Some inequalities via Ψ-Riemann-Liouville fractional integrals

1.   Introduction

It is being known that the Hermite-Hadamard inequality ^[6,7] for a convex function $\curlyvee:I\rightarrow \mathbb{R}$ on an interval $I$ is

for all $a_1, a_2\in I$ with $a_1 < a_2$. Inequality (1.1) is then revised for several generalized convex functions. For more precise data see ^{[1,3,9,10,14,16]}.

Definition 1.1 (^[8]). Let $s\in(0, 1]$. A function $\curlyvee:I\subset \mathbb{R}_{0}\rightarrow \mathbb{R}_{0}$, where $\mathbb{R}_{0} = [0, \infty)$, is called $s$-convex in the second sense, if

for all $a_1, a_2\in I$ and $r\in[0, 1]$.

Dragomir et al. ^[4] proved following inequality for $s$-convex functions.

Theorem 1.1 (^[4]). Let $s\in(0, 1)$ and $\curlyvee:\mathbb{R}_{0}\rightarrow \mathbb{R}_{0}$ is $s$-convex in the second sense. Let $a_1, a_2\in[0, \infty)$, $a_1\leq a_2$. If $\curlyvee\in L_1[a_1, a_2]$, then

In $1995$, Dragomir et al. ^[5] defined following class of functions.

Definition 1.2 (^[5]). A mapping $\curlyvee:I\rightarrow \mathbb{R}$ belongs to the class $P(I)$, if it is non-negative and satisfy the following inequality:

for all $a_1, a_2\in I$ and $r\in[0, 1]$.

Theorem 1.2 (^[5]). Let $\curlyvee\in P(I)$, $a_1, a_2\in I$ and $\curlyvee\in L_1[a_1, a_2]$, then

Fractional Hermite-Hadamard inequalities for the Riemann-Liouville, the Hadamard fractional integrals, the conformable, new conformable fractional integrals and the Katugampola fractional integrals have been studied. For examples see ^{[2,11,15,17,18,19]}. Liu et al. ^[13] established Hermite-Hadamard type inequalities for $\psi$-Riemann-Liouville fractional integrals via convex functions.

Definition 1.3 (^[12,21]) Let $(a_1, a_2)(-\infty\leq a_1 < a_2\leq \infty)$ be a finite or infinite real interval and $\beta > 0$. Let $\psi(x)$ is an increasing and positive monotone function on $(a_1, a_2]$ with continuous derivative on $(a_1, a_2)$. Then the left and right-sided $\psi$-Riemann-Liouville fractional integrals of a function $\curlyvee$ with respect to $\psi$ on $[a_1, a_2]$ are defined by

respectively; with Gamma function $\Gamma(\cdot)$.

Lemma 1.1 (^[13]). Let $\curlyvee: [a_1, a_2]\rightarrow \mathbb{R}$ be a differentiable mapping, for $0\leq a_1 < a_2$, and $\curlyvee\in L_1[a_1, a_2]$. Let $\psi(x)$ is an increasing and positive monotone function on $(a_1, a_2]$, with continuous derivative $\psi'(x)$ on $(a_1, a_2)$ and $\beta\in(0, 1)$. Then the following equality for fractional integral holds:

Lemma 1.2 (^[13]). Let $\curlyvee: [a_1, a_2]\rightarrow \mathbb{R}$ be a differentiable mapping, for $0\leq a_1 < a_2$, and $\curlyvee\in L_1[a_1, a_2]$. Let $\psi(x)$ is an increasing and positive monotone function on $(a_1, a_2]$, with continuous derivative $\psi'(x)$ on $(a_1, a_2)$ and $\beta\in(0, 1)$. Then the following equality for fractional integral holds:

where

In this paper, we studied some inequalities for functions belonging to $P(I)$ and $s$-convex functions in second sense via $\psi$-Riemann-Liouville fractional integrals.

2.   Inequalities for $s$-convex functions

First we establish the Hermite-Hadamard inequality via $\psi$-Riemann-Liouville fractional integrals.

Theorem 2.1. Let $\curlyvee: [a_1, a_2]\rightarrow \mathbb{R}$ be a positive function, for $0\leq a_1 < a_2$, and $\curlyvee\in L_1[a_1, a_2]$. Let $\psi(z)$ is an increasing and positive monotone function on $(a_1, a_2]$, with continuous derivative $\psi'(z)$ on $(a_1, a_2)$. Let $\curlyvee$ is $s$-convex function, then following inequalities for fractional integral hold:

where $B$ is Beta function defined as $B(a_1, a_2) = \int_{0}^{1}z^{a_1-1}(1-z)^{a_2-1}dz$.

Proof. Since $\curlyvee$ is $s$-convex, we have

Let $u = ra_1+(1-r)a_2$ and $v = ra_2+(1-r)a_1$, we get

Multiplying by $r^{\beta-1}$ on both sides of inequality (2.2), and then integrating with respect to $r$ over $[0, 1]$, implies

Now consider,

by using (2.3). Thus first inequality of (2.1) is proved.

For next, we again use $s$-convexity of $\curlyvee$, that is,

and

By adding

Multiplying by $r^{\beta-1}$ on both sides of inequality (2.5), and then integrating with respect to $r$ over $[0, 1]$, implies

By multiplying $\frac{\beta}{2}$ on both sides of above inequality we get,

Hence the proof is completed.

Remark 2.1. Under the similar assumptions of Theorem $2.1$.

$1.$ For $s = 1$, we get Theorem $2.1$ in ^[13].

$2.$ For $\psi(z) = z$, we get Theorem $3$ in ^[20].

$3.$ For $\psi(z) = z$ and $\beta = 1$, the inequality (2.1) becomes inequality (1.3).

$4.$ For $\psi(z) = z$ and $\beta = s = 1$, the inequality (2.1) becomes inequality (1.1).

For the next two results we use Lemma 1.1 and Lemma 1.2, respectively.

Theorem 2.2. Let $\curlyvee: [a_1, a_2]\rightarrow \mathbb{R}$ be a differentiable mapping, for $0\leq a_1 < a_2$. Let $\psi(z)$ is an increasing and positive monotone function on $(a_1, a_2]$, with continuous derivative $\psi'(z)$ on $(a_1, a_2)$ and $\beta\in(0, 1)$. If $|\curlyvee'|^q$ is $s$-convex for some fixed $s\in(0, 1)$ and $q\geq 1$, then the following inequality for fractional integral holds:

where $B_u$ is incomplete Beta function defined as:

Proof. First note that, for every $z\in(\psi^{-1}(a_1), \psi^{-1}(a_2))$, we have $a_1 < \psi(z) < a_2$. Let $r = \frac{a_2-\psi(z)}{a_2-a_1}$, then we have $\psi(z) = ra_1+(1-r)a_2$. Applying Lemma 1.1 and $s$-convexity of $|\curlyvee'|$, we obtain

Since

and

By substituting above integral values in (2.7) and after some simplification we get the required inequality (2.6) for $q = 1$.

Now consider the case when $q > 1$. Again using Lemma 1.1, power mean inequality and the $s$-convexity of $|\curlyvee'|^q$ on $[a_1, a_2]$, we get

This completes the proof.

Remark 2.2. Under the similar assumptions of Theorem $2.2$.

$1.$ For $s = 1$ and $q = 1$, we get Theorem $3.4$ in ^[13].

$2.$ For $\psi(z) = z$, we get Theorem $4$ in ^[20].

Theorem 2.3. Let $\curlyvee: [a_1, a_2]\rightarrow \mathbb{R}$ be a differentiable mapping, for $0\leq a_1 < a_2$. Let $\psi(z)$ is an increasing and positive monotone function on $(a_1, a_2]$, with continuous derivative $\psi'(z)$ on $(a_1, a_2)$ and $\beta\in(0, 1)$. If $|\curlyvee'|$ is $s$-convex for some fixed $s\in(0, 1)$, then the following inequality for fractional integral holds:

Proof. From Lemma 1.2 and the $s$-convexity of $|\curlyvee'|$, we have

where

and $k$ is defined as in Lemma 1.2. Note that

and from Theorem 2.2 for the case $q = 1$, we have

Hence by using (2.11) and (2.12) in (2.10), we get (2.9).

Remark 2.3. By taking $s = 1$ in (2.9), we get inequality $(9)$ in ^[13].

3.   Inequalities for class of non-negative functions $P(I)$

First we establish the Hermite-Hadamard inequality via $\psi$-Riemann-Liouville fractional integrals.

Theorem 3.1. Let $\curlyvee: [a_1, a_2]\rightarrow \mathbb{R}$ be a positive function, for $0\leq a_1 < a_2$, and $\curlyvee\in L_1[a_1, a_2]$. Let $\psi(z)$ is an increasing and positive monotone function on $(a_1, a_2]$, with continuous derivative $\psi'(z)$ on $(a_1, a_2)$. Let $\curlyvee\in P(I)$ is, then following inequalities for fractional integral hold:

Proof. Since the function $\curlyvee$ belongs to the class $P(I)$, we have

Let $u = ra_1+(1-r)a_2$ and $u = ra_2+(1-r)a_1$, we get

Multiplying by $r^{\beta-1}$ on both sides of inequality (2.2), and then integrating with respect to $r$ over $[0, 1]$, implies

Now consider,

by using (3.3). Thus first inequality of (3.1) is proved.

For next, again using the property of $\curlyvee\in P(I)$, that is,

and

By adding

Multiplying by $r^{\beta-1}$ on both sides of inequality (3.5), and then integrating with respect to $r$ over $[0, 1]$, implies

That is,

This completes the proof.

Remark 3.1. By taking $\beta = 1$ and $\psi(z) = z$ in (3.1), we get inequality $3.2$ of Theorem $3.1$ in ^[5].

Theorem 3.2. Let $\curlyvee: [a_1, a_2]\rightarrow \mathbb{R}$ be a differentiable mapping, for $0\leq a_1 < a_2$. Let $\psi(z)$ is an increasing and positive monotone function on $(a_1, a_2]$, with continuous derivative $\psi'(z)$ on $(a_1, a_2)$ and $\beta\in(0, 1)$. If $|\curlyvee'|\in P(I)$, then the following inequality for fractional integral holds:

Proof. Again using Lemma 1.1 and the property of $|\curlyvee'|$ on $[a_1, a_2]$, we get

Since

This completes the proof.

Theorem 3.3. Let $\curlyvee: [a_1, a_2]\rightarrow \mathbb{R}$ be a differentiable mapping, for $0\leq a_1 < a_2$. Let $\psi(z)$ is an increasing and positive monotone function on $(a_1, a_2]$, with continuous derivative $\psi'(z)$ on $(a_1, a_2)$ and $\beta\in(0, 1)$. If $|\curlyvee'|\in P(I)$, then the following inequality for fractional integral holds:

Proof. From Lemma 1.2 and the property of $|\curlyvee'|$ on $[a_1, a_2]$, we have

where

and $k$ is defined as in Lemma 1.2. Note that

and from Theorem 3.2, we have

Hence by using (3.10) and (3.11) in (3.9), we get (3.8).

4.   Application to special means

Consider the following special means of real numbers $a_1$, $a_2$ such that $a_1\neq a_2$.

${\rm (1).}$ The arithmetic mean

${\rm (2).}$ The $p$-logarithmic mean

Proposition 4.1. Let $a_1, a_2\in \mathbb{R}^+$, $a_1 < a_2$ and $0 < s < 1$, then

Proof. Apply Theorem 2.2 with $\curlyvee = z^s$, $\psi(z) = z$ and $\beta = q = 1$, we get the required result.

Proposition 4.2. Let $a_1, a_2\in \mathbb{R}^+$, $a_1 < a_2$ and $0 < s < 1$, then

Proof. Apply Theorem 2.3 with $\curlyvee = z^s$, $\psi(z) = z$ and $\beta = q = 1$, we get the required result.

Proposition 4.3. Let $a_1, a_2\in \mathbb{R}^+$, $a_1 < a_2$, then

Proof. Apply Theorem 3.2 with $\curlyvee = z^2$, $\psi(z) = z$ and $\beta = 1$, we get the required result.

Proposition 4.4. Let $a_1, a_2\in \mathbb{R}^+$, $a_1 < a_2$, then

Proof. Apply Theorem 3.3 with $\curlyvee = z^2$, $\psi(z) = z$ and $\beta = 1$, we get the required result.

Acknowledgement

This research article is supported by National University of Sciences and Technology(NUST), Islamabad, Pakistan.

Conflict of interest

The authors declare that there is no interest regarding the publication of this paper.