New Conformable Fractional Integral Inequalities of Hermite–Hadamard Type for Convex Functions

Abstract

In this work, we established new inequalities of Hermite–Hadamard type for convex functions via conformable fractional integrals. Through the conformable fractional integral inequalities, we found some new inequalities of Hermite–Hadamard type for convex functions in the form of classical integrals.

1. Introduction

A function $h:\mathcal{I}\subseteq \mathbb{R}\to \mathbb{R}$ is said to be convex on the interval $\mathcal{I}$, if the inequality

$$h(\zeta x+(1-\zeta \left)y\right)\le \zeta h\left(x\right)+(1-\zeta )h\left(y\right)$$

(1)

holds for all $x,y\in \mathcal{I}$ and $\zeta \in [0,1]$. We say that h is concave if $-h$ is convex.

For convex functions, many equalities or inequalities have been established by many authors; for example the Ostrowski type inequality [1], Hardy type inequality [2], Olsen type inequality [3], and Gagliardo–Nirenberg type inequality [4] but the most common and significant inequality is the Hermite–Hadamard type inequality [5,6], which is defined as:

$$h\left(\frac{u+v}{2}\right)\le \frac{1}{v-u}{\int }_u^{v}h\left(x\right)dx\le \frac{h\left(u\right)+h\left(v\right)}{2},$$

(2)

where the function $h:\mathcal{I}\subseteq \mathbb{R}\to \mathbb{R}$ is assumed to be a convex function with $u<v$ and $u,v\in \mathcal{I}$.

A number of mathematicians in the field of applied and pure mathematics have dedicated their efforts to extend, generalize, counterpart, and refine the Hermite–Hadamard inequality (2) for different classes of convex functions and mappings. For more recent results obtained on inequality (2), we refer the reader to References [5,7,8,9,10].

Definition 1

([11]). Suppose that $h\in L\left(\right[u,v\left]\right)$. The left and right Riemann–Liouville fractional integrals $J_{u^{+}}^{\alpha }h$ and $J_{v^{-}}^{\alpha }h$ of order $\alpha >0$ are defined by

$$J_{u^{+}}^{\alpha }h\left(x\right)=\frac{1}{\mathsf{\Gamma }\left(\alpha \right)}{\int }_u^x{(x-\zeta )}^{\alpha -1}h\left(\zeta \right)d\zeta ,x>u,$$

and

$$J_{v^{-}}^{\alpha }h\left(x\right)=\frac{1}{\mathsf{\Gamma }\left(\alpha \right)}{\int }_x^v{(\zeta -x)}^{\alpha -1}h\left(\zeta \right)d\zeta ,x<v,$$

respectively, where $\mathsf{\Gamma }\left(\alpha \right)$ is the standard gamma function defined by $\mathsf{\Gamma }\left(\alpha \right)={\int }_0^{\infty }{e}^{-\zeta }{\zeta }^{\alpha -1}d\zeta$ and $J_{u^{+}}^{0}h\left(x\right)=J_{v^{-}}^{0}h\left(x\right)=h\left(x\right)$.

As for classical integrals, many Hermite–Hadamard type inequalities have been established for the Riemann–Liouville fractional integrals; for more details and interesting applications see References [6,12,13].

Now, we give the definition of the conformable fractional derivative with its important properties which are useful in order to obtain our main results (see References [14,15,16,17,18,19,20,21]). In our study, we use the Katugampola derivative formulation of conformable derivative which is explained in the following definition:

Definition 2

([20]). Given a function $h:[0,\infty )\to \mathbb{R}$. Then, the conformable fractional derivative of h of order α of h at ζ is defined by

$$D_{\alpha }\left(h\right)\left(\zeta \right)=\underset{ϵ\to 0}{\lim }\frac{h\left(\zeta +ϵ{\zeta }^{1-\alpha }\right)-h\left(\zeta \right)}{ϵ},\alpha \in (0,1),\zeta >0.$$

(3)

If h is $\alpha$-differentiable in some $(0,\alpha ),\alpha >0$, ${\lim }_{\zeta \to 0^{+}}{h}^{\left(\alpha \right)}\left(\zeta \right)$ exist, then define

$$h^{\left(\alpha \right)}\left(0\right)=\underset{\zeta \to 0^{+}}{\lim }{h}^{\left(\alpha \right)}\left(\zeta \right).$$

Additionally, note that if h is differentiable, then

$$D_{\alpha }\left(h\right)\left(\zeta \right)={\zeta }^{1-\alpha }{h}^{\prime }\left(\zeta \right),\mathrm{where}{h}^{\prime }\left(\zeta \right)=\underset{ϵ\to 0}{\lim }\frac{h\left(\zeta +ϵ\right)-h\left(\zeta \right)}{ϵ}.$$

(4)

We can write $h^{\left(\alpha \right)}\left(\zeta \right)$ for $D_{\alpha }\left(h\right)\left(\zeta \right)$ or $\frac{d_{\alpha }}{d_{\alpha }\zeta }\left(h\left(\zeta \right)\right)$ to denote the conformable fractional derivatives of h of order $\alpha$ at $\zeta$. In addition, if the conformable fractional derivative of h of order $\alpha$ exists, then we simply say h is $\alpha$-differentiable.

Theorem 1

([20]). Let $\alpha \in (0,1]$ and $h,g$ be α-differentiable at a point $\zeta >0$. Then,

1. 

$D_{\alpha }(uf+vg)=u{D}_{\alpha }\left(h\right)+v{D}_{\alpha }\left(g\right)$ for all $u,v\in \mathbb{R}$,

2. 

$D_{\alpha }\left(hg\right)=h{D}_{\alpha }\left(g\right)+g{D}_{\alpha }\left(h\right)$,

3. 

$D_{\alpha }\left(\frac{h}{g}\right)=\frac{h{D}_{\alpha }\left(g\right)-g{D}_{\alpha }\left(h\right)}{g^2}$,

4. 

$D_{\alpha }\left(c\right)=0$ for all constant function $h\left(\zeta \right)=c$,

5. 

$D_{\alpha }\left(1\right)=0$,

6. 

$D_{\alpha }\left(\frac{1}{\alpha }{\zeta }^{\alpha }\right)=1$.

Now, we give the definition of conformable fractional integral:

Definition 3

([16]). Let $\alpha \in (0,1]$ and $0\le u\le v$. We say that a function $h:[u,v]\to \mathbb{R}$ is α-fractional integrable on $[u,v]$, if the integral

$${\int }_u^{v}h\left(\zeta \right)d_{\alpha }\zeta ={\int }_u^{v}h\left(\zeta \right){\zeta }^{\alpha -1}d\zeta$$

(5)

exists and is finite.

Remark 1.

(u) All α-fractional integrable functions on $[u,v]$ are indicated by $L_{\alpha }^1\left([u,v]\right)$.

(v) For the usual Riemann improper integral and $\alpha \in (0,1]$, we have

$$I_{\alpha }^u\left(h\right)\left(\zeta \right)=I_1^a\left({\zeta }^{\alpha -1}h\right)={\int }_u^{\zeta }{x}^{\alpha -1}h\left(x\right)dx.$$

(6)

The aim of our article is to establish some new inequalities connected with the Hermite–Hadamard inequalities (2) via conformable fractional integral.

2. Main Results

Our main results depend on the following equality:

Lemma 1.

Let $h:[u,v]\subseteq \mathbb{R}\to \mathbb{R}$ be an α-fractional differentiable mapping on $(u,v)$ with $0\le u<v$. If $D_{\alpha }\left(h\right)\in L_{\alpha }^1\left([u,v]\right)$, then the following identity for conformable fractional integral holds:

$${\mathsf{\Phi }}_{\alpha }(u,v)=\sum _{i=1}^4{\delta }_i,$$

(7)

where

$$\begin{array}{cc}\hfill {\delta }_1& =\frac{v-u}{4}{\int }_0^1\left[{\left(u\zeta +(1-\zeta )\frac{3u+v}{4}\right)}^{2\alpha -1}-{\left(\frac{3u+v}{4}\right)}^{\alpha }{\left(u\zeta +(1-\zeta )\frac{3u+v}{4}\right)}^{\alpha -1}\right]\hfill \\ \hfill & \times D_{\alpha }\left(h\right)\left(u\zeta +(1-\zeta )\frac{3u+v}{4}\right)d_{\alpha }\zeta ,\hfill \\ \hfill {\delta }_2& =\frac{v-u}{4}{\int }_0^1\left[{\left(\frac{3u+v}{4}\zeta +(1-\zeta )\frac{u+v}{2}\right)}^{2\alpha -1}-{\left(\frac{3u+v}{4}\right)}^{\alpha }{\left(\frac{3u+v}{4}\zeta +(1-\zeta )\frac{u+v}{2}\right)}^{\alpha -1}\right]\hfill \\ \hfill & \times D_{\alpha }\left(h\right)\left(\frac{3u+v}{4}\zeta +(1-\zeta )\frac{u+v}{2}\right)d_{\alpha }\zeta ,\hfill \\ \hfill {\delta }_3& =\frac{v-u}{4}{\int }_0^1\left[{\left(\frac{u+v}{2}\zeta +(1-\zeta )\frac{u+3v}{4}\right)}^{2\alpha -1}-{\left(\frac{u+3v}{4}\right)}^{\alpha }{\left(\frac{u+v}{2}\zeta +(1-\zeta )\frac{u+3v}{4}\right)}^{\alpha -1}\right]\hfill \\ \hfill & \times D_{\alpha }\left(h\right)\left(\frac{u+v}{2}\zeta +(1-\zeta )\frac{u+3v}{4}\right)d_{\alpha }\zeta ,\hfill \\ \hfill {\delta }_4& =\frac{v-u}{4}{\int }_0^1\left[{\left(\frac{u+3v}{4}\zeta +(1-\zeta )v\right)}^{2\alpha -1}-{\left(\frac{u+3v}{4}\right)}^{\alpha }{\left(\frac{u+3v}{4}\zeta +(1-\zeta )v\right)}^{\alpha -1}\right]\hfill \\ \hfill & \times D_{\alpha }\left(h\right)\left(\frac{u+3v}{4}\zeta +(1-\zeta )v\right)d_{\alpha }\zeta ,\hfill \end{array}$$

and

$$\begin{array}{cc}\hfill {\mathsf{\Phi }}_{\alpha }(u,v)& =\left[{\left(\frac{3u+v}{4}\right)}^{\alpha }-u^{\alpha }\right]h\left(u\right)+\left[v^{\alpha }-{\left(\frac{u+3v}{4}\right)}^{\alpha }\right]h\left(v\right)\hfill \\ \hfill & +\left[{\left(\frac{u+3v}{4}\right)}^{\alpha }-{\left(\frac{3u+v}{4}\right)}^{\alpha }\right]h\left(\frac{u+v}{2}\right)-\alpha {\int }_u^{v}h\left(x\right)d_{\alpha }x.\hfill \end{array}$$

Proof. 

By using the definition of the conformable fractional derivative (4), we have

$${\delta }_1=\frac{v-u}{4}{\int }_0^1\left[{\left(u\zeta +(1-\zeta )\frac{3u+v}{4}\right)}^{\alpha }-{\left(\frac{3u+v}{4}\right)}^{\alpha }\right]h^{\prime }\left(u\zeta +(1-\zeta )\frac{3u+v}{4}\right)d\zeta .$$

On integrating by parts, one can have

$${\delta }_1=\left[{\left(\frac{3u+v}{4}\right)}^{\alpha }-u^{\alpha }\right]h\left(u\right)-\alpha {\int }_0^1{\left(u\zeta +(1-\zeta )\frac{3u+v}{4}\right)}^{\alpha -1}h\left(u\zeta +(1-\zeta )\frac{3u+v}{4}\right)d\zeta .$$

Using the change of the variable $x:=u\zeta +(1-\zeta )\frac{3u+v}{4},\zeta \in [0,1]$ and definition of conformable fractional integral (5), we obtain

$$\begin{array}{cc}\hfill {\delta }_1& =\left[{\left(\frac{3u+v}{4}\right)}^{\alpha }-u^{\alpha }\right]h\left(u\right)-\alpha {\int }_u^{\frac{3u+v}{4}}{x}^{\alpha -1}h\left(x\right)dx\hfill \\ \hfill & =\left[{\left(\frac{3u+v}{4}\right)}^{\alpha }-u^{\alpha }\right]h\left(u\right)-\alpha {\int }_u^{\frac{3u+v}{4}}h\left(x\right)d_{\alpha }x.\hfill \end{array}$$

Similarly, we get

$${\delta }_2=\left[{\left(\frac{u+v}{2}\right)}^{\alpha }-{\left(\frac{3u+v}{4}\right)}^{\alpha }\right]h\left(\frac{u+v}{2}\right)-\alpha {\int }_{\frac{3u+v}{4}}^{\frac{u+v}{2}}h\left(x\right)d_{\alpha }x,$$

$${\delta }_3=\left[{\left(\frac{u+3v}{4}\right)}^{\alpha }-{\left(\frac{u+v}{2}\right)}^{\alpha }\right]h\left(\frac{u+v}{2}\right)-\alpha {\int }_{\frac{u+v}{2}}^{\frac{u+3v}{4}}h\left(x\right)d_{\alpha }x,$$

and

$${\delta }_4=\left[v^{\alpha }-{\left(\frac{u+3v}{4}\right)}^{\alpha }\right]h\left(v\right)-\alpha {\int }_{\frac{u+3v}{4}}^{v}h\left(x\right)d_{\alpha }x.$$

Adding ${\delta }_1,{\delta }_2,{\delta }_3$ and ${\delta }_4$ together, we obtain the desired identity (7). This completes the proof of Lemma 1. □

Remark 2.

With the similar assumptions of Lemma 1, if $\alpha =1$, then identity (7) reduces to the following identity:

$$\begin{array}{l}{\mathsf{\Phi }}_1(u,v)={\int }_0^1(1-\zeta )h^{\prime }\left(\frac{3u+v}{4}\zeta +(1-\zeta )\frac{u+v}{2}\right)d\zeta -{\int }_0^{1}t{h}^{\prime }\left(u\zeta +(1-\zeta )\frac{3u+v}{4}\right)d\zeta \\ +{\int }_0^1(1-\zeta )h^{\prime }\left(\frac{u+3v}{4}\zeta +(1-\zeta )v\right)d\zeta -{\int }_0^{1}t{h}^{\prime }\left(\frac{u+v}{2}\zeta +(1-\zeta )\frac{u+3v}{4}\right)d\zeta ,\hfill \end{array}$$

(8)

where

$${\mathsf{\Phi }}_1(u,v)=\frac{1}{2}\left[\frac{h\left(u\right)+h\left(v\right)}{2}+h\left(\frac{u+v}{2}\right)\right]-\frac{4}{v-u}{\int }_u^{v}h\left(x\right)dx$$

which is obtained by Shi et al. [12].

Theorem 2.

Let $h:[u,v]\subseteq \mathbb{R}\to \mathbb{R}$ be an α-fractional differentiable mapping on $(u,v)$ with $0\le u<v$. If $D_{\alpha }\left(h\right)\in L_{\alpha }^1\left([u,v]\right)$ and $\left|h^{\prime }\right|$ is convex on $[u,v]$, then the following inequality for conformable fractional integral holds:

$$\begin{array}{cc}\hfill \left|{\mathsf{\Phi }}_{\alpha }(u,v)\right|& \le \frac{v-u}{4}[d_1\left(\alpha \right)|h^{\prime }\left(u\right)|+d_2\left(\alpha \right)|h^{\prime }\left(v\right)|\hfill \\ \hfill & +d_3\left(\alpha \right)\left|h^{\prime }\left(\frac{3u+v}{4}\right)\right|+d_4\left(\alpha \right)\left|h^{\prime }\left(\frac{u+v}{2}\right)\right|+d_5\left(\alpha \right)\left|h^{\prime }\left(\frac{u+3v}{4}\right)\right|],\hfill \end{array}$$

(9)

where

$$\begin{array}{cc}\hfill d_1\left(\alpha \right)& =\frac{1}{12}\left[4u^{\alpha }+\left(\frac{3u+v}{4}\right)u^{\alpha -1}+{\left(\frac{3u+v}{4}\right)}^{\alpha -1}u-5{\left(\frac{3u+v}{4}\right)}^{\alpha }\right],\hfill \\ \hfill d_2\left(\alpha \right)& =\frac{1}{12}\left[3v^{\alpha }+\left(\frac{3u+v}{4}\right)v^{\alpha -1}+{\left(\frac{3u+v}{4}\right)}^{\alpha -1}v-5{\left(\frac{3u+v}{4}\right)}^{\alpha }\right],\hfill \\ \hfill d_3\left(\alpha \right)& =\frac{1}{12}[u^{\alpha }+\left(\frac{3u+v}{4}\right)u^{\alpha -1}+{\left(\frac{3u+v}{4}\right)}^{\alpha -1}u+\left(\frac{u+v}{2}\right){\left(\frac{3u+v}{4}\right)}^{\alpha -1}\hfill \\ \hfill & +{\left(\frac{u+v}{2}\right)}^{\alpha -1}\left(\frac{3u+v}{4}\right)+{\left(\frac{u+v}{2}\right)}^{\alpha }-5{\left(\frac{3u+v}{4}\right)}^{\alpha }],\hfill \end{array}$$

$$\begin{array}{cc}\hfill d_4\left(\alpha \right)& =\frac{1}{12}[7{\left(\frac{u+v}{2}\right)}^{\alpha }+\left(\frac{u+v}{2}\right){\left(\frac{3u+v}{4}\right)}^{\alpha -1}+{\left(\frac{u+v}{2}\right)}^{\alpha -1}\left(\frac{3u+v}{4}\right)\hfill \\ \hfill & +\left(\frac{u+v}{2}\right){\left(\frac{u+3v}{4}\right)}^{\alpha -1}+{\left(\frac{u+v}{2}\right)}^{\alpha -1}\left(\frac{u+3v}{4}\right)-5{\left(\frac{3u+v}{4}\right)}^{\alpha }-5{\left(\frac{u+3v}{4}\right)}^{\alpha }],\hfill \\ \hfill d_5\left(\alpha \right)& =\frac{1}{12}[v^{\alpha }+{\left(\frac{u+v}{2}\right)}^{\alpha }+\left(\frac{u+3v}{4}\right){\left(\frac{u+v}{2}\right)}^{\alpha -1}+{\left(\frac{u+3v}{4}\right)}^{\alpha -1}\left(\frac{u+v}{2}\right)\hfill \\ \hfill & +v{\left(\frac{u+3v}{4}\right)}^{\alpha -1}+v^{\alpha -1}\left(\frac{u+3v}{4}\right)-5{\left(\frac{u+3v}{4}\right)}^{\alpha }].\hfill \end{array}$$

Proof. 

Using Lemma 1 and the property (4), we have

$$\begin{array}{l}{\mathsf{\Phi }}_{\alpha }(u,v)=\frac{v-u}{4}\{{\int }_0^1\left[{\left(u\zeta +(1-\zeta )\frac{3u+v}{4}\right)}^{\alpha }-{\left(\frac{3u+v}{4}\right)}^{\alpha }\right]h^{\prime }\left(u\zeta +(1-\zeta )\frac{3u+v}{4}\right)d\zeta \\ +{\int }_0^1\left[{\left(\frac{3u+v}{4}\zeta +(1-\zeta )\frac{u+v}{2}\right)}^{\alpha }-{\left(\frac{3u+v}{4}\right)}^{\alpha }\right]h^{\prime }\left(\frac{3u+v}{4}\zeta +(1-\zeta )\frac{u+v}{2}\right)d\zeta \hfill \\ +{\int }_0^1\left[{\left(\frac{u+v}{2}\zeta +(1-\zeta )\frac{u+3v}{4}\right)}^{\alpha }-{\left(\frac{u+3v}{4}\right)}^{\alpha }\right]h^{\prime }\left(\frac{u+v}{2}\zeta +(1-\zeta )\frac{u+3v}{4}\right)d_{\alpha }\zeta \hfill \\ +{\int }_0^1\left[{\left(\frac{u+3v}{4}\zeta +(1-\zeta )v\right)}^{\alpha }-{\left(\frac{u+3v}{4}\right)}^{\alpha }\right]h^{\prime }\left(\frac{u+3v}{4}\zeta +(1-\zeta )v\right)d_{\alpha }\zeta \}.\hfill \end{array}$$

(10)

By using the convexity of $x^{\alpha -1}$ for $x>0,\alpha \in (0,1]$, we have

$$\begin{array}{l}{\left(u\zeta +(1-\zeta )\frac{3u+v}{4}\right)}^{\alpha }-{\left(\frac{3u+v}{4}\right)}^{\alpha }={\left(u\zeta +(1-\zeta )\frac{3u+v}{4}\right)}^{\alpha -1+1}-{\left(\frac{3u+v}{4}\right)}^{\alpha }\hfill \\ ={\left(u\zeta +(1-\zeta )\frac{3u+v}{4}\right)}^{\alpha -1}\left(u\zeta +(1-\zeta )\frac{3u+v}{4}\right)-{\left(\frac{3u+v}{4}\right)}^{\alpha }\hfill \\ \le \left[u^{\alpha -1}\zeta +(1-\zeta ){\left(\frac{3u+v}{4}\right)}^{\alpha -1}\right]\left(u\zeta +(1-\zeta )\frac{3u+v}{4}\right)-{\left(\frac{3u+v}{4}\right)}^{\alpha },\hfill \end{array}$$

(11)

$$\begin{array}{l}{\left(\frac{3u+v}{4}\zeta +(1-\zeta )\frac{u+v}{2}\right)}^{\alpha }-{\left(\frac{3u+v}{4}\right)}^{\alpha }\\ \le \left[t{\left(\frac{3u+v}{4}\right)}^{\alpha -1}+(1-\zeta ){\left(\frac{u+v}{2}\right)}^{\alpha -1}\right]\left(\frac{3u+v}{4}\zeta +(1-\zeta )\frac{u+v}{2}\right)-{\left(\frac{3u+v}{4}\right)}^{\alpha },\hfill \end{array}$$

(12)

$$\begin{array}{l}{\left(\frac{u+v}{2}\zeta +(1-\zeta )\frac{u+3v}{2}\right)}^{\alpha }-{\left(\frac{u+3v}{4}\right)}^{\alpha }\\ \le \left[t{\left(\frac{u+v}{2}\right)}^{\alpha -1}+(1-\zeta ){\left(\frac{u+3v}{4}\right)}^{\alpha -1}\right]\left(\frac{u+v}{2}\zeta +(1-\zeta )\frac{u+3v}{4}\right)-{\left(\frac{u+3v}{4}\right)}^{\alpha },\hfill \end{array}$$

(13)

and

$$\begin{array}{l}{\left(\frac{u+3v}{4}\zeta +(1-\zeta )v\right)}^{\alpha }-{\left(\frac{u+3v}{4}\right)}^{\alpha }\hfill \\ \le \left[t{\left(\frac{u+3v}{4}\right)}^{\alpha -1}+(1-\zeta )v^{\alpha -1}\right]\left(\frac{u+3v}{4}\zeta +(1-\zeta )v\right)-{\left(\frac{u+3v}{4}\right)}^{\alpha }.\hfill \end{array}$$

(14)

Using (11), (12), (13), and (14) in (10) and using the properties of modulus, we get

$$\begin{array}{l}|{\mathsf{\Phi }}_{\alpha }(u,v)|\le \frac{v-u}{4}\{{\int }_0^1\left[\left(u^{\alpha -1}\zeta +(1-\zeta ){\left(\frac{3u+v}{4}\right)}^{\alpha -1}\right)\left(u\zeta +(1-\zeta )\frac{3u+v}{4}\right)-{\left(\frac{3u+v}{4}\right)}^{\alpha }\right]\\ \times \left|h^{\prime }\left(u\zeta +(1-\zeta )\frac{3u+v}{4}\right)\right|d\zeta \hfill \\ +{\int }_0^1\left[\left({\left(\frac{3u+v}{4}\right)}^{\alpha -1}\zeta +(1-\zeta ){\left(\frac{u+v}{2}\right)}^{\alpha -1}\right)\left(\frac{3u+v}{4}\zeta +(1-\zeta )\frac{u+v}{2}\right)-{\left(\frac{3u+v}{4}\right)}^{\alpha }\right]\hfill \\ \times \left|h^{\prime }\left(\frac{3u+v}{4}\zeta +(1-\zeta )\frac{u+v}{2}\right)\right|d\zeta \hfill \\ +{\int }_0^1\left[\left({\left(\frac{u+v}{2}\right)}^{\alpha -1}\zeta +(1-\zeta ){\left(\frac{u+3v}{4}\right)}^{\alpha -1}\right)\left(\frac{u+v}{2}\zeta +(1-\zeta )\frac{u+3v}{4}\right)-{\left(\frac{u+3v}{4}\right)}^{\alpha }\right]\hfill \\ \times \left|h^{\prime }\left(\frac{u+v}{2}\zeta +(1-\zeta )\frac{u+3v}{4}\right)\right|d\zeta \hfill \\ +{\int }_0^1\left[\left({\left(\frac{u+3v}{4}\right)}^{\alpha -1}\zeta +(1-\zeta )v^{\alpha -1}\right)\left(\frac{u+3v}{4}\zeta +(1-\zeta )v\right)-{\left(\frac{u+3v}{4}\right)}^{\alpha }\right]\hfill \\ \times \left|h^{\prime }\left(\frac{u+3v}{4}\zeta +(1-\zeta )v\right)\right|d\zeta \}.\hfill \end{array}$$

(15)

Since $|h^{\prime }|$ is convex on $[u,v]$ for any $\zeta \in [0,1]$, so (17) becomes

$$\begin{array}{l}|{\mathsf{\Phi }}_{\alpha }(u,v)|\le \frac{v-u}{4}\{{\int }_0^1\left[\left(u^{\alpha -1}\zeta +(1-\zeta ){\left(\frac{3u+v}{4}\right)}^{\alpha -1}\right)\left(u\zeta +(1-\zeta )\frac{3u+v}{4}\right)-{\left(\frac{3u+v}{4}\right)}^{\alpha }\right]\\ \times \left[\zeta \left|h^{\prime }\left(u\right)\right|+(1-\zeta )\left|h^{\prime }\left(\frac{3u+v}{4}\right)\right|\right]d\zeta +{\int }_0^1[\left({\left(\frac{3u+v}{4}\right)}^{\alpha -1}\zeta +(1-\zeta ){\left(\frac{u+v}{2}\right)}^{\alpha -1}\right)\hfill \\ \times \left(\frac{3u+v}{4}\zeta +(1-\zeta )\frac{u+v}{2}\right)-{\left(\frac{3u+v}{4}\right)}^{\alpha }]\left[\zeta \left|h^{\prime }\left(\frac{3u+v}{4}\right)\right|+(1-\zeta )\left|h^{\prime }\left(\frac{u+v}{2}\right)\right|\right]d\zeta \hfill \\ +{\int }_0^1\left[\left({\left(\frac{u+v}{2}\right)}^{\alpha -1}\zeta +(1-\zeta ){\left(\frac{u+3v}{4}\right)}^{\alpha -1}\right)\left(\frac{u+v}{2}\zeta +(1-\zeta )\frac{u+3v}{4}\right)-{\left(\frac{u+3v}{4}\right)}^{\alpha }\right]\hfill \\ \times \left[\zeta \left|h^{\prime }\left(\frac{u+v}{2}\right)\right|+(1-\zeta )\left|h^{\prime }\left(\frac{u+3v}{4}\right)\right|\right]d\zeta +{\int }_0^1[\left({\left(\frac{u+3v}{4}\right)}^{\alpha -1}\zeta +(1-\zeta )v^{\alpha -1}\right)\hfill \\ \times \left(\frac{u+3v}{4}\zeta +(1-\zeta )v\right)-{\left(\frac{u+3v}{4}\right)}^{\alpha }\left]\left[\zeta \left|h^{\prime }\left(\frac{u+3v}{4}\right)\right|+(1-\zeta )\left|h^{\prime }\left(v\right)\right|\right]d\zeta \right\}.\hfill \end{array}$$

(16)

Simple calculation gives

$$\begin{array}{l}|{\mathsf{\Phi }}_{\alpha }(u,v)|\le \frac{v-u}{4}\{\left(\frac{1}{12}\left[4u^{\alpha }+\left(\frac{3u+v}{4}\right)u^{\alpha -1}+{\left(\frac{3u+v}{4}\right)}^{\alpha -1}u-5{\left(\frac{3u+v}{4}\right)}^{\alpha }\right]\right)|h^{\prime }\left(u\right)|\\ +\left(\frac{1}{12}\left[3v^{\alpha }+\left(\frac{3u+v}{4}\right)v^{\alpha -1}+{\left(\frac{3u+v}{4}\right)}^{\alpha -1}v-5{\left(\frac{3u+v}{4}\right)}^{\alpha }\right]\right)|h^{\prime }\left(v\right)|\hfill \\ +\left(\frac{1}{12}\right[u^{\alpha }+\left(\frac{3u+v}{4}\right)u^{\alpha -1}+{\left(\frac{3u+v}{4}\right)}^{\alpha -1}u+\left(\frac{u+v}{2}\right){\left(\frac{3u+v}{4}\right)}^{\alpha -1}\hfill \\ +{\left(\frac{u+v}{2}\right)}^{\alpha -1}\left(\frac{3u+v}{4}\right)+{\left(\frac{u+v}{2}\right)}^{\alpha }-5{\left(\frac{3u+v}{4}\right)}^{\alpha }\left]\right)\left|h^{\prime }\left(\frac{3u+v}{4}\right)\right|\hfill \\ +\left(\frac{1}{12}\right[7{\left(\frac{u+v}{2}\right)}^{\alpha }+\left(\frac{u+v}{2}\right){\left(\frac{3u+v}{4}\right)}^{\alpha -1}+{\left(\frac{u+v}{2}\right)}^{\alpha -1}\left(\frac{3u+v}{4}\right)\hfill \\ +\left(\frac{u+v}{2}\right){\left(\frac{u+3v}{4}\right)}^{\alpha -1}+{\left(\frac{u+v}{2}\right)}^{\alpha -1}\left(\frac{u+3v}{4}\right)-5{\left(\frac{3u+v}{4}\right)}^{\alpha }-5{\left(\frac{u+3v}{4}\right)}^{\alpha }\left]\right)\left|h^{\prime }\left(\frac{u+v}{2}\right)\right|\hfill \\ +\left(\frac{1}{12}\right[v^{\alpha }+{\left(\frac{u+v}{2}\right)}^{\alpha }+\left(\frac{u+3v}{4}\right){\left(\frac{u+v}{2}\right)}^{\alpha -1}+{\left(\frac{u+3v}{4}\right)}^{\alpha -1}\left(\frac{u+v}{2}\right)\hfill \\ +v{\left(\frac{u+3v}{4}\right)}^{\alpha -1}+v^{\alpha -1}\left(\frac{u+3v}{4}\right)-5{\left(\frac{u+3v}{4}\right)}^{\alpha }\left]\right)\left|h^{\prime }\left(\frac{u+3v}{4}\right)\right|\}.\hfill \end{array}$$

This completes the proof of Theorem 2. □

Corollary 1.

With the similar assumptions of Theorem 2, if $\alpha =1$, then

$$\begin{array}{cc}\hfill \left|{\mathsf{\Phi }}_1(u,v)\right|& \le \frac{v-u}{4}[\frac{2u-v}{12}|h^{\prime }\left(u\right)|+\frac{v-u}{12}|h^{\prime }\left(v\right)|\hfill \\ \hfill & +\frac{3u+v}{48}\left|h^{\prime }\left(\frac{3u+v}{4}\right)\right|+\frac{u+v}{24}\left|h^{\prime }\left(\frac{u+v}{2}\right)\right|+\frac{u+3v}{48}\left|h^{\prime }\left(\frac{u+3v}{4}\right)\right|].\hfill \end{array}$$

Theorem 3.

Let $h:[u,v]\subseteq \mathbb{R}\to \mathbb{R}$ be an α-fractional differentiable mapping on $(u,v)$ with $0\le u<v$. If $D_{\alpha }\left(h\right)\in L_{\alpha }^1\left([u,v]\right)$ and ${\left|h^{\prime }\right|}^q$ is convex on $[u,v]$, then the following inequality for conformable fractional integral holds:

$$\begin{array}{cc}\hfill \left|{\mathsf{\Phi }}_{\alpha }(u,v)\right|& \le \frac{v-u}{4}[{\left(A_1\left(\alpha \right)\right)}^{1-\frac{1}{q}}{\left\{A_2\left(\alpha \right){|h^{\prime }\left(u\right)|}^q+A_3\left(\alpha \right){\left|h^{\prime }\left(\frac{3u+v}{4}\right)\right|}^q\right\}}^{\frac{1}{q}}\hfill \\ \hfill & +{\left(B_1\left(\alpha \right)\right)}^{1-\frac{1}{q}}{\left\{B_2\left(\alpha \right){\left|h^{\prime }\left(\frac{3u+v}{4}\right)\right|}^q+B_3\left(\alpha \right){\left|h^{\prime }\left(\frac{u+v}{2}\right)\right|}^q\right\}}^{\frac{1}{q}}\hfill \\ \hfill & +{\left(C_1\left(\alpha \right)\right)}^{1-\frac{1}{q}}{\left\{C_2\left(\alpha \right){\left|h^{\prime }\left(\frac{u+v}{2}\right)\right|}^q+C_3\left(\alpha \right){\left|h^{\prime }\left(\frac{u+3v}{4}\right)\right|}^q\right\}}^{\frac{1}{q}}\hfill \\ \hfill & +{\left(D_1\left(\alpha \right)\right)}^{1-\frac{1}{q}}{\left\{D_2\left(\alpha \right){\left|h^{\prime }\left(\frac{u+3v}{4}\right)\right|}^q+D_3\left(\alpha \right){\left|h^{\prime }\left(v\right)\right|}^q\right\}}^{\frac{1}{q}}],\hfill \end{array}$$

(17)

where

$$\begin{array}{l}{A}_1\left(\alpha \right)=\frac{4u{(3u+v)}^{\alpha }-{\left(4u\right)}^{\alpha +1}}{4^{\alpha }(v-u)(\alpha +1)}-\frac{\alpha {(3u+v)}^{\alpha }}{4^{\alpha }},\\ A_2\left(\alpha \right)=\frac{{(3u+v)}^{\alpha +2}-{\left(4u\right)}^{\alpha +1}[\alpha (v-u)+(u+v)]}{4^{\alpha }{(v-u)}^2(\alpha +1)(\alpha +2)}-\frac{{(3u+v)}^{\alpha }}{2^{2\alpha +1}},\\ A_3\left(\alpha \right)=\frac{{\left(4u\right)}^{\alpha +2}-{(3u+v)}^{\alpha +2}(5u-v)-\alpha {(3u+v)}^{\alpha }(3u-v)(u+v)}{4^{\alpha }{(v-u)}^2(\alpha +1)(\alpha +2)}-\frac{{(3u+v)}^{\alpha }}{2^{2\alpha +1}},\end{array}$$

$$\begin{array}{l}{B}_1\left(\alpha \right)=\frac{{\left(2(u+v)\right)}^{\alpha +2}-{(3u+v)}^{\alpha +1}}{4^{\alpha }(v-u)(\alpha +1)}-\frac{{(3u+v)}^{\alpha }}{4^{\alpha }},\\ B_2\left(\alpha \right)=\frac{{\left(2(u+v)\right)}^{\alpha +2}-{(3u+v)}^{\alpha +1}(u+3v)-\alpha {(3u+v)}^{\alpha +1}(v-u)}{4^{\alpha }{(v-u)}^2(\alpha +1)(\alpha +2)}-\frac{{(3u+v)}^{\alpha }}{2^{2\alpha +1}},\\ B_3\left(\alpha \right)=\frac{{(3u+v)}^{\alpha +2}+4{\left(2(u+v)\right)}^{\alpha +1}+\alpha {\left(2(u+v)\right)}^{\alpha +1}(v-u)}{4^{\alpha }{(v-u)}^2(\alpha +1)(\alpha +2)}-\frac{{(3u+v)}^{\alpha }}{2^{2\alpha +1}},\end{array}$$

$$\begin{array}{cc}\hfill C_1\left(\alpha \right)& =\frac{4{(u+3v)}^{\alpha +1}-{\left(2(u+v)\right)}^{\alpha +2}}{4^{\alpha }(v-u)(\alpha +1)}-\frac{{(u+3v)}^{\alpha }}{4^{\alpha }},\hfill \\ \hfill C_2\left(\alpha \right)& =\frac{{\alpha }^2{(v-u)}^2{(u+3v)}^{\alpha }+16b(u+v){(u+3v)}^{\alpha }-3\alpha \left(u^2+v^2\right){(u+3v)}^{\alpha }}{2^{2\alpha +1}{(v-u)}^2(\alpha +1)(\alpha +2)}\hfill \\ \hfill & -\frac{8b{\left(2(u+v)\right)}^{\alpha +1}+2\alpha (v-u){\left(2(u+v)\right)}^{\alpha +1}}{2^{2\alpha +1}{(v-u)}^2(\alpha +1)(\alpha +2)}-\frac{{(u+3v)}^{\alpha }}{2^{2\alpha +1}},\hfill \\ \hfill C_3\left(\alpha \right)& =\frac{{\left(2(u+v)\right)}^{\alpha +2}+{(u+3v)}^{\alpha +2}+\alpha {(u+3v)}^{\alpha +2}-5(u+v){(u+3v)}^{\alpha +1}}{4^{\alpha }{(v-u)}^2(\alpha +1)(\alpha +2)},\hfill \end{array}$$

$$\begin{array}{l}{D}_1\left(\alpha \right)=\frac{{\left(4v\right)}^{\alpha +1}-{(u+3v)}^{\alpha +1}}{4^{\alpha }(\alpha +1)(v-u)}-\frac{{(u+3v)}^{\alpha }}{2^{2\alpha +1}},\\ D_2\left(\alpha \right)=\frac{{\left(4v\right)}^{\alpha +2}+{(u+3v)}^{\alpha +1}(a-5v)+\alpha {(u+3v)}^{\alpha +1(v-u)}}{4^{\alpha }{(v-u)}^2(\alpha +1)(\alpha +2)},\\ D_3\left(\alpha \right)=\frac{\alpha {\left(4v\right)}^{\alpha +2}(v-u)-2{\left(4v\right)}^{\alpha +1}(u+v)+{(u+3v)}^{\alpha +2}}{4^{\alpha }{(v-u)}^2(\alpha +1)(\alpha +2)}.\end{array}$$

Proof. 

Using Lemma 1 and the property (4), we have

$$\begin{array}{l}|{\mathsf{\Phi }}_{\alpha }(u,v)|\le \frac{v-u}{4}[{\int }_0^1\left[{\left(u\zeta +(1-\zeta )\frac{3u+v}{4}\right)}^{\alpha }-{\left(\frac{3u+v}{4}\right)}^{\alpha }\right]\left|h^{\prime }\left(u\zeta +(1-\zeta )\frac{3u+v}{4}\right)\right|d\zeta \\ +{\int }_0^1\left[{\left(\frac{3u+v}{4}\zeta +(1-\zeta )\frac{u+v}{2}\right)}^{\alpha }-{\left(\frac{3u+v}{4}\right)}^{\alpha }\right]\left|h^{\prime }\left(\frac{3u+v}{4}\zeta +(1-\zeta )\frac{u+v}{2}\right)\right|d\zeta \hfill \\ +{\int }_0^1\left[{\left(\frac{u+v}{2}\zeta +(1-\zeta )\frac{u+3v}{4}\right)}^{\alpha }-{\left(\frac{u+3v}{4}\right)}^{\alpha }\right]\left|h^{\prime }\left(\frac{u+v}{2}\zeta +(1-\zeta )\frac{u+3v}{4}\right)\right|d\zeta \hfill \\ +{\int }_0^1\left[{\left(\frac{u+3v}{4}\zeta +(1-\zeta )v\right)}^{\alpha }-{\left(\frac{u+3v}{4}\right)}^{\alpha }\right]\left|h^{\prime }\left(\frac{u+3v}{4}\zeta +(1-\zeta )v\right)\right|d\zeta \hfill \\ :=\frac{v-u}{4}(J_1+J_2+J_3+J_4).\hfill \end{array}$$

(18)

It follows from the power–mean inequality that

$$\begin{array}{cc}\hfill J_1& \le {\left({\int }_0^1\left[{\left(u\zeta +(1-\zeta )\frac{3u+v}{4}\right)}^{\alpha }-{\left(\frac{3u+v}{4}\right)}^{\alpha }\right]d\zeta \right)}^{1-\frac{1}{q}}\hfill \\ \hfill & \times {\left({\int }_0^1\left[{\left(u\zeta +(1-\zeta )\frac{3u+v}{4}\right)}^{\alpha }-{\left(\frac{3u+v}{4}\right)}^{\alpha }\right]{\left|h^{\prime }\left(u\zeta +(1-\zeta )\frac{3u+v}{4}\right)\right|}^{q}d\zeta \right)}^{\frac{1}{q}}.\hfill \end{array}$$

Since $|h^{\prime }{|}^q$ is convex on $[u,v]$ for any $\zeta \in [0,1]$, we obtain

$$\begin{array}{cc}\hfill J_1& \le {\left({\int }_0^1\left[{\left(u\zeta +(1-\zeta )\frac{3u+v}{4}\right)}^{\alpha }-{\left(\frac{3u+v}{4}\right)}^{\alpha }\right]d\zeta \right)}^{1-\frac{1}{q}}\hfill \\ \hfill & \times (|h^{\prime }{\left(u\right)|}^q{\int }_0^1\left[{\left(u\zeta +(1-\zeta )\frac{3u+v}{4}\right)}^{\alpha }-{\left(\frac{3u+v}{4}\right)}^{\alpha }\right]\zeta d\zeta \hfill \\ \hfill & +{\left|h^{\prime }\left(\frac{3u+v}{4}\right)\right|}^q{\int }_0^1\left[{\left(u\zeta +(1-\zeta )\frac{3u+v}{4}\right)}^{\alpha }-{\left(\frac{3u+v}{4}\right)}^{\alpha }\right](1-\zeta )d\zeta {)}^{\frac{1}{q}}\hfill \\ \hfill & ={\left(A_1\left(\alpha \right)\right)}^{1-\frac{1}{q}}{\left\{A_2\left(\alpha \right){|h^{\prime }\left(u\right)|}^q+A_3\left(\alpha \right){\left|h^{\prime }\left(\frac{3u+v}{4}\right)\right|}^q\right\}}^{\frac{1}{q}},\hfill \end{array}$$

where we have used the facts that

$$A_1\left(\alpha \right):={\int }_0^1\left[{\left(u\zeta +(1-\zeta )\frac{3u+v}{4}\right)}^{\alpha }-{\left(\frac{3u+v}{4}\right)}^{\alpha }\right]d\zeta =\frac{4u{(3u+v)}^{\alpha }-{\left(4u\right)}^{\alpha +1}}{4^{\alpha }(v-u)(\alpha +1)}-\frac{\alpha {(3u+v)}^{\alpha }}{4^{\alpha }},$$

$$\begin{array}{cc}\hfill A_2\left(\alpha \right)& :={\int }_0^1\left[{\left(u\zeta +(1-\zeta )\frac{3u+v}{4}\right)}^{\alpha }-{\left(\frac{3u+v}{4}\right)}^{\alpha }\right]\zeta d\zeta \hfill \\ \hfill & =\frac{{(3u+v)}^{\alpha +2}-{\left(4u\right)}^{\alpha +1}[\alpha (v-u)+(u+v)]}{4^{\alpha }{(v-u)}^2(\alpha +1)(\alpha +2)}-\frac{{(3u+v)}^{\alpha }}{2^{2\alpha +1}},\hfill \end{array}$$

and

$$\begin{array}{cc}\hfill A_3\left(\alpha \right)& :={\int }_0^1\left[{\left(u\zeta +(1-\zeta )\frac{3u+v}{4}\right)}^{\alpha }-{\left(\frac{3u+v}{4}\right)}^{\alpha }\right](1-\zeta )d\zeta \hfill \\ \hfill & =\frac{{\left(4u\right)}^{\alpha +2}-{(3u+v)}^{\alpha +2}(5u-v)-\alpha {(3u+v)}^{\alpha }(3u-v)(u+v)}{4^{\alpha }{(v-u)}^2(\alpha +1)(\alpha +2)}-\frac{{(3u+v)}^{\alpha }}{2^{2\alpha +1}}.\hfill \end{array}$$

Analogously

$$\begin{array}{l}{J}_2\le {\left(B_1\left(\alpha \right)\right)}^{1-\frac{1}{q}}{\left\{B_2\left(\alpha \right){\left|h^{\prime }\left(\frac{3u+v}{4}\right)\right|}^q+B_3\left(\alpha \right){\left|h^{\prime }\left(\frac{u+v}{2}\right)\right|}^q\right\}}^{\frac{1}{q}},\\ J_3\le {\left(C_1\left(\alpha \right)\right)}^{1-\frac{1}{q}}{\left\{C_2\left(\alpha \right){\left|h^{\prime }\left(\frac{u+v}{2}\right)\right|}^q+C_3\left(\alpha \right){\left|h^{\prime }\left(\frac{u+3v}{4}\right)\right|}^q\right\}}^{\frac{1}{q}},\end{array}$$

and

$$J_4\le {\left(D_1\left(\alpha \right)\right)}^{1-\frac{1}{q}}{\left\{D_2\left(\alpha \right){\left|h^{\prime }\left(\frac{u+3v}{4}\right)\right|}^q+D_3\left(\alpha \right){\left|h^{\prime }\left(v\right)\right|}^q\right\}}^{\frac{1}{q}}.$$

Using $J_1,J_2,J_3$, and $J_4$ in (18), we obtain the desired inequality (17). This completes the proof of Theorem 3. □

Corollary 2.

With the similar assumptions of Theorem 3, if $\alpha =1$, then

$$\begin{array}{cc}\hfill \left|{\mathsf{\Phi }}_1(u,v)\right|& \le \frac{v-u}{4}[{\left(\frac{u-v}{8}\right)}^{1-\frac{1}{q}}{\left\{\left(\frac{11u+v}{24}\right){|h^{\prime }\left(u\right)|}^q+\left(\frac{5u+v}{12}\right){\left|h^{\prime }\left(\frac{3u+v}{4}\right)\right|}^q\right\}}^{\frac{1}{q}}\hfill \\ \hfill & +{\left(\frac{v-u}{8}\right)}^{1-\frac{1}{q}}{\left\{\left(\frac{2u+v}{6}\right){\left|h^{\prime }\left(\frac{3u+v}{4}\right)\right|}^q+\left(\frac{7u+5b}{24}\right){\left|h^{\prime }\left(\frac{u+v}{2}\right)\right|}^q\right\}}^{\frac{1}{q}}\hfill \\ \hfill & +{\left(\frac{u-v}{8}\right)}^{1-\frac{1}{q}}{\left\{\left(\frac{u-v}{12}\right){\left|h^{\prime }\left(\frac{u+v}{2}\right)\right|}^q+\left(\frac{u-v}{24}\right){\left|h^{\prime }\left(\frac{u+3v}{4}\right)\right|}^q\right\}}^{\frac{1}{q}}\hfill \\ \hfill & +{\left(\frac{v-u}{8}\right)}^{1-\frac{1}{q}}{\left\{\left(\frac{u+5b}{12}\right){\left|h^{\prime }\left(\frac{u+3v}{4}\right)\right|}^q+\left(\frac{u+11b}{24}\right){\left|h^{\prime }\left(v\right)\right|}^q\right\}}^{\frac{1}{q}}].\hfill \end{array}$$

Theorem 4.

Let $h:[u,v]\subseteq \mathbb{R}\to \mathbb{R}$ be an α-fractional differentiable mapping on $(u,v)$ with $0\le u<v$. If $D_{\alpha }\left(h\right)\in L_{\alpha }^1\left([u,v]\right)$ and ${\left|h^{\prime }\right|}^q$ is concave on $[u,v]$, then the following inequality for conformable fractional integral holds:

$$\begin{array}{cc}\hfill \left|{\mathsf{\Phi }}_{\alpha }(u,v)\right|& \le \frac{v-u}{4}[A_1\left(\alpha \right)h^{\prime }\left(\frac{P_1\left(\alpha \right)}{A_1\left(\alpha \right)}\right)\hfill \\ \hfill & +B_1\left(\alpha \right)h^{\prime }\left(\frac{P_2\left(\alpha \right)}{B_1\left(\alpha \right)}\right)+C_1\left(\alpha \right)h^{\prime }\left(\frac{P_3\left(\alpha \right)}{C_1\left(\alpha \right)}\right)+D_1\left(\alpha \right)h^{\prime }\left(\frac{P_4\left(\alpha \right)}{D_1\left(\alpha \right)}\right)],\hfill \end{array}$$

(19)

where $A_1\left(\alpha \right),B_1\left(\alpha \right),C_1\left(\alpha \right)$, and $D_1\left(\alpha \right)$ are given in Theorem 3 and

$$\begin{array}{cc}{P}_1\left(\alpha \right)\hfill & =\frac{2^{1-\alpha }{(3u+v)}^{\alpha +2}-2^{2\alpha +5}-2^{2\alpha +1}(7u+6v)(v-u){(3u+v)}^{\alpha }-\alpha (7u+v)(v-u){(3u+v)}^{\alpha }}{2^{2\alpha +3}(v-u)(\alpha +2)},\hfill \\ P_2\left(\alpha \right)\hfill & =\frac{2{\left(2(u+v)\right)}^{\alpha +2}-8{(u+v)}^2{(3u+v)}^{\alpha }-\alpha (5u+3v)(v-u){(3u+v)}^{\alpha }}{2^{2\alpha +3}(v-u)(\alpha +2)},\hfill \\ P_3\left(\alpha \right)\hfill & =\frac{8{(u+v)}^2{(u+3v)}^{\alpha }-4\alpha {(u+v)}^{\alpha }{(u+3v)}^{\alpha }-\alpha {(u+3v)}^{\alpha +2}-2^{\alpha +3}{(u+v)}^{\alpha +2}}{2^{2\alpha +1}(v-u)(\alpha +2)},\hfill \\ P_4\left(\alpha \right)\hfill & =\frac{2^{2\alpha -1}\alpha (u+3v)+4^{\alpha }{(u+3v)}^3+4{\left(4v\right)}^{\alpha +2}-4^{\alpha +2}{v}^2(u+3v)-4{(u+3v)}^{\alpha +2}-2^{2\alpha +3}\alpha v^2(u+3v)}{4^{\alpha +2}(v-u)(\alpha +2)}.\hfill \end{array}$$

Proof. 

By using the power–mean inequality and the concavity of ${\left|h^{\prime }\right|}^q$ for any $\zeta \in [0,1]$, we have

$${\left|h^{\prime }\left(u\zeta +(1-\zeta )\frac{3u+v}{4}\right)\right|}^q\ge {\left(\zeta |h^{\prime }\left(u\right)|+(1-\zeta )\left|\left(\frac{3u+v}{4}\right)\right|\right)}^q.$$

This implies that

$$\left|h^{\prime }\left(u\zeta +(1-\zeta )\frac{3u+v}{4}\right)\right|\ge \zeta |h^{\prime }\left(u\right)|+(1-\zeta )\left|\left(\frac{3u+v}{4}\right)\right|.$$

(20)

This means that $|h^{\prime }|$ is also concave. Using inequality (20) in (18) and then applying the Jensen’s integral inequality, we get

$$\begin{array}{cc}\hfill J_1& \le \left({\int }_0^1\left[{\left(u\zeta +(1-\zeta )\frac{3u+v}{4}\right)}^{\alpha }-{\left(\frac{3u+v}{4}\right)}^{\alpha }\right]d\zeta \right)\hfill \\ \hfill & \times h^{\prime }\left({\displaystyle \frac{{\int }_0^1\left[{\left(u\zeta +(1-\zeta )\frac{3u+v}{4}\right)}^{\alpha }-{\left(\frac{3u+v}{4}\right)}^{\alpha }\right]\left(u\zeta +(1-\zeta )\frac{3u+v}{4}\right)d\zeta }{{\int }_0^1\left[{\left(u\zeta +(1-\zeta )\frac{3u+v}{4}\right)}^{\alpha }-{\left(\frac{3u+v}{4}\right)}^{\alpha }\right]d\zeta }}\right)\hfill \\ \hfill & =A_1\left(\alpha \right)h^{\prime }\left(\frac{P_1\left(\alpha \right)}{A_1\left(\alpha \right)}\right),\hfill \end{array}$$

where we have used the facts that

$$\begin{array}{l}{\int }_0^1\left[{\left(u\zeta +(1-\zeta )\frac{3u+v}{4}\right)}^{\alpha }-{\left(\frac{3u+v}{4}\right)}^{\alpha }\right]d\zeta =A_1\left(\alpha \right)=\frac{4u{(3u+v)}^{\alpha }-{\left(4u\right)}^{\alpha +1}}{4^{\alpha }(v-u)(\alpha +1)}-\frac{\alpha {(3u+v)}^{\alpha }}{4^{\alpha }},\\ {\int }_0^1\left[{\left(u\zeta +(1-\zeta )\frac{3u+v}{4}\right)}^{\alpha }-{\left(\frac{3u+v}{4}\right)}^{\alpha }\right]\left(u\zeta +(1-\zeta )\frac{3u+v}{4}\right)d\zeta =P_1\left(\alpha \right)\hfill \\ =\frac{2^{1-\alpha }{(3u+v)}^{\alpha +2}-2^{2\alpha +5}-2^{2\alpha +1}(7u+6v)(v-u){(3u+v)}^{\alpha }-\alpha (7u+v)(v-u){(3u+v)}^{\alpha }}{2^{2\alpha +3}(v-u)(\alpha +2)}.\hfill \end{array}$$

Similarly, we get $J_2\le B_1\left(\alpha \right)h^{\prime }\left(\frac{P_2\left(\alpha \right)}{B_1\left(\alpha \right)}\right)$, $J_3\le C_1\left(\alpha \right)h^{\prime }\left(\frac{P_3\left(\alpha \right)}{C_1\left(\alpha \right)}\right)$ and $J_4\le D_1\left(\alpha \right)h^{\prime }\left(\frac{P_4\left(\alpha \right)}{D_1\left(\alpha \right)}\right)$.

Using $J_1,J_2,J_3$, and $J_4$ in (18), we obtain the required inequality (19). This completes the proof of Theorem 4. □

Corollary 3.

With the similar assumptions of Theorem 4, if $\alpha =1$, then

$$\left|{\mathsf{\Phi }}_1(u,v)\right|\le \frac{v-u}{4}\left[A_1\left(\alpha \right)h^{\prime }\left(\frac{P_1\left(\alpha \right)}{A_1\left(\alpha \right)}\right)+B_1\left(\alpha \right)h^{\prime }\left(\frac{P_2\left(\alpha \right)}{B_1\left(\alpha \right)}\right)+C_1\left(\alpha \right)h^{\prime }\left(\frac{P_3\left(\alpha \right)}{C_1\left(\alpha \right)}\right)+D_1\left(\alpha \right)h^{\prime }\left(\frac{P_4\left(\alpha \right)}{D_1\left(\alpha \right)}\right)\right],$$

where

$$A_1\left(1\right)=\frac{u-v}{8},B_1\left(1\right)=\frac{v-u}{8},D_1\left(1\right)=\frac{u-v}{8},D_1\left(1\right)=\frac{v-u}{8},$$

and

$$\begin{array}{l}{P}_1\left(1\right)=\frac{(u-v)(11u+v)}{96},P_2\left(1\right)=\frac{(v-u)(7u+5v)}{96},\\ P_3\left(1\right)=\frac{(u-v)(5u+7v)}{96},P_4\left(1\right)=\frac{(v-u)(u+11v)}{96}.\end{array}$$

3. Conclusions

In this work, we have established new conformable fractional integral inequalities of Hermite–Hadamard type for convex functions. As a special case, if we substitute $\alpha =1$ into the general definition of conformable fractional integrals (Definition 2), we obtain the classical integrals. In view of this, we obtained some new inequalities of Hermite–Hadamard type for convex functions involving classical integrals.