF12−s,1+μk,2+μk;12+2μ/k+1μ/k+2f′bq2sμ/k+s+1−cmb−a/m22μ/k+1μ/k+42μ/k+31/q,with μ>0 and 1/p+1/q=1.Proof.
We divide the proof in two cases.Case 1.
Fixq=1. Applying Lemma 2 and strongly s,m-convexity of f′, we have(54)|2μ/k−1Γkμ+kmb−aμ/kIψ−1a+mb/2+μ,ψkf°ψψ−1mb+mμ/k+1Iψ−1a+mb/2m−μ,ψkf°ψψ−1am−12fa+mb2+mfa+mb2m|≤mb−a4∫01tμ/kf′at2+m2−t2bdt+∫01tμ/kf′am2−t2+bt2dt≤mb−a4f′a+mf′b2s∫01tμ/k+sdt+mf′b+mf′a/m22s∫012−tstμ/kdt−cmb−a2+mb−a/m224∫01tμ/k+12−tdt≤mb−a4kf′a+mf′b2sμ+sk+k+mkF12−s,1+μ/k,2+μ/k,1/2f′b+mf′a/m2k+μ−cmb−a2+mb−a/m22μ/k+44μ/k+2μ/k+3.Case 2.
Forq>1. From Lemma 2 and using power mean inequality, we get(55)|2μ/k−1Γkμ+kmb−aμ/kIψ−1a+mb/2+μ,ψkf°ψψ−1mb+mμ/k+1Iψ−1a+mb/2m−μ,ψkf°ψψ−1am−12fa+mb2+mfa+mb2m|≤mb−a4∫01tμ/kdt1−1/q∫01tμ/kf′at2+m2−t2bqdt1/q+∫01mtμ/kf′am2−t2+bt2qdt1/q≤mb−a4μ/k+11/pf′aq2s∫01ts+μ/kdt+mf′bq2s∫012−tstμ/kdt−cmb−a24∫012−ttμ/k+1dt1/q+mf′a/m22s×∫012−tstμ/kdt+f′bq2s∫01ts+μ/kdt−cmb−a/m224∫012−ttμ/k+1dt1/q≤mb−a4μ/k+11/pkf′aq2ssk+μ+k+mkf′bqF12−s,1+μ/k,2+μ/k,1/2k+μ−cmμ/k+4b−a24μ/k+2μ/k+31/q+mkf′a/m2qF12−s,1+μ/k,2+μ/k,1/2k+μ+kf′bq2ssk+μ+k−cmμ/k+4b−a/m224μ/k+2μ/k+31/q≤mb−a4μ/k+11/p12μ/k+1μ/k+21/q2kf′aqμ/k+1μ/k+22ssk+μ+k+2mμk+2f′bqF12−s,1+μk,2+μk,12−cmμ/k+4μ/k+1b−a22μ/k+31/q+2mf′am2qμk+2F12−s,1+μk,2+μk,12+2kμ/k+1μ/k+2f′bq2ssk+μ+k−cmμ/k+1μ/k+4b−a/m222μ/k+31/q.
Hence, we get (53).Remark 5.
Under the assumption of Theorem11, the following outcomes are noted.(i)
Ifs=1 and m=1 in (53), then the inequality stated in [[32], Theorem 12] is obtained.(ii)
Ifs=1, k=1, and ψ is the identity function in (53), then the inequality stated in [[31], Theorem 9] is obtained.(iii)
Ifs=1, k=1, c=0, and ψ is the identity function in (53), then the inequality stated in [[30], Theorem 2.4] is obtained.(iv)
Ifc=0, s=1,m=1, and ψ is the identity function in (53), then the inequality stated in [[23], Theorem 3.1] is obtained.(v)
Ifs=1, m=1, c=0, k=1, and ψ is the identity function in (53), then the inequality stated in [[2], Theorem 5] is obtained.(vi)
Ifq=1, s=1, m=1,c=0, k=1, μ=1, and ψ is the identity function in (53), then the inequality stated in [[36], Theorem 2.2] is obtained.Corollary 7.
Under the assumption of Theorem11 with c=0 in (53), the following inequality holds:(56)|2μ/k−1Γkμ+kmb−aμ/kIψ−1a+mb/2+μ,ψkf°ψψ−1mb+mμ/k+1Iψ−1a+mb/2m−μ,ψkf°ψψ−1am−12fa+mb2+mfa+mb2m|≤mb−a4μ/k+112μ/k+21/q2μ/k+1μ/k+2f′aq2sμ/k+s+1+2mf′bqμ/k+2×F12−s,1+μk,2+μk;121/q+2mμk+2f′am2q×F12−s,1+μk,2+μk;12+2μ/k+1μ/k+2f′bq2sμ/k+s+11/q.Corollary 8.
Under the assumption of Theorem11 with k=1, s=1, m=1,q=1, μ=1, and ψ as the identity function in (53), the following inequality holds:(57)1b−a∫abfvdv−fa+b2≤b−a8f′a+f′b−5cb−a212.Lemma 3.
(see [39]). Let p≥1 be a real number. For x1,x2,…,xn∈0,∞ and n≥2, the following inequality holds:(58)∑i=1nxip≤∑i=1nxip.Theorem 12.
Letf:I⟶ℝ be a differentiable mapping on a,b with a<b. Also, suppose that f′q is strongly s,m-convex function on a,b for q>1. Then, for k>0 and s,m∈0,12, the following k-fractional integral inequality holds for operators given in (16) and (17):(59)2μ/k−1Γkμ+kmb−aμ/kIψ−1a+mb/2+μ,ψkf°ψψ−1mb+mμ/k+1Iψ−1a+mb/2m−μ,ψkf°ψψ−1am−12fa+mb2+mfa+mb2m≤mb−a164μp/k+11/pf′a42ss+11/q+f′b4m−1+2s+12s1+s1/qq−2cmb−a231/q+mf′am24m−1+2s+12s1+s1/q+42ss+11/qf′bq−2cmb−a231/q,with μ>0 and 1/p+1/q=1.Proof.
By applying Lemma2 and using the property of modulus, we get(60)|2μ/k−1Γkμ+kmb−aμ/kIψ−1a+mb/2+μ,ψkf°ψψ−1mb+mμ/k+1Iψ−1a+mb/2m−μ,ψkf°ψψ−1am−12fa+mb2+mfa+mb2m|≤mb−a4∫01tμ/kf′at2+m2−t2bdt+∫01tμ/kf′am2−t2+bt2dt.
Now applying Hölder’s inequality for integrals, we get(61)|2μ/k−1Γkμ+kmb−aμ/kIψ−1a+mb/2+μ,ψkf°ψψ−1mb+mμ/k+1Iψ−1a+mb/2m−μ,ψkf°ψψ−1am−12fa+mb2+mfa+mb2m|≤b−a41μp/k+11/p∫01f′at2+m2−t2bqdt1/q+∫01f′am2−t2+bt2qdt1/q.
Using stronglys,m-convexity of f′q, we get(62)2μ/k−1Γkμ+kmb−aμ/kIψ−1a+mb/2+μ,ψkf°ψψ−1mb+mμ/k+1Iψ−1a+mb/2m−μ,ψkf°ψψ−1am−12fa+mb2+mfa+mb2m≤mb−a41μp/k+11/pf′aq2s∫01tsdt+mf′bq2s∫012−tsdt−cmb−a24∫01t2−tdt1/q+mf′a/m2q2s∫012−tsdt+f′bq2s∫01tsdt−cmb−a/m224∫01t2−tdt1/q=mb−a41μp/k+11/pf′aq2ss+1+mf′bq−1+2s+12s1+s−cmb−a261/q+mf′a/m2q−1+2s+12s1+s+f′bq2ss+1−cmb−a/m2261/q≤mb−a164μp/k+11/p4f′aq2ss+1+4mf′bq−1+2s+12s1+s−2cmb−a231/q+4mf′a/m2q−1+2s+12s1+s+4f′bq2ss+1−2cmb−a/m2231/q≤mb−a164μp/k+11/pf′a42ss+11/q+f′b4m−1+2s+12s1+s1/qq−2cmb−a231/q+mf′am24m−1+2s+12s1+s1/q+42ss+11/qf′bq−2cmb−a/m2231/q.
Here we have used Lemma3. This completes the proof.Remark 6.
Under the assumption of Theorem12, the following outcomes are noted.(i)
Ifs=1 and m=1 in (59), then the inequality stated in [[32], Theorem 13] is obtained.(ii)
Ifs=1 and ψ is the identity function in (53), then the inequality stated in [[31], Theorem 10] is obtained.(iii)
Ifs=1,k=1, c=0, and ψ is the identity function in (53), then the inequality stated in [[30], Theorem 2.7] is obtained.(iv)
Ifc=0, s=1, m=1, and ψ is the identity function in (59), then the inequality stated in [[23], Theorem 3.2] is obtained.(v)
Ifk=1, s=1, m=1, c=0, and ψ is the identity function in inequality (59), then the inequality stated in [[2], Theorem 6] is obtained.(vi)
Ifk=1, s=1, m=1, c=0, μ=1, and ψ is the identity function in inequality (59), then the inequality stated in [[37], Theorem 2.4] is obtained.Corollary 9.
Under the assumption of Theorem12 with c=0 in (61), the following inequality holds:(63)2μ/k−1Γkμ+kmb−aμ/kIψ−1a+mb/2+μ,ψkf°ψψ−1mb+mμ/k+1Iψ−1a+mb/2m−μ,ψkf°ψψ−1am−12fa+mb2+mfa+mb2m≤mb−a164μp/k+11/pf′a42ss+11/q+f′b4m−1+2s+12s1+s1/q+f′am24m−1+2s+12s1+s1/q+42ss+11/qf′b.Corollary 10.
Under the assumption of Theorem12 with q⟶1 and p⟶∞ in (59), the following inequality holds:(64)2μ/k−1Γkμ+kmb−aμ/kIψ−1a+mb/2+μ,ψkf°ψψ−1mb+mμ/k+1Iψ−1a+mb/2m−μ,ψkf°ψψ−1am−12fa+mb2+mfa+mb2m≤mb−a164f′a+f′b2ss+1+4m−1+2s+1f′b+f′a/m22s1+s−4cmb−a23.


## 4. Conclusion

In this article, we studied the Hadamard inequalities and their estimations for generalized Riemann–Liouville fractional integrals of stronglys,m-convex functions. These inequalities represent the generalizations and refinements of a number of well-known inequalities stated in [1, 2, 22, 23, 29–37]. The error estimations of Hadamard inequalities for differentiable strongly s,m-convex functions are better as compared to those which are obtained for convex functions, strongly convex functions, and strongly m-convex functions.



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