Hermite-Hadamard type inequalities for new conditions on h-convex functions via $\psi$-Hilfer integral operators

Abstract

We employ a new function class called B-function to create a new version of fractional Hermite–Hadamard and trapezoid type inequalities on the right-hand side that involves h-convex and $\psi$-Hilfer operators. We also provide new midpoint-type inequalities using h-convex functions.

1 Introduction

Convexity theory provides robust approaches and concepts for solving an extensive variety of problems in pure and practical mathematics. Because of their tenacity, convex functions have been applied to a wide range of mathematical disciplines, resulting in the discovery of several inequalities in the literature. One famous inequality is the Hermite–Hadamard inequality, which has found applications in convexity theory. In [16], the author introduces a novel class of functions called as h-convex functions

Definition 1

Let $h:J\rightarrow \mathbb {R}$ be a non-negative function, $h\ne 0$. We say that $f:I\rightarrow \mathbb {R}$ is an h-convex function, if f is non-negative and for all $x,y\in I$, $\alpha \in (0,1)$ we have

$$\begin{aligned} f(\alpha x+(1-\alpha )y)\le h(\alpha )f(x)+h(1-\alpha )f(y). \end{aligned}$$

(1.1)

If the inequality (1.1) is reversed, then f is said to be h-concave.

By putting $h(\alpha )=\alpha$, $h(\alpha )=1$, $h(\alpha ) =\alpha ^{s}$, $h(\alpha )=\displaystyle \frac{1}{n} \sum \nolimits _{k=1}^{n}\alpha ^{\frac{1}{k}}$ and $h(\alpha ) =\frac{1}{\alpha }$ in (1.1), Definition 1 reduces to some wel known function classes, P-functions [5, 11], s-convex functions [3], n-fractional polynomial convex functions [7] and Godunova–Levin functions [6], respectively.

2 $\psi$-Hilfer operators

Definition 2

Let $[a,b]\subseteq [ 0,+\infty ).$ Let $\beta >0$ and $\psi$ be a positive, strictly increasing differentiable function such that $\psi \,^{\prime }(\tau )\ne 0$ for all $\tau \in [ a,b]$. The left and right sided $\psi$-Hilfer fractional integral of a function f with respect to the function $\psi$ on [a, b] are defined respectively as follows.

$$\begin{aligned}&{}^{\psi }\mathfrak {J}\,_{a^{+}}^{\beta }f(x)=\displaystyle \frac{1}{\Gamma (\beta )}\int _{a}^{x} \left( \psi (x)-\psi (t)\right) ^{\beta -1} \psi ^{\prime }(t)f(t)dt,\quad a<x\le b. \end{aligned}$$

(2.1)

$$\begin{aligned}&{}^{\psi }\mathfrak {J}\,_{b^{-}}^{\beta }f(x) =\displaystyle \frac{1}{\Gamma (\beta )}\int _{x}^{b}\left( \psi (t)-\psi (x)\right) ^{\beta -1} \psi ^{\prime }(t)f(t)dt,\quad a\le x<b, \end{aligned}$$

(2.2)

where the gamma function verified

$$\begin{aligned} \Gamma (\beta )=\displaystyle \int _{0}^{\infty }t^{\beta -1} e^{-t}dt\,,\quad \beta \,\Gamma (\beta )=\Gamma (\beta +1). \end{aligned}$$

For these operators, consider the following space

$$\begin{aligned} X[a, b]=\left\{ f: \ \parallel f\parallel _{X}=\left( \displaystyle \int _{a}^{b} \mid f(x)\mid \psi \,^{\prime }(x)dx\right) <\infty \right\} . \end{aligned}$$

One essential property of $\psi$-Hilfer operators is that they are dependent on the function $\psi$ and produce a particular type of fractional integrals.

• (1) Taking $\psi (\tau )=\tau$, we get Riemann-Liouville fractional operator of order $\beta >0$

  $$\begin{aligned}{} & {} \mathcal{R}\mathcal{L}_{a^{+}}^{\beta }f(x)=\displaystyle \frac{1}{\Gamma (\beta )} \int _{a}^{x}(x-s)^{\beta -1}f(s)ds,\quad x>a, \\{} & {} \mathcal{R}\mathcal{L}_{b^{-}}^{\beta }f(x)=\displaystyle \frac{1}{\Gamma (\beta )} \int _{x}^{b}(s-x)^{\beta -1}f(s)ds,,\quad x<b. \end{aligned}$$

• (2) Using $\psi (\tau )=\ln \tau$, we deduce Hadamard fractional operator of order $\beta >0$

  $$\begin{aligned}{} & {} \mathcal {H}_{a^{+}}^{\beta }f(x)=\displaystyle \frac{1}{\Gamma (\beta )} \int _{a}^{x}\left( \ln \frac{x}{s}\right) ^{\beta -1}f(s)\frac{ds}{s}, \quad x>a>1, \\{} & {} \mathcal {H}_{a^{+}}^{\beta }f(x)=\displaystyle \frac{1}{\Gamma (\beta )} \int _{x}^{b}\left( \ln \frac{s}{x}\right) ^{\beta -1} f(s)\frac{ds}{s},\quad 1<x<b. \end{aligned}$$

• (3) Putting $\psi (\tau )=\frac{\tau ^{\rho }}{\rho }$ where $\rho >0$, we obtain Katugompola fractional operators of order $\beta >0$.

  $$\begin{aligned}{} & {} \mathcal {K}_{a^{+}}^{\beta }f(x)=\displaystyle \frac{(\rho )^{1-\beta }}{\Gamma (\beta )}\int _{a}^{x}\left( x^{\rho }-s^{\rho }\right) ^{\beta -1} f(s)s^{\rho -1}ds,\ x>a, \\{} & {} \mathcal {K}_{b^{-}}^{\beta }f(x)=\displaystyle \frac{(\rho )^{1-\beta }}{ \Gamma (\beta )}\int _{x}^{b}\left( s^{\rho }-x^{\rho }\right) ^{\beta -1}f(s)s^{\rho -1}ds,\ x<b. \end{aligned}$$

In the literature, papers devoted to fractional integral inequalities. For some of them, please refer to [1, 2, 4, 9, 12,13,14].

3 B-function

We now give a definition of the B-function.

Definition 3

Let $a<b$ and $h:(a,b)\subset \mathbb {R}\rightarrow \mathbb {R}$ be a non-negative function. The function h is a B-function, or that h belongs to the class B(a, b), if for all $x\in (a,b)$, we have

$$\begin{aligned} h(x-a)+h(b-x)\le 2\,h\left( \frac{a+b}{2}\right) . \end{aligned}$$

(3.1)

If the inequality (3.1) is reversed, h is called A-function, or that h belongs to the class A(a, b).

If we have the equality in (3.1), h is called AB-function, or that h belongs to the class AB(a, b).

Corollary 1

Let $h: (0, 1) \rightarrow \mathbb {R}$ be a non-negative function.

• (1) The function $h(\alpha )=1$ is AB-function, B-function and A-function. And the function $h(\alpha )=\alpha$ is AB-function, B-function and A-function under the conditions $a=0$, $a<0$ and $a>0$, respectively.

• (2) The function $h(\alpha )=\alpha ^{s},\ s\in (0,1]$ is B-function.

• (3) The function $h(\alpha ) =\displaystyle \frac{1}{n}\sum \nolimits _{k=1}^{n} \alpha ^{\frac{1}{k}},\ n,k\in \mathbb {N}$ is B-function.

• (4) The function $h(\alpha )=\frac{1}{\alpha }$ is A-function.

Proof

• (1) The first case is obvious.

  The following result (3.2) is required to prove the next case.

  Let $g(x)=\ln (x+1)$, $g(0)=0$, $g(1)=\ln 2$ and $M(1,\ln 2)$. On the interval [0, 1], the graph of the function g appears over the line (OM). This gives us, for all $x\in (0,1]$

  $$\begin{aligned} \ln (x+1)\ge x\ln 2\Leftrightarrow (1+x)^{\frac{1}{x}}\ge 2, \end{aligned}$$

  taking $x=\frac{1}{p}$, we have

  $$\begin{aligned} \hbox {for}\ \hbox {all}\ p\ge 1, \quad \left( 1+\frac{1}{p}\right) ^{p}\ge 2\,. \end{aligned}$$

  (3.2)

• (2) Let $h(\alpha )=\alpha ^{s},\ s\in (0,1]$ and taking $s=\frac{1}{p}$. The function h is a B-function, this means

  $$\begin{aligned} \alpha ^{\frac{1}{p}}+(1-\alpha )^{\frac{1}{p}} \le \left( \frac{1}{2} \right) ^{\frac{1}{p}-1}. \end{aligned}$$

  (3.3)

  We use absurdity to demonstrate inequality (3.3), suppose that exist $p\ge 1$ verified

  $$\begin{aligned} \alpha ^{\frac{1}{p}}+(1-\alpha )^{\frac{1}{p}} >\left( \frac{1}{2}\right) ^{\frac{1}{p}-1}, \end{aligned}$$

  we get

  $$\begin{aligned} \displaystyle \int _{0}^{1}\left[ \alpha ^{\frac{1}{p}} +(1-\alpha )^{\frac{1}{p}}\right] d\alpha >\left( \frac{1}{2}\right) ^{\frac{1}{p}-1}, \end{aligned}$$

  thus

  $$\begin{aligned} 2\displaystyle \left( \frac{p}{p+1}\right) >\left( \frac{1}{2}\right) ^{\frac{1}{p}-1}. \end{aligned}$$

  This gives

  $$\begin{aligned} \displaystyle \left( \frac{p}{p+1}\right) ^{p}>\frac{1}{2}, \end{aligned}$$

  hence

  $$\begin{aligned} \left( 1+\frac{1}{p}\right) ^{p}<2\ \hbox {which}\ \hbox {is}\ \hbox {absurd}. \end{aligned}$$

• (3) Let $h(\alpha )=\displaystyle \frac{1}{n}\sum \nolimits _{k=1}^{n} \alpha ^{\frac{1}{k}},\ n,k\in \mathbb {N}$, we have

  $$\begin{aligned} \alpha ^{\frac{1}{k}}+(1-\alpha )^{\frac{1}{k}}\le \left( \frac{1}{2}\right) ^{\frac{1}{k}-1}, \end{aligned}$$

  then

  $$\begin{aligned} \displaystyle \frac{1}{n}\sum \limits _{k=1}^{n}\alpha ^{\frac{1}{k}} +\displaystyle \frac{1}{n}\sum \limits _{k=1}^{n}(1-\alpha )^{\frac{1}{k}} \le \displaystyle \frac{2}{n}\sum \limits _{k=1}^{n}\left( \frac{1}{2} \right) ^{\frac{1}{k}}. \end{aligned}$$

  This yields

  $$\begin{aligned} h(\alpha )+h(1-\alpha )\le 2\,h\left( \frac{1}{2}\right) . \end{aligned}$$

• (4) Let $h(\alpha )=\frac{1}{\alpha }$ with $\alpha \in (0, 1)$, we have

  $$\begin{aligned} \frac{(2\alpha -1)^{2}}{\alpha (1-\alpha )}\ge 0 \Leftrightarrow \frac{1}{\alpha } +\frac{1}{1-\alpha }\ge 4, \end{aligned}$$

  therefore

  $$\begin{aligned} h(\alpha ) +h(1-\alpha )\ge 2 \,h\left( \frac{1}{2}\right) . \end{aligned}$$

$\square$

Motivated by previous literature, we use $\psi$-Hilfer operators to establish a new version of Hermite–Hadamard and trapezoid inequalities based on h-convex functions where h belongs to the class B(0, 1). We also provide new midpoint inequalities using h-convex functions.

4 Hermite–Hadamard inequalities

We now present the first result of the Hermite–Hadamard inequality for h-convexity of the function f involving $\psi$-Hilfer operators.

Theorem 4.1

Let $\beta >0$, h be a B-function. Let $f\in X[a,b]$ be a h-convex function and $\psi$ be a positive, strictly increasing differentiable function such that $\psi \,^{\prime }(\tau )\ne 0$ for all $\tau \in [ a,b]$. Then the following inequalities hold

$$\begin{aligned} \displaystyle \frac{1}{2\,h\left( \frac{1}{2}\right) }f\left( \displaystyle \frac{a+b}{2}\right)&\le \displaystyle \frac{\Gamma (\beta +1)}{4\,\left( \psi (b)-\psi (a)\right) ^{\beta }}\left[ {\,^{\psi } \mathfrak {J}_{b^{-}}^{\beta }}F(a)+{\,^{\psi } \mathfrak {J}_{a^{+}}^{\beta }}F(b)\right] \nonumber \\&\le 2\,h\left( \frac{1}{2}\right) \left[ \displaystyle \frac{f(b)+f(a)}{2}\right] , \end{aligned}$$

(4.1)

where

$$\begin{aligned} F(t)=f(t)+f\left( a+b-t\right) . \end{aligned}$$

(4.2)

Proof

For any $t\in [ a,b]$, we have

$$\begin{aligned} f\left( \displaystyle \frac{a+b}{2}\right)&=f\left( \displaystyle \frac{1}{2}(a+b-t)+\frac{1}{2}t\right) \\&\le h\left( \frac{1}{2}\right) \,f\left( a+b-t\right) +h\left( \frac{1}{2}\right) \,f(t), \end{aligned}$$

then

$$\begin{aligned} f\left( \displaystyle \frac{a+b}{2}\right) \le h\left( \frac{1}{2}\right) \,F(t). \end{aligned}$$

(4.3)

Multiplying (4.3) by $\beta \left( \psi (b)-\psi (t)\right) ^{\beta -1}\psi ^{^{\prime }}(t)$ and integrating over $t\in [ a,b]$, we result

$$\begin{aligned} f\left( \displaystyle \frac{a+b}{2}\right) \left( \psi (b) -\psi (a)\right) ^{\beta }\le \Gamma (\beta +1)h \left( \frac{1}{2}\right) \,{\,^{\psi } \mathfrak {J}\,_{a^{+}}^{\beta }}F(b). \end{aligned}$$

(4.4)

Multiplying (4.3) by $\beta \left( \psi (t)-\psi (a)\right) ^{\beta -1}\psi ^{^{\prime }}(t)$ and integrating over $t\in [ a,b]$, we get

$$\begin{aligned} f\left( \displaystyle \frac{a+b}{2}\right) \left( \psi (b) -\psi (a)\right) ^{\beta }\le \Gamma (\beta +1)h \left( \frac{1}{2}\right) \,{\,^{\psi } \mathfrak {J}\,_{b^{-}}^{\beta }}F(a). \end{aligned}$$

(4.5)

By adding the inequalities (4.4) and (4.5), we deduce

$$\begin{aligned} \displaystyle \frac{1}{h\left( \frac{1}{2}\right) }f \left( \displaystyle \frac{a+b}{2}\right) \le \displaystyle \frac{ \Gamma (\beta +1)}{2\,\left( \psi (b)-\psi (a)\right) ^{\beta }} \left[ {\,^{\psi }\mathfrak {J}\,_{b^{-}}^{\beta }}F(a)+{\,^{\psi } \mathfrak {J}\,_{a^{+}}^{\beta }}F(b)\right] . \end{aligned}$$

(4.6)

Put $t=(1-s)\,a+s\,b$ in (4.2) for $s\in (0,1)$ and using the h-convexity of f, we get

$$\begin{aligned} F(t)&=f\left( (1-s)\,b+s\,a\right) +f\left( (1-s)\,a+s\,b\right) \\&\le \displaystyle \,h(1-s)\left[ f(b)+f(a)\right] +\,h(s) \left[ f(b)+f(a)\right] \\&=\left( \,h(s)+\,h(1-s)\right) \left[ f(b)+f(a)\right] , \end{aligned}$$

given that h is a B-function, we conclude

$$\begin{aligned} F(t)\le 2\,h\left( \frac{1}{2}\right) \,\left[ f(b)+f(a)\right] . \end{aligned}$$

(4.7)

Using the same method as previously on (4.7), we obtain

$$\begin{aligned} \displaystyle \frac{\Gamma (\beta +1)}{2\,\left( \psi (b) -\psi (a)\right) ^{\beta }}\left[ {\,^{\psi }\mathfrak {J} \,_{b^{-}}^{\beta }}F(a)+{\,^{\psi } \mathfrak {J}\,_{a^{+}}^{\beta }}F(b)\right] \le 2\,h\left( \frac{1}{2}\right) \,\left[ f(b)+f(a)\right] . \end{aligned}$$

(4.8)

Combining the inequality (4.6) with the inequality (4.8) yields the desired outcome. $\square$

Taking $\psi (x)=x$ and $\beta =1$, we conclude the following new version of the Hermite–Hadamard inequality, see (Theorem.1 [15]).

Corollary 2

Let h be a B-function. Let $f\in L[a,b]$ be a h-convex function. Then the following inequalities hold

$$\begin{aligned} \displaystyle \frac{1}{2 \,h\left( \frac{1}{2}\right) }f\left( \displaystyle \frac{a+b}{2}\right) \le \displaystyle \frac{1}{b-a}\int _{a}^{b}f(t)dt \le \left[ f(b)+f(a)\right] \,h\left( \frac{1}{2}\right) . \end{aligned}$$

(4.9)

The following results are dependent on the function h presented in Theorem 4.1. First, assuming $h(\alpha )=\alpha$, we may obtain the result using the convex functions described in [8].

Corollary 3

Let $f\in X[a,b]$ be a convex function and $\psi$ be a positive, strictly increasing differentiable function such that $\psi \,^{\prime }(\tau )\ne 0$ for all $\tau \in [ a,b]$, $\beta >0$. Then the following inequalities hold

$$\begin{aligned} f\left( \displaystyle \frac{a+b}{2}\right) \le \displaystyle \frac{\Gamma (\beta +1)}{4\,\left( \psi (b)-\psi (a)\right) ^{\beta }}\left[ {\,^{\psi } \mathfrak {J}\,_{b^{-}}^{\beta }}F(a)+{\,^{\psi }\mathfrak {J} \,_{a^{+}}^{\beta }}F(b)\right] \le \left[ \displaystyle \frac{f(b)+f(a)}{2}\right] , \end{aligned}$$

(4.10)

where $F(t)=f(t)+f(a+b-t)$.

By setting $h(\alpha )=1$, we get the following result using $\psi$-Hilfer operators with function f belongs to class P-function.

Corollary 4

Let $\beta >0$, $f\in X[a,b]$ be a P-function and $\psi$ be a positive, strictly increasing differentiable function such that $\psi \,^{\prime }(\tau )\ne 0$ for all $\tau \in [ a,b]$. Then the following inequalities hold

$$\begin{aligned} f\left( \displaystyle \frac{a+b}{2}\right) \le \displaystyle \frac{\Gamma (\beta +1)}{2\,\left( \psi (b)-\psi (a)\right) ^{\beta }}\left[ {\,^{\psi } \mathfrak {J}\,_{b^{-}}^{\beta }}F(a)+{\,^{\psi }\mathfrak {J} \,_{a^{+}}^{\beta }}F(b)\right] \le 2\,\left[ f(b)+f(a)\right] , \end{aligned}$$

(4.11)

where $F(t)=f(t)+f(a+b-t)$.

Using $h(\alpha )=\alpha ^{s}$, we obtain the following result through $\psi$-Hilfer operators and s-convex functions.

Corollary 5

Let $\beta >0$, $s\in (0,1]$ and $f\in X[a,b]$ be a s-convex function and let $\psi$ be a positive, strictly increasing differentiable function such that $\psi \,^{\prime }(\tau )\ne 0$ for all $\tau \in [ a,b]$. Then the following inequalities hold

$$\begin{aligned} 2^{s-1}\,f\left( \displaystyle \frac{a+b}{2}\right)&\le \displaystyle \frac{ \Gamma (\beta +1)}{4\,\left( \psi (b)-\psi (a)\right) ^{\beta }}\left[ { \,^{\psi }\mathfrak {J}\,_{b^{-}}^{\beta }}F(a)+{\,^{\psi }\mathfrak {J} \,_{a^{+}}^{\beta }}F(b)\right] \nonumber \\&\le \left[ \displaystyle \frac{f(b)+f(a)}{ 2^{s}\,}\right] , \end{aligned}$$

(4.12)

where $F(t)=f(t)+f(a+b-t)$.

Taking $h(\alpha )=\frac{1}{n}\sum \nolimits _{k=1}^{n}\alpha ^{\frac{1}{k}}$, we deduce the following result through $\psi$-Hilfer operators and n-fractional polynomial convex functions.

Corollary 6

Let $\beta >0$ and $f\in X [a,b]$ be a n-fractional polynomial convex function and $\psi$ be a positive, strictly increasing differentiable function such that $\psi \,^{\prime }(\tau )\ne 0$ for all $\tau \in [a, b]$. Then the following inequalities hold

$$\begin{aligned} \displaystyle \frac{1}{C_{n}}f\left( \displaystyle \frac{a+b}{2}\right)&\le \displaystyle \frac{\Gamma (\beta +1)}{2\, \left( \psi (b) -\psi (a) \right) ^{\beta } }\left[ {\,^{\psi }\mathfrak {J}\,^{\alpha }_{b^{-}}}F(a) + { \,^{\psi }\mathfrak {J}\,^{\beta }_{a^{+}}}F(b)\right] \nonumber \\&\le C_{n}\left[ \displaystyle \frac{f(b)+f(a)}{2}\right] , \end{aligned}$$

(4.13)

where $F( t) = f(t) + f( a + b - t)$ and $C_{n}=\frac{2}{n} \sum \nolimits _{k=1}^{n}\left( \frac{1}{2}\right) ^{\frac{1}{k}}$.

Remark 1

If we choose $\psi (\tau )=\tau$, $\psi (\tau )=\ln \tau$ and $\psi (\tau )=\frac{ \tau ^{\rho }}{\rho }$, respectively, in Corollaries 4, 5 and 6, we obtain Hermite–Hadamard inequality for P-function, s-convex functions and n-fractional polynomial convex functions, respectively, involving Riemann-Liouville fractional operator, Hadamard fractional operator, Katugompola fractional operators, respectively.

5 Trapezoid type inequalities

In this section, we will explore certain trapezoid-like inequalities for h-convexity functions utilizing $\psi$-Hilfer operators, as well as their particular results, where h belongs to the class B(0, 1). To accomplish this, we must first establish equality in the lemma that follows.

Lemma 5.1

If $\beta ,\psi$ are defined as in Theorem 4.1 and $f:[a,b]\rightarrow \mathbb {R}$ is a differentiable mapping to (a, b), then the following identity holds.

$$\begin{aligned}&\displaystyle \frac{f(a)+f(b)}{2}-\displaystyle \frac{\Gamma (\beta +1)}{ 4\,\left( \psi (b)-\psi (a)\right) ^{\beta }}\left[ {\,^{\psi }\mathfrak {J} \,_{b^{-}}^{\beta }}F(a)+{\,^{\psi }\mathfrak {J}\,_{a^{+}}^{\beta }}F(b) \right] =\displaystyle \frac{b-a}{4\, \left( \psi (b)-\psi (a)\right) ^{\beta }} \nonumber \\&\displaystyle \times \int _{0}^{1}\left( 2\,\left( \psi (b)-\psi (a)\right) ^{\beta }-A_{\psi ,\beta }(s)\right) \left[ f^{\prime }\left( sa+(1-s)b\right) -f^{\prime }\left( (1-s)a+sb\right) \right] ds, \end{aligned}$$

(5.1)

where

$$\begin{aligned} A_{\psi ,\beta }(s)=\left( \psi (b)-\psi \left( sa+(1-s)b\right) \right) ^{\beta }+\left( \psi \left( (1-s)a+sb\right) -\psi (a)\right) ^{\beta }. \end{aligned}$$

(5.2)

Proof

Let

$$\begin{aligned} J_{1}=\int _{a}^{b}\left[ \left( \psi (b)-\psi (a)\right) ^{\beta } -\left( \psi (b)-\psi (t)\right) ^{\beta }\right] F^{\prime }(t)dt. \end{aligned}$$

(5.3)

Integrating by parts (5.3) and using (4.2), we get

$$\begin{aligned} J_{1}=\left[ \left( \psi (b)-\psi (a)\right) ^{\beta }-\left( \psi (b)-\psi (t)\right) ^{\beta }\right] F(t)\Bigg |_{a}^{b}-{\beta }\displaystyle \int _{a}^{b}\left( \psi (b)-\psi (t)\right) ^{\beta -1}\psi ^{\prime }(t)F(t)dt, \end{aligned}$$

therefore

$$\begin{aligned} J_{1}=\left( \psi (b)-\psi (a)\right) ^{\beta }F(b)-\Gamma (\beta +1) \ {\,^{\psi }\mathfrak {J}\,_{a^{+}}^{\beta }}F(b). \end{aligned}$$

(5.4)

Similarly, let

$$\begin{aligned} J_{2}=\int _{a}^{b}\left[ \left( \psi (b)-\psi (a)\right) ^{\beta } -\left( \psi (t)-\psi (a)\right) ^{\beta }\right] F^{\prime }(t)dt. \end{aligned}$$

(5.5)

Integrating by parts (5.5), we deduce

$$\begin{aligned} J_{2}=-\left( \psi (b)-\psi (a)\right) ^{\beta }F(a)+\Gamma (\beta +1) \ { \,^{\psi }\mathfrak {J}\,_{b^{-}}^{\beta }}F(a). \end{aligned}$$

(5.6)

Since $F(a)=F(b)=f(a)+f(b)$, from (5.4) and (5.6), we obtain

$$\begin{aligned} J_{1}-J_{2}=2\,\left( \psi (b)-\psi (a)\right) ^{\beta } \left( f(a)+f(b)\right) -\Gamma (\beta +1)\left[ {\,^{\psi }\mathfrak {J} \,_{b^{-}}^{\beta }}F(a)+{\,^{\psi }\mathfrak {J}\,_{a^{+}}^{\beta }}F(b) \right] , \end{aligned}$$

thus

$$\begin{aligned}&\displaystyle \frac{f(a)+f(b)}{2}-\displaystyle \frac{\Gamma (\beta +1)}{ 4\,\left( \psi (b)-\psi (a)\right) ^{\beta }}\left[ {\,^{\psi }\mathfrak {J} \,_{b^{-}}^{\beta }}F(a)+{\,^{\psi }\mathfrak {J}\,_{a^{+}}^{\beta }}F(b) \right] \nonumber \\&\quad =\displaystyle \frac{1}{4\,\left( \psi (b)-\psi (a)\right) ^{\beta }} \left( J_{1}-J_{2}\right) . \end{aligned}$$

(5.7)

On the other hand, given that $F^{\prime }(t)=f^{\prime }(t)-f^{\prime }(a+b-t)$, then from (5.3), we get

$$\begin{aligned} J_{1}=\displaystyle \int _{a}^{b}\left[ \left( \psi (b)-\psi (a)\right) ^{\beta }-\left( \psi (b)-\psi (t)\right) ^{\beta }\right] \left( f^{\prime }(t)-f^{\prime }(a+b-t)\right) dt. \end{aligned}$$

Changing the variable $t=sa+(1-s)b$ for $s\in [ 0,1]$ produces

$$\begin{aligned} J_{1}= & {} (b-a)\,\displaystyle \int _{0}^{1} \left[ \left( \psi (b)-\psi (a)\right) ^{\beta }-\left( \psi (b)-\psi (sa+(1-s)b)\right) ^{\beta }\right] \\{} & {} \displaystyle \times \left[ f^{\prime }\left( sa+(1-s)b\right) -f^{\prime }\left( (1-s)a+sb\right) \right] ds. \end{aligned}$$

Similarly, from (5.5), we get

$$\begin{aligned} J_{2}= & {} \displaystyle \int _{a}^{b}\left[ \left( \psi (b)-\psi (a)\right) ^{\beta }-\left( \psi (t)-\psi (a)\right) ^{\beta }\right] \left( f^{\prime }(t)-f^{\prime }(a+b-t)\right) dt \\= & {} (b-a)\,\displaystyle \int _{0}^{1}\left[ \left( \psi (b)-\psi (a)\right) ^{\beta }-\left( \psi ((1-s)a+sb)-\psi (a)\right) ^{\beta }\right] \\{} & {} \displaystyle \times \left[ f^{\prime }\left( (1-s)a+sb\right) -f^{\prime }\left( sa+(1-s)b\right) \right] ds. \end{aligned}$$

As a result,

$$\begin{aligned} J_{1}-J_{2}&=(b-a)\,\int _{0}^{1}\left( 2\,\left( \psi (b)-\psi (a)\right) ^{\beta }-A_{\psi ,\beta }(s)\right) \nonumber \\&\quad \left[ f^{\prime }\left( sa+(1-s)b\right) -f^{\prime }\left( (1-s)a+sb\right) \right] ds. \end{aligned}$$

(5.8)

To get the desired equality (5.1), simply replace (5.8) with (5.7). $\square$

Theorem 5.1

Assume h is a B-function and $\beta , \psi$ are defined according to Lemma 5.1. If $|f^{\prime }|$ is a h-convex mapping on [a, b], then the trapezoid type inequality is obtained as

$$\begin{aligned}&\Bigg | \displaystyle \frac{ f(a) + f(b) }{2} - \displaystyle \frac{ \Gamma (\beta +1) }{4\,\Omega (\psi ,\beta ) } \left[ {\,^{\psi }\mathfrak {J} \,^{\beta }_{b^{-}}} F(a) + {\,^{\psi }\mathfrak {J}\,^{\beta }_{a^{+}}} F(b) \right] \Bigg | \nonumber \\&\quad \le \displaystyle \frac{ (b-a)\,h\left( \frac{1}{2}\right) }{2\,\Omega (\psi ,\beta )}\Big [ |f^{\prime }(a) | + | f^{\prime }(b) | \Big ]\int _0^1 \big | 2\,\Omega (\psi ,\beta ) -A_{\psi , \beta }(s) \big | ds, \end{aligned}$$

(5.9)

where $\Omega (\psi ,\beta )= \left( \psi (b)-\psi (a)\right) ^{\beta }$.

Proof

Using the absolute value of identity (5.1), the h-convexity of the function $|f^{\prime }|$ and h is a B-function, we get

$$\begin{aligned}&\Bigg | \displaystyle \frac{ f(a) + f(b) }{2} - \displaystyle \frac{ \Gamma (\beta +1) }{4\, \Omega (\psi ,\beta ) } \left[ {\,^{\psi }\mathfrak {J} \,^{\beta }_{b^{-}}} F(a) + {\,^{\psi }\mathfrak {J}\,^{\beta }_{a^{+}}} F(b) \right] \Bigg | \\&\quad \le \displaystyle \frac{ b-a }{4\,\Omega (\psi ,\beta ) }\int _{0}^{1}\big | 2\,\Omega (\psi ,\beta )-A_{\psi ,\beta }(s)\big |\Big |f^{\prime }\left( sa+(1-s)b\right) -f^{\prime }\left( (1-s)a+sb\right) \Big |ds \\&\quad \le \displaystyle \frac{b-a}{4\,\Omega (\psi ,\beta )}\int _{0}^{1}\big | 2\,\Omega (\psi ,\beta )-A_{\psi ,\beta }(s)\big | \left( h(s) +h(1-s)\right) \left[ \big |f^{\prime }(a)\big |+\big |f^{\prime }(b)\big |\right] ds \\&\quad \le \displaystyle \frac{b-a}{4\,\Omega (\psi ,\beta )}\left( 2 \,h\left( \frac{1}{2}\right) \left[ \big |f^{\prime }(a)\big |+\big |f^{\prime }(b)\big | \right] \right) \int _{0}^{1}\big |2\,\Omega (\psi ,\beta )-A_{\psi ,\beta }(s) \big | ds. \end{aligned}$$

This completes the proof. $\square$

Using $\psi (x)=x$ and $\beta =1$, we derive the following new version of the trapezoidal inequality, see ([10, Corollary.1]).

Corollary 7

Let h be a B-function. If $|f^{\prime }|$ is a h-convex mapping on [a, b], then the following inequalities hold

$$\begin{aligned} \Bigg | \displaystyle \frac{ f(a) + f(b) }{2} - \displaystyle \frac{1}{b-a} \int _{a}^{b}f(t)dt\Bigg | \le \displaystyle \frac{ (b-a)\,h\left( \frac{1}{2} \right) }{4}\Big [ |f^{\prime }(a) | + | f^{\prime }(b) | \Big ]. \end{aligned}$$

(5.10)

The following Corollaries depend on the function h given in Theorem 5.1. First, Let’s put $h(\alpha )=\alpha$ to get a new version using the convex functions proven in [8].

Corollary 8

Assume that $\beta ,\psi$ are defined as in Lemma 5.1. If $|f^{\prime }|$ is a convex mapping on [a, b], then

$$\begin{aligned}&\Bigg |\displaystyle \frac{f(a)+f(b)}{2}-\displaystyle \frac{\Gamma (\beta +1)}{ 4\,\Omega (\psi ,\beta )}\left[ {\,^{\psi }\mathfrak {J}\,_{b^{-}}^{\beta }} F(a)+{\,^{\psi }\mathfrak {J}\,_{a^{+}}^{\beta }}F(b)\right] \Bigg | \nonumber \\&\quad \le \displaystyle \frac{b-a}{4\,\Omega (\psi ,\beta )}\Big [|f^{\prime }(a)|+|f^{\prime }(b)|\Big ]\int _{0}^{1}\big |2\,\Omega (\psi ,\beta )-A_{\psi ,\beta }(s)\big |ds, \end{aligned}$$

(5.11)

where $\Omega (\psi ,\beta )=\left( \psi (b)-\psi (a)\right) ^{\beta }$.

Put $h(\alpha )=1$, we get the following result via $\psi$-Hilfer operators with function f belongs to the class P-function.

Corollary 9

Assume that $\beta , \psi$ are defined as in Lemma 5.1. If $|f^{\prime }|$ is a P-function on [a, b], thus

$$\begin{aligned}&\Bigg | \displaystyle \frac{ f(a) + f(b) }{2} - \displaystyle \frac{ \Gamma (\beta +1) }{4\,\Omega (\psi ,\beta ) } \left[ {\,^{\psi }\mathfrak {J} \,^{\beta }_{b^{-}}} F(a) + {\,^{\psi }\mathfrak {J}\,^{\beta }_{a^{+}}} F(b) \right] \Bigg | \nonumber \\&\quad \le \displaystyle \frac{b-a}{2\,\Omega (\psi ,\beta )}\Big [ |f^{\prime }(a) | + | f^{\prime }(b) | \Big ]\int _0^1 \big | 2\,\Omega (\psi ,\beta ) -A_{\psi , \beta }(s) \big | ds, \end{aligned}$$

(5.12)

where $\Omega (\psi ,\beta )= \left( \psi (b)-\psi (a)\right) ^{\beta }$.

Put $h(\alpha )=\alpha ^{s}$, we get the following result via $\psi$-Hilfer operators with s-convex functions.

Corollary 10

Assume that $\beta , \psi$ are defined as in Lemma 5.1. If $|f^{\prime }|$ is a s-convex mapping on [a, b], then the trapezoid type inequality is obtained as

$$\begin{aligned}&\Bigg | \displaystyle \frac{ f(a) + f(b) }{2} - \displaystyle \frac{ \Gamma (\beta +1) }{4\,\Omega (\psi ,\beta ) } \left[ {\,^{\psi }\mathfrak {J} \,^{\beta }_{b^{-}}} F(a) + {\,^{\psi }\mathfrak {J}\,^{\beta }_{a^{+}}} F(b) \right] \Bigg | \nonumber \\&\quad \le \displaystyle \frac{b-a}{2^{s+1}\,\Omega (\psi ,\beta )}\Big [ |f^{\prime }(a) | + | f^{\prime }(b) | \Big ]\int _0^1 \big | 2\,\Omega (\psi ,\beta ) -A_{\psi , \beta }(s) \big | ds, \end{aligned}$$

(5.13)

where $\Omega (\psi ,\beta )= \left( \psi (b)-\psi (a)\right) ^{\beta }$.

Put $h(\alpha )=\frac{1}{n}\sum \nolimits _{k=1}^{n} \alpha ^{\frac{1}{k}}$, we get the following result via $\psi$-Hilfer operators with n-fractional polynomial convex functions.

Corollary 11

Assume that $\beta , \psi$ are defined as in Lemma 5.1. If $|f^{\prime }|$ is a n-fractional polynomial convex mapping on [a, b], yields

$$\begin{aligned}&\Bigg | \displaystyle \frac{ f(a) + f(b) }{2} - \displaystyle \frac{ \Gamma (\beta +1) }{4\,\Omega (\psi ,\beta ) } \left[ {\,^{\psi }\mathfrak {J} \,^{\beta }_{b^{-}}} F(a) + {\,^{\psi }\mathfrak {J}\,^{\beta }_{a^{+}}} F(b) \right] \Bigg | \nonumber \\&\quad \le \displaystyle \frac{ (b-a)\,C_{n}}{4\,\Omega (\psi ,\beta )}\Big [ |f^{\prime }(a) | + | f^{\prime }(b) | \Big ]\int _0^1 \big | 2\,\Omega (\psi ,\beta ) -A_{\psi , \beta }(s) \big | ds, \end{aligned}$$

(5.14)

where $\Omega (\psi ,\beta )= \left( \psi (b)-\psi (a)\right) ^{\beta }$ and $C_{n}=\frac{2}{n}\sum \nolimits _{k=1}^{n} \left( \frac{1}{2}\right) ^{\frac{1}{k}}$.

Remark 2

If we choose $\psi (\tau )=\tau$, $\psi (\tau )=\ln \tau$ and $\psi (\tau )=\frac{ \tau ^{\rho }}{\rho }$, respectively, in the corollaries 8, 9, 10 and 11, we establish trapezoid-type inequalities for the convex function, the function P, s-convex functions, and polynomial n-fractional convex functions, respectively, involving the Riemann-Liouville fractional operator, the Hadamard fractional operator, and the Katugompola fractional operators, respectively.

Theorem 5.2

Let h be a B-function, $p>1$ and $\frac{1}{p^{\prime }}+ \frac{1}{p}=1$. Assume that $\beta ,\psi$ are defined as in Lemma 5.1. If $\left| f^{\prime }\right| ^{p}$ is a h-convex mapping on [a, b], then the following trapezoid type inequality is obtained

$$\begin{aligned}&\Bigg |\displaystyle \frac{f(a)+f(b)}{2}-\displaystyle \frac{\Gamma (\beta +1)}{ 4\,\Omega (\psi ,\beta )}\left[ {\,^{\psi }\mathfrak {J}\,_{b^{-}}^{\beta }} F(a)+{\,^{\psi }\mathfrak {J}\,_{a^{+}}^{\beta }}F(b)\right] \Bigg | \nonumber \\&\quad \le \displaystyle \frac{(b-a)\left( 2\,h\left( \frac{1}{2}\right) \right) ^{ \frac{1}{p}}}{4\,\Omega (\psi ,\beta )}\left( 2\int _{0}^{1}\big |2\,\Omega (\psi ,\beta )-A_{\psi ,\beta }(s)\big |^{p^{\prime }}ds\right) ^{\frac{1}{ p^{\prime }}}\left( \Big |f^{\prime }(a)\Big |^{p}+\Big |f^{\prime }(b)\Big | ^{p}\right) ^{\frac{1}{p}} \nonumber \\&\quad \le \displaystyle \frac{(b-a)\left( 2\,h\left( \frac{1}{2}\right) \right) ^{ \frac{1}{p}}}{4\,\Omega (\psi ,\beta )}\left( 2\int _{0}^{1}\big |2\,\Omega (\psi ,\beta )-A_{\psi ,\beta }(s)\big |^{p^{\prime }}ds\right) ^{\frac{1}{ p^{\prime }}}\left( \Big |f^{\prime }(a)\Big |+\Big |f^{\prime }(b)\Big |\right) , \end{aligned}$$

(5.15)

where $\Omega (\psi ,\beta )=\left( \psi (b)-\psi (a)\right) ^{\beta }$.

Proof

By using the absolute value on equality in Lemma 5.1, we get

$$\begin{aligned}&\Bigg |\displaystyle \frac{f(a)+f(b)}{2}-\displaystyle \frac{\Gamma (\beta +1)}{ 4\,\Omega (\psi ,\beta )}\left[ {\,^{\psi }\mathfrak {J}\,_{b^{-}}^{\beta }} F(a)+{\,^{\psi }\mathfrak {J}\,_{a^{+}}^{\beta }}F(b)\right] \Bigg | \\&\quad \le \displaystyle \frac{b-a}{4\,\Omega (\psi ,\beta )}\int _{0}^{1}\Big | 2\,\Omega (\psi ,\beta )-A_{\psi ,\beta }(s)\Big |\,\Big |f^{\prime }\left( sa+(1-s)b\right) \Big |ds \\&\qquad \; +\displaystyle \frac{b-a}{4\,\Omega (\psi ,\beta )}\int _{0}^{1}\Big | 2\,\Omega (\psi ,\beta )-A_{\psi ,\beta }(s)\Big |\,\Big |f^{\prime }\left( (1-s)a+sb\right) \Big |ds, \end{aligned}$$

Utilizing Hölder’s inequality, we get

$$\begin{aligned}&\Bigg |\displaystyle \frac{f(a)+f(b)}{2}-\displaystyle \frac{\Gamma (\beta +1)}{ 4\,\Omega (\psi ,\beta )}\left[ {\,^{\psi }\mathfrak {J}\,_{b^{-}}^{\beta }} F(a)+{\,^{\psi }\mathfrak {J}\,_{a^{+}}^{\beta }}F(b)\right] \Bigg | \\&\quad \le \displaystyle \frac{b-a}{4\,\Omega (\psi ,\beta )}\left( \int _{0}^{1} \big |2\,\Omega (\psi ,\beta )-A_{\psi ,\beta }(s)\big |^{p^{\prime }}ds\right) ^{\frac{1}{p^{\prime }}}\left( \int _{0}^{1}\Big |f^{\prime }\left( sa+(1-s)b\right) \Big |^{p}ds\right) ^{\frac{1}{p}} \\&\qquad \; +\displaystyle \frac{b-a}{4\,\Omega (\psi ,\beta )}\left( \int _{0}^{1}\big | 2\,\Omega (\psi ,\beta )-A_{\psi ,\beta }(s)\big |^{p^{\prime }}ds\right) ^{ \frac{1}{p^{\prime }}}\left( \int _{0}^{1}\Big |f^{\prime }\left( (1-s)a+sb\right) \Big |^{p}ds\right) ^{\frac{1}{p}}. \end{aligned}$$

Notice that for $p>1$ and $A,B\ge 0$: $\displaystyle A^{\frac{1}{p}}+B^{ \frac{1}{p}}\le 2^{1-\frac{1}{p}}(A+B)^{\frac{1}{p}}$, then

$$\begin{aligned}&\Bigg |\displaystyle \frac{f(a)+f(b)}{2}-\displaystyle \frac{\Gamma (\beta +1)}{ 4\,\Omega (\psi ,\beta )}\left[ {\,^{\psi }\mathfrak {J}\,_{b^{-}}^{\beta }} F(a)+{\,^{\psi }\mathfrak {J}\,_{a^{+}}^{\beta }}F(b)\right] \Bigg | \\&\quad \le \displaystyle \frac{b-a}{4\,\Omega (\psi ,\beta )}\left( 2\,\int _{0}^{1} \big |2\,\Omega (\psi ,\beta )-A_{\psi ,\beta }(s)\big |^{p^{\prime }}ds\right) ^{\frac{1}{p^{\prime }}} \\&\qquad \; \times \displaystyle \left[ \int _{0}^{1}\Big |f^{\prime }\left( sa+(1-s)b\right) \Big |^{p}ds+\int _{0}^{1}\Big |f^{\prime }\left( (1-s)a+sb\right) \Big |^{p}ds\right] ^{\frac{1}{p}}. \end{aligned}$$

Given that $\left| f^{\prime }\right| ^{p}$ is h-convex function and h is B-function, we get

$$\begin{aligned}&\displaystyle \int _{0}^{1}\Big |f^{\prime }\left( sa+(1-s)b\right) \Big |^{p}ds +\displaystyle \int _{0}^{1}\Big |f^{\prime }\left( (1-s)a+sb\right) \Big |^{p}ds \\&\quad \le \left( \Big |f^{\prime }(a)\Big |^{p}+\Big |f^{\prime }(b)\Big |^{p}\right) \displaystyle \int _{0}^{1}\left( h(s)+h(1-s)\right) ds \\&\quad \le 2\,h\left( \frac{1}{2}\right) \left( \Big |f^{\prime }(a)\Big |^{p}+\Big | f^{\prime }(b)\Big |^{p}\right) , \end{aligned}$$

therefore

$$\begin{aligned}&\Bigg |\displaystyle \frac{f(a)+f(b)}{2}-\displaystyle \frac{\Gamma (\beta +1)}{ 4\,\Omega (\psi ,\beta )}\left[ {\,^{\psi }\mathfrak {J}\,_{b^{-}}^{\beta }} F(a)+{\,^{\psi }\mathfrak {J}\,_{a^{+}}^{\beta }}F(b)\right] \Bigg | \\&\quad \le \displaystyle \frac{(b-a)\left( 2\,h\left( \frac{1}{2}\right) \right) ^{ \frac{1}{p}}}{4\,\Omega (\psi ,\beta )}\left( 2\int _{0}^{1}\big |2\,\Omega (\psi ,\beta )-A_{\psi ,\beta }(s)\big |^{p^{\prime }}ds\right) ^{\frac{1}{ p^{\prime }}}\left( \Big |f^{\prime }(a)\Big |^{p}+\Big |f^{\prime }(b)\Big | ^{p}\right) ^{\frac{1}{p}}. \end{aligned}$$

This accomplishes the first inequality in (6.14). Notice that $A^{p}+B^{p}\le (A+B)^{p}\ \hbox {for}\ p>1$, it results the second inequality in (6.14). $\square$

6 Midpoint type inequalities

This section presents new results on midpoint inequality involving h-convex functions via $\psi$-Hilfer operators, based on the identity presented in the following lemma.

Lemma 6.1

Let $\beta >0$, $f\in X [a,b]$ be a h-convex function and $\psi$ be a positive, strictly increasing differentiable function such that $\psi \,^{\prime }(\tau )\ne 0$ for all $\tau \in [a, b]$. The following identity holds

$$\begin{aligned}&\displaystyle \frac{\Gamma (\beta +1)}{4\,\Omega (\psi ,\beta )}\left[ { \,^{\psi }\mathfrak {J}\,^{\beta }_{b^{-}}} F(a) + {\,^{\psi }\mathfrak {J} \,^{\beta }_{a^{+}}} F(b) \right] -\displaystyle f\left( \frac{a+b}{2}\right) \nonumber \\&\quad =\displaystyle \frac{b-a}{4\,\Omega (\psi ,\beta )}\int _{0}^{1}A_{\psi ,\beta }(s)\left[ f^{\prime }\Big (sa+(1-s)b\Big )-f^{\prime }\Big ((1-s)a+sb\Big ) \right] ds \nonumber \\&\qquad \;\;-\displaystyle \frac{b-a}{2}\int _{\frac{1}{2}}^{1}\left[ f^{\prime }\Big ( sa+(1-s)b\Big )-f^{\prime }\Big ((1-s)a+sb\Big )\right] ds. \end{aligned}$$

(6.1)

Proof

Let

$$\begin{aligned} K_{1}=-\int _{a}^{b}\left( \psi (b)-\psi (t)\right) ^{\beta }F\,^{\prime }(t)dt+\left( \psi (b)-\psi (a)\right) ^{\beta }\int _{a}^{\frac{a+b}{2} }F\,^{\prime }(t)dt. \end{aligned}$$

(6.2)

Integrating by parts (6.2), we get

$$\begin{aligned} K_{1}&=-\left( \psi (b)-\psi (t)\right) ^{\beta }F(t)\Big \vert _{a}^{b}-{ \beta }\displaystyle \int _{a}^{b}\left( \psi (b)-\psi (t)\right) ^{\beta -1}\,F(t)dt \\&\quad \; +\left( \psi (b)-\psi (a)\right) ^{\beta }F(t)\Big \vert _{a}^{\frac{a+b}{2}}. \end{aligned}$$

This gives

$$\begin{aligned} K_{1}=\left( \psi (b)-\psi (a)\right) ^{\beta }F\left( \frac{a+b}{2}\right) -\Gamma (\beta +1)\ {\,^{\psi } \mathfrak {J}\,_{a^{+}}^{\beta }}F(b). \end{aligned}$$

(6.3)

Consider the following integral

$$\begin{aligned} K_{2}=\int _{a}^{b}\left( \psi (t)-\psi (a)\right) ^{\beta }F\,^{\prime }(t)dt-\left( \psi (b)-\psi (a)\right) ^{\beta }\int _{\frac{a+b}{2} }^{b}F\,^{\prime }(t)dt. \end{aligned}$$

(6.4)

Integrating by parts yields

$$\begin{aligned} K_{2}&=\left( \psi (t)-\psi (a)\right) ^{\beta }F(t)\Big \vert _{a}^{b}-{ \beta }\displaystyle \int _{a}^{b}\left( \psi (t)-\psi (a)\right) ^{\beta -1}\,F(t)dt \\&\quad \; -\left( \psi (b)-\psi (a)\right) ^{\beta }F(t)\Big \vert _{\frac{a+b}{2} }^{b}, \end{aligned}$$

which gives

$$\begin{aligned} K_{2}=\left( \psi (b)-\psi (a)\right) ^{\beta }F \left( \frac{a+b}{2}\right) -\Gamma (\beta +1)\grave{u}{\,^{\psi } \mathfrak {J}\,_{b^{-}}^{\beta }}F(a). \end{aligned}$$

(6.5)

By adding (6.3) and (6.5), then using (4.2) we obtain

$$\begin{aligned} K_{1}+K_{2}=4\,\Omega (\psi ,\beta )\,f\left( \frac{a+b}{2}\right) -\Gamma (\beta +1) \left[ {\,^{\psi }\mathfrak {J}\,_{b^{-}}^{\beta }}F(a)+{\,^{\psi } \mathfrak {J }\,_{a^{+}}^{\beta }}F(b)\right] . \end{aligned}$$

(6.6)

Furthermore, by changing the variable $t=sa+(1-s)b$ in (6.2) and using (4.2), we get

$$\begin{aligned} K_{1}&=-(b-a)\,\displaystyle \int _{0}^{1}\left( \psi (b)-\psi (sa+(1-s)b)\right) ^{\beta }\displaystyle \\&\quad \left[ f^{\prime }\left( sa+(1-s)b\right) -f^{\prime }\left( (1-s)a+sb\right) \right] ds \\&\quad \; +(b-a)\,\left( \psi (b)-\psi (a)\right) ^{\beta }\displaystyle \int _{\frac{1}{2}}^{1}\left[ f^{\prime }\left( sa+(1-s)b\right) -f^{\prime } \left( (1-s)a+sb\right) \right] ds. \end{aligned}$$

Changing the variable $t=(1-s)a+sb$ in (6.4) and using (4.2), we obtain

$$\begin{aligned} K_{2}&=(b-a)\,\displaystyle \int _{0}^{1}\left( \psi ((1-s)a+sb)-\psi (a)\right) ^{\beta }\displaystyle \nonumber \\&\quad \left[ f^{\prime }\left( (1-s)a+sb\right) -f^{\prime }\left( sa+(1-s)b\right) \right] ds \\&\quad \; -(b-a)\,\left( \psi (b)-\psi (a)\right) ^{\beta }\displaystyle \int _{ \frac{1}{2}}^{1}\left[ f^{\prime }\left( (1-s)a+sb\right) -f^{\prime }\left( sa+(1-s)b\right) \right] ds, \end{aligned}$$

therefore

$$\begin{aligned} K_{1}+K_{2}&=-(b-a)\,\displaystyle \int _{0}^{1}A_{\psi ,\beta }(s)\left[ f^{\prime }\left( sa+(1-s)b\right) -f^{\prime }\left( (1-s)a+sb\right) \right] ds \nonumber \\&\quad \; +2\,(b-a)\,\Omega (\psi ,\beta )\displaystyle \int _{\frac{1}{2}}^{1} \left[ f^{\prime }\left( sa+(1-s)b\right) -f^{\prime }\left( (1-s)a+sb\right) \right] ds. \end{aligned}$$

(6.7)

Replacing (6.7) in (6.6), we get the desired equality (6.1). $\square$

Theorem 6.1

Assume h is a B-function and $\beta , \psi$ are defined according to Lemma 5.1. If $|f^{\prime }|$ is a h-convex mapping on [a, b], then the midpoint type inequality is achieve.

$$\begin{aligned}&\Bigg | \displaystyle \frac{ \Gamma (\beta +1) }{4\, \Omega (\psi , \beta ) } \left[ {\,^{\psi }\mathfrak {J}\,^{\beta }_{b^{-}}} F(a) + {\,^{\psi }\mathfrak {J} \,^{\beta }_{a^{+}}} F(b) \right] - \displaystyle f\left( \frac{ a +b}{2} \right) \Bigg | \nonumber \\&\quad \le \displaystyle \frac{b-a }{4} \left( 1+ \displaystyle \frac{1}{\Omega (\psi ,\beta ) }\int _0^1 \Big | A_{\psi , \beta }(s)\Big | ds\right) \Big [ |f^{\prime }(a) | + | f^{\prime }(b) | \Big ] . \end{aligned}$$

(6.8)

Proof

Using the absolute value of the identity (6.1) and the h-convexity of the function $|f^{\prime }|$, we get

$$\begin{aligned}&\Bigg |\displaystyle \frac{\Gamma (\beta +1)}{4\,\Omega (\psi ,\beta )}\left[ { \,^{\psi }\mathfrak {J}\,_{b^{-}}^{\beta }}F(a)+{\,^{\psi }\mathfrak {J} \,_{a^{+}}^{\beta }}F(b)\right] -\displaystyle f\left( \frac{a+b}{2}\right) \Bigg | \\&\quad \le \displaystyle \frac{b-a}{4\,\Omega (\psi ,\beta )}\int _{0}^{1}\Big | A_{\psi ,\beta }(s)\Big |\left[ \Big |f^{\prime }\big (sa+(1-s)b\big )\Big |+\Big | f^{\prime }\big ((1-s)a+sb\big )\Big |\right] ds \\&\qquad \; +\displaystyle \frac{b-a}{2}\int _{\frac{1}{2}}^{1} \left[ \Big |f^{\prime }\big (sa+(1-s)b\big )\Big |+\Big |f^{\prime } \big ((1-s)a+sb\big )\Big |\right] ds\\&\quad \le \displaystyle \frac{b-a}{4\,\Omega (\psi ,\beta )}\int _{0}^{1}\Big | A_{\psi ,\beta }(s)\Big |\left( h(s)+h(1-s)\right) \left[ \big |f^{\prime }(a) \big |+\big |f^{\prime }(b)\big |\right] ds \\&\qquad \; +\displaystyle \frac{b-a}{2}\int _{\frac{1}{2}}^{1}\left( h(s)+h(1-s)\right) \left[ \big |f^{\prime }(a)\big | +\big |f^{\prime }(b)\big |\right] ds. \end{aligned}$$

Given that h is a B-function, we get

$$\begin{aligned}&\Bigg |\displaystyle \frac{\Gamma (\beta +1)}{4\,\Omega (\psi ,\beta )}\left[ { \,^{\psi }\mathfrak {J}\,_{b^{-}}^{\beta }}F(a)+{\,^{\psi }\mathfrak {J} \,_{a^{+}}^{\beta }}F(b)\right] -\displaystyle f\left( \frac{a+b}{2}\right) \Bigg | \\&\quad \le \displaystyle \frac{b-a}{4\,\Omega (\psi ,\beta )}\int _{0}^{1}\Big | A_{\psi ,\beta }(s)\Big |\left( 2\,h\left( \frac{1}{2}\right) \left[ \big | f^{\prime }(a)\big |+\big |f^{\prime }(b)\big |\right] \right) ds \\&\qquad \; +\displaystyle \frac{b-a}{2}\int _{\frac{1}{2}}^{1}\left( 2\,h\left( \frac{1}{2}\right) \left[ \big |f^{\prime }(a)\big |+\big |f^{\prime }(b)\big | \right] \right) ds \\&\quad =\displaystyle \frac{(b-a)\,h\left( \frac{1}{2}\right) }{2}\left( 1+ \displaystyle \frac{1}{\Omega (\psi ,\beta )}\int _{0}^{1}\Big |A_{\psi ,\beta }(s)\Big |ds\right) \Big [|f^{\prime }(a)|+|f^{\prime }(b)|\Big ]. \end{aligned}$$

This end the proof. $\square$

Using $\psi (x)=x$ and $\beta =1$, we get the following midpoint inequality.

Corollary 12

Assume h is a B-function. If $|f^{\prime }|$ is a h-convex mapping on [a, b], then the midpoint type inequality holds.

$$\begin{aligned} \Bigg | \displaystyle \frac{1}{b-a}\int _{a}^{b}f(t)dt - \displaystyle f\left( \frac{ a +b}{2}\right) \Bigg | \le ( b-a)\,h\left( \frac{1}{2}\right) \Big [ |f^{\prime }(a) | + | f^{\prime }(b) | \Big ] . \end{aligned}$$

(6.9)

The following Corollaries depend on the function h given in Theorem 6.1. First, Let’s put $h(\alpha )=\alpha$ to get midpoint inequality with convex functions.

Corollary 13

Assume that $\beta ,\psi$ are defined as in Lemma 5.1. If $|f^{\prime }|$ is a convex mapping on [a, b], then

$$\begin{aligned}&\Bigg |\displaystyle \frac{\Gamma (\beta +1)}{4\,\Omega (\psi ,\beta )}\left[ { \,^{\psi }\mathfrak {J}\,_{b^{-}}^{\beta }}F(a)+{\,^{\psi }\mathfrak {J} \,_{a^{+}}^{\beta }}F(b)\right] -\displaystyle f\left( \frac{a+b}{2}\right) \Bigg | \nonumber \\&\quad \le \displaystyle \frac{b-a}{4}\left( 1+\displaystyle \frac{1}{\Omega (\psi ,\beta )}\int _{0}^{1}\Big |A_{\psi ,\beta }(s)\Big |ds\right) \Big [|f^{\prime }(a)|+|f^{\prime }(b)|\Big ], \end{aligned}$$

(6.10)

where $\Omega (\psi ,\beta )=\left( \psi (b)-\psi (a)\right) ^{\beta }$.

Put $h(\alpha )=1$, we get midpoint type inequality via $\psi$-Hilfer operators with function f belongs to the class P-function.

Corollary 14

Assume that $\beta ,\psi$ are defined as in Lemma 5.1. If $|f^{\prime }|$ is a P-function on [a, b], thus

$$\begin{aligned}&\Bigg |\displaystyle \frac{\Gamma (\beta +1)}{4\,\Omega (\psi ,\beta )} \left[ {\,^{\psi }\mathfrak {J}\,_{b^{-}}^{\beta }}F(a)+{\,^{\psi } \mathfrak {J}\,_{a^{+}}^{\beta }}F(b)\right] -\displaystyle f \left( \frac{a+b}{2}\right) \Bigg | \nonumber \\&\quad \le \displaystyle \frac{b-a}{2}\left( 1+\displaystyle \frac{1}{\left( \psi (b)-\psi (a)\right) ^{\beta }}\int _{0}^{1}\Big |A_{\psi ,\beta }(s)\Big | ds\right) \Big [|f^{\prime }(a)|+|f^{\prime }(b)|\Big ], \end{aligned}$$

(6.11)

where $\Omega (\psi ,\beta )=\left( \psi (b)-\psi (a)\right) ^{\beta }$.

Put $h(\alpha )=\alpha ^{s}$, we get the following midpoint type inequality involving $\psi$-Hilfer operators with s-convex functions.

Corollary 15

Assume that $\beta ,\psi$ are defined as in Lemma 5.1. If $|f^{\prime }|$ is a s-convex mapping on [a, b], then the trapezoid type inequality is obtained as

$$\begin{aligned}&\Bigg |\displaystyle \frac{\Gamma (\beta +1)}{4\,\Omega (\psi ,\beta )}\left[ { \,^{\psi }\mathfrak {J}\,_{b^{-}}^{\beta }}F(a)+{\,^{\psi }\mathfrak {J} \,_{a^{+}}^{\beta }}F(b)\right] -\displaystyle f\left( \frac{a+b}{2}\right) \Bigg | \nonumber \\&\quad \le \displaystyle \frac{b-a}{2^{s+1}}\left( 1+\displaystyle \frac{1}{\left( \psi (b)-\psi (a)\right) ^{\beta }}\int _{0}^{1}\Big |A_{\psi ,\beta }(s)\Big | ds\right) \Big [|f^{\prime }(a)|+|f^{\prime }(b)|\Big ], \end{aligned}$$

(6.12)

where $\Omega (\psi ,\beta )=\left( \psi (b)-\psi (a)\right) ^{\beta }$.

Put $h(\alpha )=\frac{1}{n}\sum \nolimits _{k=1}^{n}\alpha ^{\frac{1}{k}}$, we get midpoint type inequality via $\psi$-Hilfer operators with n-fractional polynomial convex functions.

Corollary 16

Assume that $\beta ,\psi$ are defined as in Lemma 5.1. If $|f^{\prime }|$ is a h-fractional polynomial convex mapping on [a, b], yields

$$\begin{aligned}&\Bigg |\displaystyle \frac{\Gamma (\beta +1)}{4\,\Omega (\psi ,\beta )}\left[ { \,^{\psi }\mathfrak {J}\,_{b^{-}}^{\beta }}F(a)+{\,^{\psi }\mathfrak {J} \,_{a^{+}}^{\beta }}F(b)\right] -\displaystyle f \left( \frac{a+b}{2}\right) \Bigg | \nonumber \\&\quad \le \displaystyle \frac{(b-a)\,C_{n}}{4}\left( 1+\displaystyle \frac{1}{ \left( \psi (b)-\psi (a)\right) ^{\beta }}\int _{0}^{1}\Big |A_{\psi ,\beta }(s)\Big |ds\right) \Big [|f^{\prime }(a)|+|f^{\prime }(b)|\Big ], \end{aligned}$$

(6.13)

where $\Omega (\psi ,\beta )=\left( \psi (b)-\psi (a)\right) ^{\beta }$ and $C_{n}=\frac{2}{n}\sum \nolimits _{k=1}^{n}\left( \frac{1}{2}\right) ^{\frac{1}{k} }$.

Remark 3

If we choose $\psi (\tau )=\tau$, $\psi (\tau )=\ln \tau$ and $\psi (\tau )=\frac{ \tau ^{\rho }}{\rho }$, respectively, in the corollaries 13, 14, 15 and 16, we establish midpoint type inequalities for the convex function, the function P, s-convex functions, and polynomial n-fractional convex functions, respectively, involving the Riemann-Liouville fractional operator, the Hadamard fractional operator, and the Katugompola fractional operators, respectively.

Theorem 6.2

Let $\ p>1$, $\frac{1}{p^{\prime }}+\frac{1}{p}=1$ and h be a B-function. Assume that $\beta ,\psi$ are defined as in Lemma 6.1. If $\left| f^{\prime }\right| ^{p}$ is a h-convex mapping on [a, b], we get the following midpoint type inequality

$$\begin{aligned}&\Bigg |\displaystyle \frac{\Gamma (\beta +1)}{4\,\Omega (\psi ,\beta )}\left[ { \,^{\psi }\mathfrak {J}\,_{b^{-}}^{\beta }}F(a)+{\,^{\psi }\mathfrak {J} \,_{a^{+}}^{\beta }}F(b)\right] -\displaystyle f\left( \frac{a+b}{2}\right) \Bigg | \nonumber \\&\quad \le \displaystyle \frac{(b-a)\left( 2\, h\left( \frac{1}{2}\right) \right) ^{ \frac{1}{p}}}{2}\left( 1+\frac{1}{\Omega \,^{p^{\prime }}(\psi ,\beta )} \int _{0}^{1}\big |A_{\psi ,\beta }(s)\big |^{p^{\prime }}ds\right) ^{\frac{1}{ p^{\prime }}}\left[ \big |f^{\prime }(a)\big |^{p}+\big |f^{\prime }(b)\big |^{p} \right] ^{\frac{1}{p}} \nonumber \\&\quad \le \displaystyle \frac{(b-a)\left( 2\, h\left( \frac{1}{2}\right) \right) ^{ \frac{1}{p}}}{2}\left( 1+\frac{1}{\Omega \,^{p^{\prime }}(\psi ,\beta )} \int _{0}^{1}\big |A_{\psi ,\beta }(s)\big |^{p^{\prime }}ds\right) ^{\frac{1}{ p^{\prime }}}\left( \Big |f^{\prime }(a)\Big |+\Big |f^{\prime }(b)\Big |\right) . \end{aligned}$$

(6.14)

Proof

Apply Lemma 5.1 and the absolute value, we get

$$\begin{aligned}&\Bigg |\displaystyle \frac{\Gamma (\beta +1)}{4\,\Omega (\psi ,\beta )}\left[ { \,^{\psi }\mathfrak {J}\,_{b^{-}}^{\beta }}F(a)+{\,^{\psi }\mathfrak {J} \,_{a^{+}}^{\beta }}F(b)\right] -\displaystyle f\left( \frac{a+b}{2}\right) \Bigg | \\&\quad \le \displaystyle \frac{b-a}{4\,\Omega (\psi ,\beta )}\int _{0}^{1}\Big | A_{\psi ,\beta }(s)\Big |\left[ \Big |f^{\prime }\big (sa+(1-s)b\big )\Big |+\Big | f^{\prime }\big ((1-s)a+sb\big )\Big |\right] ds \\&\qquad \; +\displaystyle \frac{b-a}{2}\int _{\frac{1}{2}}^{1}\left[ \Big |f^{\prime }\big ( sa+(1-s)b\big )\Big |+\Big |f^{\prime }\big ((1-s)a+sb\big )\Big |\right] ds. \end{aligned}$$

Using Hölder inequality and $\displaystyle A^{\frac{1}{p}}+B^{\frac{1}{p} }\le 2^{1-\frac{1}{p}}(A+B)^{\frac{1}{p}}$, we get

$$\begin{aligned}&\Bigg |\displaystyle \frac{\Gamma (\beta +1)}{4\,\Omega (\psi ,\beta )}\left[ { \,^{\psi }\mathfrak {J}\,_{b^{-}}^{\beta }}F(a)+{\,^{\psi }\mathfrak {J} \,_{a^{+}}^{\beta }}F(b)\right] -\displaystyle f\left( \frac{a+b}{2}\right) \Bigg | \\&\quad \le \displaystyle \frac{b-a}{4\,\Omega (\psi ,\beta )}\left( \int _{0}^{1} \big |A_{\psi ,\beta }(s)\big |^{p^{\prime }}ds\right) ^{\frac{1}{p^{\prime }}}\\&\qquad \left[ \left( \int _{0}^{1}\Big |f^{\prime }\left( sa+(1-s)b\right) \Big | ^{p}ds\right) ^{\frac{1}{p}}+\left( \int _{0}^{1}\Big |f^{\prime }\left( (1-s)a+sb\right) \Big |^{p}ds\right) ^{\frac{1}{p}}\right] \\&\qquad \; +\displaystyle \frac{b-a}{2}\left( \int _{\frac{1}{2}}^{1}ds\right) ^{\frac{1}{ p^{\prime }}}\\&\qquad \left[ \left( \int _{\frac{1}{2}}^{1}\Big |f^{\prime }\left( sa+(1-s)b\right) \Big |^{p}ds\right) ^{\frac{1}{p}}+\left( \int _{\frac{1}{2} }^{1}\Big |f^{\prime }\left( (1-s)a+sb\right) \Big |^{p}ds\right) ^{\frac{1}{p} }\right] \\&\quad \le \displaystyle \frac{b-a}{4\,\Omega (\psi ,\beta )}\left( 2\,\int _{0}^{1} \big |A_{\psi ,\beta }(s)\big |^{p^{\prime }}ds\right) ^{\frac{1}{p^{\prime }}}\\&\qquad \left[ \int _{0}^{1}\Big |f^{\prime }\left( sa+(1-s)b\right) \Big | ^{p}ds+\int _{0}^{1}\Big |f^{\prime }\left( (1-s)a+sb\right) \Big |^{p}ds\right] ^{\frac{1}{p}} \\&\qquad \; +\displaystyle \frac{b-a}{2}\left[ \int _{\frac{1}{2}}^{1}\Big |f^{\prime }\left( sa+(1-s)b\right) \Big |^{p}ds+\int _{\frac{1}{2}}^{1}\Big |f^{\prime }\left( (1-s)a+sb\right) \Big |^{p}ds\right] ^{\frac{1}{p}}. \end{aligned}$$

Given that $\left| f^{\prime }\right| ^{p}$ is a h-convex and h is B-function, we obtain

$$\begin{aligned}&\Bigg |\displaystyle \frac{\Gamma (\beta +1)}{4\,\Omega (\psi ,\beta )}\left[ { \,^{\psi }\mathfrak {J}\,_{b^{-}}^{\beta }}F(a)+{\,^{\psi }\mathfrak {J} \,_{a^{+}}^{\beta }}F(b)\right] -\displaystyle f\left( \frac{a+b}{2}\right) \Bigg | \\&\quad \le \displaystyle \frac{b-a}{4\,\Omega (\psi ,\beta )}\left( 2\,\int _{0}^{1} \big |A_{\psi ,\beta }(s)\big |^{p^{\prime }}ds\right) ^{\frac{1}{p^{\prime }}}\\&\qquad \left[ \int _{0}^{1}\left( h(s)+h(1-s)\right) \left[ \big |f^{\prime }(a)\big | ^{p}+\big |f^{\prime }(b)\big |^{p}\right] ds\right] ^{\frac{1}{p}} \\&\qquad +\displaystyle \frac{b-a}{2}\left[ \int _{\frac{1}{2}}^{1}\left( h(s)+h(1-s)\right) \left[ \big |f^{\prime }(a)\big |^{p}+\big |f^{\prime }(b) \big |^{p}\right] ds\right] ^{\frac{1}{p}} \\&\quad \le \displaystyle \frac{b-a}{4\,\Omega (\psi ,\beta )}\left( 2\,\int _{0}^{1} \big |A_{\psi ,\beta }(s)\big |^{p^{\prime }}ds\right) ^{\frac{1}{p^{\prime }}} \left[ 2\,h\left( \frac{1}{2}\right) \left[ \big |f^{\prime }(a)\big |^{p}+ \big |f^{\prime }(b)\big |^{p}\right] \right] ^{\frac{1}{p}} \\&\qquad +\displaystyle \frac{b-a}{2}\left[ h\left( \frac{1}{2}\right) \left[ \big | f^{\prime }(a)\big |^{p}+\big |f^{\prime }(b)\big |^{p}\right] \right] ^{\frac{1 }{p}} \\&\quad =\displaystyle \frac{(b-a)\left( h\left( \frac{1}{2}\right) \right) ^{\frac{1 }{p}}}{2}\\&\qquad \left[ 1+\left( \frac{1}{\Omega \,^{p^{\prime }}(\psi ,\beta )} \int _{0}^{1}\big |A_{\psi ,\beta }(s)\big |^{p^{\prime }}ds\right) ^{\frac{1}{ p^{\prime }}}\right] \left[ \big |f^{\prime }(a)\big |^{p}+\big |f^{\prime }(b) \big |^{p}\right] ^{\frac{1}{p}} \\&\quad \le \displaystyle \frac{(b-a)\left( h\left( \frac{1}{2}\right) \right) ^{ \frac{1}{p}}}{2}2^{1-\frac{1}{p^{\prime }}}\\&\qquad \left( 1+\frac{1}{\Omega \,^{p^{\prime }}(\psi ,\beta )}\int _{0}^{1}\big |A_{\psi ,\beta }(s)\big | ^{p^{\prime }}ds\right) ^{\frac{1}{p^{\prime }}}\left[ \big |f^{\prime }(a) \big |^{p}+\big |f^{\prime }(b)\big |^{p}\right] ^{\frac{1}{p}} \\&\quad =\displaystyle \frac{(b-a)\left( 2\, h\left( \frac{1}{2}\right) \right) ^{ \frac{1}{p}}}{2}\left( 1+\frac{1}{\Omega \,^{p^{\prime }}(\psi ,\beta )} \int _{0}^{1}\big |A_{\psi ,\beta }(s)\big |^{p^{\prime }}ds\right) ^{\frac{1}{ p^{\prime }}}\left[ \big |f^{\prime }(a)\big |^{p}+\big |f^{\prime }(b)\big |^{p} \right] ^{\frac{1}{p}}. \end{aligned}$$

This accomplishes the first inequality in (6.14).

Notice that for $p>1$ and $A,B\ge 0$, we have $A^{p}+B^{p}\le (A+B)^{p}$, which yields to the second inequality in (6.14). $\square$