# New Hermite–Hadamard Type Inequalities forψ-Riemann–Liouville Fractional Integral via Convex Functions

Authors: Yining Sun; Run Xu
Journal: Discrete Dynamics in Nature and Society
 (2020)

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## Abstract

This paper established some new Hermite–Hadamard type inequalities forψ-Riemann–Liouville fractional integrals via convex functions. As applications, we applied the inequalities to special means of real numbers and constructed inequalities for the beta function.


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## Body

## 1. Introduction

Hermite–Hadamard integral inequality provides the estimate for the integral average of any convex function defined on a compact interval. In recent years, many mathematicians extended it tos-convex functions, quasi-convex functions, ψ-Riemann–Liouville convex functions, η-convex functions, r-convex functions, α,m-convex functions, and Lipschitzian functions, see [1–7]. The most famous are trapezoid inequality and midpoint inequality [8, 9]: suppose that f:a,b⟶ℝ is a differentiable function on a,b with a<b. If f′ is convex on a,b, then(1)fa+fb2−1b−a∫abfxdx≤b−a8f′a+f′b,1b−a∫abfxdx−fa+b2≤b−a8f′a+f′b.In [10], Mehrez and Agarwal extended the interval of f′ and established the following result: let f:a,b⟶ℝ be a differentiable function on a,b, and f′:3a−b/2,3b−a/2⟶ℝ be a continuous function on 3a−b/2,3b−a/2. q≥1. If f′q is convex on 3a−b/2,3b−a/2, then the following inequality holds:(2)1b−a∫abfxdx−fa+b2≤b−a8f′3a−b2q+f′3b−a2q1/q.The classical integro-differential equation has been studied deeply, and a rigorous and systematic theoretical system has been established. In recent decades, the theoretical research of fractional integro-differential equation has been constantly improved. Many theoretical achievements have been obtained [2–4, 7, 11–15]. The research of Hermite–Hadamard type inequalities for fractional integrals has also been deepened.In [3], Liu et al. established the Hermite–Hadamard type inequalities for ψ-Riemann–Liouville fractional integrals via convex functions: let g:a,b⟶ℝ be a positive convex integrable function, 0≤a<b. ψx is an increasing positive monotone function and have continuous derivation on a,b, and α∈0,1. Then, the following inequality holds:(3)ga+b2≤Γα+12b−aαIψ−1a+α:ψg∘ψψ−1b+Iψ−1b−α:ψg∘ψψ−1a≤ga+gb2.Letg:a,b⟶ℝ be a differentiable mapping on a,b, a<b, g′ is convex on a,b. ψx is an increasing positive monotone function on a,b and have continuous derivation on a,b, and α∈0,1. Then, the following inequalities hold:(4)ga+gb2−Γα+12b−aαIψ−1a+α:ψg∘ψψ−1bIψ−1b−α:ψg∘ψψ−1a≤b−a2α+11−12αg′a+g′b,(5)Γα+12b−aαIψ−1a+α:ψg∘ψψ−1b+Iψ−1b−α:ψg∘ψψ−1a−ga+b2≤gb−ga2+b−a2α+11−12αg′a+g′b,where Iψ−1a+α:ψ and Iψ−1b−α:ψ are the left-sided and right-sided ψ-Riemann–Liouville fractional integral operators of order α, which are listed in Section 2.The aim of this paper is to extend the range of argument of the functiong′ and establish some new Hermite–Hadamard type inequalities for ψ-Riemann–Liouville fractional integrals via convex functions on the basis of (3)–(5). Then, we apply them to special means of real numbers and construct inequalities for the beta function.


## 2. Preliminaries

In this section, we recall the definition ofψ-Riemann–Liouville fractional integral and some lemmas about Γα+1/2b−aαIψ−1a+α:ψf∘ψψ−1b+Iψ−1b−α:ψf∘ψψ−1a.Definition 1.
(see [16]). Let a,b−∞≤a<b≤∞ be a finite or infinite interval of the real line ℝ and α>0. Also, let ψx be an increasing and positive monotone function on a,b, having a continuous derivative ψ′x on a,b. The left-sided and right-sided ψ-Riemann–Liouville fractional integrals of order α for function f with respect to another function ψ on a,b are defined by(6)Ia+α:ψfx=1Γα∫axψ′tψx−ψtα−1ftdt,x>a,Ib−α:ψfx=1Γα∫xbψ′tψt−ψxα−1ftdt,x<b,respectively. Here, Γα is the gamma function.Lemma 1.
(see [3]). Let f:a,b⟶ℝ be a differentiable mapping on a,b; suppose that g′∈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, we have(7)ga+gb2−Γα+12b−aαIψ−1a+α:ψg∘ψψ−1b+Iψ−1b−α:ψg∘ψψ−1a=12b−aα∫ψ−1aψ−1bψt−aα−b−ψtαg′∘ψtψ′tdt.Lemma 2.
(see [3]). Under the assumptions of Lemma 1, we have(8)Γα+12b−aαIψ−1a+α:ψg∘ψψ−1b+Iψ−1b−α:ψg∘ψψ−1a−ga+b2=∫ψ−1aψ−1bkg′∘ψtψ′tdt+12b−aα∫ψ−1aψ−1bb−ψtα−ψt−aαg′∘ψtψ′tdt,where(9)k=12,ψ−1a+b2≤t<ψ−1b,−12,ψ−1a<t<ψ−1a+b2.Lemma 3.
(see [17]). Assume that a≥0,p≥q≥0 and p≠0. Then,(10)aq/p≤qpKq−p/pa+p−qpKq/p,for any K>0.


## 3. Main Results

First, we give an estimate forΓα+1/2b−aαIψ−1a+α:ψg∘ψψ−1b+Iψ−1b−α:ψg∘ψψ−1a.Lemma 4.
Letg:a,b⟶ℝ be a positive function, g∈L1a,b, ψx is an increasing and positive monotone function on a,b, with a continuous derivative ψ′x on a,b, 0≤a<b, and α∈0,1. If g is a convex function on a,b, then the following inequalities hold:(11)ga+b2≤Γα+12b−aαIψ−1a+α:ψg∘ψψ−1b+Iψ−1b−α:ψg∘ψψ−1a≤g3a−b/2+g3b−a/2+2ga+b/24.Proof.
From the left side of (3), we can directly get the left side of (11). Then, we prove the right side of (11).(12)Γα+12b−aαIψ−1a+α:ψg∘ψψ−1b+Iψ−1b−α:ψg∘ψψ−1a=α2b−aα∫ψ−1aψ−1bψ′vb−ψvα−1g∘ψvdv+∫ψ−1aψ−1bψ′vψv−aα−1g∘ψvdvletψv=34t+a+b4=3α8b−aα∫3a−b/33b−a/33b−a4−34tα−1g34t+a+b4dt+∫3a−b/33b−a/3b−3a4+34tα−1g34t+a+b4dt≤3α8b−aα∫3a−b/33b−a/33b−a4−34tα−112g32t+12ga+b2dt+∫3a−b/33b−a/3b−3a4+34tα−112g32t+12ga+b2dt=3α16b−aα∫3a−b/33b−a/33b−a4−34tα−1+b−3a4+34tα−1g32tdt+∫3a−b/33b−a/33b−a4−34tα−1+b−3a4+34tα−1ga+b2dt.
Let(13)N1=∫3a−b/33b−a/33b−a4−34tα−1+b−3a4+34tα−1g32tdt,N2=∫3a−b/33b−a/33b−a4−34tα−1+b−3a4+34tα−1ga+b2dt,we have(14)N1=12α−1∫3a−b/33b−a/33b−a2−32tα−1+b−3a2+32tα−1g32tdtletψv=32t=12α−2⋅3∫ψ−13a−b/2ψ−13b−a/23b−a2−ψvα−1+ψv−3a−b2α−1g∘ψvψ′vdv=Γα2α−2⋅3Iψ−13a−b/2+α:ψg∘ψψ−13b−a2+Iψ−13a−b/2−α:ψg∘ψψ−13a−b2,(15)N2=ga+b2∫3a−b/33b−a/33b−a4−34tα−1+b−3a4+34tα−1dt=43ga+b2∫3a−b/43b−a/43b−a4−xα−1+b−3a4+xα−1dx=8b−aα3αga+b2.
From (12)–(15), we have the inequality(16)Γα+12b−aαIψ−1a+α:ψg∘ψψ−1b+Iψ−1b−α:ψg∘ψψ−1a≤3α16b−aαN1+N2=Γα+12α+2b−aαIψ−13a−b/2+α:ψg∘ψψ−13b−a2+Iψ−13a−b/2−α:ψg∘ψψ−13a−b2+12ga+b2.
From (3), we have(17)0<Γα+12α+2b−aαIψ−13a−b/2+α:ψg∘ψψ−13b−a2+Iψ−13b−a/2−α:ψg∘ψψ−13a−b2≤g3a−b/2+g3b−a/24.
From (16) and (17), we obtain the right-hand side of inequality (11).
This completes the proof.Remark 1.
Based on (16), we have the inequality(18)Γα+12b−aαIψ−1a+α:ψg∘ψψ−1b+Iψ−1b−α:ψg∘ψψ−1a−ga+b/22≤Γα+12α+2b−aαIψ−13a−b/2+α:ψg∘ψψ−13b−a2+Iψ−13b−a/2−α:ψg∘ψψ−13a−b2.
From (17) and (18), we obtain a new inequality(19)0<Γα+12b−aαIψ−1a+α:ψg∘ψψ−1b+Iψ−1b−α:ψg∘ψψ−1a−ga+b/22≤g3a−b/2+g3b−a/24.Theorem 1.
Letg:a,b⟶ℝ be a differentiable mapping on a,b with 0≤a<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. g′:3a−b/2,3b−a/2⟶ℝ is a continuous function on 3a−b/2,3b−a/2. q≥1. If g′q is a convex function on 3a−b/2,3b−a/2, then the following inequalities hold:(20)Γα+12b−aαIψ−1a+α:ψg∘ψψ−1b+Iψ−1b−α:ψg∘ψψ−1a−ga+b2≤b−a21α+1+12g′3b−a2+g′3a−b2.Proof.
By Lemma2, substituting 3a−b/2 for a and 3b−a/2 for b, we have(21)Γα+12α+1b−aαIψ−13a−b/2+α:ψg∘ψψ−13b−a2+Iψ−13b−a/2−α:ψg∘ψψ−13a−b2=ga+b2+∫ψ−13a−b/2ψ−1a+b/2−12g′∘ψvψ′vdv+∫ψ−1a+b/2ψ−13b−a/212g∘ψvψ′vdv+12α+1b−aα∫ψ−13a−b/2ψ−13b−a/23b−a2−ψvα−ψv−3a−b2αg′∘ψvψ′vdv.
From (16) and (21), we have(22)Γα+12b−aαIψ−1a+α:ψg∘ψψ−1b+Iψ−1b−α:ψg∘ψψ−1a−ga+b2≤Γα+12α+2b−aαIψ−13a−b/2+α:ψg∘ψψ−13b−a2+Iψ−13b−a/2−α:ψg∘ψψ−13a−b2−12ga+b2=12∫ψ−13a−b/2ψ−1a+b/2−12g′∘ψvψ′vdv+∫ψ−1a+b/2ψ−13b−a/212g′∘ψvψ′vdv+12α+1b−aα∫ψ−13a−b/2ψ−13b−a/23b−a2−ψvα−ψv−3a−b2αg′∘vψ′vdv≤∫ψ−13a−b/2ψ−1a+b/214g′∘ψvψ′vdv+∫ψ−1a+b/2ψ−13b−a/214g′∘ψvψ′vdv+12α+2b−aα∫ψ−13a−b/2ψ−13b−a/23b−a2−ψvα−ψv−3a−b2αg′∘ψvψ′vdv≤∫ψ−13a−b/2ψ−13b−a/214g′∘ψvψ′vdv+12α+2b−aα∫ψ−13a−b/2ψ−13b−a/23b−a2−ψvαg′∘ψvψ′vdv+∫ψ−13a−b/2ψ−13b−a/2ψv−3a−b2αg′∘ψvψ′vdvletψv=3b−a2−2tb−a≤b−a2∫01g′3b−a21−t+3a−b2tdt+b−a2∫01tαg′3b−a21−t+3a−b2tdt+b−a2∫011−tαg′3b−a21−t+3a−b2tdt.
Ifq=1 and the function g′ is a convex function on 3a−b/2,3b−a/2, then we have(23)g′t3a−b2+1−t3b−a2≤tg′3a−b2+1−tg′3b−a2,for any t∈0,1.
Therefore, we obtain(24)∫01g′t3a−b2+1−t3b−a2dt≤∫01tg′3a−b2dt+∫011−tg′3b−a2dt=g′3a−b/2+g′3b−a/22,(25)∫011−tαg′3b−a21−t+3a−b2tdt+∫01tαg′3b−a21−t+3a−b2tdt≤∫011−tαtg′3a−b2+1−tg′3b−a2dt+∫01tαtg′3a−b2+1−tg′3b−a2dt=g′3a−b2∫011−tαt+tα+1dt+g′3b−a2∫011−tα+1+tα1−tdt=1α+1g′3a−b2+1α+1g′3b−a2=1α+1g′3a−b2+g′3b−a2.
From (22), (24), and (25), if q=1, we have(26)Γα+12b−aαIψ−1a+α:ψg∘ψψ−1b+Iψ−1b−α:ψg∘ψψ−1a−ga+b2≤b−a2⋅α+32α+2g′3a−b2+g′3b−a2.
Suppose thatq>1, g′q is convex on 3a−b/2,3b−a/2; using Ho¨lder’s inequality q>1,p=q/q−1 and Jensen inequality, we have(27)∫01g′t3a−b2+1−t3b−a2dt≤∫011q/q−1dt1−1/q∫01g′t3a−b2+1−t3b−a2qdt1/q≤∫01tg′3a−b2q+1−tg′3b−a2qdt1/q=g′3a−b/2q2+g′3b−a/2q21/q≤12g′3a−b2+g′3b−a2,(28)∫01tαg′3b−a21−t+3a−b2tdt=∫01tα1−1/qtα1/qg′3b−a21−t+3a−b2tdt≤∫01tαdt1−1/q∫01tαg′3b−a21−t+3a−b2tqdt1/q≤1α+11−1/q∫01tαtg′3a−b2q+1−tg′3b−a2qdt1/q=1α+11−1/q1α+2g′3a−b2q+Bα+1,2g′3b−a2q1/q,(29)∫011−tαg′3b−a21−t+3a−b2tdt=∫011−tα1−1/q1−tα1/qg′3b−a21−t+3a−b2tdt≤∫011−tαdt1−1/q∫011−tαg′3b−a21−t+3a−b2tqdt1/q≤1α+11−1/q∫011−tαtg′3a−b2q+1−tg′3b−a2qdt1/q=1α+11−1/qB2,α+1g′3a−b2q+1α+2g′3b−a2q1/q,where B is the beta function.
From (27)–(29), using Jensen inequality, we have(30)∫01g′3b−a21−t+3a−b2tdt+∫01tαg′3b−a21−t+3a−b2tdt+∫011−tαg′3b−a21−t+3a−b2tdt≤12g′3a−b2+g′3b−a2+1α+11−1/q2121α+2+B2,α+1g′3b−a2q+g′3a−b2q1/q≤12g′3a−b2+g′3b−a2+1α+11−1/q121/q−11α+2+B2,α+11/q21/q−1g′3b−a2+g′3a−b2=1α+11−1/q1α+2+B2,α+11/q+12g′3b−a2+g′3a−b2=1α+11−1/q1α+11/q+12g′3b−a2+g′3a−b2=1α+1+12g′3b−a2+g′3a−b2.
From (22) and (30), if q>1, we have(31)Γα+12b−aαIψ−1a+α:ψg∘ψψ−1b+Iψ−1b−α:ψg∘ψψ−1a−ga+b2≤b−a21α+1+12g′3b−a2+g′3a−b2.
In conclusion, we get (20). The proof is complete.
In the context ofq=1, comparing (5) with (20), the conclusion of (20) is easier to get. If we get the value of g′3b−a/2,g′3a−b/2, we get the estimation for(32)Γα+12b−aαIψ−1a+α:ψg∘ψψ−1b+Iψ−1b−α:ψg∘ψψ−1a−ga+b2.
But (5) needs to calculate the value of g′a,g′b,g′a,g′b.
Based on Theorem1, we give Theorem 2.Theorem 2.
Under the assumptions of Theorem1, if q>1, then the following inequalities hold:(33)Γα+12b−aαIψ−1a+α:ψg∘ψψ−1b+Iψ−1b−α:ψg∘ψψ−1a−ga+b2≤b−a21+αqq−11−1/q+12g′3b−a2+g′3a−b2.Proof.
By Ho¨lder’s inequality 1/p+1/q=1,p=q/q−1 and Jensen inequality, we have(34)∫01tαg′3b−a21−t+3a−b2tdt≤∫01tαq/q−1dt1−1/q∫01g′3b−a21−t+3a−b2tqdt1/q≤1+αqq−11−1/q∫01tg′3a−b2q+1−tg′3b−a2qdt1/q=1+αqq−11−1/q12g′3a−b2q+12g′3b−a2q1/q,(35)∫011−tαg′3b−a21−t+3a−b2tdt≤∫011−tαq/q−1dt1−1/q∫01g′3b−a21−t+3a−b2tqdt1/q≤1+αqq−11−1/q∫01tg′3a−b2q+1−tg′3b−a2qdt1/q=1+αqq−11−1/q12g′3a−b2q+12g′3b−a2q1/q.
From (27), (34), and (35), using Jensen inequality, we have(36)∫01g′3b−a21−t+3a−b2tdt+∫01tαg′3b−a21−t+3a−b2tdt+∫011−tαg′3b−a21−t+3a−b2tdt≤1+αqq−11−1/q+12g′3b−a2+g′3a−b2.
From (22) and (36), we have(37)Γα+12b−aαIψ−1a+α:ψg∘ψψ−1b+Iψ−1b−α:ψg∘ψψ−1a−ga+b2≤b−a21+αqq−11−1/q+12g′3b−a2+g′3a−b2.
The proof is complete.Corollary 1.
From Theorems1‐2, we get the following inequalities for q>1:(38)Γα+12b−aαIψ−1a+α:ψg∘ψψ−1b+Iψ−1b−α:ψg∘ψψ−1a−ga+b2≤b−a2minN1,N2g′3b−a2+g′3a−b2,where N1=1/α+1+1/2andN2=1+αq/q−11−1/q+1/2.
Now, we get the estimation for(39)Γα+12b−aαIψ−1a+α:ψg∘ψψ−1b+Iψ−1b−α:ψg∘ψψ−1a−g3a−b/2+g3b−a/2+2ga+b/24,in different ways from above.Theorem 3.
Under the assumptions of Theorem1, if g′q is a convex function on 3a−b/2,3b−a/2, q≥1, then the following inequalities hold:(40)Γα+12b−aαIψ−1a+α:ψg∘ψψ−1b+Iψ−1b−α:ψg∘ψψ−1a−g3a−b/2+g3b−a/2+2ga+b/24≤b−a2α+2g′3a−b2+g′3b−a2.Proof.
By Lemma1, substituting 3a−b/2 for a and 3b−a/2 for b, we have(41)Γα+12α+1b−aαIψ−13a−b/2+α:ψg∘ψψ−13b−a2+Iψ−13b−a/2−α:ψg∘ψψ−13a−b2=g3a−b/2+g3b−a/22−12α+1b−aα∫ψ−13a−b/2ψ−13b−a/2ψv−3a−b2α−3b−a2−ψvαg′∘ψvψ′vdv.
From (16) and (41), we have(42)Γα+12b−aαIψ−1a+α:ψg∘ψψ−1b+Iψ−1b−α:ψg∘ψψ−1a−g3a−b/2+g3b−a/2+2ga+b/24≤Γα+12α+2b−aαIψ−13a−b/2+α:ψg∘ψψ−13b−a2+Iψ−13b−a/2−α:ψg∘ψψ−13a−b2−g3a−b/2+g3b−a/24=−12α+1b−aα∫ψ−13a−b/2ψ−13b−a/2ψv−3a−b2α−3b−a2−ψvαg′∘ψvψ′vdv≤−12α+1b−aα∫ψ−13a−b/2ψ−13b−a/2ψv−3a−b2α−3b−a2−ψvαg′∘ψvψ′vdvletψv=3a−b2+2tb−a=b−a2∫01tα−1−tαg′3a−b21−t+3b−a2tdt≤b−a2∫01tαg′3a−b21−t+3b−a2tdt+∫011−tαg′3a−b21−t+3b−a2tdt.
Suppose thatq=1; the function g′ is a convex function on 3a−b/2,3b−a/2, so we have(43)Γα+12b−aαIψ−1a+α:ψg∘ψψ−1b+Iψ−1b−α:ψg∘ψψ−1a−g3a−b/2+g3b−a/2+2ga+b/24≤b−a2∫01tα1−tg′3a−b2+tg′3b−a2dt+∫011−tα1−tg′3a−b2+tg′3b−a2dt≤b−a2∫01tα1−tg′3a−b2dt+∫01tα+1g′3b−a2dt+∫011−tα+1g′3a−b2dt+∫011−tαtg′3b−a2dt=b−a2B2,α+1+1α+2g′3a−b2+g′3b−a2.
Then, we suppose thatq>1; the function g′|q is a convex function on 3a−b/2,3b−a/2, so we have(44)∫01tαg′3a−b21−t+3b−a2tdt≤∫01tα1−1/qtα1/qg′3a−b21−t+3b−a2tdt≤∫01tαdt1−1/q∫01tαg′t3b−a2+1−t3a−b2qdt1/q≤1α+11−1/q∫01tαtg′3b−a2q+1−tg′3a−b2qdt1/q≤1α+11−1/q1α+2g′3b−a2q+B2,α+1f′3a−b2q1/q,(45)∫011−tαg′3a−b21−t+3b−a2tdt≤∫011−tα1−1/q1−tα1/qg′3a−b21−t+3b−a2tdt≤∫011−tαdt1−1/q∫011−tαg′t3b−a2+1−t3a−b2qdt1/q≤1α+11−1/q∫011−tαtg′3b−a2q+1−tg′3a−b2qdt1/q≤1α+11−1/qB2,α+1g′3b−a2q+1α+2g′3a−b2q1/q.
From (42), (44), and (45), using Jensen inequality, for q>1, we have(46)Γα+12b−aαIψ−1a+α:ψg°ψψ−1b+Iψ−1b−α:ψg°ψψ−1a−g3a−b/2+g3b−a/2+2ga+b/24≤121/qb−a1α+11−1/q1α+2+B2,α+11/qg′3a−b2q+g′3b−a2q1/q≤b−a2α+2g′3a−b2+g′3b−a2.
In conclusion, we get (40). The proof is complete.
Based on Theorem3, we give Theorems 4 and 5.Theorem 4.
Under the assumptions of Theorem1, if g′q is a convex function on 3a−b/2,3b−a/2, q>1, then the following inequalities hold:(47)Γα+12b−aαIψ−1a+α:ψg∘ψψ−1b+Iψ−1b−α:ψg∘ψψ−1a−g3a−b/2+g3b−a/2+2ga+b/24≤b−a21+αqq−11−1/qg′3b−a2+g′3a−b2.Proof.
From the convex function ofg′q and Holder inequality, we have(48)∫01tαg′3a−b21−t+3b−a2tdt≤∫01tαq/q−1dt1−1/q∫01g′t3b−a2+1−t3a−b2qdt1/q≤1+αqq−11−1/q∫01tg′3b−a2q+1−tg′3a−b2qdt1/q≤1+αqq−11−1/q12g′3b−a2q+12g′3a−b2q1/q,(49)∫011−tαg′3a−b21−t+3b−a2tdt≤∫011−tαq/q−1dt1−1/q∫01g′t3b−a2+1−t3a−b2qdt1/q≤1+αqq−11−1/q∫01tg′3b−a2q+1−tg′3a−b2qdt1/q≤1+αqq−11−1/q12g′3b−a2q+12g′3a−b2q1/q.
From (42), (47), and (49), using Jensen inequality, we have(50)Γα+12b−aαIψ−1a+α:ψg∘ψψ−1b+Iψ−1b−α:ψg∘ψψ−1a−g3a−b/2+g3b−a/2+2ga+b/24≤b−a1+αqq−11−1/q121/qg′3b−a2q+g′3a−b2q1/q≤b−a21+αqq−11−1/qg′3b−a2+g′3a−b2.
The proof is complete.Theorem 5.
Under the assumptions of Theorem1, if g′q is a convex function on 3a−b/2,3b−a/2, q>1, then the following inequalities hold:(51)Γα+12b−aαIψ−1a+α:ψg∘ψψ−1b+Iψ−1b−α:ψg∘ψψ−1a−g3a−b/2+g3b−a/2+2ga+b/24≤b−a42pα+1+1−p1/pg′3b−a2+g′3a−b2,where 1/p+1/q=1.Proof.
By (42), using Holder inequality, Jensen inequality, and Lemma 3 (let K=1,q/p=p), we have(52)Γα+12b−aαIψ−1a+α:ψg∘ψψ−1b+Iψ−1b−α:ψg∘ψψ−1a−g3a−b/2+g3b−a/2+2ga+b/24≤b−a2∫01tα−1−tαg′3a−b21−t+3b−a2tdt≤b−a2∫01tα−1−tαdt∫01g′3a−b21−t+3b−a2tdt≤b−a2∫01tα+1−tαpdt1/p∫01g′3a−b21−t+3b−a2tqdt1/q≤b−a2∫01p1−tα+ptα+1−pdt1/p∫011−tg′3a−b2q+tg′3b−a2qdt1/q≤b−a42pα+1+1−p1/pg′3b−a2+g′3a−b2.
The proof is complete.Corollary 2.
From Theorems3–5, we get the following inequality for q>1:(53)Γα+12b−aαIψ−1a+α:ψg∘ψψ−1b+Iψ−1b−α:ψg∘ψψ−1a−g3a−b/2+g3b−a/2+2ga+b/24≤minN3,N4,N5g′3b−a2+g′3a−b2,where N3=b−a/2α+2,N4=b−a/21+αq/q−11−1/q,N5=b−a/42p/α+1+1−p1/p.


## 4. Applications

Example 1.
We consider the following the arithmetic means, geometric means, logarithmic means, and generalized logarithmic means for positive real numbersa,b,anda≠b:(54)Aa,b=a+b2,a,b∈R,Ga,b=ab,a,b∈R,La,b=b−alnb−lna,a≠b,ab≠0,Lna,b=bn+1−an+1n+1b−a1/n,n∈Z∖−1,0,a,b∈R,a≠b.
Now, using the results of Section3, we have some conclusions to the special means of real numbers.(1)
Leta,b∈ℝ,0<a<b, then we have(55)L−1a,b−A−1a,b≤b−aminN1,N2A3b−a2−2,3a−b2−2,whereN1=1,N2=1+q/q−11−1/q+1/2.In fact, applying Corollary1 with gx=1/x,ψx=x,andα=1, we have(56)L−1a,b−A−1a,b=lnb−lnab−a−2a+b=12b−a∫ab1vdv+∫ab1vdv−2a+b=α2b−aα∫ψ−1aψ−1bψ′vb−ψvα−1g∘ψvdv+∫ψ−1aψ−1bψ′vψv−aα−1g∘ψvdv−ga+b2=Γα+12b−aαIψ−1a+α:ψg∘ψψ−1b+Iψ−1b−α:ψg∘ψψ−1a−ga+b2≤b−a2minN1,N2g′3b−a2+g′3a−b2=b−a2minN1,N23b−a2−2+3a−b2−2=b−aminN1,N2A3b−a2−2,3a−b2−2.Then, we obtain the result.(2)
Leta,b∈ℝ,0<a<b, then we have(57)Lnna,b−Ana,b≤nb−aminN1,N2A3b−a2n−1,3a−b2n−1,whereN1=1,N2=1+q/q−11−1/q+1/2.Applying Corollary1 with gx=xn,ψx=x,andα=1, by similar calculation, we obtain the result.(3)
Leta,b∈ℝ,0<a<b, then we have(58)G−2a,b−A−2a,b≤2b−aminN1,N2A3b−a2−3,3a−b2−3,where N1=1,andN2=1+q/q−11−1/q+1/2.
Applying Corollary1 with gx=1/x2,ψx=x,α=1, by similar calculation, we obtain the result.
From Corollary2, we get the estimate for Bα,ρ.Example 2.
Letα>0,a=0,b=1,gx=xρ−1x∈0,1,ρ≥3,ψx=x, Γα is the gamma function, and Bα,ρ is the beta function. Then, g′ is convex on −1/2,3/2. So, we have(59)Γα+12b−aαIψ−1a+α:ψg∘ψψ−1b=α2∫011−xα−1xρ−1dx=α2Bα,ρ,Γα+12b−aαIψ−1b−α:ψg∘ψψ−1a=α2∫01xα+ρ−2dx=α2α+ρ−1.
From Corollary2, the following inequality holds:(60)Bα,ρ+1α+ρ−1−−1/2ρ−1+3/2ρ−1+21/2ρ−12α≤2αρ−1minN3,N4,N512ρ−2+32ρ−2,
whereN3=1/2α+2,N4=1/21+αq/q−11−1/q,andN5=1/42p/α+1+1−p1/p.


## 5. Conclusion

This paper extends the range of argument of the functiong′ and establishes some new Hermite–Hadamard type inequalities for ψ-Riemann–Liouville fractional integrals via convex functions. If g is a convex function on a,b, we obtain an estimate for(61)Γα+12b−aαIψ−1a+α:ψg∘ψψ−1b+Iψ−1b−α:ψg∘ψψ−1a.Ifg′q is a convex function on 3a−b/2,3b−a/2, we obtain estimates for(62)Γα+12b−aαIψ−1a+α:ψg∘ψψ−1b+Iψ−1b−α:ψg∘ψψ−1a−ga+b2,Γα+12b−aαIψ−1a+α:ψg∘ψψ−1b+Iψ−1b−α:ψg∘ψψ−1a−g3a−b/2+g3b−a/2+2ga+b/24.In the context ofq=1, both (5) and (20) estimate for(63)Γα+12b−aαIψ−1a+α:ψg∘ψψ−1b+Iψ−1b−α:ψg∘ψψ−1a−ga+b2.In Theorem1, we extend the range of argument of the function g′, and get a different conclusion. g′ is convex on 3a−b/2,3b−a/2. If we get the value of g′3b−a/2,g′3a−b/2, we get the conclusion of (20). In [3], g′ is convex on a,b, and if we have the value of g′a, g′b, ga and gb, we have the conclusion of (5).



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