Fractional version of the Jensen-Mercer and Hermite-Jensen-Mercer type inequalities for strongly h-convex function

1.   Introduction

Convex functions and mathematical inequalities play vital role in the advancement of several fields of pure and applied sciences. Various inequalities for convex and related functions have been studied in recent decades, also their variants in fractional calculus are analyzed, see ^[1,2,3,4,5].

The employment of fractional integral operators for establishing the generalized versions of classical inequalities become a fashion in the study of mathematical inequalities. In this regard the most celebrated inequality called Hadamard inequality has been studied extensively for fractional integral operators. The presence of Mittag-Leffler function in fractional integral operators produces interesting fractional integral inequalities.

The purpose of this paper is to present new Hadamard type inequalities for $k$-fractional integrals of strongly $h$-convex functions.

Let $f \in[\phi, \varphi]$ then the Riemann-Liouville fractional integrals $J_{\phi+}^{\varepsilon} f$ and $J_{\varphi-}^{\varepsilon} f$ of order $\varepsilon$ with $\varepsilon > 0$, 0 $\leq\phi < \varphi$ are defined by ^[6]:

and

respectively, where $\Gamma(\varepsilon) = \int_{0}^{\infty}e^{- t } t ^{\phi-1}d t$ is the Gamma function and $\left(J_{\phi+}^{0}\right) f \left(v\right) = \left(J_{\varphi-}^{0}\right) f \left(v\right) = f \left(v\right).$

In ^[7], Jarad et al. defined the new fractional integral operator as follows:

and

Note that, by taking $\phi$ = 0 and $\varepsilon\rightarrow 0$, the new conformable fractional integral operator coincides with the generalized fractional integrals ^[8], by taking $\phi$ = 0 and $\varepsilon$ = 1, the (1.3) becomes Riemann-Liouville fractional operator (1.1). Furthermore, by taking $\varphi$ = 0 and $\varepsilon$ = 1, the (1.4) reduces to the Riemann-Liouville.

The generalized k-fractional conformable integrals ^[9] are defined as follows:

and

If k $>$ 0, then the k-Gamma function $\Gamma_{k}$ is defined as ^[10]:

If Re ($\varepsilon > 0$), then k-Gamma function in integral form is defined as

with $\varepsilon\Gamma_{k}(\varepsilon) = \Gamma_{k}(\varepsilon+k)$.

Let $f$ : $I^{\circ} \subseteq$ R $\rightarrow$ R be a convex function. Then

holds $\forall$ $\phi, \varphi\in I^{\circ}$ with $\phi\neq\varphi$ and is known as Hermite-Hadamard inequality. The above inequality is reversed, if $f$ is a concave function on $I^{\circ}$, see ^{[11,12,13,14,15]}.

In ^[16], the following generalization of Hermite-Hadamard inequality for strongly $h$-convex function is established:

Let $f : I^{\circ} \subseteq R \rightarrow R$ is Lebesgue integrable and strongly $h$-convex function with modulus c $> 0$. If $h$ : (0, 1)$\rightarrow (0, \infty)$ be a given function. Then

holds $\forall$ u, v $\in I^{\circ}$, u $<$ v.

Some further interesting extensions and refinements of the Hermite-Hadamard inequalities for strongly $h$-convex mappings have been widely studied in the literature, see ^[17]. The other well known inequality is Jensen inequality and different variants of Jensen-Mercer inequalities are present in literature, see ^{[18,19,20,21]}.

In this paper we find further versions of generalized Hadamard type fractional integral inequality for $k$-fractional integrals. For this purpose we utilize the definition of $h$-convex function.

2.   Definitions

Next, we recall some basic definitions which will be used in the present article:

The gamma function is denoted by $\Gamma(.)$ and is defined as follows:

The beta function is denoted by $\beta$ and is defined as follows:

where $x, y, q$ are positive real numbers.

The $L_{1}[a, b]$ is a class of all real-valued functions integrable in $[a, b]$.

Further, we give definitions of convex and related functions which are useful to understand the inequalities established in this work.

Definition 2.1. ^[23] A function $f: [a, b]\rightarrow \mathbb{R}$ is said to be convex if

holds. If the inequality in (2.1) is reversed, then $f$ is called concave function.

Definition 2.2. ^[24,25] Let $I\subset (0, \infty)$ be an interval and $p\in \mathbb{R}\setminus\{0\}$. Then a function $f: I\rightarrow \mathbb{R}$ is said to be $p$-convex, if

holds for $a, b\in I$ and $t\in [0, 1]$. If the inequality in (2.2) is reversed, then $f$ is called $p$-concave function.

Definition 2.3. ^[23] Let $I, J$ be intervals in $\mathbb{R}$, $(0, 1)\subseteq J$ and $h: J\rightarrow \mathbb{R}$ be a non-negative function and $h\neq 0$. A function $f: I\rightarrow \mathbb{R}$ is said to be an $h$-convex, if

holds for $x, y\in I$ and $t\in (0, 1)$. If the inequality in (2.3) is reversed, then $f$ is called $h$-concave function.

Definition 2.4. ^[26] Let $h: J\supseteq (0, 1)\rightarrow \mathbb{R}$ be a positive function. Let $I\subset (0, \infty)$ be an interval and $p\in \mathbb{R}\setminus\{0\}$. A function $f: I\rightarrow \mathbb{R}$ is said to be ($p$, $h$)-convex, if

holds for $a, b\in I$ and $t\in [0, 1]$. If the inequality in (2.4) is reversed, then $f$ is called ($p$, $h$)-concave function.

Definition 2.5. A function h : $I^{\circ}\subseteq R \rightarrow$ R is called superadditive function if

$\forall$ u, v $\in I^{\circ}$.

Definition 2.6. ^[27] Let $h: J\supseteq (0, 1)\rightarrow \mathbb{R}$ be a positive function and $I\subset (0, \infty)$ be an interval. A function $f: I^{\circ} \subseteq R \rightarrow$ R is said to be strongly convex function with modulus c, u, v $\in I^{\circ}$ and $t\in\left[0, 1\right]$ if the following inequality holds

Definition 2.7. ^[16] Let $h: J\supseteq (0, 1)\rightarrow \mathbb{R}$ be a positive function and $I\subset (0, \infty)$ be an interval. A function $f: I^{\circ} \subseteq R \rightarrow$ R is said to be strongly convex function with modulus c, u, v $\in I^{\circ}$ and $t\in\left[0, 1\right]$ if the following inequality holds

3.   Main results

3.1.   Jensen-Mercer type inequality

Our main purpose of this section is to prove the Jensen-Mercer inequality for strongly $h$-convex function. To present Jensen-Mercer inequality first we will prove the following lemma:

Let $0 < u_{1} \leq u_{2} \leq...\leq u_{n}$ be real numbers and let $t _{j} (1 \leq j \leq n)$ be positive weights associated with these $u_{j}$ and $\sum_{J = 1}^{n} u_{j}$ = 1. let $h : I^{\circ} \rightarrow$ R be a non negative function defined over an interval $I^{\circ} \subset R$ such that $(0, 1)\in I^{\circ}$.

Lemma 3.1. Let $f : I^{\circ} \rightarrow$ R be a strongly h-convex function with modulus c $>$ 0, then

for any $u_{j}\in I^{\circ}, (1 \leq j \leq n)$ and for all $t \in(0, 1)$, where $M^{'} = \sup \{h(t) : t \in(0, 1)\}$.

Proof. Write $v_{j} = u_{1} + u_{n} - u_{j}$. Then $u_{1}+u_{n} = u_{j}+v_{j}$ so that the pairs $u_{1}$, $u_{n}$ and $u_{j}$, $v_{j}$ posses the same mid point. Since that is the case, there exist a $t$ such that

where $0 \leq t \leq 1$ and $1\leq j \leq n$. Then, applying strongly h-convexity of $f$

where $M^{'} = \sup\{h(t) : t \in(0, 1)\}$ and since $v_{j} = u_{1}+u_{n}-u_{j}$, then

This completes the proof.

Remark 3.2. 1. Taking $M^{'}$ = sup $\{h(t) : t \in(0, 1)\}$ = 1 and $c$ = 0 in Lemma 3.1, we get Lemma 1.3 of ^[20].

2. Taking $c$ = 0 in Lemma 3.1, we get Lemma 3.1 of ^[18].

3. Taking $M^{'}$ = sup $\{h(t) : t \in(0, 1)\}$ = 1 in Lemma 3.1, we get Lemma 2.2 of ^[19].

Theorem 3.3. Let $f : I^{\circ} \subset R \rightarrow R$ be a strongly h-convex function with modulus $c > 0$ and $0 < u_{1}\leq u_{2} \leq...\leq u_{n}$ be real numbers in $I^{\circ}$. If $t _{j}(1 \leq j \leq n)$ be positive numbers such that $\sum_{j = 1}^{n}t_j = 1$ and $\bar{u} = \sum_{j = 1}^{n} t _{j}u_{j}$. Then

where $M^{'} = \sup \{h(t) : t \in(0, 1)\}$.

Proof. Since $\sum_{j = 1}^{n} t _{j}$ = 1 we have

By using Lemma 3.1, we have

This completes the proof.

Remark 3.4. 1. Taking $c = 0$ in Theorem 3.3, we get Theorem 3.2 of ^[18].

2. Taking $h(t) = t$ and $M^{'}$ = sup $\{ h(t) : t \in(0, 1)\}$ = 1 in Theorem 3.3, we get Theorem 2.2 of ^[19].

3. Taking $c = 0$, $h(t) = t$ and $M^{'}$ = sup $\{ h(t) : t \in(0, 1)\}$ = 1 in Theorem 3.3, we get Theorem 1.2 of ^[20]).

3.2.   Hermite-Jensen-Mercer type inequalities

Theorem 3.5. Assume $f : I^{\circ} \subseteq R \rightarrow$ R be a strongly h-convex function with modulus c $>$ 0 and $\phi, \varphi\in I^{\circ}$ with $\phi < \varphi$. If h is a super-additive function and take $\varepsilon, \delta >$ 0. Then

holds $\forall$ u, v $\in[\phi, \varphi]$, where $M^{'}$ = sup $\left\lbrace h\left(t \right) : t \in \left(0, 1 \right) \right\rbrace$.

Proof. Since $f$ is strongly h-convex function then we write

$\forall$ $u_{1}, v_{1} \in [\phi, \varphi]$.

If we take $u_{1} = \frac{ t }{2}u + \frac{2- t }{2} v$ and $v_{1} = \frac{2- t }{2} u + \frac{ t }{2} v$. Then for u, v $\in[\phi, \varphi]$ and $t \in[0, 1]$, we have

Multiplying (3.4) by $\left(\frac{1-(1- t)^{\varepsilon}}{\gamma} \right)^{\frac{\delta}{k}-1} (1- t)^{\varepsilon-1}$ and integrating the resulting inequality with respect to $t$ over [0, 1] and then combining the obtained inequality with the definition of integral operator gives

Note that

By simple calculations, we have

which completes the proof of first inequality of (3.3).

For second part of inequality (3.3) take the definition of strongly h-convex function and by similar discussion, we get

and

Adding the above two inequalities give

By using super-additivity of function, we have

Multiplying (3.7) by $\left(\frac{1-(1- t)^{\gamma}}{\gamma} \right)^{\frac{\delta}{k}-1} (1- t)^{\gamma-1}$ and then integrating the resulting inequality with respect to $t$ over [0, 1], we obtain

By computing the above integrals, we have

that is,

Multiplying both sides of (3.8) by $h\left(\frac{1}{2}\right)$ and then subtract $c\left(\frac{1}{4}\right) \left(\frac{\delta}{k}\beta\left(\frac{\delta}{k}, \frac{2}{\varepsilon}+1 \right)\right) (u-v)^{2}$ leads to the conclusion that

Combining (3.6) and (3.9) lead to (3.3).

Remark 3.6. 1. Taking $h\left(t \right)$ = $t$, $M^{'}$ = sup $\left\lbrace h\left(t \right) : t \in \left(0, 1 \right) \right\rbrace$ = 1 and c = 0 in Theorem 3.5, we get Theorem 2.1 of ^[28].

2. Taking $h\left(t \right)$ = $t$, $M^{'}$ = sup $\left\lbrace h\left(t \right) : t \in \left(0, 1 \right) \right\rbrace$ = 1, c = 0, k = 1, u = $\phi$, and v = $\varphi$ in Theorem 3.5, we get Theorem 2.1 of ^[29].

3. Taking $\left(t \right)$ = $t$, $M^{'}$ = sup $\left\lbrace h\left(t \right) : t \in \left(0, 1 \right) \right\rbrace$ = 1, c = 0, $\varepsilon = k = 1$, $u = \phi$ and $v = \varphi$ in Theorem 3.5, we get Theorem 2 of ^[30].

Theorem 3.7. Let $f : I^{\circ} \subseteq R \rightarrow$ R be a strongly h-convex function with modulus c $>$ 0 and $\phi, \varphi\in I^{\circ}$ with $f < \varphi$. If h is a super-additive function and take $\varepsilon, \delta >$ 0. Then

and

holds $\forall$ u, v $\in [\phi, \varphi]$, where M $^{'}$ = sup $\left\lbrace h\left(t \right) : t \in \left(0, 1 \right) \right\rbrace$.

Proof. Employing Jensen-Mercer inequality (3.3), we have

$\forall$ $u_{1}, v_{1}\in[\phi, ]$.

By changing variables $u_{1} = t u + (1- t)v$ and $v_{1} = (1- t)u + t v$ for all u, v $\in [\phi, \varphi]$ and $t \in[0, 1]$ in (3.12), we get

Multiplying (3.13) by $\left(\frac{1-(1- t)^{\varepsilon}}{\varepsilon}\right) ^{\frac{\delta}{k}-1}(1- t)^{\varepsilon-1}$ and integrating the obtained inequality with respect to $t$ over [0, 1] gives

that is,

this completes the proof of first inequality of (3.10).

Now for second part of inequality (3.10) take the definition of strongly h-convex function, we have

Multiplying (3.15) by $\left(\frac{1-(1- t)^{\varepsilon}}{\varepsilon} \right)^{\frac{\delta}{k}-1} (1- t)^{\varepsilon-1}$ and then integrating the resulting inequality with respect to $t$ over [0, 1], we obtain

By some calculations, we get

Multiplying (3.16) by $h(1),$ we have

Adding $h\left(\frac{1}{2}\right) 2M^{'} \left[ f (\phi)+ f (\varphi)\right]$ to both sides of $(3.17)$, we get

Adding $-c\Bigg[\frac{1}{2}h\left(\frac{1}{2}\right) \left(1- \frac{\delta}{k}\beta\left(\frac{\delta}{k}, \frac{2}{\varepsilon}+1\right) \right)\left(1+4\frac{\delta}{k}\left(\beta\left(\frac{\delta}{k}, \frac{2}{\varepsilon}+1\right) -\beta\left(\frac{\delta}{k}, \frac{1}{\varepsilon}+1\right)\right) \right) (v-u)^{2}+\frac{1}{4}\left(\frac{\delta}{k}\beta\left(\frac{\delta}{k}, \frac{2}{\varepsilon}+1\right) \right)\left(1+4\frac{\delta}{k}\left(\beta\left(\frac{\delta}{k}, \frac{2}{\varepsilon}+1\right) -\beta\left(\frac{\delta}{k}, \frac{1}{\varepsilon}+1\right)\right) \right) (v-u)^{2}\Bigg]$ to both sides of above inequality, we have

Combining $(3.14)$ and $(3.18)$ yields $(3.10)$.

Next, we prove inequality $(3.11)$ by using strong h-convexity of $f$, we have

for all $u_{1}, v_{1}\in[\phi, \varphi]$.

For $u_{1} = t u+ (1- t)v$ and $v_{1} = (1- t)u+ t v$. Then $(3.19)$ leads to the conclusion that

Multiplying $(3.20)$ by $\left(\frac{1-(1- t)^{\varepsilon}}{\varepsilon}\right) ^{\frac{\delta}{k}-1}(1- t)^{\varepsilon-1}$ and then integrating the resulting inequality with respect to $t$ over [0, 1], we have

By computing the above integrals, we get

Employing strong h-convexity of $f$ gives

and

Adding above two inequalities and employing supper-additivity of function, we have

Now employing the Jensen-Mercer inequality, we get

Multiplying $(3.22)$ by $\left(\frac{1-(1- t)^{\varepsilon}}{\varepsilon}\right) ^{\frac{\delta}{k}-1}(1- t)^{\varepsilon-1}$ and integrate the resulting inequality with respect to $t$ over [0, 1], we have

Adding $-c\left(1+4\frac{\delta}{k}\left(\beta\left(\frac{\delta}{k}, \frac{2}{\varepsilon}+1\right) -\beta\left(\frac{\delta}{k}, \frac{1}{\varepsilon}+1\right) \right) \right)(u-v)^{2}$ to both sides of above inequality and combining the result with $(3.21)$ yields $(3.11)$.

Remark 3.8. 1. Taking $h\left(t \right)$ = $t$, c = 0 and $M^{'}$ = sup $\left\lbrace h\left(t \right) : t \in \left(0, 1 \right) \right\rbrace$ = 1 in Theorem 3.7, we get Theorem 2.3 of ^[28].

2. Taking $\varepsilon = \delta = k = 1$, c = 0, $h\left(t \right)$ = $t$ and $M^{'}$ = sup $\left\lbrace h\left(t \right) : t \in \left(0, 1 \right) \right\rbrace$ = 1 in Theorem $3.7$, we get Theorem 2.1 of ^[31].

Lemma 3.9. ^[28] Let $\varepsilon, \delta > 0, \phi < \varphi$ and $f : [\phi, \varphi] \rightarrow$ R be a differentiable mapping such that $f ^{'}\in L[\phi, \varphi]$. Then the inequality

holds for all $u, v \in[\phi, \varphi]$.

Theorem 3.10. Let $f : I^{\circ} \subseteq R \rightarrow$ R ba a differentiable function on $I^{\circ}$, $\phi, \varphi \in I^{\circ}$ with $\phi < \varphi$ and $f ^{'}\in L[\phi, \varphi]$. If $\lvert f ^{'}\rvert$ is a strongly $h$-convex function with modulus $c >$ 0 on $[\phi, \varphi]$ and take $\varepsilon, \delta >$ 0. Then

holds $\forall$ $u, v \in[\phi, \varphi]$, where $\beta_{1}\left(h, \dfrac{2- t }{2}\right) = \int_{0}^{1}\left(\dfrac{1-(1- t)^{\varepsilon}}{\varepsilon}\right)^{\frac{\delta}{k}}h\left(\dfrac{2- t }{2}\right) d t,$ $\beta_{2}\left(h, \dfrac{ t }{2}\right) = \int_{0}^{1}\left(\dfrac{1-(1- t)^{\varepsilon}}{\varepsilon}\right)^{\frac{\delta}{k}}h\left(\dfrac{ t }{2}\right) d t$ and $M ^{'} = \sup \left\lbrace h\left(t \right) : t \in \left(0, 1 \right) \right\rbrace$.

Proof. Employing Lemma $3.9$, inequality $(3.2)$ and definition of strong h-convexity of $\lvert f ^{'}\rvert$ yield the desired result.

Remark 3.11. 1. Taking $h\left(t \right)$ = $t$, $M^{'}$ = sup $\left\lbrace h\left(t \right) : t \in \left(0, 1 \right) \right\rbrace$ = 1 and c = 0 in Theorem $3.10$, we get Theorem 2.10 of ^[28].

2. Taking $h\left(t \right)$ = $t$, $M^{'}$ = sup$\left\lbrace h\left(t \right) : t \in \left(0, 1 \right) \right\rbrace$ = 1, c = 0, k = 1, u = $\phi$ and v = $\varphi$ in Theorem 3.10, we get Theorem 3.1 of ^[29].

3. Taking $\varepsilon = k = 1$, $h\left(t \right)$ = $t$, $M^{'}$ = sup $\left\lbrace h\left(t \right) : t \in \left(0, 1 \right) \right\rbrace$ = 1, c = 0, u = $\phi$ and v = $\varphi$ in Theorem 3.10, we get Theorem 5 of ^[32] in the case of q = 1.

Theorem 3.12. Let $f : I^{\circ} \subseteq R \rightarrow$ R be a differentiable function on $I^{\circ}$, $\phi, \varphi \in I^{\circ}$ with $\phi < \varphi$ and $f ^{'} \in L [\phi, \varphi]$. If $\lvert f ^{'}\rvert^{q}$ is a strongly $h$-convex function with modulus $c >$ 0 on $[\phi, \varphi]$ for $q >$ 1 and take $\varepsilon, \delta >$ 0. Then

holds $\forall$ u, v $\in[\phi, \varphi]$, where $\beta_{1}\left(h, \dfrac{2- t }{2}\right),$ $\beta_{2}\left(h, \dfrac{ t }{2}\right)$ and $M^{'}$ are same as in Theorem 3.10.

Proof. Employing Lemma $3.9$, inequality $(3.2)$, power-mean inequality and definition of strongly $h$-convex function of $\lvert f ^{'}\rvert$ yield the desired result.

Remark 3.13. Taking $h\left(t \right)$ = $t$, $M^{'}$ = sup $\left\lbrace h\left(t \right) : t \in \left(0, 1 \right) \right\rbrace$ = 1 and c = 0 in Theorem $3.12$, we get Theorem 2.12 of ^[28].

Theorem 3.14. Let $f : I^{\circ} \subseteq R \rightarrow$ R be a differentiable function on $I^{\circ}$, $\phi, \varphi \in I^{\circ}$ with $\phi < \varphi$ and $f ^{'} \in L [\phi, \varphi]$. If $\lvert f ^{'}\rvert^{q}$ is a strongly $h$-convex function with modulus $c >$ 0 on $[\phi, \varphi]$ for $p, q >$ 1 with $\frac{1}{p}+\frac{1}{q} = 1$ and take $\varepsilon, \delta >$ 0. Then

holds $\forall$ $u, v \in[\phi, \varphi]$, where $C_{1}\left(h, \dfrac{2- t }{2}\right) = \int_{0}^{1}h\left(\dfrac{2- t }{2}\right) d t,$ $C_{2}\left(h, \dfrac{ t }{2}\right) = \int_{0}^{1}h\left(\dfrac{ t }{2}\right) d t$ and $M^{'} = \sup \left\lbrace h\left(t \right) : t \in \left(0, 1 \right) \right\rbrace$.

Proof. From applying Lemma $3.9$, inequality $3.2$, Hölder integral inequalities and definition of strongly h-convex function simultaneously give the desired result.

Remark 3.15. Taking $h\left(t \right)$ = $t$, $M^{'}$ = sup $\left\lbrace h\left(t \right) : t \in \left(0, 1 \right) \right\rbrace$ = 1 and c = 0 in Theorem $3.14$, we get Theorem 2.14 of ^[28].

Theorem 3.16. Assume $f : I^{\circ} \subseteq R \rightarrow$ R be a differentiable function on $I^{\circ}$, $\phi, \varphi \in I^{\circ}$ with $\phi < \varphi$ and $f ^{'} \in L [\phi, \varphi]$. If $\lvert f ^{'}\rvert^{q}$ is a strongly h-convex function with modulus $c >$ 0 on $[\phi, \varphi]$ for $p, q >$ 1 with $\frac{1}{p}+\frac{1}{q} = 1$ and take $\varepsilon, \delta >$ 0. Then

holds $\forall$ $u, v \in[\phi, \varphi]$, where $\beta_{1}\left(h, \dfrac{2- t }{2}\right),$ $\beta_{2}\left(h, \dfrac{ t }{2}\right)$ and $M^{'}$ are same as in Theorem 3.10.

Proof. Using Lemma $3.9$, inequality $(3.2)$, Hölder integral inequalities and definition of strongly h-convex function, we obtain our desired result.

Remark 3.17. Taking $h\left(t \right)$ = $t$, $M^{'}$ = sup $\left\lbrace h\left(t \right) : t \in \left(0, 1 \right) \right\rbrace$ = 1 and c = 0 in Theorem $3.16$, we get Theorem 2.16 of ^[28].

3.3.   New inequalities in the sense of improved Hölder inequality

Theorem 3.18. Let $f :I^{\circ} \subseteq R \rightarrow$ R be a differentiable function on $I^{\circ}$, $\phi, \varphi \in I^{\circ}$ with $\phi < \varphi$ and $f ^{'} \in L [\phi, \varphi]$. If $\lvert f ^{'}\rvert^{q}$ is a strongly $h$-convex function with modulus $c >$ 0 on $[\phi, \varphi]$ for $p, q >$ 1 with $\frac{1}{p}+\frac{1}{q} = 1$ and take $\varepsilon, \delta >$ 0. Then

holds for all u, v $\in [\phi, \varphi]$, where $\beta_{1}\left((1- t)h, \dfrac{2- t }{2}\right) = \int_{0}^{1}(1- t) h\left(\dfrac{2- t }{2}\right) d t,$ $\beta_{2}\left((1- t)h, \dfrac{ t }{2}\right) = \int_{0}^{1}(1- t) h\left(\dfrac{ t }{2}\right) d t,$ $\beta_{3}\left(t h, \dfrac{2- t }{2}\right) = \int_{0}^{1} t h\left(\dfrac{2- t }{2}\right) d t,$ $\beta_{4}\left(t h, \dfrac{ t }{2}\right) = \int_{0}^{1} t h\left(\dfrac{ t }{2}\right) d t$ and $M^{'} = \sup \left\lbrace h\left(t \right) : t \in \left(0, 1 \right) \right\rbrace$.

Proof. Using Lemma $3.9$, inequality $(3.2)$, Hölder-Íscan integral inequality given in Theorem 1.4 of ^[33] and definition of strongly $h$-convex function of $\lvert f ^{'}\rvert^{q}$ yield the desired result.

Remark 3.19. Taking $h\left(t \right)$ = $t$, c = 0 and $M^{'} = \sup \left\lbrace h\left(t \right) : t \in \left(0, 1 \right) \right\rbrace$ = 1 in Theorem 3.18, we get Theorem 3.1 of ^[28].

Theorem 3.20. Let $f :I^{\circ} \subseteq R \rightarrow$ R be a differentiable function on $I^{\circ}$, $\phi, \varphi \in I^{\circ}$ with $\phi < \varphi$ and $f ^{'} \in L [\phi, \varphi]$. If $\lvert f ^{'}\rvert^{q}$ is a strongly $h$-convex function with modulus c on $[\phi, \varphi]$, where $p, q >$ 1 with $\frac{1}{p}+\frac{1}{q} = 1$ and take $\varepsilon, \delta >$ 0. Then

holds for all u, v $\in [\phi, \varphi]$, where $\beta_{1}\left((1- t)h, \dfrac{2- t }{2}\right) = \int_{0}^{1}(1- t)\left(\dfrac{1-(1- t)^{\varepsilon}}{\varepsilon}\right) ^{\frac{\delta}{k}} h\left(\dfrac{2- t }{2}\right) d t,$ $\beta_{2}\left((1- t)h, \dfrac{ t }{2}\right) = \int_{0}^{1}(1- t)\left(\dfrac{1-(1- t)^{\varepsilon}}{\varepsilon}\right) ^{\frac{\delta}{k}} h\left(\dfrac{ t }{2}\right) d t,$ $\beta_{3}\left(t h, \dfrac{2- t }{2}\right) = \int_{0}^{1} t \left(\dfrac{1-(1- t)^{\varepsilon}}{\varepsilon}\right) ^{\frac{\delta}{k}} h\left(\dfrac{2- t }{2}\right) d t,$ $\beta_{4}\left(t h, \dfrac{ t }{2}\right) = \int_{0}^{1} t \left(\dfrac{1-(1- t)^{\varepsilon}}{\varepsilon}\right) ^{\frac{\delta}{k}} h\left(\dfrac{ t }{2}\right) d t$ and $M^{'} = \sup \left\lbrace h\left(t \right) : t \in \left(0, 1 \right) \right\rbrace$.

Proof. Employing Lemma $3.9$, inequality $(3.2)$, improved power-mean integral inequality given in Theorem 1.5 of ^[33] and definition of strong h-convexity of $\lvert f ^{'}\rvert^{q}$ yield the required result.

Remark 3.21. Taking $h\left(t \right)$ = $t$, c = 0 and M = sup $\left\lbrace h\left(t \right) : t \in \left(0, 1 \right) \right\rbrace$ = 1 in Theorem 3.20, we get Theorem 3.2 of ^[28].

4.   Conclusions

The convex functions play an important role in applied mathematics as well as in optimization theory. In past years, researchers paid a huge attention to establish properties of different variants of convex functions. In this paper, we studied $h$-convex functions in the setting of $k$-fractional integrals. We established various important versions of Hermite-Hadamard type inequalities for $h$-convex. Our results generalize and extend many existing results in literature.

Acknowledgement

The study was supported by Fundamental Research Funds for the Central Universities (Grant No. 2020RC14) and Young teachers undergraduate teaching top talents Project of Shandong University of Science and Technology (No. BJRC20190508).

Conflict of interest

The author declares no conflict of interests.