Hadamard and Fejér-Hadamard inequalities for generalized $k$-fractional integrals involving further extension of Mittag-Leffler function

1.   Introduction

We first give the definitions of fractional integral operators and some theorems which establish the basis for the second and third part of the paper.

Definition 1. (see ^[1]) Let $w, \alpha, l, \gamma, c\in \mathbb{C}, \Re(\alpha), \Re(l) > 0, \Re(c) > \Re(\gamma) > 0$ with $\tilde{p}\geq 0, \mu, \delta > 0$ and $0 < v\leq \delta+\mu$. Let $f\in L_{1}[a, b]$ and $x\in [a, b]$. Then, the generalized fractional integral operators $F^{\gamma, \delta, v, c}_{\mu, \alpha, l, w, a+}f$ and $F^{\gamma, \delta, v, c}_{\mu, \alpha, l, w, b-}f$ are defined by:

where

is the extended generalized Mittag-Leffler function and $\beta_{\tilde{p}}$ is the extension of beta function is defined as follows:

where $\Re(x), \Re(y), \Re(\tilde{p}) > 0$.

Definition 2. (see ^[2]) Let $f, g:[a, b]\rightarrow \mathbb{R}, 0 < a < b$, be the functions such that $f$ be positive and $f\in L_{1}[a, b]$ and $g$ be differentiable and strictly increasing. Also let $\frac{\phi}{x}$ be an increasing function on $[a, \infty)$ and $w, \alpha, l, \gamma, c\in \mathbb{C}, \Re(\alpha), \Re(l) > 0, \Re(c) > \Re(\gamma) > 0$ with $\tilde{p}\geq 0, \mu, \delta > 0$ and $0 < v\leq \delta+\mu$. Then for $x\in [a, b]$ the fractional integral operators are defined by:

Definition 3. (see ^[3]) Let $f, g:[a, b]\rightarrow \mathbb{R}, 0 < a < b$, be the functions such that $f$ be positive and $f\in L_{1}[a, b]$ and $g$ be differentiable and strictly increasing. Also, let $w, \alpha, l, \gamma, c\in \mathbb{C}, \Re(\alpha), \Re(l) > 0, \Re(c) > \Re(\gamma) > 0$ with $\tilde{p}\geq 0, \mu, \delta > 0$ and $0 < v\leq \delta+\mu$. Then for $x\in [a, b]$ the unified integral operators are defined by:

Definition 4. (see ^[4]) A function $f:[a, b]\rightarrow \mathbb{R}$ is said to be convex, if

holds for all $x, y\in [a, b]$ and $t\in [0, 1]$.

Definition 5. (see ^[5]) Let $I\subset(0, \infty)$ be a real interval and $p\in\mathbb{R}\setminus\{0\}$. A function $f:I\rightarrow \mathbb{R}$ is said to be $p$-convex function, if

holds for all $x, y\in I$ and $t\in [0, 1]$.

Definition 6. (see ^[6]) Let $p\in\mathbb{R}\setminus\{0\}$. Then a function $f:[a, b]\subset (0, \infty)\rightarrow \mathbb{R}$ is said to be $p$-symmetric with respect to $\left[\frac{a^{p}+b^{p}}{2}\right]^{\frac{1}{p}}$, if

for $t\in [0, 1]$.

Following inequality is the well-known Hadamard inequality.

Theorem 1. (see ^[7]) Let $f:[a, b]\rightarrow \mathbb{R}$ be a convex function for $a < b$. Then the following inequality holds:

The Fejér-Hadamard inequality is a weighted version of the Hadamard inequality given by Fejér in ^[8].

Theorem 2. Let $f:[a, b]\rightarrow \mathbb{R}$ be a convex function and $g:[a, b]\rightarrow \mathbb{R}$ be non-negative, integrable and symmetric with respect to $\frac{a+b}{2}$. Then the following inequality holds:

The aim of this paper is to study the Hadamard and the Fejér-Hadamard inequalities for generalized $k$-fractional integrals containing Mittag-Leffler functions. In the subsequent section we deduce $k$-fractional integral operators containing Mittag-Leffler functions. In Section 3, we utilize these $k$-fractional integral operators to construct the $k$-fractional Hadamard and Fejér-Hadamard type inequalities.

2.   $k$-fractional integral operators containing Mittag-Leffler function

We define $k$-fractional integral operators containing Mittag-Leffler function by setting $\phi(x) = x^{\alpha/k}, \, \alpha > k > 0$ in (1.4). Here we consider all involved parameters as real numbers.

Definition 7. Let $f, g:[a, b]\rightarrow \mathbb{R}, 0 < a < b$, be the functions such that $f$ be positive and $f\in L_{1}[a, b]$ and $g$ be differentiable and strictly increasing. Also, let $\alpha > k > 0$ and $w, l, \gamma, c\in \mathbb{R}, c > \gamma > 0$ with $\tilde{p}\geq 0, \mu, \delta, l > 0$ and $0 < v\leq \delta+\mu$. Then for $x\in [a, b]$ the left and right generalized $k$-fractional integral operators $\left(_{g}^{k}F^{\gamma, \delta, v, c}_{\mu, \alpha, l, w, a+}f\right)$ and $\left(_{g}^{k}F^{\gamma, \delta, v, c}_{\mu, \alpha, l, w, b-}f\right)$ are defined by

and

Remark 1. The following $k$-fractional integrals can be deduced from (2.1) and (2.2).

(i) If we set $\tilde{p} = 0$ and $g(x) = x$ in Eqs (2.1) and (2.2), then we have

and

(ii) If we set $l = \delta = 1$ and $g(x) = x$ in Eqs (2.1) and (2.2), then we have

and

(iii) If we set $\tilde{p} = 0$, $l = \delta = 1$ and $g(x) = x$ in Eqs (2.1) and (2.2), then we have

and

(iv) If we set $\tilde{p} = 0$, $l = \delta = v = 1$ and $g(x) = x$ in Eqs (2.1) and (2.2), then we have

and

(v) For $\tilde{p} = w = 0$, $g(x) = x$ and $k = 1$ the Eqs (2.1) and (2.2) reduce to classical Riemann–Liouville fractional integral operators.

(vi) If we set $\gamma = \delta = l = v = 1$ and $w = \tilde{p} = 0$ in Eqs (2.1) and (2.2), Definition 1 of ^[9] is obtained.

Remark 2. Some more definitions of integral operators are composed as follows:

(i) In Remark 1 (i), if we take $w = 0$, (8) of ^[10] is obtained.

(ii) In Remark 1 (iii), if we take $\gamma = k = 1$, (2.2) of ^[11] is obtained.

(iii) In Remark 1 (iv), if we take $w = 0$ and $k = 1$, we get classical Riemann-Liouville fractional integral.

Remark 3.

(i) In Remark 2, if we take $w = 0$ and $c = 1$, Definition 4 of ^[12] is obtained.

3.   $k$-fractional integral inequalities of Hadamard and Fejér-Hadamard type for $p$-convex function

First, we give the following generalized $k$-fractional Hadamard inequality.

Theorem 3. Let $f, g:[a, b]\rightarrow \mathbb{R}$, $0 < a < b$, be the functions such that $f$ positive and $f\in L_{1}[a, b]$ and $g$ be differentiable and strictly increasing. If $f$ is $p$-convex, $p\in {\rm \mathbb{R}}\backslash \left\{0\right\}$, then the following inequalities for k-fractional integral operators (2.1) and (2.2) hold:

(i) If $p > 0$, then

where $\bar{w} = \frac{w}{\left(g^{p} \left(b\right)-g^{p} \left(a\right)\right)^{\mu } }$ and $\theta \left(t\right) = g^{\frac{1}{p}} \left(t\right)$ for all $t\in \left[a^{p}, b^{p} \right]$.

(ii) If $p < 0$, then

where $\bar{w} = \frac{w}{\left(g^{p} \left(a\right)-g^{p} \left(b\right)\right)^{\mu } }$ and $\theta \left(t\right) = g^{\frac{1}{p}} \left(t\right)$ for all $t\in \left[b^{p}, a^{p} \right]$.

Proof. We prove the assertion (i) as follows:

(i) Since $f$ is $p$-convex on $\left[a, b\right]$, for all $x, y\in I$, we have

Setting $g\left(x\right) = \left(tg^{p} \left(a\right)+(1-t)g^{p} \left(b\right)\right)^{\frac{1}{p}}$ and $g\left(y\right) = \left(tg^{p} \left(b\right)+(1-t)g^{p} \left(a\right)\right)^{\frac{1}{p}}$ in above inequality, we have

Multiplying both sides of (3.4) by $2t^{\frac{\alpha}{k}-1} E_{\mu, \alpha, l}^{\gamma, \delta, v, c} \left(wt^{\mu }; \tilde{p}\right)$ and then integrating over $\left[0, 1\right]$, we have

Setting $g\left(m\right) = tg^{p} \left(a\right)+(1-t)g^{p} \left(b\right)$ and $g\left(n\right) = tg^{p} \left(b\right)+(1-t)g^{p} \left(a\right)$ in (3.5), we have

by using k-fractional integral operators (2.1) and (2.2), the first inequality of (3.1) is obtained.

Now to prove the second inequality of (3.1), again from $p$-convexity of $f$ over $\left[0, 1\right]$ and for $t\in \left[0, 1\right]$, we have

Multiplying both sides of (3.7) by $t^{\frac{\alpha}{k}-1}E_{\mu, \alpha, l}^{\gamma, \delta, v, c}\left(wt^{\mu}; \tilde{p}\right)$ and then integrating over $\left[0, 1\right]$, we have

Setting $g\left(m\right) = tg^{p} \left(a\right)+(1-t)g^{p} \left(b\right)$ and $g\left(n\right) = tg^{p} \left(b\right)+(1-t)g^{p} \left(a\right)$ in (3.8), then by using k-fractional integral operators (2.1) and (2.2), the second inequality of (3.1) is obtained.

(ii) Proof is similar to the proof of (i).

Remark 4. By setting $\tilde{p} = w = 0$ in (3.1) and (3.2), Corollary 23 of ^[13] is obtained.

Remark 5. By using (3.1) and (3.2), some more $k$-fractional inequalities are presented as follows:

(i) By setting $\tilde{p} = w = 0$ and $g = I$, we get

(ii) By setting $\tilde{p} = 0{\rm, \; }p = -1$ and $g = I$, we get

(iii) By setting $g = I$ and $p = -1$, we get

(iv) By setting $w = \tilde{p} = 0; p = -1$ and $g = I$, we get

Corollary 1. By using (3.1) and (3.2), some more $k$-fractional inequalities are presented as follows:

(i) By setting $p = 1$, we get

(ii) By setting $p = -1$, we get

(iii) By setting $p = -2$, we get

Corollary 2. The aforementioned $k$-fractional inequalities are further connected with already known results as follows:

(i) By setting $k = 1$ in Remark 5 (i), Theorem 9 of ^[6] is obtained.

(ii) By setting $k = 1$ in Remark 5 (ii), Theorem 2.1 of ^[14] is obtained.

(iii) By setting $k = 1$ in Remark 5 (iii), Theorem 2.1 of ^[15] is obtained.

(iv) By setting $k = 1$ in Remark 5 (iv), Theorem 4 of ^[16] is obtained.

(v) By setting $k = 1$ in Remark 5 (v), Theorem 2.1 of ^[17] is obtained.

The following lemma is useful to present the Fejér-Hadamard inequality for generalized $k$-fractional integrals.

Lemma 1. Let $f, g:[a, b]\rightarrow \mathbb{R}$, $0 < a < b$, be the functions such that $f$ positive and $f\in L_{1} [a, b]$ and $g$ be differentiable and strictly increasing. If $f$ is $p$-convex, $p\in {\rm \mathbb{R}}\backslash \left\{0\right\}$ and $f\left(g^{\frac{1}{p}} \left(x\right)\right) = f\left(\left[g^{p} \left(a\right)+g^{p} \left(b\right)-g\left(x\right)\right]^{\frac{1}{p}} \right)$, then for generalized k-fractional integral operators (2.1) and (2.2), we have:

(i) If $p > 0$, then

with $\theta \left(t\right) = g^{\frac{1}{p}} \left(t\right)$ for all $t\in \left[a^{p}, b^{p} \right]$.

(ii) If $p < 0$, then

with $\theta \left(t\right) = g^{\frac{1}{p}} \left(t\right)$ for all $t\in \left[b^{p}, a^{p} \right]$.

Proof. We prove the assertion (i) as follows:

(i) By definition of generalized k-fractional integral operators (2.1) and (2.2), we have

Setting $g\left(t\right) = g^{p} \left(a\right)+g^{p} \left(b\right)-g\left(x\right)$ in the above equation and using $f\left(g^{\frac{1}{p}}\left(x\right)\right) = f\left(\left[g^{p}\left(a\right)+g^{p} \left(b\right)-g\left(x\right)\right]^{\frac{1}{p}}\right)$, we have:

This implies

By adding $\left({}_{g}^{k} F_{\mu, \alpha, l, \bar{w}, g^{-1} \left(g^{p} \left(a\right)\right)+}^{\gamma, \delta, v, c} f\circ \theta \right)\left(g^{-1} \left(g^{p} \left(b\right)\right); \tilde{p}\right)$ on both sides of (3.13), we have

From Eqs (3.13) and (3.14), the result (3.9) can be obtained.

(ii) Proof is similar to the proof of (i).

The first version of the Fejér-Hadamard inequality for generalized $k$-fractional integrals is given as follows:

Theorem 4. Let $f, g:[a, b]\rightarrow \mathbb{R}$, $0 < a < b$, be the functions such that $f$ positive and $f\in L_{1} [a, b]$ and $g$ be differentiable and strictly increasing, $h$ is a non-negative and integrable function. If $f$ is $p$-convex, $p\in {\rm R}\backslash \left\{0\right\}$ and $f\left(g^{\frac{1}{p}} \left(x\right)\right) = f\left(\left[g^{p} \left(a\right)+g^{p} \left(b\right)-g\left(x\right)\right]^{\frac{1}{p}} \right)$, then the following inequalities for generalized k-fractional integral operators (2.1) and (2.2) hold:

(i) If $p > 0$, then

where $\bar{w} = \frac{w}{\left(g^{p} \left(b\right)-g^{p} \left(a\right)\right)^{\mu } }$ and $\theta \left(t\right) = g^{\frac{1}{p}} \left(t\right)$ for all $t\in \left[a^{p}, b^{p} \right]$.

(ii) If $p < 0$, then

where $\bar{w} = \frac{w}{\left(g^{p} \left(a\right)-g^{p} \left(b\right)\right)^{\mu } }$ and $\theta \left(t\right) = g^{\frac{1}{p}} \left(t\right)$ for all $t\in \left[b^{p}, a^{p} \right]$.

Proof. We prove the assertion (i) as follows:

(i) Multiplying both sides of (3.4) by $2t^{\frac{\alpha}{k}-1} E_{\mu, \alpha, l}^{\gamma, \delta, v, c} \left(wt^{\mu }; \tilde{p}\right)h\left(\left[tg^{p} \left(a\right)+\left(1-t\right)g^{p} \left(b\right)\right]^{\frac{1}{p}} \right)$ and then integrating over $\left[0, 1\right]$, we have

Setting $g(x) = tg^{p}(a)+(1-t)g^{p}(b)$, that is, $g^{p}(a)+g^{p}(b)-g(x) = tg^{p}(b)+(1-t)g^{p}(a)$, in (3.17) and using $f\left(g^{\frac{1}{p}} \left(x\right)\right) = f\left(\left[g^{p} \left(a\right)+g^{p} \left(b\right)-g\left(x\right)\right]^{\frac{1}{p}} \right)$, we have

This implies

Using Lemma 3.1 (i) in above inequality, then first inequality of (3.15) is obtained.

Now to prove second inequality of (3.15), multiplying both sides of (3.7) by $t^{\frac{\alpha}{k}-1} E_{\mu, \alpha, l}^{\gamma, \delta, v, c} \left(wt^{\mu }; \tilde{p}\right) h\left(\left[tg^{p} \left(a\right)+\left(1-t\right)g^{p}\left(b\right)\right]^{\frac{1}{p}} \right)$ and then integrating over $\left[0, 1\right]$, we have

Setting $g\left(x\right) = tg^{p} \left(a\right)+\left(1-t\right)g^{p} \left(b\right)$ and using $f\left(g^{\frac{1}{p}} \left(x\right)\right) = f\left(\left[g^{p} \left(a\right)+g^{p} \left(b\right)-g\left(x\right)\right]^{\frac{1}{p}} \right)$ in (3.20), we have

Using Lemma 3.1 (i) in above inequality, then second inequality of (3.15) is obtained.

(ii) Proof is similar to the proof of (i).

Remark 6. By using (3.15) and (3.16), some more $k$-fractional inequalities are presented as follows:

(i) By setting $\tilde{p} = 0$ and $g = I$, we get

(ii) By setting $w = \tilde{p} = 0$ and $g = I$, we get

(iii) By setting $\tilde{p} = 0$, $h\left(x\right) = 1$, $p = -1$ and $g = I$, we get

(iv) By setting $g = I$, $h\left(x\right) = 1$ and $p = -1$, we get

(v) By setting $w = \tilde{p} = 0$, $h\left(x\right) = 1$, $p = -1$ and $g = I$, we get

Corollary 3. By using (3.15) and (3.16), some more $k$-fractional inequalities are presented as follows:

(i) By setting $p = 1$, we get

(ii) By setting $p = -1$, we get

(iii) By setting $p = -2$, we get

Corollary 4. The aforementioned $k$-fractional inequalities are further connected with already known results as follows:

(i) By setting $k = 1$ in Remark 6 (i), Theorem 2.2 of ^[18] is obtained.

(ii) By setting $k = 1$ in Remark 6 (ii), Theorem 9 of ^[6] is obtained.

(iii) By setting $k = 1$ in Remark 6 (iii), Theorem 2.1 of ^[14] is obtained.

(iv) By setting $k = 1$ in Remark 6 (iv), Theorem 2.1 of ^[15] is obtained.

(v) By setting $k = 1$ in Remark 6 (v), Theorem 4 of ^[16] is obtained.

(vi) By setting $k = 1$ in Remark 6 (vi), Theorem 2.5 of ^[17] is obtained.

In the next theorem we present another version of the Hadamard inequality.

Theorem 5. Let $f, g:[a, b]\rightarrow \mathbb{R}$, $0 < a < b$, be the functions such that $f$ positive and $f\in L_{1} [a, b]$ and $g$ be differentiable and strictly increasing. If $f$ is $p$-convex, $p\in {\rm \mathbb{R}}\backslash \left\{0\right\}$, then for generalized k-fractional integral operators (2.1) and (2.2), we have:

(i) If $p > 0$, then

where $\bar{\bar{w}} = \frac{2^{\mu } w}{\left(g^{p} \left(b\right)-g^{p} \left(a\right)\right)^{\mu } }$ and $\theta \left(t\right) = g^{\frac{1}{p}} \left(t\right)$ for all $t\in \left[a^{p}, b^{p} \right]$.

(ii) If $p < 0$, then

where $\bar{\bar{w}} = \frac{2^{\mu } w}{\left(g^{p} \left(b\right)-g^{p} \left(a\right)\right)^{\mu } }$ and $\theta \left(t\right) = g^{\frac{1}{p}} \left(t\right)$ for all $t\in \left[b^{p}, a^{p} \right]$.

Proof. We prove the assertion (i) as follows:

(i) Setting $g(x) = \big[\big(\frac{t}{2}\big)g^{p}(a)+\big(\frac{2-t}{2}\big)g^{p}(b)\big]^{\frac{1}{p}}$ and $g(y) = \big[\big(\frac{t}{2}\big)g^{p}(b)+\big(\frac{2-t}{2}\big)g^{p}(a)\big]^{\frac{1}{p}}$ in (3.3), we have

Multiplying both sides of (3.24) by $2t^{\frac{\alpha}{k}-1} E_{\mu, \alpha, l}^{\gamma, \delta, v, c} \left(wt^{\mu }; \tilde{p}\right)$ and then integrating over $\left[0, 1\right]$, we have

Setting $g(x) = \big[\big(\frac{t}{2}\big)g^{p}(a)+\big(\frac{2-t}{2}\big)g^{p}(b)\big]^{\frac{1}{p}}$ and $g(y) = \big[\big(\frac{t}{2}\big)g^{p}(b)+\big(\frac{2-t}{2}\big)g^{p}(a)\big]^{\frac{1}{p}}$ in (3.25), then by using k-fractional integral operators (2.1) and (2.2), the first inequality of (3.22) is obtained.

Now to prove the second inequality of (3.22), again from $p$-convexity of $f$ over $\left[a, b\right]$ and for $t\in \left[0, 1\right]$, we have

Multiplying both sides of (3.26) by $t^{\frac{\alpha}{k}-1} E_{\mu, \alpha, l}^{\gamma, \delta, v, c} \left(wt^{\mu }; \tilde{p}\right)$ and then integrating over $\left[0, 1\right]$, we have

Setting $g(m) = \big(\frac{t}{2}\big)g^{p}(a)+\big(\frac{2-t}{2}\big)g^{p}(b)$ and $g(n) = \big(\frac{t}{2}\big)g^{p}(b)+\big(\frac{2-t}{2}\big)g^{p}(a)$ in (3.27), then by using k-fractional integral operators (2.1) and (2.2), the second inequality of (3.22) is obtained.

(ii) Proof is similar to the proof of (i).

Remark 7. By using (3.22) and (3.23), some more $k$-fractional inequalities are presented as follows: (i) By setting $\tilde{p} = 0{\rm, \; }p = -1$ and $g = I$, we get

(ii) By setting $g = I$ and $p = -1$, we get

Corollary 5. By using (3.22) and (3.23), some more $k$-fractional inequalities are presented as follows:

(i) By setting $p = 1$, we get

(ii) By setting $p = -1$, we get

(iii) By setting $p = -2$, we get

Corollary 6. The aforementioned $k$-fractional inequalities are further connected with already known results as follows:

(i) By setting $k = 1$ in Remark 7 (i), Theorem 2.3 of ^[14] is obtained.

(ii) By setting $k = 1$ in Remark 7 (ii), Theorem 2.3 of ^[15] is obtained.

(iii) By setting $k = 1$ in Remark 7 (iii), Theorem 2.7 of ^[17] is obtained.

The second version of the Fejér-Hadamard inequality for generalized $k$-fractional integrals is given as follows:

Theorem 6. Let $f, g:[a, b]\rightarrow \mathbb{R}$, $0 < a < b$, be the functions such that $f$ positive and $f\in L_{1} [a, b]$ and $g$ be differentiable and strictly increasing, $h$ is a non-negative and integrable function. If $f$ is $p$-convex, $p\in {\rm \mathbb{R}}\backslash \left\{0\right\}$ and $f\left(g^{\frac{1}{p}} \left(x\right)\right) = f\left(\left[g^{p} \left(a\right)+g^{p} \left(b\right)-g\left(x\right)\right]^{\frac{1}{p}} \right)$, then the following inequalities for generalized k-fractional integral operators (2.1) and (2.2) hold:

(i) If $p > 0$, then

where $\bar{\bar{w}} = \frac{2^{\mu } w}{\left(g^{p} \left(b\right)-g^{p} \left(a\right)\right)^{\mu } }$ and $\theta \left(t\right) = g^{\frac{1}{p}} \left(t\right)$ for all $t\in \left[a^{p}, b^{p} \right]$.

(ii) If $p < 0$, then

where $\bar{\bar{w}} = \frac{2^{\mu } w}{\left(g^{p} \left(a\right)-g^{p} \left(b\right)\right)^{\mu } }$ and $\theta \left(t\right) = g^{\frac{1}{p}} \left(t\right)$ for all $t\in \left[b^{p}, a^{p} \right]$.

Proof. We prove the first assertion as follows:

(i) Multiplying (3.24) by $2t^{\frac{\alpha}{k}-1} E_{\mu, \alpha, l}^{\gamma, \delta, v, c} \left(wt^{\mu }; \tilde{p}\right)h\left(\big[\big(\frac{t}{2}\big)g^{p}(a)+\big(\frac{2-t}{2}\big)g^{p}(b)\big]^{\frac{1}{p}} \right)$ and then integrating over $\left[0, 1\right]$, we have

Setting $g(x) = \big(\frac{t}{2}\big)g^{p}(a)+\big(\frac{2-t}{2}\big)g^{p}(b)$ and $g(y) = \big(\frac{t}{2}\big)g^{p}(b)+\big(\frac{2-t}{2}\big)g^{p}(a)$, that is, $g^{p} \left(a\right)+g^{p} \left(b\right)-g\left(x\right) = \left(\frac{t}{2} \right)g^{p} \left(b\right)+\left(\frac{2-t}{2}\right)g^{p} \left(a\right)$ in (3.30), then by using $f\left(g^{\frac{1}{p}} \left(x\right)\right) = f\left(\left[g^{p} \left(a\right)+g^{p} \left(b\right)-g\left(x\right)\right]^{\frac{1}{p}} \right)$ and k-fractional integral operators (2.1) and (2.2), the first inequality of (3.28) is obtained.

Now to prove the second inequality of (3.28), multiplying both sides of (3.26) by $2t^{\frac{\alpha}{k}-1} E_{\mu, \alpha, l}^{\gamma, \delta, v, c} \left(wt^{\mu }; \tilde{p}\right) h\left(\big[\big(\frac{t}{2}\big)g^{p}(a)+\big(\frac{2-t}{2}\big)g^{p}(b)\big]^{\frac{1}{p}} \right)$ and then integrating over $\left[0, 1\right]$, we have

Setting $g(x) = \big(\frac{t}{2}\big)g^{p}(a)+\big(\frac{2-t}{2}\big)g^{p}(b)$ and $g(y) = \big(\frac{t}{2}\big)g^{p}(b)+\big(\frac{2-t}{2}\big)g^{p}(a)$, that is, $g^{p} \left(a\right)+g^{p} \left(b\right)-g\left(x\right) = \left(\frac{t}{2} \right)g^{p} \left(b\right)+\left(\frac{2-t}{2}\right)g^{p} \left(a\right)$ in (3.31), then by using $f\left(g^{\frac{1}{p}} \left(x\right)\right) = f\left(\left[g^{p} \left(a\right)+g^{p} \left(b\right)-g\left(x\right)\right]^{\frac{1}{p}} \right)$ and k-fractional integral operators (2.1) and (2.2), the second inequality of (3.28) is obtained.

(ii) Proof is similar to the proof of (i).

Remark 8. By setting $\tilde{p} = w = 0$ in (3.28) and (3.29), Corollary 31 of ^[13] is obtained.

Remark 9. By using (3.28) and (3.29), some more $k$-fractional inequalities are presented as follows:

(i) By setting $\tilde{p} = 0{\rm, \; }p = -1$ and $g = I$, we get

(ii) By setting $g = I$ and $p = -1$, we get

Corollary 7. By using (3.28) and (3.29), some more $k$-fractional inequalities are presented as follows:

(i) By setting $p = 1$, we get

(ii) By setting $p = -1$, we get

(iii) By setting $p = -2$, we get

Corollary 8. The aforementioned $k$-fractional inequalities are further connected with already known results as follows:

(i) By setting $k = 1$ in Remark 9 (i), Theorem 2.6 of ^[14] is obtained.

(ii) By setting $k = 1$ in Remark 9 (ii), Theorem 2.6 of ^[15] is obtained.

(iii) By setting $k = 1$ in Remark 9 (iii), Theorem 2.10 of ^[17] is obtained.

Corollary 9. When we set $w = \tilde{p} = 0{\rm, \; }p = -1$ and $g = I$ in Theorem 2.4, then we get the following inequalities.