New extensions of Chebyshev-Pólya-Szegö type inequalities via conformable integrals

1.   Introduction and preliminaries

This article is based on the well known Chebyshev functional ^[1]:

where $f$ ve $g$ are two integrable functions which are synchronous on $\left[a, b\right]$, i.e.

for any $x, y\in \left[a, b\right]$, then the Chebyshev inequality states that $T\left(f, g\right) \geq 0$.

The functional (1.1) has attracted many researchers attention due to diverse applications in numerical quadrature, transform theory, probability and statistical problems. Among those applications, the functional (1.1) has also been employed to yield a number of integral inequalities. For some recent counterparts, generalizations of Chebyshev inequality, the reader may refer to ^[2,3,9] and ^[10].

Another important inequality which will be useful to prove our main results is Pólya and Szegö inequality: (see ^[4])

In ^[5], Dragomir and Diamond obtained the following Grüss type inequality by using the Pólya-Szegö inequality:

Theorem 1.1. Let $f, g:\left[a, b\right] \rightarrow \mathbb{R} _{+}$ be two integrable functions so that

for $x\in \left[a, b\right]$. Then we have

The constant $\frac{1}{4}$ is best possible in (1.2) in the sense it can not be replaced by a smaller constant.

Let us recall some well-known concepts. We note that the beta function $B\left(\alpha, \beta \right)$ is defined (see ^[11])

where $\Gamma$ is the familiar Gamma function. Here and in the following, let $\mathbb{C}, \mathbb{R}, \mathbb{R} ^{+}$ and $\mathbb{Z} _{0}^{-}$ be the sets of complex numbers, real numbers, positive real numbers and non-positive integers, respectively and let $\mathbb{R} _{0}^{+} = \mathbb{R} ^{+}\cup \left\{ 0\right\}.$

Definition 1.1. (See ^[6,7]) Let $f\in L\left[a, b\right]$. The Riemann-Liouville integrals $J_{a^{+}}^{\alpha }f$ and $J_{b^{-}}^{\alpha }f$ of order $\alpha > 0$ with $a\geq 0$ are defined by

and

where $\Gamma \left(\alpha \right) = \int_{0}^{\infty }e^{-x}x^{\alpha -1}dx$ is the Gamma function.

Definition 1.2. (See ^[8]) Let $\alpha \in \left(n, n+1\right]$, $n = 0, 1, 2, ...$ and set $\beta = \alpha -n$. Then the left conformable fractional integral of order $\alpha > 0$ is defined by

the right conformable fractional integral of order $\alpha > 0$ is defined by

Notice that if $\alpha = n+1$ then $\beta = \alpha -n = n+1-n = 1$ and hence $\left(I_{\alpha }^{a}f\right) \left(t\right) = \left(J_{n+1}^{a}f\right) \left(t\right).$

Several recent results related to different kinds of fractional integral operators can be found in ^{[12,13,14,15,16,17,18,19,20,21,22,23,24]}.

The main aim of this present paper is to prove certain new Pólya-Szeg ö and Chebyshev types integral inequalities involving conformable fractional integral operator. We also give some special cases of our results.

2.   Main results

In this section, we establish certain Pólya-Szegö type integral inequalities for positive integral functions involving conformable fractional integral operator.

Lemma 2.1. Let $f$ and $g$ be two positive integrable functions on $\left[0, \infty \right)$. Also let $\alpha \in \left(n, n+1\right]$, $n = 0, 1, 2, ...$, set $\beta = \alpha -n$. Assume that there exist four positive integrable functions $v_{1}, v_{2}, w_{1}$ and $w_{2}$ such that:

Then the following inequality holds:

Proof. From (2.1), for $\tau \in \left[0, x\right]$, $x > 0$, we can write

and

multiplying (2.3) and (2.4), we get

From the above inequality, we can write

Multiplying both sides of (2.5) by $\frac{1}{n!}\left(x-\tau \right) ^{n}\tau ^{\alpha -n-1}$ and integrating the resulting inequality with respect to $\tau$ over $\left(0, x\right)$, we get

applying the AM-GM inequality, i.e. ($a+b\geq 2\sqrt{ab},$ $a, b\in \mathbb{R} ^{+}$), we have

which implies that

So, we get the desired result.

Corollary 2.1. If $v_{1} = m$, $v_{2} = M$, $w_{1} = n$ and $w_{2} = N$, then we have

Lemma 2.2. Let $f$ and $g$ be two positive integrable functions on $\left[0, \infty \right)$. Also let $\alpha \in \left(n, n+1\right]$, $\theta \in \left(k, k+1\right]$, $n, k = 0, 1, 2, ...$. Assume that there exist four positive integrable functions $v_{1}, v_{2}, w_{1}$ and $w_{2}$ satisfying condition (2.1). Then the following inequality holds:

Proof. From (2.1), we get

and

which leads to

Multiplying both sides of (2.8) by $w_{1}\left(\xi \right) w_{2}\left(\xi \right) g^{2}\left(\xi \right),$ we have

Multiplying both sides (2.9) by $\left(\frac{1}{n!}\right) \left(\frac{1}{k!}\right) \left(x-\tau \right) ^{n}\left(x-\xi \right) ^{k}\tau ^{\alpha -n-1}\xi ^{\theta -k-1}$and integrating the resulting inequality with respect to $\tau$ and $\xi$ over $\left(0, x\right) ^{2}$, we get

Applying the AM-GM inequality, we obtain

which leads to the desired inequality in (2.7). The proof is completed.

Corollary 2.2. If $v_{1} = m$, $v_{2} = M$, $w_{1} = n$ and $w_{2} = N$, then we have

Lemma 2.3. Let $f$ and $g$ be two positive integrable functions on $\left[0, \infty \right)$. Also let $\alpha \in \left(n, n+1\right]$, $\theta \in \left(k, k+1\right]$, $n, k = 0, 1, 2, ...$. Assume that there exist four positive integrable functions $v_{1}, v_{2}, w_{1}$ and $w_{2}$ satisfying condition (2.1). Then the following inequality holds:

Proof. Using the condition (2.1), we get

Multiplying both sides of (2.11) by $\left(\frac{1}{n!}\right) \left(x-\tau \right) ^{n}\tau ^{\alpha -n-1}$and integrating the resulting inequality with respect to $\tau$ over $\left(0, x\right)$, we obtain

which leads to

Similarly, we can write

By a similar argument, we have

which implies

Multiplying (2.12) and (2.13), we get the (2.10). The proof is completed.

Corollary 2.3. If $v_{1} = m$, $v_{2} = M$, $w_{1} = n$ and $w_{2} = N$, then we have

Theorem 2.1. Let $f$ and $g$ be two positive integrable functions on $\left[0, \infty \right)$. Also let $\alpha \in \left(n, n+1\right]$, $\theta \in \left(k, k+1\right]$, $n, k = 0, 1, 2, ...$. Assume that there exist four positive integrable functions $v_{1}, v_{2}, w_{1}$ and $w_{2}$ satisfying condition (2.1). Then the following inequality holds:

where

and

Proof. Let $f$ and $g$ be two positive integrable functions on $\left[0, \infty \right)$. For $\tau, \xi \in \left(0, x\right)$ with $x > 0$, we define $H\left(\tau, \xi \right)$ as

namely

Multiplying both sides of (2.15) by

and double integrating the resulting inequality with respect to $\tau$ and $\xi$ over $\left(0, x\right) ^{2},$ we get

Applying the Cauchy-Schwarz inequality for double integrals, we can write

As a consequence

Applying Lemma 2.1 with $w_{1}\left(\tau \right) = w_{2}\left(\tau \right) = g\left(\tau \right) = 1$, we get

This implies that

and

Similarly, applying Lemma 2.1 with $v_{1}\left(\tau \right) = v_{2}\left(\tau \right) = f\left(\tau \right) = 1$, we have

and

Using (2.16)-(2.19), we conclude the result.

Theorem 2.2. Let $f$ and $g$ be two positive integrable functions on $\left[0, \infty \right)$. Also let $\alpha \in \left(n, n+1\right]$, $\theta \in \left(k, k+1\right]$, $n, k = 0, 1, 2, ...$. Assume that there exist four positive integrable functions $v_{1}, v_{2}, w_{1}$ and $w_{2}$ satisfying condition (2.1). Then the following inequality holds:

where

Proof. Setting $\alpha = \theta$ in (2.14), we obtain (2.20).

Corollary 2.4. If $v_{1} = m$, $v_{2} = M$, $w_{1} = n$ and $w_{2} = N$, then we have

Conflict of interest

All authors declare no conflicts of interest in this paper.