## 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 [ab] 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. (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. (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. (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(ab), 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(ab).

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

### Corollary 1

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

1. (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. (2)

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

3. (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. (4)

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

### Proof

1. (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. (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. (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. (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.

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 (ab), 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 [ab], 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 [ab], 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 [ab], 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 [ab], 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 [ab], 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 [ab], 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 [ab], 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 [ab], 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 [ab], 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 [ab], 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 [ab], 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 [ab], 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 [ab], 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 [ab], 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). $$\square$$