Riemann-Liouville fractional Hermite-Hadamard inequalities. Part I: for once differentiable geometric-arithmetically s-convex functions

Abstract

By using the definition of geometric-arithmetically s-convex functions in (Analysis 33:197-208, 2013) and first-order fractional integral identities in (Math. Comput. Model. 57:2403-2407, 2013; J. Appl. Math. Stat. Inform. 8:21-28, 2012; Comput. Math. Appl. 63:1147-1154, 2012), we present some interesting Riemann-Liouville fractional Hermite-Hadamard inequalities for differentiable geometric-arithmetically s-convex functions. Both beta function and incomplete beta function are used in the desired estimations.

MSC:26A33, 26A51, 26D15.

1 Introduction

Fractional integrals and derivatives arise in many engineering and scientific disciplines as the mathematical modeling of systems and processes in the fields of physics, chemistry, aerodynamics, electrodynamics of a complex medium. In particular, the subject of fractional calculus has become a rapidly growing area and has found applications in diverse fields ranging from physical sciences and engineering to biological sciences and economics [1–8].

Fractional Hermite-Hadamard inequalities involving all kinds of fractional integrals have attracted by many researchers. Many authors paid attention to the study of fractional Hermite-Hadamard inequalities according to the first-order integral equalities and convex functions of different classes. For instance, the readers can refer to [9–11] for convex functions and [12] for nondecreasing functions, [13–15] for m-convex functions and [16] for $(s,m)$-convex functions, [17, 18] for functions satisfying s-e-condition, [19] for $(\alpha ,m)$-logarithmically convex functions and the references therein.

Very recently, the authors [20] introduced the new concept of geometric-arithmetically s-convex functions and established some interesting Hermite-Hadamard type inequalities for integer integrals of such functions. However, to our knowledge, fractional Hermite-Hadamard inequalities for geometric-arithmetically s-convex functions have not been reported. Motivated by [9, 10, 13, 20], we study Riemann-Liouville fractional Hermite-Hadamard type inequalities for geometric-arithmetically s-convex functions by means of first-order fractional integral equalities.

2 Preliminaries

In this section, we introduce notations, definitions and preliminary facts.

Definition 2.1 (see [3]) Let $f\in L[a,b]$. The symbols $J_{a^{+}}^{\alpha }f$ and $J_{b^{-}}^{\alpha }f$ denote the left-sided and right-sided Riemann-Liouville fractional integrals of the order $\alpha \in R^{+}$ and are defined by

$$\left(J_{a^{+}}^{\alpha }f\right)(x)=\frac{1}{\mathrm{\Gamma }(\alpha )}{\int }_a^x{(x-t)}^{\alpha -1}f(t)dt(0\le a<x\le b)$$

and

$$\left(J_{b^{-}}^{\alpha }f\right)(x)=\frac{1}{\mathrm{\Gamma }(\alpha )}{\int }_x^b{(t-x)}^{\alpha -1}f(t)dt(0\le a\le x<b),$$

respectively, here $\mathrm{\Gamma }(\cdot )$ is the gamma function.

Definition 2.2 (see [20])

Let $f:I\subseteq R^{+}\to R^{+}$ and $s\in (0,1]$. A function $f(x)$ is said to be geometric-arithmetically s-convex on I if for every $x,y\in I$ and $t\in [0,1]$, we have

$$f\left(x^t{y}^{1-t}\right)\le t^s(f(x))+{(1-t)}^{s}f(y).$$

Definition 2.3 (see [21])

The incomplete beta function is defined as follows:

$$B_x(a,b)={\int }_0^x{t}^{a-1}{(1-t)}^{b-1}dt,$$

where $x\in [0,1]$, $a,b>0$.

The following inequality will be used in the sequel.

Lemma 2.4 (see [19])

For $t\in [0,1]$, we have

$$\begin{array}{c}{(1-t)}^n\le 2^{1-n}-t^n\mathit{\text{for }}n\in [0,1],\hfill \\ {(1-t)}^n\ge 2^{1-n}-t^n\mathit{\text{for }}n\in [1,\mathrm{\infty }).\hfill \end{array}$$

The following elementary inequality was used in the proof directly in [20]. Here, we revisit this inequality from the point of our view and give a proof.

Lemma 2.5 For $t\in [0,1]$, $x,y>0$, we have

$$tx+(1-t)y\ge y^{1-t}{x}^t.$$

Proof If $x\ge y$, then we consider the function $f(z)=tz+1-t-z^t$, $z\ge 1$. If $y\ge x$, then we consider the function $g(z)=t+(1-t)z-z^{1-t}$, $z\ge 1$. Clearly $f(\cdot )$ and $g(\cdot )$ are increasing functions for all $z\ge 1$, then $f(z)\ge f(1)=0$ and $g(z)\ge g(1)=0$. So, $f(\frac{x}{y})\ge 0$ and $g(\frac{y}{x})\ge 0$, i.e., the statement holds. □

We collect the following first-order fractional integrals identities.

Lemma 2.6 ([[9], Lemma 2])

Let $f:[a,b]\to R$ be a differentiable mapping on $(a,b)$ with $a<b$. If $f^{\prime }\in L[a,b]$, then the following equality for fractional integrals holds:

$$\begin{array}{r}\frac{f(a)+f(b)}{2}-\frac{\mathrm{\Gamma }(\alpha +1)}{2{(b-a)}^{\alpha }}[J_{a^{+}}^{\alpha }f(b)+J_{b^{-}}^{\alpha }f(a)]\\ =\frac{b-a}{2}{\int }_0^1[{(1-t)}^{\alpha }-t^{\alpha }]f^{\prime }(ta+(1-t)b)dt.\end{array}$$

Lemma 2.7 ([[10], Lemma 2.1])

Let $f:[a,b]\to \mathbb{R}$ be a differentiable mapping on $(a,b)$. If $f^{\prime }\in L[a,b]$, then the following equality for fractional integrals holds:

$$\begin{array}{r}\frac{\mathrm{\Gamma }(\alpha +1)}{2{(b-a)}^{\alpha }}[J_{a+}^{\alpha }f(b)+J_{b-}^{\alpha }f(a)]-f\left(\frac{a+b}{2}\right)\\ =\frac{b-a}{2}{\int }_0^1[h(t)-{(1-t)}^{\alpha }+t^{\alpha }]f^{\prime }(ta+(1-t)b)dt,\end{array}$$

where

$$h(t)=\{\begin{array}{cc}1,\hfill & 0<t<\frac{1}{2},\hfill \\ -1,\hfill & \frac{1}{2}<t<1.\hfill \end{array}$$

Lemma 2.8 ([[13], Lemma 2])

Let $f:[a,b]\to R$ be a differentiable mapping on $(a,b)$ with $a<b$. If $f^{\prime }\in L[a,b]$, then the following equality for fractional integrals holds:

$$\begin{array}{r}\frac{{(x-a)}^{\alpha }+{(b-x)}^{\alpha }}{b-a}f(x)-\frac{\mathrm{\Gamma }(\alpha +1)}{b-a}[J_{x^{-}}^{\alpha }f(a)+J_{x^{+}}^{\alpha }f(b)]\\ =\frac{{(x-a)}^{\alpha +1}}{b-a}{\int }_0^1{t}^{\alpha }{f}^{\prime }(tx+(1-t)a)dt\\ -\frac{{(b-x)}^{\alpha +1}}{b-a}{\int }_0^1{t}^{\alpha }{f}^{\prime }(tx+(1-t)b)dt.\end{array}$$

3 Main results

In this section, we use Lemmas 2.6, 2.7 and 2.8 via geometric-arithmetically s-convex functions to derive the main results in this paper.

3.1 The first results

By using Lemma 2.6, we can obtain the main results in this section.

Theorem 3.1 Let $f:[0,b]\to R$ be a differentiable mapping. If $|f^{\prime }|$ is measurable and $|f^{\prime }|$ is decreasing and geometric-arithmetically s-convex on $[0,b]$ for some fixed $\alpha \in (0,\mathrm{\infty })$, $s\in (0,1]$, $0\le a<b$, then the following inequality for fractional integrals holds:

$$\begin{array}{r}|\frac{f(a)+f(b)}{2}-\frac{\mathrm{\Gamma }(\alpha +1)}{2{(b-a)}^{\alpha }}[J_{a^{+}}^{\alpha }f(b)+J_{b^{-}}^{\alpha }f(a)]|\\ \le \frac{(b-a)(2^{\alpha +s}|f^{\prime }(b)|-|f^{\prime }(a)|-|f^{\prime }(b)|)}{(\alpha +s+1)2^{\alpha +s+1}}\\ +(b-a)|f^{\prime }(a)|[0.5B(s+1,\alpha +1)-B_{0.5}(\alpha +1,s+1)]\\ +(b-a)|f^{\prime }(b)|[B_{0.5}(s+1,\alpha +1)-0.5B(s+1,\alpha +1)],\end{array}$$

where

$$B(s+1,\alpha +1)={\int }_0^1{t}^s{(1-t)}^{\alpha }dt,$$

and

$$B_{0.5}(s+1,\alpha +1)={\int }_0^{0.5}{t}^s{(1-t)}^{\alpha }dt.$$

Proof By using Definition 2.2, Definition 2.3, Lemma 2.5 and Lemma 2.6, we have

$$\begin{array}{r}|\frac{f(a)+f(b)}{2}-\frac{\mathrm{\Gamma }(\alpha +1)}{2{(b-a)}^{\alpha }}[J_{a^{+}}^{\alpha }f(b)+J_{b^{-}}^{\alpha }f(a)]|\\ \le \frac{b-a}{2}{\int }_0^1|{(1-t)}^{\alpha }-t^{\alpha }|\left|f^{\prime }(ta+(1-t)b)\right|dt\\ \le \frac{b-a}{2}{\int }_{\frac{1}{2}}^1(t^{\alpha }-{(1-t)}^{\alpha })\left|f^{\prime }(ta+(1-t)b)\right|dt\\ +\frac{b-a}{2}{\int }_0^{\frac{1}{2}}({(1-t)}^{\alpha }-t^{\alpha })\left|f^{\prime }(ta+(1-t)b)\right|dt\\ \le \frac{b-a}{2}{\int }_{\frac{1}{2}}^1(t^{\alpha }-{(1-t)}^{\alpha })\left|f^{\prime }\left(a^t{b}^{1-t}\right)\right|dt\\ +\frac{b-a}{2}{\int }_0^{\frac{1}{2}}({(1-t)}^{\alpha }-t^{\alpha })\left|f^{\prime }\left(a^t{b}^{1-t}\right)\right|dt\\ \le \frac{b-a}{2}{\int }_{\frac{1}{2}}^1(t^{\alpha }-{(1-t)}^{\alpha })[t^s|f^{\prime }(a)|+{(1-t)}^s|f^{\prime }(b)|]dt\\ +\frac{b-a}{2}{\int }_0^{\frac{1}{2}}({(1-t)}^{\alpha }-t^{\alpha })[t^s|f^{\prime }(a)|+{(1-t)}^s|f^{\prime }(b)|]dt\\ \le \frac{(b-a)|f^{\prime }(a)|}{2}{\int }_{\frac{1}{2}}^1{t}^{\alpha +s}dt-\frac{(b-a)|f^{\prime }(a)|}{2}{\int }_{\frac{1}{2}}^1{t}^s{(1-t)}^{\alpha }dt\\ -\frac{(b-a)|f^{\prime }(b)|}{2}{\int }_{\frac{1}{2}}^1{(1-t)}^{\alpha +s}dt+\frac{(b-a)|f^{\prime }(b)|}{2}{\int }_{\frac{1}{2}}^1{t}^{\alpha }{(1-t)}^{s}dt\\ -\frac{(b-a)|f^{\prime }(a)|}{2}{\int }_0^{\frac{1}{2}}{t}^{\alpha +s}dt+\frac{(b-a)|f^{\prime }(a)|}{2}{\int }_0^{\frac{1}{2}}{t}^s{(1-t)}^{\alpha }dt\\ +\frac{(b-a)|f^{\prime }(b)|}{2}{\int }_0^{\frac{1}{2}}{(1-t)}^{\alpha +s}dt-\frac{(b-a)|f^{\prime }(b)|}{2}{\int }_0^{\frac{1}{2}}{t}^{\alpha }{(1-t)}^{s}dt\\ \le (b-a)|f^{\prime }(a)|{\int }_{\frac{1}{2}}^1{t}^{\alpha +s}dt-(b-a)|f^{\prime }(a)|{\int }_{\frac{1}{2}}^1{t}^s{(1-t)}^{\alpha }dt\\ -(b-a)|f^{\prime }(b)|{\int }_{\frac{1}{2}}^1{(1-t)}^{\alpha +s}dt+(b-a)|f^{\prime }(b)|{\int }_{\frac{1}{2}}^1{t}^{\alpha }{(1-t)}^{s}dt\\ -(b-a)|f^{\prime }(a)|{\int }_0^1{t}^{\alpha +s}dt+\frac{(b-a)|f^{\prime }(a)|}{2}{\int }_0^1{t}^s{(1-t)}^{\alpha }dt\\ +\frac{(b-a)|f^{\prime }(b)|}{2}{\int }_0^1{(1-t)}^{\alpha +s}dt-\frac{(b-a)|f^{\prime }(b)|}{2}{\int }_0^1{t}^{\alpha }{(1-t)}^{s}dt\\ \le -(b-a)|f^{\prime }(a)|\frac{1}{(\alpha +s+1)2^{\alpha +s+1}}\\ +(b-a)|f^{\prime }(a)|\frac{1}{\alpha +s+1}-(b-a)|f^{\prime }(a)|B_{0.5}(\alpha +1,s+1)\\ -(b-a)|f^{\prime }(b)|\frac{1}{(\alpha +s+1)2^{\alpha +s+1}}+(b-a)|f^{\prime }(b)|B_{0.5}(s+1,\alpha +1)\\ -(b-a)|f^{\prime }(a)|\frac{1}{\alpha +s+1}+\frac{(b-a)|f^{\prime }(a)|}{2}B(s+1,\alpha +1)\\ +\frac{(b-a)|f^{\prime }(b)|}{2}\frac{1}{\alpha +s+1}-\frac{(b-a)|f^{\prime }(b)|}{2}B(s+1,\alpha +1)\\ \le \frac{(b-a)(2^{\alpha +s}|f^{\prime }(b)|-|f^{\prime }(a)|-|f^{\prime }(b)|)}{(\alpha +s+1)2^{\alpha +s+1}}\\ +(b-a)|f^{\prime }(a)|[0.5B(s+1,\alpha +1)-B_{0.5}(\alpha +1,s+1)]\\ +(b-a)|f^{\prime }(b)|[B_{0.5}(s+1,\alpha +1)-0.5B(s+1,\alpha +1)].\end{array}$$

The proof is done. □

Theorem 3.2 Let $f:[0,b]\to R$ be a differentiable mapping and $1<q<\mathrm{\infty }$. If ${|f^{\prime }|}^q$ is measurable and ${|f^{\prime }|}^q$ is decreasing and geometric-arithmetically s-convex on $[0,b]$ for some fixed $\alpha \in (0,\mathrm{\infty })$, $s\in (0,1]$, $0\le a<b$, then the following inequality for fractional integrals holds:

$$\begin{array}{r}|\frac{f(a)+f(b)}{2}-\frac{\mathrm{\Gamma }(\alpha +1)}{2{(b-a)}^{\alpha }}[J_{a^{+}}^{\alpha }f(b)+J_{b^{-}}^{\alpha }f(a)]|\\ \le \frac{b-a}{2}{\left(\frac{{|f^{\prime }(a)|}^q+{|f^{\prime }(b)|}^q}{s+1}\right)}^{\frac{1}{q}}{\left(\frac{2-2^{1-p\alpha }}{p\alpha +1}\right)}^{\frac{1}{p}},\end{array}$$

where $\frac{1}{p}+\frac{1}{q}=1$.

Proof By using Definition 2.2, Lemma 2.5, Hölder’s inequality and Lemma 2.6, we have

$$\begin{array}{r}|\frac{f(a)+f(b)}{2}-\frac{\mathrm{\Gamma }(\alpha +1)}{2{(b-a)}^{\alpha }}[J_{a^{+}}^{\alpha }f(b)+J_{b^{-}}^{\alpha }f(a)]|\\ \le \frac{b-a}{2}{\int }_0^1|{(1-t)}^{\alpha }-t^{\alpha }|\left|f^{\prime }(ta+(1-t)b)\right|dt\\ \le \frac{b-a}{2}{\left({\int }_0^1{|{(1-t)}^{\alpha }-t^{\alpha }|}^{p}dt\right)}^{\frac{1}{p}}{\left({\int }_0^1{\left|f^{\prime }(ta+(1-t)b)\right|}^{q}dt\right)}^{\frac{1}{q}}\\ \le \frac{b-a}{2}{\left({\int }_0^1{|{(1-t)}^{\alpha }-t^{\alpha }|}^{p}dt\right)}^{\frac{1}{p}}{\left({\int }_0^1{\left|f^{\prime }\left(a^t{b}^{1-t}\right)\right|}^{q}dt\right)}^{\frac{1}{q}}\\ \le \frac{b-a}{2}{\left({\int }_0^1{|{(1-t)}^{\alpha }-t^{\alpha }|}^{p}dt\right)}^{\frac{1}{p}}{\left({\int }_0^1[t^s{|f^{\prime }(a)|}^q+{(1-t)}^s{|f^{\prime }(b)|}^q]dt\right)}^{\frac{1}{q}}\\ \le \frac{b-a}{2}{\left(\frac{{|f^{\prime }(a)|}^q+{|f^{\prime }(b)|}^q}{s+1}\right)}^{\frac{1}{q}}{\left({\int }_0^1{|{(1-t)}^{\alpha }-t^{\alpha }|}^{p}dt\right)}^{\frac{1}{p}}\\ \le \frac{b-a}{2}{\left(\frac{{|f^{\prime }(a)|}^q+{|f^{\prime }(b)|}^q}{s+1}\right)}^{\frac{1}{q}}{({\int }_{\frac{1}{2}}^1{(t^{\alpha }-{(1-t)}^{\alpha })}^{p}dt+{\int }_0^{\frac{1}{2}}{({(1-t)}^{\alpha }-t^{\alpha })}^{p}dt)}^{\frac{1}{p}}\\ \le \frac{b-a}{2}{\left(\frac{{|f^{\prime }(a)|}^q+{|f^{\prime }(b)|}^q}{s+1}\right)}^{\frac{1}{q}}{({\int }_{\frac{1}{2}}^1(t^{p\alpha }-{(1-t)}^{p\alpha })dt+{\int }_0^{\frac{1}{2}}({(1-t)}^{p\alpha }-t^{p\alpha })dt)}^{\frac{1}{p}}\\ \le \frac{b-a}{2}{\left(\frac{{|f^{\prime }(a)|}^q+{|f^{\prime }(b)|}^q}{s+1}\right)}^{\frac{1}{q}}{\left(\frac{2-2^{1-p\alpha }}{p\alpha +1}\right)}^{\frac{1}{p}}.\end{array}$$

The proof is done. □

3.2 The second results

By using Lemma 2.7, we can obtain the main results in this section.

Theorem 3.3 Let $f:[0,b]\to R$ be a differentiable mapping. If $|f^{\prime }|$ is measurable and $|f^{\prime }|$ is decreasing and geometric-arithmetically s-convex on $[0,b]$ for some fixed $\alpha \in (0,\mathrm{\infty })$, $s\in (0,1]$, $0\le a<b$, then the following inequality for fractional integrals holds:

$$\begin{array}{r}\frac{\mathrm{\Gamma }(\alpha +1)}{2{(b-a)}^{\alpha }}[J_{a+}^{\alpha }f(b)+J_{b-}^{\alpha }f(a)]-f\left(\frac{a+b}{2}\right)\\ \le 0.5(b-a)(|f^{\prime }(a)|+|f^{\prime }(b)|)\\ \times (B_{0.5}(\alpha +1,s+1)-B_{0.5}(s+1,\alpha +1)+\frac{2^{-\alpha -1}}{\alpha +s+1}+\frac{1}{s+1}).\end{array}$$

Proof By using Definition 2.2, Lemma 2.5 and Lemma 2.7, we have

$$\begin{array}{r}|\frac{\mathrm{\Gamma }(\alpha +1)}{2{(b-a)}^{\alpha }}[J_{a+}^{\alpha }f(b)+J_{b-}^{\alpha }f(a)]-f\left(\frac{a+b}{2}\right)|\\ \le \frac{b-a}{2}{\int }_0^1|h(t)-{(1-t)}^{\alpha }+t^{\alpha }|\left|f^{\prime }(ta+(1-t)b)\right|dt\\ \le \frac{b-a}{2}{\int }_{\frac{1}{2}}^1|-1-{(1-t)}^{\alpha }+t^{\alpha }|\left|f^{\prime }(ta+(1-t)b)\right|dt\\ +\frac{b-a}{2}{\int }_0^{\frac{1}{2}}|1-{(1-t)}^{\alpha }+t^{\alpha }|\left|f^{\prime }(ta+(1-t)b)\right|dt\\ \le \frac{b-a}{2}{\int }_{\frac{1}{2}}^1(1+{(1-t)}^{\alpha }-t^{\alpha })\left|f^{\prime }(ta+(1-t)b)\right|dt\\ +\frac{b-a}{2}{\int }_0^{\frac{1}{2}}(1-{(1-t)}^{\alpha }+t^{\alpha })\left|f^{\prime }(ta+(1-t)b)\right|dt\\ \le \frac{b-a}{2}{\int }_{\frac{1}{2}}^1(1+{(1-t)}^{\alpha }-t^{\alpha })\left|f^{\prime }\left(a^t{b}^{1-t}\right)\right|dt\\ +\frac{b-a}{2}{\int }_0^{\frac{1}{2}}(1-{(1-t)}^{\alpha }+t^{\alpha })\left|f^{\prime }\left(a^t{b}^{1-t}\right)\right|dt\\ \le \frac{b-a}{2}{\int }_{\frac{1}{2}}^1(1+{(1-t)}^{\alpha }-t^{\alpha })[t^s|f^{\prime }(a)|+{(1-t)}^s|f^{\prime }(b)|]dt\\ +\frac{b-a}{2}{\int }_0^{\frac{1}{2}}(1-{(1-t)}^{\alpha }+t^{\alpha })[t^s|f^{\prime }(a)|+{(1-t)}^s|f^{\prime }(b)|]dt\\ \le -\frac{(b-a)|f^{\prime }(a)|}{2}{\int }_{\frac{1}{2}}^1{t}^{\alpha +s}dt+\frac{(b-a)|f^{\prime }(a)|}{2}{\int }_{\frac{1}{2}}^1{t}^{s}dt+\frac{(b-a)|f^{\prime }(a)|}{2}{\int }_{\frac{1}{2}}^1{t}^s{(1-t)}^{\alpha }dt\\ +\frac{(b-a)|f^{\prime }(b)|}{2}{\int }_{\frac{1}{2}}^1{(1-t)}^{\alpha +s}dt-\frac{(b-a)|f^{\prime }(b)|}{2}{\int }_{\frac{1}{2}}^1{t}^{\alpha }{(1-t)}^{s}dt\\ +\frac{(b-a)|f^{\prime }(b)|}{2}{\int }_{\frac{1}{2}}^1{(1-t)}^{s}dt+\frac{(b-a)|f^{\prime }(a)|}{2}{\int }_0^{\frac{1}{2}}{t}^{\alpha +s}dt\\ -\frac{(b-a)|f^{\prime }(a)|}{2}{\int }_0^{\frac{1}{2}}{t}^s{(1-t)}^{\alpha }dt+\frac{(b-a)|f^{\prime }(a)|}{2}{\int }_0^{\frac{1}{2}}{t}^{s}dt\\ -\frac{(b-a)|f^{\prime }(b)|}{2}{\int }_0^{\frac{1}{2}}{(1-t)}^{\alpha +s}dt+\frac{(b-a)|f^{\prime }(b)|}{2}{\int }_0^{\frac{1}{2}}{t}^{\alpha }{(1-t)}^{s}dt\\ +\frac{(b-a)|f^{\prime }(b)|}{2}{\int }_0^{\frac{1}{2}}{(1-t)}^{s}dt\\ \le -\frac{(b-a)|f^{\prime }(a)|}{2}\frac{1-2^{-\alpha -s-1}}{\alpha +s+1}+\frac{(b-a)|f^{\prime }(a)|}{2}\frac{1-2^{-s-1}}{s+1}\\ +\frac{(b-a)|f^{\prime }(a)|}{2}{B}_{0.5}(\alpha +1,s+1)+\frac{(b-a)|f^{\prime }(b)|}{2}\frac{2^{-\alpha -s-1}}{\alpha +s+1}\\ -\frac{(b-a)|f^{\prime }(b)|}{2}{B}_{0.5}(s+1,\alpha +1)+\frac{(b-a)|f^{\prime }(b)|}{2}\frac{2^{-s-1}}{s+1}\\ +\frac{(b-a)|f^{\prime }(a)|}{2}\frac{2^{-\alpha -s-1}}{\alpha +s+1}\\ -\frac{(b-a)|f^{\prime }(a)|}{2}{B}_{0.5}(s+1,\alpha +1)+\frac{(b-a)|f^{\prime }(a)|}{2}\frac{2^{-s-1}}{s+1}\\ -\frac{(b-a)|f^{\prime }(b)|}{2}\frac{1-2^{-\alpha -s-1}}{\alpha +s+1}\\ +\frac{(b-a)|f^{\prime }(b)|}{2}\frac{1-2^{-s-1}}{s+1}+\frac{(b-a)|f^{\prime }(b)|}{2}{B}_{0.5}(\alpha +1,s+1)\\ \le \frac{(b-a)|f^{\prime }(a)|}{2}\frac{2^{-\alpha -s}-1}{\alpha +s+1}+\frac{(b-a)|f^{\prime }(a)|}{2}[B_{0.5}(\alpha +1,s+1)-B_{0.5}(s+1,\alpha +1)]\\ +\frac{(b-a)|f^{\prime }(a)|}{2(s+1)}+\frac{(b-a)|f^{\prime }(b)|}{2}\frac{2^{-\alpha -s}-1}{\alpha +s+1}\\ +\frac{(b-a)|f^{\prime }(b)|}{2}[B_{0.5}(\alpha +1,s+1)-B_{0.5}(s+1,\alpha +1)]+\frac{(b-a)|f^{\prime }(b)|}{2(s+1)}\\ \le 0.5(b-a)(|f^{\prime }(a)|+|f^{\prime }(b)|)\\ \times (B_{0.5}(\alpha +1,s+1)-B_{0.5}(s+1,\alpha +1)+\frac{2^{-\alpha -1}}{\alpha +s+1}+\frac{1}{s+1}).\end{array}$$

The proof is done. □

Theorem 3.4 Let $f:[0,b]\to R$ be a differentiable mapping and $1<q<\mathrm{\infty }$. If ${|f^{\prime }|}^q$ is measurable and ${|f^{\prime }|}^q$ is decreasing and geometric-arithmetically s-convex on $[0,b]$ for some fixed $\alpha \in (0,\mathrm{\infty })$, $s\in (0,1]$, $0\le a<b$, then the following inequality for fractional integrals holds:

$$\begin{array}{r}|\frac{\mathrm{\Gamma }(\alpha +1)}{2{(b-a)}^{\alpha }}[J_{a+}^{\alpha }f(b)+J_{b-}^{\alpha }f(a)]-f\left(\frac{a+b}{2}\right)|\\ \le max\{\frac{b-a}{2}{\left(\frac{{|f^{\prime }(a)|}^q+{|f^{\prime }(b)|}^q}{s+1}\right)}^{\frac{1}{q}}{(\frac{{(1+2^{1-\alpha })}^p}{2}-\frac{2^p(1-2^{-p\alpha })}{(p\alpha +1)})}^{\frac{1}{p}},\\ (b-a){\left(\frac{{|f^{\prime }(a)|}^q+{|f^{\prime }(b)|}^q}{s+1}\right)}^{\frac{1}{q}}{(2^{p-1}-\frac{2^p(1-2^{-p\alpha -1})}{p\alpha +1})}^{\frac{1}{p}}\},\end{array}$$

where $\frac{1}{p}+\frac{1}{q}=1$.

Proof To achieve our aim, we divide our proof into two cases.

Case 1: $\alpha \in (0,1)$. By using Definition 2.2, Lemma 2.4, Lemma 2.5, Hölder’s inequality and Lemma 2.7, we have

$$\begin{array}{r}|\frac{\mathrm{\Gamma }(\alpha +1)}{2{(b-a)}^{\alpha }}[J_{a+}^{\alpha }f(b)+J_{b-}^{\alpha }f(a)]-f\left(\frac{a+b}{2}\right)|\\ \le \frac{b-a}{2}{\int }_0^1|h(t)-{(1-t)}^{\alpha }+t^{\alpha }|\left|f^{\prime }(ta+(1-t)b)\right|dt\\ \le \frac{b-a}{2}{\left({\int }_0^1{|h(t)-{(1-t)}^{\alpha }+t^{\alpha }|}^{p}dt\right)}^{\frac{1}{p}}{\left({\int }_0^1{\left|f^{\prime }(ta+(1-t)b)\right|}^{q}dt\right)}^{\frac{1}{q}}\\ \le \frac{b-a}{2}{\left({\int }_0^1{|h(t)-{(1-t)}^{\alpha }+t^{\alpha }|}^{p}dt\right)}^{\frac{1}{p}}{\left({\int }_0^1{\left|f^{\prime }\left(a^t{b}^{1-t}\right)\right|}^{q}dt\right)}^{\frac{1}{q}}\\ \le \frac{b-a}{2}{\left({\int }_0^1{|h(t)-{(1-t)}^{\alpha }+t^{\alpha }|}^{p}dt\right)}^{\frac{1}{p}}{\left({\int }_0^1[t^s{|f^{\prime }(a)|}^q+{(1-t)}^s{|f^{\prime }(b)|}^q]dt\right)}^{\frac{1}{q}}\\ \le \frac{b-a}{2}{\left(\frac{{|f^{\prime }(a)|}^q+{|f^{\prime }(b)|}^q}{s+1}\right)}^{\frac{1}{q}}{\left({\int }_0^1{|h(t)-{(1-t)}^{\alpha }+t^{\alpha }|}^{p}dt\right)}^{\frac{1}{p}}\\ \le \frac{b-a}{2}{\left(\frac{{|f^{\prime }(a)|}^q+{|f^{\prime }(b)|}^q}{s+1}\right)}^{\frac{1}{q}}({\int }_{\frac{1}{2}}^1{(1-t^{\alpha }+{(1-t)}^{\alpha })}^{p}dt\\ {+{\int }_0^{\frac{1}{2}}{(1-{(1-t)}^{\alpha }+t^{\alpha })}^{p}dt)}^{\frac{1}{p}}\\ \le \frac{b-a}{2}{\left(\frac{{|f^{\prime }(a)|}^q+{|f^{\prime }(b)|}^q}{s+1}\right)}^{\frac{1}{q}}({\int }_{\frac{1}{2}}^1{[1-t^{\alpha }+2^{1-\alpha }-t^{\alpha }]}^{p}dt\\ {+{\int }_0^{\frac{1}{2}}{[(1+t^{\alpha })-(1-t^{\alpha })]}^{p}dt)}^{\frac{1}{p}}\\ \le \frac{b-a}{2}{\left(\frac{{|f^{\prime }(a)|}^q+{|f^{\prime }(b)|}^q}{s+1}\right)}^{\frac{1}{q}}{({\int }_{\frac{1}{2}}^1[{(1+2^{1-\alpha })}^p-2^p{t}^{p\alpha }]dt+{\int }_0^{\frac{1}{2}}{2}^p{t}^{p\alpha }dt)}^{\frac{1}{p}}\\ \le \frac{b-a}{2}{\left(\frac{{|f^{\prime }(a)|}^q+{|f^{\prime }(b)|}^q}{s+1}\right)}^{\frac{1}{q}}{(\frac{{(1+2^{1-\alpha })}^p}{2}-\frac{2^p(1-2^{-p\alpha })}{(p\alpha +1)})}^{\frac{1}{p}}.\end{array}$$

Case 2: $\alpha \in [1,\mathrm{\infty })$. By using Definition 2.2, Lemma 2.4 and Lemma 2.7, we have

$$\begin{array}{r}|\frac{\mathrm{\Gamma }(\alpha +1)}{2{(b-a)}^{\alpha }}[J_{a+}^{\alpha }f(b)+J_{b-}^{\alpha }f(a)]-f\left(\frac{a+b}{2}\right)|\\ \le \frac{b-a}{2}{\left(\frac{{|f^{\prime }(a)|}^q+{|f^{\prime }(b)|}^q}{s+1}\right)}^{\frac{1}{q}}({\int }_{\frac{1}{2}}^1{(1-t^{\alpha }+{(1-t)}^{\alpha })}^{p}dt\\ {+{\int }_0^{\frac{1}{2}}{(1-{(1-t)}^{\alpha }+t^{\alpha })}^{p}dt)}^{\frac{1}{p}}\\ \le (b-a){\left(\frac{{|f^{\prime }(a)|}^q+{|f^{\prime }(b)|}^q}{s+1}\right)}^{\frac{1}{q}}{\left({\int }_{\frac{1}{2}}^1{(1-t^{\alpha }+1-t^{\alpha })}^{p}dt\right)}^{\frac{1}{p}}\\ \le (b-a){\left(\frac{{|f^{\prime }(a)|}^q+{|f^{\prime }(b)|}^q}{s+1}\right)}^{\frac{1}{q}}{\left({\int }_{\frac{1}{2}}^1{2}^p(1-t^{p\alpha })dt\right)}^{\frac{1}{p}}\\ \le (b-a){\left(\frac{{|f^{\prime }(a)|}^q+{|f^{\prime }(b)|}^q}{s+1}\right)}^{\frac{1}{q}}{(2^{p-1}-\frac{2^p(1-2^{-p\alpha -1})}{p\alpha +1})}^{\frac{1}{p}}.\end{array}$$

The proof is done. □

3.3 The third results

By using Lemma 2.8, we can obtain the main results in this section.

Theorem 3.5 Let $f:[0,b]\to R$ be a differentiable mapping. If $|f^{\prime }|$ is measurable and $|f^{\prime }|$ is decreasing and geometric-arithmetically s-convex on $[0,b]$ for some fixed $\alpha \in (0,\mathrm{\infty })$, $s\in (0,1]$, $0\le a<x<b$, then the following inequality for fractional integrals holds:

$$\begin{array}{r}|\frac{{(x-a)}^{\alpha }+{(b-x)}^{\alpha }}{b-a}f(x)-\frac{\mathrm{\Gamma }(\alpha +1)}{b-a}[J_{x^{-}}^{\alpha }f(a)+J_{x^{+}}^{\alpha }f(b)]|\\ \le \frac{|{(x-a)}^{\alpha +1}-{(b-x)}^{\alpha +1}|}{b-a}[\frac{|f^{\prime }(a)|}{\alpha +s+1}+|f^{\prime }(b)|B(\alpha +1,s+1)],\end{array}$$

where

$$B(s+1,\alpha +1)={\int }_0^1{t}^s{(1-t)}^{\alpha }dt.$$

Proof By using Definition 2.2, Lemma 2.5 and Lemma 2.8, we have

$$\begin{array}{r}|\frac{{(x-a)}^{\alpha }+{(b-x)}^{\alpha }}{b-a}f(x)-\frac{\mathrm{\Gamma }(\alpha +1)}{b-a}[J_{x^{-}}^{\alpha }f(a)+J_{x^{+}}^{\alpha }f(b)]|\\ \le \frac{|{(x-a)}^{\alpha +1}-{(b-x)}^{\alpha +1}|}{b-a}{\int }_0^1{t}^{\alpha }\left|f^{\prime }(tx+(1-t)a)\right|dt\\ \le \frac{|{(x-a)}^{\alpha +1}-{(b-x)}^{\alpha +1}|}{b-a}{\int }_0^1{t}^{\alpha }\left|f^{\prime }\left(x^t{a}^{1-t}\right)\right|dt\\ \le \frac{|{(x-a)}^{\alpha +1}-{(b-x)}^{\alpha +1}|}{b-a}{\int }_0^1{t}^{\alpha }[t^s|f^{\prime }(a)|+{(1-t)}^s|f^{\prime }(b)|]dt\\ \le \frac{|{(x-a)}^{\alpha +1}-{(b-x)}^{\alpha +1}|}{b-a}{\int }_0^1[t^{\alpha +s}|f^{\prime }(a)|+t^{\alpha }{(1-t)}^s|f^{\prime }(b)|]dt\\ \le \frac{|{(x-a)}^{\alpha +1}-{(b-x)}^{\alpha +1}|}{b-a}[\frac{|f^{\prime }(a)|}{\alpha +s+1}+|f^{\prime }(b)|B(\alpha +1,s+1)].\end{array}$$

The proof is done. □

Theorem 3.6 Let $f:[0,b]\to R$ be a differentiable mapping and $1<q<\mathrm{\infty }$. If ${|f^{\prime }|}^q$ is measurable and ${|f^{\prime }|}^q$ is decreasing and geometric-arithmetically s-convex on $[0,b]$ for some fixed $\alpha \in (0,\mathrm{\infty })$, $s\in (0,1]$, $0\le a<b$, then the following inequality for fractional integrals holds:

$$\begin{array}{r}|\frac{{(x-a)}^{\alpha }+{(b-x)}^{\alpha }}{b-a}f(x)-\frac{\mathrm{\Gamma }(\alpha +1)}{b-a}[J_{x^{-}}^{\alpha }f(a)+J_{x^{+}}^{\alpha }f(b)]|\\ \le \frac{b-a}{2(p\alpha +1)}{\left(\frac{{|f^{\prime }(a)|}^q+{|f^{\prime }(b)|}^q}{s+1}\right)}^{\frac{1}{q}}.\end{array}$$

Proof By using Definition 2.2, Lemma 2.5, Hölder’s inequality and Lemma 2.8, we have

$$\begin{array}{r}|\frac{{(x-a)}^{\alpha }+{(b-x)}^{\alpha }}{b-a}f(x)-\frac{\mathrm{\Gamma }(\alpha +1)}{b-a}[J_{x^{-}}^{\alpha }f(a)+J_{x^{+}}^{\alpha }f(b)]|\\ \le \frac{|{(x-a)}^{\alpha +1}-{(b-x)}^{\alpha +1}|}{b-a}{\int }_0^1{t}^{\alpha }\left|f^{\prime }(tx+(1-t)a)\right|dt\\ \le \frac{b-a}{2}{\left({\int }_0^1{t}^{p\alpha }dt\right)}^{\frac{1}{p}}{\left({\int }_0^1{\left|f^{\prime }(ta+(1-t)b)\right|}^{q}dt\right)}^{\frac{1}{q}}\\ \le \frac{b-a}{2(p\alpha +1)}{\left({\int }_0^1{\left|f^{\prime }(ta+(1-t)b)\right|}^{q}dt\right)}^{\frac{1}{q}}\\ \le \frac{b-a}{2(p\alpha +1)}{\left({\int }_0^1{\left|f^{\prime }\left(a^t{b}^{1-t}\right)\right|}^{q}dt\right)}^{\frac{1}{q}}\\ \le \frac{b-a}{2(p\alpha +1)}{\left({\int }_0^1[t^s{|f^{\prime }(a)|}^q+{(1-t)}^s{|f^{\prime }(b)|}^q]dt\right)}^{\frac{1}{q}}\\ \le \frac{b-a}{2(p\alpha +1)}{\left(\frac{{|f^{\prime }(a)|}^q+{|f^{\prime }(b)|}^q}{s+1}\right)}^{\frac{1}{q}},\end{array}$$

where $\frac{1}{p}+\frac{1}{q}=1$. The proof is done. □

4 Applications to special means

Consider the following special means (see Pearce and Pec̆arić [22]) for arbitrary real numbers α, β, $\alpha \ne \beta$ as follows:

• (i) $H(\alpha ,\beta )=\frac{2}{\frac{1}{\alpha }+\frac{1}{\beta }}$, $\alpha ,\beta \in R\setminus \{0\}$;

• (ii) $A(\alpha ,\beta )=\frac{\alpha +\beta }{2}$, $\alpha ,\beta \in R$;

• (iii) $L(\alpha ,\beta )=\frac{\beta -\alpha }{ln|\beta |-ln|\alpha |}$, $|\alpha |\ne |\beta |$, $\alpha \beta \ne 0$;

• (iv) $L_n(\alpha ,\beta )={[\frac{{\beta }^{n+1}-{\alpha }^{n+1}}{(n+1)(\beta -\alpha )}]}^{\frac{1}{n}}$, $n\in \mathbb{Z}\setminus \{-1,0\}$, $\alpha ,\beta \in R$, $\alpha \ne \beta$.

Next, we apply the obtained results to give some applications to special means of real numbers.

Theorem 4.1 For some $s\in (0,1]$, $n\in \mathbb{Z}\setminus \{-1,0\}$, $0\le a<b$, the following inequality for fractional integrals holds:

$$|A(a^n,b^n)-L_n^n(a^n,b^n)|\le \frac{n(b-a)}{2}{\left(\frac{a^{(n-1)q}+b^{(n-1)q}}{s+1}\right)}^{\frac{1}{q}}{\left(\frac{2-2^{1-p}}{p+1}\right)}^{\frac{1}{p}},$$

where

$$\frac{1}{p}+\frac{1}{q}=1,1<q<\mathrm{\infty }.$$

Proof Applying Theorem 3.2 for $f(x)=x^n$, $\alpha =1$, one can obtain the result immediately. □

Theorem 4.2 For some $s\in (0,1]$, $n\in \mathbb{Z}\setminus \{-1,0\}$, $0\le a<b$, the following inequality for fractional integrals holds:

$$\begin{array}{r}|L_n^n(a^n,b^n)-A^n(a,b)|\\ \le (b-a){\left(\frac{a^{(n-1)q}+b^{(n-1)q}}{s+1}\right)}^{\frac{1}{q}}max\{{(2^{p-2}-\frac{2^{p-1}-0.5}{p+1})}^{\frac{1}{p}},{(2^{p-1}-\frac{2^p-0.5}{p+1})}^{\frac{1}{p}}\},\end{array}$$

where

$$\frac{1}{p}+\frac{1}{q}=1,1<q<\mathrm{\infty }.$$

Proof Applying Theorem 3.4 for $f(x)=x^n$, $\alpha =1$, one can obtain the result immediately. □

Theorem 4.3 For some $s\in (0,1]$, $n\in \mathbb{Z}\setminus \{-1,0\}$, $1<q<\mathrm{\infty }$, $0\le a<x<b$, the following inequality for fractional integrals holds:

$$|L_n^n(a^n,b^n)-x^n|\le \frac{b-a}{2(p+1)}{\left(\frac{{|f^{\prime }(a)|}^q+{|f^{\prime }(b)|}^q}{s+1}\right)}^{\frac{1}{q}}.$$

Proof Applying Theorem 3.6 for $f(x)=x^n$, $\alpha =1$, one can obtain the result immediately. □

Theorem 4.4 For some $s\in (0,1]$, $0\le a<b$, the following inequality for fractional integrals holds:

$$|\frac{1}{H(a,b)}-\frac{1}{L(a,b)}|\le \frac{b-a}{2}{\left(\frac{a^{2q}+b^{2q}}{(s+1)a^{2q}{b}^{2q}}\right)}^{\frac{1}{q}}{\left(\frac{2-2^{1-p}}{p+1}\right)}^{\frac{1}{p}},$$

where

$$\frac{1}{p}+\frac{1}{q}=1,1<q<\mathrm{\infty }.$$

Proof Applying Theorem 3.2 for $f(x)=x^{-1}$, $\alpha =1$, one can obtain the result immediately. □

Theorem 4.5 For some $s\in (0,1]$, $0\le a<b$, the following inequality for fractional integrals holds:

$$\begin{array}{r}|\frac{1}{H(a,b)}-\frac{1}{A(a,b)}|\\ \le (b-a){\left(\frac{a^{2q}+b^{2q}}{(s+1)a^{2q}{b}^{2q}}\right)}^{\frac{1}{q}}max\{{(\frac{{(1+2^{1-\alpha })}^p}{4}-\frac{2^{p-1}(1-2^{-p\alpha })}{(p\alpha +1)})}^{\frac{1}{p}},\\ {(2^{p-1}-\frac{2^p(1-2^{-p\alpha -1})}{p\alpha +1})}^{\frac{1}{p}}\},\end{array}$$

where

$$\frac{1}{p}+\frac{1}{q}=1,1<q<\mathrm{\infty }.$$

Proof Applying Theorem 3.4 for $f(x)=x^{-1}$, $\alpha =1$, one can obtain the result immediately. □

Theorem 4.6 For some $s\in (0,1]$, $0\le a<x<b$, the following inequality for fractional integrals holds:

$$|\frac{1}{H(a,b)}-\frac{1}{x}|\le \frac{b-a}{2(p+1)}{\left(\frac{a^{2q}+b^{2q}}{(s+1)a^{2q}{b}^{2q}}\right)}^{\frac{1}{q}},$$

where

$$\frac{1}{p}+\frac{1}{q}=1,1<q<\mathrm{\infty }.$$

Proof Applying Theorem 3.6 for $f(x)=x^{-1}$, $\alpha =1$, one can obtain the result immediately. □