# Refinements and Generalizations of Some Fractional Integral Inequalities via Strongly Convex Functions

Authors: Ghulam Farid; Hafsa Yasmeen; Chahn Yong Jung; Soo Hak Shim; Gaofan Ha
Journal: Mathematical Problems in Engineering
 (2021)

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## Abstract

In this article, we have established the Hadamard inequalities for strongly convex functions using generalized Riemann–Liouville fractional integrals. The findings of this paper provide refinements of some fractional integral inequalities. Furthermore, the error bounds of these inequalities are given by using two generalized integral identities.


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## Body

## 1. Introduction

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.The above inequality is well-known as the Hadamard inequality. This inequality provides lower and upper estimates for integral average of a convex function. Since the appearance of this result in literature, it has drawn attention of many mathematicians of recent age and it is one of the most extensively studied results for convex functions. In [1, 2], Sarikaya et al. have studied it via Riemann–Liouville fractional integrals of convex functions. After these versions of Hadamard inequality, many researchers were motivated and elegantly produced fractional inequalities using different types of fractional integrals. Also, many new classes of functions have been introduced in the establishment of fractional Hadamard inequalities; for details, we refer the readers to [3–11].Fractional calculus studies the integrals and derivatives of any arbitrary order, real or complex. Its history begins at the end of seventeenth century, when G. W. Leibniz and Marquis de l’Hospital in 1695 introduced it for first time by discussing the differentiation of functions of order1/2. However, it experienced a rapid growth over the short span of time. For example, Lagrange, Laplace, Lacroix, Fourier, Abel, Liouville, Riemann, Green, Holmgren, Grunwald, Letnikov, Sonin, Laurent, Nekrassov, Krug, and Weyl made their major contributions to establish a solid foundation of fractional calculus (see [12–14] and references there in). Fractional integral and derivative operators are the key factors in the development of fractional calculus. Recently, the generalizations [15–17], extensions [18–20], and applications [21–23] for fractional operators have been made by many researchers in mathematics, fluid mechanics [24–26], biological population models [27], and numerical methods [28].Our aim in this paper is to utilize generalized Riemann–Liouville fractional integrals with monotonically increasing function. The Hadamard inequality is studied for these integral operators of strongly convex functions, and also, by using some integral identities, error bounds are established. Next, we give the definition of strongly convex function introduced by Polyak [29] (see also [30]).Definition 1.
LetD be a convex subset of X, X,. be a normed space. A function f:D⊂X⟶ℝ will be called strongly convex function with modulus C≥0 if(2)fxt+1−ty≤tfx+1−tfy−Ct1−ty−x2,holds ∀x,y∈D⊆X,t∈0,1. For C=0, (2) gives the definition of convex function.In the following, we give the definition of Riemann–Liouville fractional integrals.Definition 2.
Letf∈L1a,b. Then, left-sided and right-sided Riemann–Liouville fractional integrals of a function f of order μ where ℜμ>0 are defined as follows:(3)Ia+μfx=1Γμ∫axx−tμ−1ftdt,x>a,(4)Ib−μfx=1Γμ∫xbt−xμ−1ftdt,x<b.The fractional versions of Hadamard inequality by Riemann–Liouville fractional integrals are given in the following theorems.Theorem 1.
(see [1]). Let f:a,b⟶ℝ be a positive function with 0≤a<b and f∈L1a,b. If f is a convex function on a,b, then the following fractional integral inequalities hold:(5)fa+b2≤Γα+12b−aαIa+αfb+Ib−αfa≤fa+fb2,with α>0.Theorem 2.
(see [2]). Let f:a,b⟶ℝ be a positive function with 0≤a<b and f∈L1a,b. If f is a convex function on a,b, then the following fractional integral inequalities hold:(6)fa+b2≤2α−1Γα+1b−aαIa+b/2+αfb+Ia+b/2−αfa≤fa+fb2,with α>0.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−Γα+12b−aαIa+αfb+Ib−αfa≤b−a2α+11−12αf′a+f′b.In the following, refinements of Theorem1–3 are given.Theorem 4.
(see [31]). Let f:a,b⟶ℝ be a positive function with 0≤a<b and f∈L1a,b. If f is strongly convex function on a,b with modulus C≥0, then the following fractional integral inequalities hold:(8)fa+b2+Cb−a2α2−α+24α+1α+2≤Γα+12b−aαIa+αfb+Ib−αfa≤fa+fb2−Cαb−a2α+1α+2,with α.Theorem 5.
(see [32]). Let f:a,b⟶ℝ be a positive function with 0≤a<b and f∈L1a,b. If f is strongly convex function on a,b with modulus C≥0, then the following fractional integral inequalities hold:(9)fa+b2+Cb−a22α+1α+2≤2α−1Γα+1b−aαIa+b/2+αfb+Ia+b/2−αfa≤fa+fb2−Cαb−a2α+34α+1α+2,with α.Theorem 6.
(see [32]). Let f:a,b⟶ℝ be a differentiable mapping on a,b with a<b and f∈L1a,b. If f is strongly convex function on a,b with modulus C≥0, then the following fractional integral inequalities hold:(10)fa+fb2−Γα+12b−aαIa+αfb+Ib−αfa≤b−a2α+11−12αf′a+f′b−Cb−a3α+2α+31−α+42α+2,with α.In [33], k-fractional Riemann–Liouville integrals are defined as follows.Definition 3.
Letf∈L1a,b. Then, k-fractional Riemann–Liouville integrals of order μ, where ℜμ>0, k>0, are defined by(11)Ika+μfx=1kΓkμ∫axx−tμ/k−1ftdt,x>a,(12)Ikb−μfx=1kΓkμ∫xbt−xμ/k−1ftdt,x<b,where Γk. is defined as follows [34]:(13)Γkμ=∫0∞tα−1e−tk/kdt,ℜμ>0.Ifk=1, (11) and (12) coincide with (3) and (4).Farid et al. [35, 36] proved the following k-fractional Hadamard inequalities.Theorem 7.
(see [35]). 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 inequalities for k-fractional integrals hold:(14)fa+b2≤Γkα+k2b−aα/kIka+αfb+Ikb−αfa≤fa+fb2.Theorem 8.
(see [36]). 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 inequalities for k-fractional integrals hold:(15)fa+b2≤2α/k−1Γkα+kb−aα/kIka+b/2+αfb+Ika+b/2−αfa≤fa+fb2.Theorem 9.
(see [35]). 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 hold:(16)fa+fb2−Γkα+k2b−aα/kIka+αfb+Ikb−αfa≤b−a2α/k+11−12α/kf′a+f′b.In the following, we give the definition of generalized Riemann–Liouville fractional integrals by a monotonically increasing function:Definition 4.
(see [37]). 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 defined by(17)Ia+μ,ψfx=1Γμ∫axψ′tψx−ψtμ−1ftdt,x>a,(18)Ib−μ,ψfx=1Γμ∫xbψ′tψt−ψxμ−1ftdt,x<b.Ifψ is identity function, then (17) and (18) coincide with (3) and (4).Thek-analogue of generalized Riemann–Liouville fractional integrals are defined as follows:Definition 5.
(see [38]). 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(19)Ika+μ,ψfx=1kΓkμ∫axψ′tψx−ψtμ/k−1ftdt,x>a,(20)Ikb−μ,ψfx=1kΓkμ∫xbψ′tψt−ψxμ/k−1ftdt,x<b.Ifk=1, (19) and (20) coincide with (17) and (18). If ψ is taken as identity function, (19) and (20) coincide with (11) and (12). If ψ is taken as identity function along with k=1, (19) and (20) coincide with (3) and (4). For more details of above defined fractional integrals, one can see [13, 39].In Section2, we establish Hadamard inequalities for generalized Riemann–Liouville fractional integrals of strongly convex functions. The particular cases are given as consequences of these inequalities which are connected with already published results. In Section 3, by using two integral identities for generalized fractional integrals, the error bounds of fractional Hadamard inequalities are established. The findings of this paper are connected with results that are explicitly proved in [1, 2, 31, 35, 36, 40–44].


## 2. Main Results

Theorem 10.
Letf:a,b⟶ℝ be a positive function with 0≤a<b and f∈L1a,b. Also, suppose that f is strongly convex function on a,b with modulus C≥0, ψ is an increasing and positive monotone function on a,b, having a continuous derivative ψ′x on a,b. Then, for k>0, the following k-fractional integral inequalities hold:(21)fa+b2+Cb−a2α2−kα+2k24α+kα+2k≤Γkα+k2b−aα/kIkψ−1a+α,ψf∘ψψ−1b+Ikψ−1b−α,ψf∘ψψ−1a≤fa+fb2−Ckαb−a2α+kα+2k,with α>0.Proof.
Since the functionf is strongly convex function, so for x,y∈a,b, we have(22)fx+y2≤fx+fy2−C4x−y2.
Letx=at+1−tb, y=1−ta+tb for t∈0,1 in (22) and multiplying the resulting inequality with tα/k−1 on both sides, we get(23)2fa+b2tα/k−1≤fat+1−tbtα/k−1+f1−ta+bttα/k−1−C2b−a21−2t2tα/k−1.
Integrating (23) over the interval 0,1, we get(24)2fa+b2∫01tα/k−1dt≤∫01fat+1−tbtα/k−1dt+∫01f1−ta+bttα/k−1dt−C2b−a2∫011−2t2tα/k−1dt,2kαfa+b2≤∫01fat+1−tbtα/k−1dt+∫01f1−ta+bttα/k−1dt−Ckb−a2α2−kα+2k22αα+kα+2k.
Multiplying (24) by α/2k, we get(25)fa+b2≤α2k∫01fat+1−tbtα/k−1dt+∫01f1−ta+bttα/k−1dt−Cb−a2α2−kα+2k24α+kα+2k.
Takingu∈a,b so that ψu=at+b1−t, that is, t=b−ψu/b−a, and v∈a,b so that ψv=a1−t+bt, that is, t=ψv−a/b−a, in (25), then by applying Definition 5, we get the following inequality:(26)fa+b2+Cb−a2α2−kα+2k24α+kα+2k≤Γkα+k2b−aα/kIkψ−1a+α,ψf∘ψψ−1b+Ikψ−1b−α,ψf∘ψψ−1a.
Sincef is strongly convex function, for t∈0,1, we also have following inequality:(27)fta+1−tb+f1−ta+tb≤fa+fb−2Ct1−tb−a2.
Multiplying (27) with tα/k−1 and then integrating over the interval 0,1, we get(28)∫01tα/k−1fta+1−tbdt+∫01tα/k−1f1−ta+tbdt≤fa+fb∫01tα/k−1dt−2Cb−a2∫01tα/k1−tdt,∫01tα/k−1fta+1−tbdt+∫01tα/k−1f1−ta+tbdt≤kfa+fbα−2Ck2b−a2α+kα+2k.
Multiplying (28) by α/2k, we get(29)α2k∫01tα/k−1fta+1−tbdt+∫01tα/k−1f1−ta+tbdt≤fa+fb2−Ckαb−a2α+kα+2k.
Again takingψu=at+b1−t, that is, t=b−ψu/b−a, and ψv=a1−t+bt, that is, t=ψv−a/b−a, in (29), then by applying Definition 5, we get the following inequality:(30)Γkα+k2b−aα/kIkψ−1a+α,ψf∘ψψ−1b+Ikψ−1b−α,ψf∘ψψ−1a≤fa+fb2−Ckαb−a2α+kα+2k.
Combining (26) and (30), we get (21).Remark 1.
Under the assumption of Theorem10, one can get the following results:(i)
IfC=0, k=1, and ψ is identity function in (21), then Theorem 1 is obtained.(ii)
IfC=0 and ψ is identity function in (21), then Theorem 7 is obtained.(iii)
Ifk=1 and ψ is identity function in (21), then Theorem 4 is obtained.(iv)
Ifα=1, k=1, C=0, and ψ is identity function in (21), then Hadamard inequality is obtained.(v)
IfC=0 in (21), then the inequality (Theorem 1) stated in [41] is obtained.(vi)
IfC=0 and k=1 in (21), then the inequality (Theorem 2.1) stated in [40] is obtained.(vii)
Ifk=1, α=1, and ψ is identity function in (21), then the inequality (Theorem 6) stated in [44] is obtained.Corollary 1.
Under the assumption of Theorem10 with k=1 in (21), the following inequality holds:(31)fa+b2+Cb−a2α2−α+24α+1α+2≤Γα+12b−aαIψ−1a+α,ψf∘ψψ−1b+Iψ−1b−α,ψf∘ψψ−1a≤fa+fb2−Cαb−a2α+1α+2.Corollary 2.
Under the assumption of Theorem10 with ψ as identity function in (21), the following inequality holds:(32)fa+b2+Cb−a2α2−kα+2k24α+kα+2k≤Γkα+k2b−aα/kIka+αfb+Ikb−αfa≤fa+fb2−Ckαb−a2α+kα+2k.Theorem 11.
Letf:a,b⟶ℝ be a positive function with 0≤a<b and f∈L1a,b. Also, suppose that f is strongly convex function on a,b with modulus C≥0 and ψ is an increasing and positive monotone function on a,b, having a continuous derivative ψ′x on a,b. Then, for k>0, the following k-fractional integral inequalities hold:(33)fa+b2+k2Cb−a22α+kα+2k≤2α/k−1Γkα+kb−aα/kIkψ−1a+b/2+α,ψf∘ψψ−1b+Ikψ−1a+b/2−α,ψf∘ψψ−1a≤fa+fb2−Cαb−a2α+3k4α+kα+2k,with α>0.Proof.
Letx=at/2+2−t/2b and y=2−t/2a+bt/2 in (22), and multiplying the resulting inequality with tα/k−1, we get(34)fa+b2tα/k−1≤12fat2+2−t2btα/k−1+f2−t2a+bt2tα/k−1−C4b−a21−t2tα/k−1.
Integrating (34) over 0,1, we get(35)2kαfa+b2≤∫01fat2+2−t2btα/k−1dt+∫01fa2−t2+bt2tα/k−1dt−k3Cb−a2αα+kα+2k.
Multiplying above inequality withα/2k, we get(36)fa+b2+Ck2b−a22α+kα+2k≤α2k∫01fat2+2−t2btα/k−1dt+∫01fa2−t2+bt2tα/k−1dt.
Takingu∈a,b so that ψu=at/2+b2−t/2, that is, t=2b−ψu/b−a, and v∈a,b so that ψv=a2−t/2+bt/2, that is, t=2ψv−a/b−a, in (36), then by applying Definition 5, we get following inequality:(37)fa+b2+Ck2b−a22α+kα+2k≤2α/k−1Γkα+kb−aα/kIkψ−1a+b/2+α,ψf∘ψψ−1b+Ikψ−1a+b/2−α,ψf∘ψψ−1a.
Sincef is strongly convex function on a,b, for t∈0,1, we have the following inequality:(38)fat2+2−t2b+fa2−t2+bt2≤fa+fb−Ct2−tb−a22.
Multiplying (38) with tα/k−1 on both sides and integrating over 0,1, we get(39)∫01fat2+2−t2btα/k−1dt+∫01fa2−t2+bt2tα/k−1dt≤fa+fb∫01tα/k−1dt−Cb−a22∫012−ttα/kdt.
Multiplying (39) with α/2k on both sides, we get(40)α2k∫01fat2+2−t2btα/k−1dt+∫01f2−t2a+bt2tα/k−1dt≤fa+fb2−Cαb−a2α+3k4α+kα+2k.
Again takingψu=at/2+b2−t/2, that is, t=2b−ψv/b−a, and ψv=a2−t/2+bt/2, that is, t=2ψv−a/b−a, in (40), then by applying Definition 5, we get the following inequality:(41)2α/kΓkα+k2b−aα/kIkψ−1a+b/2+α,ψf∘ψψ−1b+Ikψ−1a+b/2−α,ψf∘ψψ−1a≤fa+fb2−Cαb−a2α+3k4α+kα+2k.
Combining (37) and (41), (33) is obtained.Remark 2.
Under the assumption of Theorem11, one can get the following results:(i)
Ifk=1, C=0, and ψ is identity function in (33), then Theorem 2 is obtained.(ii)
Ifk=1 and ψ is identity function in (33), then Theorem 5 is obtained.(iii)
IfC=0 and ψ is identity function in (33), then inequality (Theorem 2.1) stated in [36] is obtained.(iv)
Ifk=1, C=0, α=1, and ψ is identity function in (33), then Hadamard inequality is obtained.(v)
Ifk=1, α=1, and ψ is identity function in (33), then the inequality (Theorem 6) stated in [44] is obtained.Corollary 3.
Under the assumption of Theorem11 with C=0 in (33), the following inequality holds:(42)fa+b2≤2α/k−1Γkα+kb−aα/kIkψ−1a+b/2+α,ψf∘ψψ−1b+Ikψ−1a+b/2−α,ψf∘ψψ−1a≤fa+fb2.Corollary 4.
Under the assumption of Theorem11 with k=1 in (33), the following inequality holds:(43)fa+b2+Cb−a22α+1α+2≤2α−1Γα+1b−aαIψ−1a+b/2+α,ψf∘ψψ−1b+Iψ−1a+b/2−α,ψf∘ψψ−1a≤fa+fb2−Cαb−a2α+34α+1α+2.Corollary 5.
Under the assumption of Theorem11 with ψ as identity function in (33), the following inequality holds:(44)fa+b2+Ck2b−a22α+kα+2k≤2α/k−1Γkα+kb−aα/kIka+b/2+αfb+Ika+b/2−αfa≤fa+fb2−Cαb−a2α+3k4α+kα+2k.


## 3. Error Bounds of Hadamard Inequalities for Strongly Convex Functions

In this section, we provide the error bounds of fractional Hadamard inequalities using generalized Riemann–Liouville fractional integrals via strongly convex functions. Estimations here are further refined as compared to those already established for convex functions. The following lemma is useful to prove the next result.Lemma 1.
(see [41]). Let a<b and f:a,b⟶ℝ be a differentiable mapping on a,b. Also, suppose that f′∈La,b, ψ 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:(45)fa+fb2−Γkα+k2b−aα/kIkψ−1a+α,ψf∘ψψ−1b+Ikψ−1b−α,ψf∘ψψ−1a=b−a2∫011−tα/k−tα/kf′ta+1−tbdt.Theorem 12.
Letf:a,b⟶ℝ be a differentiable mapping on a,b with a<b. Also, suppose that f′ is strongly convex function on a,b with modulus C≥0 and ψx is an increasing and positive monotone function on a,b, having a continuous derivative ψ′x on a,b. Then, for k>0, the following k-fractional integral inequalities hold:(46)fa+fb2−Γkα+k2b−aα/kIkψ−1a+α,ψf∘ψψ−1b+Ikψ−1b−α,ψf∘ψψ−1a≤b−a2α/k+11−12α/kf′a+f′b−Cb−a3α/k+2α/k+31−α/k+42α/k+2,with α>0.Proof.
From Lemma1 and strongly convexity of f′, we have(47)fa+fb2−Γkα+k2b−aα/kIkψ−1a+α,ψf∘ψψ−1b+Ikψ−1b−αf∘ψψ−1b≤b−a2∫011−tα/k−tα/kf′ta+1−tdt≤b−a2∫011−tα/k−tα/ktf′a+1−tf′b−Ct1−tb−a2dt=b−a2∫01/21−tα/k−tα/ktf′a+1−tf′b−Ct1−tb−a2dt+∫1/21tα/k−1−tα/ktf′a+1−tf′b−Ct1−tb−a2dt.
It can be noted that(48)∫01/21−tα/k−tα/ktf′a+1−tf′b−Ct1−tb−a2dt=f′a∫01/2t1−tα/k−tα/k+1dt+f′b∫01/21−tα/k+1−1−ttα/kdt−Cb−a2∫01/2t1−tα/k+1dt−∫01/21−ttα/k+1dt=f′a1α/k+1α/k+2−1/2α/k+1α/k+1+f′b1α/k+2−1/2α/k+1α/k+1−Cb−a2α/k+2α/k+31−α/k+42α/k+2.
By similar evaluation, one can have(49)∫1/21tα/k−1−tα/ktf′a+1−tf′b−Ct1−tb−a2dt=f′a1α/k+2−1/2α/k+1α/k+1+f′b1α/k+1α/k+2−1/2α/k+1α/k+1−Cb−a2α/k+2α/k+31−α/k+42α/k+2.
Therefore, (47) implies(50)fa+fb2−Γkα+k2b−aα/kIkψ−1a+α,ψf∘ψψ−1b+Ikψ−1b−αf∘ψψ−1b≤b−a2f′a1α/k+1α/k+2−1/2α/k+1α/k+1+f′b1α/k+2−1/2α/k+1α/k+1−Cb−a2α/k+2α/k+31−α/k+42α/k+2+f′a1α/k+2−1/2α/k+1α/k+1+f′b1α/k+1α/k+2−1/2α/k+1α/k+1−Cb−a2α/k+2α/k+31−α/k+42α/k+2.
From which after a little computation, one can get (46).Remark 3.
Under the assumption of Theorem12, one can get the following results:(i)
Ifk=1 and ψ is identity function in (46), then Theorem 6 is obtained.(ii)
IfC=0 and ψ is identity function in (46), then Theorem 9 is obtained.(iii)
Ifk=1, C=0, and ψ is identity function in (46), then Theorem 3 is obtained.(iv)
Ifk=1, C=0, α=1, and ψ is identity function in (46), then Theorem 2.2 in [42] is obtained.(v)
Ifk=1 and ψ is identity function in (46), then Theorem 6 is obtained.Corollary 6.
Under the assumption of Theorem12 with C=0 in (46), the following inequality holds:(51)fa+fb2−Γkα+k2b−aα/kIkψ−1a+α,ψf∘ψψ−1b+Ikψ−1b−α,ψf∘ψψ−1a≤b−a2α/k+11−12α/kf′a+f′b.Corollary 7.
Under the assumption of Theorem12 with k=1 in (46), the following inequality holds:(52)fa+fb2−Γα+12b−aαIψ−1a+α,ψf∘ψψ−1b+Iψ−1b−α,ψf∘ψψ−1a≤b−a2α+11−12αf′a+f′b−Cb−a3α+2α+31−α+42α+2.Corollary 8.
Under the assumption of Theorem12 with ψ as identity function in (46), the following inequality holds:(53)fa+fb2−Γkα+k2b−aα/kIka+αfb+Ikb−αfa≤b−a2α/k+11−12α/kf′a+f′b−Cb−a3α/k+2α/k+31−α/k+42α/k+2.We now derive a new fractional integral identity for fractional integrals (19) and (20).Lemma 2.
Leta<b and f:a,b⟶ℝ be a differentiable mapping on a,b. Also, suppose that f′∈La,b and ψ is an increasing and positive monotone function on a,b, having a continuous derivative ψ′x on a,b. Then, for k>0, the following identity holds:(54)2α/k−1Γkα+kb−aα/kIkψ−1a+b/2+α,ψf∘ψψ−1b+Ikψ−1a+b/2−α,ψf∘ψψ−1a−fa+b2=b−a4∫01tα/kf′at2+2−t2bdt−∫01tα/kf′2−t2a+bt2dt,with α>0.Proof.
Let(55)I1=2α/k−1Γkα+kb−aα/kIkψ−1a+b/2+α,ψf∘ψψ−1b,I2=2α/k−1Γkα+kb−aα/kIkψ−1a+b/2−α,ψf∘ψψ−1a.
Then, we have(56)I1=2α/k−1αkb−aα/k∫ψ−1a+b/2ψ−1bψ′vb−ψvα/k−1f∘ψvdv=−2α/k−1b−aα/k∫ψ−1a+b/2ψ−1bfψvdb−ψvα/k=12fa+b2+12∫ψ−1a+b/2ψ−1bf′ψv2b−ψvb−aα/kψ′vdv.
By substitutingt=2b−ψv/b−a, we will get(57)I1=12fa+b2+b−a4∫01tα/kf′at2+2−t2bdt.
We also have(58)I2=2α/k−1αkb−aα/k∫ψ−1aψ−1a+b/2ψ′vψv−aα/k−1f∘ψvdv=2α/k−1b−aα/k∫ψ−1aψ−1a+b/2dψv−aα/kfψv=12fa+b2−12∫ψ−1aψ−1a+b/2f′ψv2ψv−ab−aα/kψ′vdv.
By substitutings=2ψv−a/b−a, we will get(59)I2=12fa+b2−b−a4∫01f′a2−s2+bs2sα/kds.
By summing (57) and (59), we get (54).Remark 4.
Under the assumption of Lemma2, one can get the following results:(i)
Ifk=1 and ψ is identity function in (54), then the identity (Lemma 3) stated in [2] is obtained.(ii)
Ifk=1, α=1, and ψ is identity function in (54), then the identity (Corollary 1) stated in [2] is obtained.(iii)
Ifψ is identity function in (54), then the identity (Lemma 3.1) stated in [36] is obtained.Corollary 9.
Under the assumption of Lemma2 with k=1 in (54), the following identity holds:(60)2α−1Γα+1b−aαIψ−1a+b/2+α,ψf∘ψψ−1b−Iψ−1a+b/2−α,ψf∘ψψ−1a−fa+b2=b−a4∫01tαf′at2+2−t2bdt+∫01tαf′2−t2a+bt2dt.Using above lemma, we give the following error bounds of thek-fractional Hadamard inequality.Theorem 13.
Letf:I⟶ℝ be a differentiable mapping on a,b with a<b. Also, suppose that f′q is strongly convex function on a,b with modulus C≥0 for q≥1, and ψ is an increasing and positive monotone function on a,b, having a continuous derivative ψ′x on a,b. Then, for k>0, the following k-fractional integral inequalities hold:(61)2α/k−1Γkα+kb−aα/kIkψ−1a+b/2+α,ψf∘ψψ−1b+Ikψ−1a+b/2−α,ψf∘ψψ−1a−fa+b2≤b−a4α/k+112α/k+21/qαk+1f′aq+αk+3f′bq−Cb−a2α/k+1α/k+42α/k+31/q+αk+3f′aq+αk+1f′bq−Cb−a2α/k+1α/k+42α/k+31/q,with α>0.Proof.
From Lemma2 and strongly convexity of f′, let q=1, we have(62)2α/k−1Γkα+kb−aα/kIkψ−1a+b/2+α,ψf∘ψψ−1b+Ikψ−1a+b/2−α,ψf∘ψψ−1a−fa+b2≤b−a4∫01tα/kf′at2+2−t2bdt+∫01tα/kf′a2−t2+bt2dt≤b−a4|f′a|+f′b∫01tα/kdt−C2b−a2∫01tα/k+12−tdt≤b−a4α/k+1|f′a|+f′b−Cb−a2α/k+4α/k+1α/k+2α/k+k.
Now, forq>1, we proceed as follows.
From Lemma2 and using power mean inequality, we get(63)2α/k−1Γkα+kb−aα/kIkψ−1a+b/2+α,ψf∘ψψ−1b+Ikψ−1a+b/2−α,ψf∘ψψ−1a−fa+b2≤b−a4∫01tα/kdt1−1/q∫01tα/kf′at2+2−t2bqdt1/q+∫01tα/kf′2−t2a+bt2qdt1/q.
Strongly convexity off′q gives(64)2α/k−1Γα+kb−aα/kIkψ−1a+b/2+α,ψf∘ψψ−1b+Ikψ−1a+b/2−α,ψf∘ψψ−1a−fa+b2≤b−a4α/k+11/pf′aq∫01tα/k+12dt+f′bq∫012tα/k−tα/k+12dt−Cb−a24∫01tα/k+12−tdt1/q+f′aq∫012tα/k−tα/k+12dt+f′bq∫01tα/k+12dt−Cb−a24∫01tα/k+12−tdt1/q≤b−a4α/k11/pf′aq2α/k+2+f′bqα/k+32α/k+1α/k+2−Cb−a2α/k+44α/k+2α/k+31/q+f′aqα/k+32α/k+1α/k+2+f′bq2α/k+2−Cb−a2α/k+44α/k+2α/k+31/q,which after a little computation gives the required result.Remark 5.
Under the assumption of Theorem13, one can get the following results:(i)
IfC=0 and ψ is identity function in (61), then the inequality (Theorem 3.1) stated in [36] is obtained.(ii)
IfC=0, k=1, and ψ is identity function in (61), then the inequality (Theorem 5) stated in [2] is obtained.(iii)
Ifq=1, C=0, k=1, α=1, and ψ is identity function in (61), then the inequality (Theorem 2.2) stated in [43] is obtained.Corollary 10.
Under the assumption of Theorem13 with C=0 in (61), the following inequality holds:(65)2α/k−1Γkα+kb−aα/kIkψ−1a+b/2+α,ψf∘ψψ−1b+Ikψ−1a+b/2−α,ψf∘ψψ−1a−fa+b2≤b−a4α/k+112α/k+21/qαk+1f′aq+αk+3f′bq1/q+αk+3f′aq+αk+1f′bq1/q.Corollary 11.
Under the assumption of Theorem13 with k=1 in (61), the following inequality holds:(66)2α−1Γα+1b−aαIψ−1a+b/2+α,ψf∘ψψ−1b+Iψ−1a+b/2−α,ψf∘ψψ−1a−fa+b2≤b−a4α+112α+21/qα+1f′aq+α+3f′bq−Cb−a2α/k+1α+42α+31/q+α+3f′aq+α+1f′bq−Cb−a2α+1α+42α+31/q.Corollary 12.
Under the assumption of Theorem13 with ψ is identity function in (61), the following inequality holds:(67)2α/k−1Γkα+kb−aα/kIka+b/2+αfb−Ika+b/2−αfa−fa+b2≤b−a4α/k+112α/k+21/qαk+1f′aq+αk+3f′bq−Cb−a2α/k+1α/k+42α/k+31/q+αk+3f′aq+αk+1f′bq−Cb−a2α/k+1α/k+42α/k+31/q.Corollary 13.
Under the assumption of Theorem13 with k=1, q=1, α=1, and ψ as identity function in (61), the following inequality holds:(68)1b−a∫abfvdv−fa+b2≤b−a8f′a+f′b−5Cb−a212.Theorem 14.
Letf:I⟶ℝ be a differentiable mapping on a,b with a<b. Also, suppose that f′q is strongly convex function on a,b for q>1, and ψ is an increasing and positive monotone function on a,b, having a continuous derivative ψ′x on a,b. Then, for k>0, the following k-fractional integral inequalities hold:(69)2α/k−1Γkα+kb−aα/kIkψ−1a+b/2+α,ψf∘ψψ−1b+Ikψ−1a+b/2−α,ψf∘ψψ−1a−fa+b2≤b−a164αp/k+11/pf′a+31/qf′bq−2Cb−a231/q+31/qf′a+f′bq−2Cb−a231/q,with α>0.Proof.
From Lemma2 and using the property of modulus, we get(70)2α/k−1Γkα+kb−aα/kIkψ−1a+b/2+α,ψfψ−1b+Ikψ−1a+b/2−α,ψf∘ψψ−1a−fa+b2≤b−a4∫01tα/kf′t2a+2−t2bdt+∫01tα/kf′2−t2a+t2bdt.
Now applyingHölder’s inequality for integrals, we get(71)2α/k−1Γkα+kb−aα/kIkψ−1a+b/2+α,ψf∘ψψ−1b+Ikg−1a+b/2−α,ψf∘ψψ−1a−fa+b2≤b−a4∫01tαp/kdt1/p∫01f′t2a+2−t2bqdt1/q+∫01tαp/kdt1/p∫01f′2−t2a+t2bqdt1/q.
Using strongly convexity off′q, we get(72)2α/k−1Γkα+kb−aα/kIkψ−1a+b/2+α,ψf∘ψψ−1b+Ikψ−1a+b/2−α,a+b/2f∘ψψ−1a−fa+b2≤b−a41αp/k+11/pf′aq∫01t2dt+f′bq∫012−t2dt−Cb−a24∫01t2−tdt1/q+f′aq∫012−t2dt+f′bq∫01t2dtCb−a24∫01t2−tdt1/q=b−a41αp/k+11/pf′aq+3f′bq−2Cb−a231/q+3f′aq+f′bq−2Cb−a231/q≤b−a41αp/k+11/pf′a+31/qf′bq−2Cb−a231/q+31/qf′a+f′bq−2Cb−a231/q.
Here, we have used the factaq+bq≤a+bq, for q>1, a,b≥0. This completes the proof.Remark 6.
Under the assumption of Theorem14, one can get the following results:(i)
IfC=0 and ψ is identity function in (69), then the inequality (Theorem 3.2) stated in [36] is obtained.(ii)
Ifk=1, C=0, α=1 and ψ is identity function in (69), then the inequality (Theorem 2.3) stated in [43] is obtained.Corollary 14.
Under the assumption of Theorem14 with C=0 in (69), the following inequality holds:(73)2α/k−1Γkα+kb−aα/kIkψ−1a+b/2+α,ψf∘ψψ−1b+Ikψ−1a+b/2−α,ψf∘ψψ−1a−fa+b2≤b−a44αp/k+11/pf′a+f′b.Corollary 15.
Under the assumption of Theorem14 with k=1 in (69), the following inequality holds:(74)2α−1Γα+1b−aαIψ−1a+b/2+α,ψf∘ψψ−1b+Iψ−1a+b/2−α,ψψ−1a−fa+b2≤b−a164αp+11/pf′a+31/qf′bq−2Cb−a231/q+31/qf′a+f′bq−2Cb−a231/q.Corollary 16.
Under the assumption of Theorem14 with C=0 and k=1 in (69), the following inequality holds:(75)2α−1Γα+1b−aαIψ−1a+b/2+α,ψf∘ψψ−1b+Iψ−1a+b/2−α,ψf∘ψψ−1a−fa+b2≤b−a44αp+11/pf′a+f′b.Corollary 17.
Under the assumption of Theorem14 with ψ is identity function in (69), the following inequality holds:(76)2α/k−1Γkα+kb−aα/kIka+b/2+αfb+Ika+b/2−αfa−fa+b2≤b−a164αp/k+11/pf′a+31/qf′bq−2Cb−a231/q+31/qf′a+f′bq−2Cb−a231/q.Corollary 18.
Under the assumption of Theorem14 with k=1, α=1, and ψ is identity function in (69), the following inequality holds:(77)1b−a∫abftdt−fa+b2≤b−a164p+11/pf′a+31/qf′bq−2Cb−a231/q+31/qf′a+f′bq−2Cb−a231/q.


## 4. Conclusion

In this paper, we have studied Hadamard inequalities and their error estimations using generalized Riemann–Liouville fractional integrals of strongly convex functions. The Hadamard inequalities obtained in this work are refinements as well as generalizations of many well-known inequalities. The error estimations of the Hadamard inequalities for differentiable strongly convex functions are better as compared to those which are obtained for convex functions. The authors are analyzing other well-known fractional integral operators for several kinds of functions in their future work.



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