Inequalities for different type of functions via Caputo fractional derivative

1.   Introduction

Inequalities develop tools for analyzing the problems in pure and applied mathematics. Thus, inequalities have become the centre of interest for many branches of science. A large number of extensions, various generalizations have been investigated in the theory of mathematical inequalities. We refer the reader to the references ^{[1,2,3,4,5,6,7]}. Fractional calculus is the generalization of derivatives and integrals of arbitrary non-integer order. In the last few years, some new definitions of fractional operators have been introduced to provide the best method for fractional calculus. Such type of studies promotes future researches to obtain new ideas to unify the fractional derivatives and fractional inequalities. For a review of this topic we direct the reader to the monographes ^[8,9].

In ^[10], Samraiz et al. developed a new class of trapezium-type inequalities up to twice differentiable h-convex mappings for fractional integrals of Riemann-type. Ali et al. established a new version of generalized fractional Hadamard and Fejér-Hadamard type integral inequalities in ^[11]. Rahman et al. investigated some novel inequalities for a class of differentiable functions related to Chebyshev's functionals in ^[12]. In ^[13], Mohammed et al. considered some Hermite-Hadamard-Fejér fractional integral inequalities and related results for the weighted fractional operators.

Some authors extended fractional inequalities using time scale calculus. For instance, Mohammed et al. introduced new time scales on $\mathbb{Z}$ and investigated the discrete inequality of Hermite-Hadamard type for discrete convex functions in ^[14]. A variety of various types of some classical inequalities have been established utilizing the fractional derivative and integral operators and their extensions are found in ^{[15,16,17,18]}.

In ^[19], authors established several inequalities connected with the Riemann-Liouville fractional integrals. Our aim in this paper is to extend and establish some new inequalities using Caputo fractional derivative for different classes of convex functions and mappings. Now we give the definitions of different classes of functions and Caputo fractional derivative to obtain our main results.

Definition 1.1. A function $f:\left[ a, b\right] \rightarrow \mathbb{R}$ is said to be convex if the inequality

holds for all $\; x, y\in \left[ a, b\right]$ and $\lambda \in \left[ 0, 1\right]$. If the inequality holds in the reverse direction, the function $f$ is called concave.

Definition 1.2. A function $f:I\subseteq \mathbb{R}\; \rightarrow \mathbb{R\; }$is said to belong to the class of $Q\left(I\right)$, if it is nonnegative and for all $x, y\in \; I\;$and $\lambda \in \left(0, 1\right) \;$satisfies the following inequality:

Definition 1.3. A function $f:I\subseteq \mathbb{R}\; \rightarrow \mathbb{R}$ is $P$ function or that $f$ belongs to the class of $P\left(I\right)$, if it is nonnegative and for all $x, y\in \; I\;$and $\lambda \in \left[ 0, 1\right]$ satisfies the following inequality:

The power mean $M_{r}\left(x, y;\lambda \right)$ of order $r$ of positive numbers $x, y$ is defined by

In ^[1], Pearce et al. generalized this inequality to r-convex positive function $f$ which is defined on an interval $\left[ a, b \right]$, for all $x, y\in$ $\left[ a, b\right]$ and $\lambda \in \left[ 0, 1 \right]$,

Definition 1.4. Let $h:J\subseteq \mathbb{R}\; \rightarrow \mathbb{R\; }$ be a positive function. We say that $f:I\subseteq \mathbb{R}\; \rightarrow \mathbb{R\; }$is h-convex or that $f$ belongs to the class of $SX\left(h, I\right)$. If $f$ is nonnegative and for all$\; x, y\in I\;$and $\lambda \in \left(0, 1\right)$, $\;$we have

If the above inequality is reversed, then $f$ is said to be h-concave, i.e. $f\in SV\left(h, I\right)$.

Definition 1.5. $f:I\rightarrow \mathbb{R}$ is a convex function on the interval I of real numbers and $u, v\in I$ with $u < v$, the inequality

is known as the Hermite-Hadamard inequality.

Definition 1.6. Suppose that $\alpha > 0, \; t > a, \; \alpha, a, t\in \mathbb{R}$. The fractional operator

is called the Caputo fractional derivative or Caputo fractional differential operator of order $\alpha.$ This operator is introduced by the Italian mathematician Caputo in 1967.

Definition 1.7. Let $\alpha > 0\;$ and $\alpha \notin \{1, 2, 3, ...\},$ $n =$ $\left[ \alpha\right] +1,$ $f\in C^{n}\left[ a, b\right]$ be a continuous $n$ order differentiable function. Right-sided and left-sided Caputo fractional derivatives are defined as follows:

and

Definition 1.8. Let $p\geq 1, \; 0 < \int\limits_{a}^{b}f^{p}\left(x\right) dx < \infty$ and $0 < \int\limits_{a}^{b}g^{p}\left(x\right) dx < \infty.$ Then Minkowski integral inequality is given as follow:

Lemma 1.1. Let $f:\left[ a, b\right] \rightarrow \mathbb{R}$ be a differentiable mapping on $\left(a, b\right)$ with $a < b$. If $f^{\left(n+1\right) }\in L\left[ a, b\right], \;$then the following fractional equalityholds:

2.   Results

In this section, we establish some new inequalities for different type of functions using Caputo derivatives.

Theorem 2.1. Let $f:\left[ a, b\right] \rightarrow \mathbb{R}$ be a differentiable mappingon $\left(a, b\right)$. If $\left\vert f^{\left(n+1\right)}\right\vert$ belongs to the class of $Q\left(I\right)$, $a, b \in I$ with$0\leq a < b.$ Then we have the following inequality:

Proof. By using the Lemma 1 and the Definition 1.2, we have

This ends the proof.

Theorem 2.2. Let $f:\left[ a, b\right] \rightarrow \mathbb{R}$ be a differentiable mappingon $\left(a, b\right)$ with $a < b$. If the function $\left\vert f^{\left(n+1\right) }\right\vert$ belongs to the class of $P\left(I\right)$, then $f$the following inequality holds:

Proof. As in the proof of the Theorem 2.1, using the definition of $P\left(I\right),$ we have

Theorem 2.3. Let $f:\left[ a, b\right] \rightarrow \mathbb{R}$ be a differentiable mappingon $\left(a, b\right)$. If $\left\vert f^{\left(n+1\right)}\right\vert$ is a r-convex positive function, then we have the following inequality:

Proof. Using the definition of r-convex function and the Minkowski integral inequality, we write

Theorem 2.4. Let $f:\left[ a, b\right] \rightarrow \mathbb{R}$ be a differentiable mappingon $\left(a, b\right)$. If the function $\left\vert f^{\left(n+1\right)}\right\vert$ belongs to the class of $SX\left(h, I \right)$, then we have the following inequality:

Proof. From Lemma 1.1, we have

Theorem 2.5. Let $f$ be a function belongs to the class $Q\left(I\right)$, $a, b\in I$ with $0\leq a < b$ and $f\in L_{1}\left[ a, b\right]$. Then we have

for $\alpha > 0$.

Proof. Since $f\in$ $Q\left(I\right)$, using Definition 1.2, we have

for all $x, y\in I$ and $\lambda = \frac{1}{2}$. Substituting $x = ta+\left(1-t\right)b$, $y = \left(1-t\right)a+tb$, we obtain

Then multiplying both sides of (2.2) by $t^{n-\alpha -1}$ and integrating the inequality with respect to $t$ over $\left[0, 1\right]$, we obtain

This completes the proof.

Theorem 2.6. Let $f$ be a function belongs to the class $P\left(I\right)$, $\; a, b\in I\;$with$\; 0\leq a < b\;$and $f\in L_{1}\left[ a, b\right]$. Then we have

for $\alpha > 0$.

Proof. Using Definition 1.3 and substituting with $x = ta+\left(1-t\right) b, \; y = \left(1-t\right)a+tb$ and $\lambda = \frac{1}{2}$, we have

for all $t\in \left[ 0, 1\right]$.

As in the proof of Theorem 2.5, multiplying both sides of inequality (2.4) by $t^{n-\alpha -1}$ and integrating the inequality with respect to $t$ over $\left[ 0, 1\right]$, we have

which completes the first part of the proof of inequality (2.3).

Since $f \in P\left(I\right)$, we have

and

Using above inequalities, we obtain

If we multiply both sides of inequality (2.5) by $t^{n-\alpha -1}$ and integrate with respect to $t$ from 0 to 1, we get

and thus the second part of the proof of inequality (2.3) is proved.

Theorem 2.7. Let $f:\left[ a, b\right] \rightarrow \left(0, \infty \right)$ be r-convex$\;$function on$\; \left[ a, b\right]$ with $a < b\;$and $0 < r\leq 1$. Then thefollowing inequality for fractional derivative holds:

where $\beta$ is the beta function.

Proof. Since $f$ is r-convex and $r > 0$, we have

and

for all $t\in \left[ 0, 1\right]$. By using these inequalities, we have

If we multiply both sides of above inequality by $t^{n-\alpha -1}\;$and integrate with respect to $t$ from 0 to 1, we obtain

We know that

By using the Minkowski inequality, we have

and similarly we obtain

Thus, we have

This completes the proof.

Theorem 2.8. Let $f\in SX\left(h, I\right)$, $a, b\in I$ with $a < b$ and $L_{1}\left[ a, b\right]$. Then we have inequality for h-convex functions via fractional derivative:

Proof. According to inequality (1.3), if we replace $x = ta+\left(1-t\right)b$, $y = \left(1-t\right)a+tb$ and $n-\alpha = \frac{1}{2}$, we have

If we multiply both sides of inequality (2.7) by $t^{n-\alpha -1}$ and integrate with respect to $t$ from 0 to 1, we get

and the first part of the inequality (2.6) is proved.

Since $f\in SX\left(h, I\right)$, we have

and

Using these inequalities, we get

If we replace $x = a$ and $y = b$ in inequality (2.8), we find

Then multiplying both sides of the inequality (2.9) by $t^{n-\alpha -1}\; \;$ and integrating the inequality with respect to $t$ over $\left[ 0, 1\right],$ we have

and

so that the second part of the inequality (2.6) is proved. Hence, the proof is complete.

3.   Conclusions

Convex functions play an important role in the advancement of many inequalities. Most of the well-known inequalities are the consequences of convex functions. In this work, we have motivated by different type of convex functions which are integrable and we have obtained new results for Caputo fractional derivative. Fractional inequalities have gained considerable importance and popularity. Thus, this paper presents a new approach to new versions of different inequalities. We conclude that the results presented in this study would be a guide for further investigations concerning inequalities for various kinds of fractional calculus.

Conflict of interest

The author declares no conflict of interest.