Hermite–Hadamard-type inequalities for generalized s-convex functions on real linear fractal set $\mathbb {R}^{\alpha }(0<\alpha <1)$

Abstract

In this paper, we establish the Hermite–Hadamard-type inequalities for the generalized s-convex functions in the second sense on real linear fractal set $\mathbb {R}^{\alpha }(0<\alpha <1).$

Introduction

The convex function plays an important role in the class mathematical analysis course and other fields. In [1], Hudzik and Maligranda introduced two kinds of s-convex functions in the space of European space $\mathbb {R}$. In addition, many important inequalities are established for the s-convex functions in $\mathbb {R}$. For example, the Hermite–Hadamard's inequality is one of the best known results in the literature, see [2,3,4] and so on.

In recent years, the fractal theory has received significantly remarkable attention [5]. The calculus on fractal set can lead to better comprehension for the various real-world models from the engineering and science [6].

On the fractal set, Mo et al. [7, 8] introduced the definition of the generalized convex function and established Hermite–Hadamard-type inequality. In [9], the authors introduced two kinds of generalized s-convex functions on fractal sets $\mathbb {R}^{\alpha }(0<\alpha <1).$

The definitions of the generalized s-convex functions are as follows:

Definition 1.1

[9] Suppose that $\mathbb {R_+}=[0,\infty ).$ If the function $f:\mathbb {R_+}\rightarrow \mathbb {R^{\alpha }}$ satisfies the following inequality:

$$\begin{aligned} f(\lambda _1u+\lambda _2v)\le \lambda _1^{\alpha s}f(u)+\lambda _2^{\alpha s}f(v), \end{aligned}$$

(1.1)

for all $u,v\in \mathbb {R_+}$ and all $\lambda _1,\lambda _2\ge 0$ with $\lambda _1^ s+\lambda _2^s=1$ and $0<s<1,$ then f is said to be a generalized s-convex function in the first sense. We denote this by $f\in GK_s^1.$

Definition 1.2

[9] Suppose that $\mathbb {R_+}=[0,\infty ).$ If the function $f:\mathbb {R_+}\rightarrow \mathbb {R^{\alpha }}$ satisfies the following inequality:

$$ f(\lambda _1u+\lambda _2v)\, \le\, \lambda _1^{\alpha s}f(u)+\lambda _2^{\alpha s}f(v), $$

(1.2)

for all $u,v\in \mathbb {R_+}$ and all $\lambda _1,\lambda _2\ge 0$ with $\lambda _1+\lambda _2=1$ and $0<s<1,$ then f is said to be a generalized s-convex function in the second sense. We denote this by $f\in GK_s^2.$

Note that the generalized s-convex function in both sense is generalized convex function [9] for $s=1$.

Inspired by [2, 3, 8], in this paper, we will establish the Hermite–Hadamard-type inequalities for the generalized s-convex functions.

Preliminaries

Now, let us review the operations with real line number on fractal space. In addition, we will use the Gao–Yang–Kang’s idea to describe the definitions of the local fractional derivative and local fractional integral [10,11,12,13,14].

Let $a^\alpha ,b^\alpha$ and $c^\alpha$ belong to the set $\mathbb {R}^\alpha (0<\alpha <1)$ of real line numbers, then

• 1. $a^\alpha b^\alpha$ and $a^\alpha +b^\alpha$ belong to the set $\mathbb {R}^\alpha$;

• 2. $a^\alpha +b^\alpha =(a+b)^\alpha =b^\alpha +a^\alpha =(b+a)^\alpha$;

• 3. $a^\alpha +(b^\alpha +c^\alpha )=(a^\alpha +b^\alpha )+c^\alpha$;

• 4. $a^\alpha b^\alpha =(ab)^\alpha =b^\alpha a^\alpha =(ba)^\alpha$;

• 5. $a^\alpha (b^\alpha c^\alpha )=(a^\alpha b^\alpha )c^\alpha$;

• 6. $a^\alpha (b^\alpha +c^\alpha )=a^\alpha b^\alpha +a^\alpha c^\alpha$;

• 7. $0^\alpha +a^\alpha =a^\alpha +0^\alpha =a^\alpha$ and $1^\alpha \cdot a^\alpha =a^\alpha \cdot 1^\alpha =a^\alpha$.

Definition 2.1

([10]) If the function $f:[a,b]\rightarrow \mathbb {R^\alpha }$ satisfies the inequality

$$\begin{aligned} |f(x)-f(y)|<c|x-y|^\alpha , \quad (x,y\in [a,b]), \end{aligned}$$

for $c>0$ and $\alpha (0<\alpha \le 1),$ then f is called a Hölder continuous function. In this case, we think that f is in the space $C_\alpha [a,b].$

Definition 2.2

[10] Let $\triangle ^\alpha (f(x)-f(x_0))\cong \Gamma (1+a)(f(x)-f(x_0)).$ Then, the local fractional derivative of f of order $\alpha$ at $x=x_0$ is defined by

$$\begin{aligned} f^{(\alpha )}(x_0)=\frac{d^\alpha f(x)}{dx^\alpha }|_{x=x_0}=\lim \limits _{x\rightarrow x_0}\frac{\triangle ^\alpha (f(x)-f(x_0))}{(x-x_0)^\alpha }. \end{aligned}$$

If there exists $f^{((k+1)\alpha )}(x)=\mathop {\overbrace{D_x^{\alpha } \ldots D_x^\alpha }}\limits ^{k+1\; {times}}f(x)$ for any $x\in I\subseteq \mathbb {R},$ then we denoted $f\in D_{(k+1)\alpha }(I)$, where $k=0,1,2\dots .$

Definition 2.3

[10] For $f\in C_{\alpha }[a,b],$ the local fractional integral of the function f is defined by

$$\begin{aligned} \begin{array}{cl} &{}_aI_b^{(\alpha )}f(x)\\ &{}=\frac{1}{\Gamma (1+a)}\int _a^bf(t)(\mathrm{d}t)^\alpha \\ &{}=\frac{1}{\Gamma (1+a)}\lim \limits _{\triangle t\rightarrow 0}\sum \limits _{j=0}^{N}f(t_j)(\bigtriangleup t_j)^\alpha , \end{array} \end{aligned}$$

where $\triangle t_j=t_{j+1}-t_j,\;\triangle t=\max \{\triangle t_1,\triangle t_2,\triangle t_j,\ldots \}$ and $[t_j,t_j+1],$ $j=0,\ldots ,N-1,$ $t_0=a,$ $t_N=b,$ is a partition of the interval [a, b].

Lemma 2.1

[10] Let $f\in C_\alpha [g(a), g(b)]$ and $g\in C_1[a,b],$ then

$$\begin{aligned} _{g(a)}I_{g(b)}^{(\alpha )}f(x)=_{a}I_{b}^{(\alpha )}f(g)(s)[g'(s)]^{\alpha }. \end{aligned}$$

Lemma 2.2

[10]

• 1. Let $f(x)=g^{(\alpha )}(x)\in C_{\alpha }[a,b]$, then we have

  $$\begin{aligned} _{a}I_{b}^{(\alpha )}f(x)=g(b)-g(a). \end{aligned}$$

• 2. Let $f(x), g(x)\in D_{\alpha }[a,b]$ and $f^{(\alpha )}(x), g^{(\alpha )}(x)\in C_{\alpha }[a,b]$, then we have

  $$\begin{aligned} _{a}I_{b}^{\alpha }f(x)g^{(\alpha )}(x)=f(x)g(x)\bigg |_{a}^{b}-_aI_{b}^{(\alpha )}f^{(\alpha )}(x)g(x). \end{aligned}$$

Lemma 2.3

[10]

$$\begin{aligned} \frac{d^\alpha x^{k\alpha }}{dx^\alpha }=\frac{\Gamma (1+k\alpha )}{\Gamma (1+(k-1)\alpha )}x^{(k-1)\alpha }. \end{aligned}$$

From the above formula and Lemma 2.2 , we have

$$\begin{aligned} \frac{1}{\Gamma (1+\alpha )}\int _a^bx^{k\alpha }(dx)^{\alpha }=\frac{\Gamma (1+k\alpha )}{\Gamma (1+(k+1)\alpha )}(b^{(k+1)\alpha }-a^{(k+1)\alpha }),k\in R. \end{aligned}$$

Lemma 2.4

[10] (Generalized Hölder’s inequality) Let $f,g\in C_{\alpha }[a,b]$ and $p,q>1$ with $1/p+1/q=1.$ Then, it follows that

$$\begin{aligned}&\frac{1}{\Gamma (1+\alpha )}\int _a^b|f(x)g(x)|(\mathrm{d}x)^{\alpha }\le \left( \frac{1}{\Gamma (1+\alpha )}\int _a^b|f(x)|^p(\mathrm{d}x)^{\alpha }\right) ^{1/p}\\&\times \left( \frac{1}{\Gamma (1+\alpha )}\int _a^b|g(x)|^q(\mathrm{d}x)^{\alpha }\right) ^{1/q}. \end{aligned}$$

Main results

Theorem 3.1

Let $f:\mathbb {R_+}\rightarrow \mathbb {R^{\alpha }}$ be a generalized s-convex function in the second sense for $0<s<1$ and $a,b\in [0,\infty )$ with $a<b.$ Then, for $f\in C_{\alpha }[a,b],$ the following inequalities hold:

$$\begin{aligned} \begin{array}{cl}\frac{2^{(s-1)\alpha }}{\Gamma (1+\alpha )}f\left( \frac{a+b}{2}\right) \le \frac{_{a}I_{b}^{\alpha }f(x)}{(b-a)^\alpha }\le \frac{\Gamma (1+s\alpha )}{\Gamma (1+(s+1)\alpha )}\bigl (f(a)+f(b)\bigr ). \end{array} \end{aligned}$$

(3.1)

Proof

Let $x=a+b-t.$ Then, from Lemma 2.1, we have $_{\frac{a+b}{2}}I_b^{(\alpha )}f(x)={_a}I_{\frac{a+b}{2}}^{(\alpha )}f(a+b-t).$

Since f is a generalized s-convex function in the second sense, then

$$\begin{aligned} \begin{array}{cl}_aI_b^{(\alpha )}f(x) &{} ={_a}I_{\frac{a+b}{2}}^{(\alpha )}\bigl [f(x)+f(a+b-x)\bigr ]\\ &{} \ge 2^{\alpha s} {_a}I_{\frac{a+b}{2}}^{(\alpha )}f\left( \frac{a+b}{2}\right) \\ &{} =\frac{2^{(s-1)\alpha }}{\Gamma (1+\alpha )}(b-a)^\alpha f\left( \frac{a+b}{2}\right) . \end{array} \end{aligned}$$

In the other hand, let $x=b-(b-a)t$, $0\le t\le 1,$ then we get

$$\begin{aligned} _aI_b^{(\alpha )}f(x)&=(b-a)^\alpha {_0}I_1^{(\alpha )}f\bigl [ta+(1-t)b\bigr ]\\ & \le (b-a)^\alpha {_0}I_1^{(\alpha )}\bigl [t^{\alpha s}f(a)+(1-t)^{\alpha s}f(b)\bigr ]\\ & =(b-a)^\alpha \bigl [f(a) {_0}I_1^{(\alpha )}t^{\alpha s}+f(b) {_0}I_1^{(\alpha )}(1-t)^{\alpha s}\bigr ]. \end{aligned}$$

From Lemma 2.3, it is easy to see that

$$\begin{aligned} {_0}I_1^{(\alpha )}t^{\alpha s}=\frac{\Gamma (1+s\alpha )}{\Gamma (1+(s+1)\alpha )}, \end{aligned}$$

and

$$\begin{aligned} {_0}I_1^{(\alpha )}(1-t)^{\alpha s}=\frac{\Gamma (1+s\alpha )}{\Gamma (1+(s+1)\alpha )}. \end{aligned}$$

Therefore

$$\begin{aligned} _aI_b^{(\alpha )}f(x)\le (b-a)^\alpha \frac{\Gamma (1+s\alpha )}{\Gamma (1+(s+1)\alpha )}\bigl (f(a)+f(b)\bigr ). \end{aligned}$$

Combining the above estimates, we obtain

$$\begin{aligned} \frac{2^{(s-1)\alpha }}{\Gamma (1+\alpha )}f\left( \frac{a+b}{2}\right) \le \frac{_{a}I_{b}^{\alpha }f(x)}{(b-a)^\alpha }\le \frac{\Gamma (1+s\alpha )}{\Gamma (1+(s+1)\alpha )}\bigl (f(a)+f(b)\bigr ). \end{aligned}$$

$\square$

Theorem 3.2

Let $I\subset \mathbb {R}$ be an interval, and $I^0$ be the interior of I. Suppose that $f: I\rightarrow \mathbb {R^{\alpha }}$ is a differentiable function on $I^0$ such that $f^{(\alpha )}\in C_{\alpha }[a,b]$ , where $a,b\in I^0$ with $a<b.$ If $|f^{(\alpha )}|^q$ is a generalized s-convex function in the second sense on [a, b] for some fixed $s\in (0,1)$ and $q\ge 1$, then

$$\begin{aligned} & \bigg |\frac{f(a)+f(b)}{2^\alpha }-\frac{\Gamma (1+\alpha )}{(b-a)^\alpha }{_a}I_b^{(\alpha )}f(x)\bigg |\\ &\quad \le \frac{(b-a)^\alpha }{2^\alpha }\bigg (\frac{\Gamma (1+\alpha )}{\Gamma (1+2\alpha )}\bigg )^{\frac{q-1}{q}}\bigg [\frac{\Gamma (1+s\alpha )}{\Gamma (1+(s+1)\alpha )}+\frac{\Gamma (1+\alpha )\Gamma (1+s\alpha )}{\Gamma (1+(s+2)\alpha )} \left( \left( \frac{1}{2}\right) ^{\alpha s}-2^\alpha \right) \bigg ]^{\frac{1}{q}}\\ &\qquad \times \bigg [\big |f^{(\alpha )}(a)\big |^q+\big |f^{(\alpha )}(b)\big |^q\bigg ]^{\frac{1}{q}}.\\ \end{aligned}$$

To show Theorem 3.2 is right, we need the following Lemma.

Lemma 3.1

([8]) Let f :  $I\rightarrow \mathbb {R^{\alpha }},$ $I\subset [0,\infty ).$ If $f\in D_\alpha (I^0)$ and $f^{(\alpha )}\in C_\alpha [a,b]$ for $a,b\in I^0$ with $a<b,$ then the following equality holds:

$$\begin{aligned}&\frac{f(a)+f(b)}{2^\alpha }-\frac{\Gamma (1+\alpha )}{(b-a)^\alpha }{_a}I_b^{(\alpha )}f(x)\\&=\frac{(b-a)^\alpha }{2^\alpha }\frac{1}{\Gamma (1+\alpha )} \int _0^1(1-2t)^\alpha f^{(\alpha )}\big (ta+(1-t)b\big )(\mathrm{d}t)^\alpha . \end{aligned}$$

Now, let us give the proof of Theorem 3.2.

Proof

From Lemma 3.1, it is obvious that

$$\begin{aligned} & \bigg |\frac{f(a)+f(b)}{2^\alpha }-\frac{\Gamma (1+\alpha )}{(b-a)^\alpha }{_a}I_b^{(\alpha )}f(x)\bigg |\\ &\quad \le \frac{(b-a)^\alpha }{2^\alpha }\frac{1}{\Gamma (1+\alpha )}\int _0^1|1-2t|^\alpha \big |f^{(\alpha )}(ta+(1-t)b)\big |(\mathrm{d}t)^\alpha . \end{aligned}$$

(3.2)

Let us estimate

$$\begin{aligned} \frac{1}{\Gamma (1+\alpha )}\int _0^1|1-2t|^\alpha \big |f^{(\alpha )}(ta+(1-t)b)\big |(\mathrm{d}t)^\alpha , \end{aligned}$$

for $q=1$ and $q>1.$

Case I. $q=1.$

Since $|f^{(\alpha )}|$ is generalized sconvex on [a, b] in the second sense, we can know that for any $t\in [0,1]$

$$\begin{aligned} \big |f^{(\alpha )}(ta+(1-t)b)\big |\le t^{\alpha s}|f^{(\alpha )}(a)|+(1-t)^{\alpha s}|f^{(\alpha )}(b)|. \end{aligned}$$

Then, we have

$$\begin{aligned} \begin{array}{cl} &{}\frac{1}{\Gamma (1+\alpha )}\int _0^1|1-2t|^\alpha \left| f^{(\alpha )}(ta+(1-t)b)\right| (\mathrm{d}t)^\alpha \\ &{}\le \frac{1}{\Gamma (1+\alpha )}\int _0^1|1-2t|^\alpha \left[ t^{\alpha s}|f^{(\alpha )}(a)|+(1-t)^{\alpha s}|f^{(\alpha )}(b)|\right] (\mathrm{d}t)^\alpha \\ &{}=\left[ |f^{(\alpha )}(a)|\frac{1}{\Gamma (1+\alpha )}\int _0^1t^{\alpha s}|1-2t|^\alpha (\mathrm{d}t)^\alpha +|f^{(\alpha )}(b)|\frac{1}{\Gamma (1+\alpha )}\int _0^1(1-t)^{\alpha s}|1-2t|^{\alpha }(\mathrm{d}t)^\alpha \right] .\\ \end{array} \end{aligned}$$

(3.3)

From Lemmas 2.2 and 2.3, it is easy to see that

$$\begin{aligned} &\frac{1}{\Gamma (1+\alpha )}\int _0^1t^{\alpha s}|1-2t|^\alpha (\mathrm{d}t)^\alpha \\ &\quad=\frac{1}{\Gamma (1+\alpha )}\left[ \int _0^{\frac{1}{2}}t^{\alpha s}(1-2t)^\alpha (\mathrm{d}t)^\alpha +\int _{\frac{1}{2}}^1t^{\alpha s}(2t-1)^\alpha (\mathrm{d}t)^\alpha \right] \\ &\quad =\left[ \frac{\Gamma (1+s\alpha )}{\Gamma (1+(s+1)\alpha )}+\frac{\Gamma (1+\alpha )\Gamma (1+s\alpha )}{\Gamma (1+(s+2)\alpha )}\left( \left( \frac{1}{2}\right) ^{\alpha s}-2^\alpha \right) \right] . \end{aligned}$$

(3.4)

In addition, let $1-t=x,$ then by Lemma 2.1 and (3.4), we have

$$\begin{aligned} & \frac{1}{\Gamma (1+\alpha )}\int _0^1(1-t)^{\alpha s}|1-2t|^\alpha (\mathrm{d}t)^\alpha \\ &\quad =\frac{1}{\Gamma (1+\alpha )}\int _0^1 x^{\alpha s}|1-2x|^\alpha (\mathrm{d}x)^\alpha \\ &\quad =\left[ \frac{\Gamma (1+s\alpha )}{\Gamma (1+(s+1)\alpha )}+\frac{\Gamma (1+\alpha )\Gamma (1+s\alpha )}{\Gamma (1+(s+2)\alpha )}\left( \left( \frac{1}{2}\right) ^{\alpha s}-2^\alpha \right) \right] . \end{aligned}$$

(3.5)

Thus, substituting (3.4) and (3.5) into (3.3), we have

$$\begin{aligned} &\frac{1}{\Gamma (1+\alpha )}\int _0^1|1-2t|^\alpha \big |f^{(\alpha )}(ta+(1-t)b)\big |(\mathrm{d}t)^\alpha \\ &\quad \le \frac{\Gamma (1+\alpha )}{\Gamma (1+2\alpha )} \bigg [\frac{\Gamma (1+s\alpha )}{\Gamma (1+(s+1)\alpha )}+\frac{\Gamma (1+\alpha )\Gamma (1+s\alpha )}{\Gamma (1+(s+2)\alpha )} \left( \left( \frac{1}{2}\right) ^{\alpha s}-2^\alpha \right) \bigg ]\\ &\qquad \times \bigg [\big |f^{(\alpha )}(a)\big |^q+\big |f^{(\alpha )}(b)\big |\bigg ].\\ \end{aligned}$$

(3.6)

Thus, from (3.2), we obtain

$$\begin{aligned} \begin{array}{cl} &{}\bigg |\frac{f(a)+f(b)}{2^\alpha }-\frac{\Gamma (1+\alpha )}{(b-a)^\alpha }{_a}I_b^{(\alpha )}f(x)\bigg |\\ &{}\le \frac{(b-a)^\alpha }{2^\alpha }\frac{\Gamma (1+\alpha )}{\Gamma (1+2\alpha )} \bigg [\frac{\Gamma (1+s\alpha )}{\Gamma (1+(s+1)\alpha )}+\frac{\Gamma (1+\alpha )\Gamma (1+s\alpha )}{\Gamma (1+(s+2)\alpha )} \left( \left( \frac{1}{2}\right) ^{\alpha s}-2^\alpha \right) \bigg ]\\ &{}\quad \times \bigg [\big |f^{(\alpha )}(a)\big |^q+\big |f^{(\alpha )}(b)\big |\bigg ].\\ \end{array} \end{aligned}$$

Case II. $q>1.$

Using the generalized Hölder’s inequality (Lemma 2.4), we obtain

$$\begin{aligned} &\frac{1}{\Gamma (1+\alpha )}\int _0^1|1-2t|^\alpha \big |f^{(\alpha )}(ta+(1-t)b)\big |(\mathrm{d}t)^\alpha \\ &\quad =\frac{1}{\Gamma (1+\alpha )}\int _0^1|1-2t|^{\alpha \frac{q-1}{q}}|1-2t|^{\alpha \frac{1}{q}}\big |f^{(\alpha )}(ta+(1-t)b)\big |(\mathrm{d}t)^\alpha \\ &\quad \le \left( \frac{1}{\Gamma (1+\alpha )}\int _0^1|1-2t|^\alpha (\mathrm{d}t)^\alpha \right) ^{\frac{q-1}{q}}\\ &\qquad \times \left( \frac{1}{\Gamma (1+\alpha )}\int _0^1|1-2t|^\alpha |f^{(\alpha )}(ta+(1-t)b)|^q(\mathrm{d}t)^\alpha \right) ^{\frac{1}{q}}. \end{aligned}$$

(3.7)

It is obvious that

$$\begin{aligned} &\frac{1}{\Gamma (1+\alpha )}\int _0^1|1-2t|^\alpha (\mathrm{d}t)^\alpha \\ &\quad=\frac{1}{\Gamma (1+\alpha )}\int _0^{\frac{1}{2}}(1-2t)^\alpha (\mathrm{d}t)^\alpha + \frac{1}{\Gamma (1+\alpha )}\int _{\frac{1}{2}}^1(2t-1)^\alpha (\mathrm{d}t)^\alpha =\frac{\Gamma (1+\alpha )}{\Gamma (1+2\alpha )}. \end{aligned}$$

(3.8)

Moreover, since $|f^{(\alpha )}|^q$ is generalized s convex in the second sense on [a, b],  then

$$\begin{aligned} &\frac{1}{\Gamma (1+\alpha )}\int _0^1|1-2t|^\alpha \big |f^{(\alpha )}(ta+(1-t)b)\big |^q(\mathrm{d}t)^\alpha \\ &\quad \le \frac{1}{\Gamma (1+\alpha )}\int _0^1|1-2t|^\alpha \bigg (t^{\alpha s}|f^{(\alpha )}(a)|^q+(1-t)^{\alpha s}|f^{(\alpha )}(b)|^q\bigg )(\mathrm{d}t)^\alpha \\ &\quad=|f^{(\alpha )}(a)|^q\frac{1}{\Gamma (1+\alpha )}\int _0^1|1-2t|^\alpha t^{\alpha s}(\mathrm{d}t)^\alpha \\ &\qquad +|f^{(\alpha )}(b)|^q\frac{1}{\Gamma (1+\alpha )}\int _0^1|1-2t|^\alpha (1-t)^{\alpha s}(\mathrm{d}t)^\alpha .\\ \end{aligned}$$

From (3.3) and (3.4), it is easy to see that

$$\begin{aligned} &\frac{1}{\Gamma (1+\alpha )}\int _0^1|1-2t|^\alpha t^{\alpha s}(\mathrm{d}t)^\alpha \\ &\quad = \frac{1}{\Gamma (1+\alpha )}\int _0^1|1-2t|^\alpha (1-t)^{\alpha s}(\mathrm{d}t)^\alpha \\ &\quad = \bigg [\frac{\Gamma (1+s\alpha )}{\Gamma (1+(s+1)\alpha )}+\frac{\Gamma (1+\alpha )\Gamma (1+s\alpha )}{\Gamma (1+(s+2)\alpha )}\left( \left( \frac{1}{2}\right) ^{\alpha s}-2^\alpha \right) \bigg ]. \end{aligned}$$

Therefore

$$\begin{aligned} & \frac{1}{\Gamma (1+\alpha )}\int _0^1|1-2t|^\alpha \big |f^{(\alpha )}(ta+(1-t)b)\big |^q(\mathrm{d}t)^\alpha \\ &\quad =\bigg [\frac{\Gamma (1+s\alpha )}{\Gamma (1+(s+1)\alpha )}+\frac{\Gamma (1+\alpha )\Gamma (1+s\alpha )}{\Gamma (1+(s+2)\alpha )} \left( \left( \frac{1}{2}\right) ^{\alpha s}-2^\alpha \right) \bigg ]^{\frac{1}{q}}\bigg [\big |f^{(\alpha )}(a)\big |^q+\big |f^{(\alpha )}(b)\big |^q\bigg ]^{\frac{1}{q}}. \end{aligned}$$

(3.9)

Thus, substituting (3.8) and (3.9) into (3.7), we have

$$\begin{aligned} \begin{array}{cl} &{}\frac{1}{\Gamma (1+\alpha )}\int _0^1|1-2t|^\alpha \big |f^{(\alpha )}(ta+(1-t)b)\big |(\mathrm{d}t)^\alpha \\ \le &{}\bigg (\frac{\Gamma (1+\alpha )}{\Gamma (1+2\alpha )}\bigg )^{\frac{q-1}{q}}\bigg [\frac{\Gamma (1+s\alpha )}{\Gamma (1+(s+1)\alpha )}+\frac{\Gamma (1+\alpha )\Gamma (1+s\alpha )}{\Gamma (1+(s+2)\alpha )} \left( \left( \frac{1}{2}\right) ^{\alpha s}-2^\alpha \right) \bigg ]^{\frac{1}{q}} \bigg [\big |f^{(\alpha )}(a)\big |^q+\big |f^{(\alpha )}(b)\big |^q\bigg ]^{\frac{1}{q}}. \end{array} \end{aligned}$$

Therefore, from (3.2), it follows that

$$\begin{aligned} \begin{array}{cl} &{}\bigg |\frac{f(a)+f(b)}{2^\alpha }-\frac{\Gamma (1+\alpha )}{(b-a)^\alpha }{_a}I_b^{(\alpha )}f(x)\bigg |\\ &{}\le \frac{(b-a)^\alpha }{2^\alpha }\bigg (\frac{\Gamma (1+\alpha )}{\Gamma (1+2\alpha )}\bigg )^{\frac{q-1}{q}}\bigg [\frac{\Gamma (1+s\alpha )}{\Gamma (1+(s+1)\alpha )}+\frac{\Gamma (1+\alpha )\Gamma (1+s\alpha )}{\Gamma (1+(s+2)\alpha )} \left( \left( \frac{1}{2}\right) ^{\alpha s}-2^\alpha \right) \bigg ]^{\frac{1}{q}}\\ &{}\quad \times \bigg [\big |f^{(\alpha )}(a)\big |^q+\big |f^{(\alpha )}(b)\big |^q\bigg ]^{\frac{1}{q}}.\\ \end{array} \end{aligned}$$

Thus, we complete the proof of Theorem 3.2. $\square$

Theorem 3.3

Suppose that f :  $I\rightarrow \mathbb {R^{\alpha }},$ $I\subset [0,\infty )$ is a differentiable function on $I^0$, such that $f^{(\alpha )}\in C_{\alpha }[a,b]$ , where $a,b\in I$ with $a<b$. If $|f^{(\alpha )}|^q$ is a generalized s-convex function in the second sense on [a, b] for some fixed $s\in (0,1)$ and $q>1$, then

$$\begin{aligned} \begin{array}{cl} &{}\bigg |\frac{f(a)+f(b)}{2^\alpha }-\frac{\Gamma (1+\alpha )}{(b-a)^\alpha }{_a}I_b^{(\alpha )}f(x)\bigg |\\ &{}\le \frac{(b-a)^\alpha }{2^\alpha }\bigg [\frac{\Gamma (1+\frac{q}{q-1}\alpha )}{2^\alpha \Gamma (1+\frac{2q-1}{q-1}\alpha )}\bigg ]^{\frac{q-1}{q}} \bigg (\frac{\Gamma (1+s\alpha )}{2^\alpha \Gamma (1+(s+1)\alpha )}\bigg )^{\frac{1}{q}}\\ {} &{} \quad \times \left[ \left( |f^{(\alpha )}(a)|^q+\left| f^{(\alpha )}\left( \frac{a+b}{2}\right) \right| ^q\right) ^{\frac{1}{q}}+\left( \left| f^{(\alpha )}\left( \frac{a+b}{2}\right) \right| ^q+|f^{(\alpha )}(b)|^q\right) ^{\frac{1}{q}}\right] . \end{array} \end{aligned}$$

Proof

From Lemma 3.1, we have

$$\begin{aligned} \begin{array}{cl} &{}\bigg |\frac{f(a)+f(b)}{2^\alpha }-\frac{\Gamma (1+\alpha )}{(b-a)^\alpha }{_a}I_b^{(\alpha )}f(x)\bigg |\\ &{}\le \frac{(b-a)^\alpha }{2^\alpha }\frac{1}{\Gamma (1+\alpha )}\int _0^1|1-2t|^\alpha |f^{(\alpha )}(ta+(1-t)b)|(\mathrm{d}t)^\alpha \\ &{}\le \frac{(b-a)^\alpha }{2^\alpha }\bigg [\frac{1}{\Gamma (1+\alpha )}\int _0^{\frac{1}{2}}(1-2t)|f^{(\alpha )}(ta+(1-t)b)|(\mathrm{d}t)^\alpha \\ &{}\quad +\frac{1}{\Gamma (1+\alpha )}\int _{\frac{1}{2}}^1(2t-1)|f^{(\alpha )}(ta+(1-t)b)|(\mathrm{d}t)^\alpha \bigg ]. \end{array} \end{aligned}$$

(3.10)

Let us estimate

$$\begin{aligned} \frac{1}{\Gamma (1+\alpha )}\int _0^{\frac{1}{2}}(1-2t)|f^{(\alpha )}(ta+(1-t)b)|(\mathrm{d}t)^\alpha \end{aligned}$$

and

$$\begin{aligned} \frac{1}{\Gamma (1+\alpha )}\int _{\frac{1}{2}}^1(2t-1)|f^{(\alpha )}(ta+(1-t)b)|(\mathrm{d}t)^\alpha , \end{aligned}$$

respectively.

Using the generalized Hölder’s inequality(Lemma 2.4), we obtain

$$\begin{aligned} \begin{array}{cl} &{}\frac{1}{\Gamma (1+\alpha )}\int _0^{\frac{1}{2}}(1-2t)|f^{(\alpha )}(ta+(1-t)b)|(\mathrm{d}t)^\alpha \\ &{}\le \left( \frac{1}{\Gamma (1+\alpha )}\int _0^{\frac{1}{2}}(1-2t)^{\alpha \frac{q}{q-1}}(\mathrm{d}t)^\alpha \right) ^{\frac{q-1}{q}} \bigg (\frac{1}{\Gamma (1+\alpha )}\int _0^{\frac{1}{2}}|f^{(\alpha )}(ta+(1-t)b)|^{\alpha q}(\mathrm{d}t)^\alpha \bigg )^{\frac{1}{q}}. \end{array} \end{aligned}$$

(3.11)

It is easy to see that

$$\begin{aligned} \begin{array}{cl} \frac{1}{\Gamma (1+\alpha )}\int _0^{\frac{1}{2}}(1-2t)^{\alpha \frac{q}{q-1}}(\mathrm{d}t)^\alpha =\frac{1}{\Gamma (1+\alpha )}\int _{\frac{1}{2}}^1(1-2t)^{\alpha \frac{q}{q-1}}(\mathrm{d}t)^\alpha =\frac{\Gamma (1+\frac{q}{q-1}\alpha )}{2^\alpha \Gamma (1+\frac{2q-1}{q-1}\alpha )}. \end{array} \end{aligned}$$

(3.12)

Let $|f^{(\alpha )}(ta+(1-t)b)|^q=U(t).$ It is easy to see that U(t) is a generalized sconvex function in the second sense. Thus, from the right-hand side of (3.1), it follows that

$$\begin{aligned} &\frac{1}{\Gamma (1+\alpha )}\int _0^{\frac{1}{2}}|f^{(\alpha )}(ta+(1-t)b)|^q(\mathrm{d}t)^\alpha \\ &\quad= \frac{1}{\Gamma (1+\alpha )}\int _0^{\frac{1}{2}}U(t)(\mathrm{d}t)^\alpha \\ &\quad \le \left( \frac{1}{2}-0\right) ^\alpha \frac{\Gamma (1+s\alpha )}{\Gamma (1+(s+1)\alpha )}\left(U(0)+U(\frac{1}{2})\right)\\ &\quad =\frac{\Gamma (1+s\alpha )}{2^\alpha \Gamma (1+(s+1)\alpha )}\left( \left| f^{(\alpha )}\left( \frac{a+b}{2}\right) \right| ^q+|f^{(\alpha )}(b)|^q\right) . \end{aligned}$$

(3.13)

Thus, substituting (3.12) and (3.13) into (3.11), we get

$$\begin{aligned} \begin{array}{cl} &{}\frac{1}{\Gamma (1+\alpha )}\int _0^{\frac{1}{2}}(1-2t)|f^{(\alpha )}(ta+(1-t)b)|(\mathrm{d}t)^\alpha \\ &{}\le \bigg (\frac{\Gamma (1+\frac{q}{q-1}\alpha )}{2^\alpha \Gamma (1+\frac{2q-1}{q-1}\alpha )}\bigg )^{\frac{q-1}{q}} \bigg (\frac{\Gamma (1+s\alpha )}{2^\alpha \Gamma (1+(s+1)\alpha )}\bigg )^{\frac{1}{q}} \left( |f^{(\alpha )}(a)|^q+\left| f^{(\alpha )}\left( \frac{a+b}{2}\right) \right| ^q\right) ^{\frac{1}{q}}. \end{array} \end{aligned}$$

(3.14)

Moreover

$$\begin{aligned} \begin{array}{cl} \frac{1}{\Gamma (1+\alpha )}\int _{\frac{1}{2}}^1(2t-1)^{\alpha \frac{q}{q-1}}(\mathrm{d}t)^\alpha =\frac{1}{\Gamma (1+\alpha )}\int _{\frac{1}{2}}^1(1-2t)^{\alpha \frac{q}{q-1}}(\mathrm{d}t)^\alpha =\frac{\Gamma (1+\frac{q}{q-1}\alpha )}{2^\alpha \Gamma (1+\frac{2q-1}{q-1}\alpha )}. \end{array} \end{aligned}$$

In addition, similar to the estimate of (3.13), we have

$$ \frac{1}{\Gamma (1+\alpha )}\int _{\frac{1}{2}}^1|f^{(\alpha )}(ta+(1-t)b)|^q(\mathrm{d}t)^\alpha \le \frac{\Gamma (1+s\alpha )}{2^\alpha \Gamma (1+(s+1)\alpha )}\left( \left| f^{(\alpha )}\left( \frac{a+b}{2}\right) \right| ^q+|f^{(\alpha )}(b)|^q\right) . $$

Therefore, it is analogues to the estimate of (3.11), we have

$$\begin{aligned} \begin{array}{cl} &{}\frac{1}{\Gamma (1+\alpha )}\int _0^{\frac{1}{2}}(1-2t)|f^{(\alpha )}(ta+(1-t)b)|(\mathrm{d}t)^\alpha \\ &{}\le \bigg (\frac{1}{\Gamma (1+\alpha )}\int _{\frac{1}{2}}^1(2t-1)^{\alpha \frac{q}{q-1}}(\mathrm{d}t)^\alpha \bigg )^{\frac{q-1}{q}} \bigg (\frac{1}{\Gamma (1+\alpha )}\int _{\frac{1}{2}}^1|f^{(\alpha )}(ta+(1-t)b)|^{q\alpha }(\mathrm{d}t)^\alpha \bigg )^{\frac{1}{q}}\\ &{}\le \bigg (\frac{\Gamma (1+\frac{q}{q-1}\alpha )}{2^\alpha \Gamma (1+\frac{2q-1}{q-1}\alpha )}\bigg )^{\frac{q-1}{q}} \bigg (\frac{\Gamma (1+s\alpha )}{2^\alpha \Gamma (1+(s+1)\alpha )}\bigg )^{\frac{1}{q}} \left( \left| f^{(\alpha )}\left( \frac{a+b}{2}\right) \right| ^q+|f^{(\alpha )}(b)|^q)|^q\right) ^{\frac{1}{q}}. \end{array} \end{aligned}$$

(3.15)

Thus, combining (3.10), (3.14), and (3.15), we obtain

$$\begin{aligned} \begin{array}{cl} &{}\bigg |\frac{f(a)+f(b)}{2^\alpha }-\frac{\Gamma (1+\alpha )}{(b-a)^\alpha }{_a}I_b^{(\alpha )}f(x)\bigg |\\ &{}\le \frac{(b-a)^\alpha }{2^\alpha }\bigg [\frac{\Gamma (1+\frac{q}{q-1}\alpha )}{2^\alpha \Gamma (1+\frac{2q-1}{q-1}\alpha )}\bigg ]^{\frac{q-1}{q}} \bigg (\frac{\Gamma (1+s\alpha )}{2^\alpha \Gamma (1+(s+1)\alpha )}\bigg )^{\frac{1}{q}}\\ {} &{} \quad \times \left[ \left( |f^{(\alpha )}(a)|^q+\left| f^{(\alpha )}\left( \frac{a+b}{2}\right) \right| ^q\right) ^{\frac{1}{q}}+\left( \left| f^{(\alpha )}\left( \frac{a+b}{2}\right) \right| ^q+|f^{(\alpha )}(b)|^q\right) ^{\frac{1}{q}}\right] . \end{array} \end{aligned}$$

$\square$

Therefore, we complete the proof of Theorem 3.3.