Boundedness of fractional integral operators containing Mittag-Leffler functions via (s,m)-convexity

1.   Introduction

Convex functions are useful in various aspects in diverse fields of mathematical sciences. They produce an elegant theory of convex analysis, see ^[22,24,27].

Definition 1. ^[27] A function $f: I\rightarrow\mathbb{R}$ is said to be convex function, if the following inequality holds:

for all $a, b \in I$ and $t\in \left[0, 1\right]$.

Convex functions have been extended and generalized from their analytical interpretations. A generalization of convex function defined on right half of real line is called $s$-convex function given as follows:

Definition 2. ^[16] Let $s\in \left[0, 1\right]$. A function $f:[0, \infty) \rightarrow\mathbb{R}$ is said to be $s$-convex function in the second sense if

holds for all $a, b \in[0, \infty)$ and $t\in \left[0, 1\right]$.

Another generalization of convex function defined on right half of real line is called $m$-convex function given as follows:

Definition 3. ^[2] A function $f:[0, b] \rightarrow\mathbb{R}$ is said to be $m$-convex function, where $m\in\left[0, 1\right]$ and $b > 0$, if for every $x, y\in[0, b]$ and $t\in[0, 1]$ we have

Aforementioned functions can be generalized by $(s, m)$-convex functions defined as follows:

Definition 4. ^[2] A function $f:[0, b] \rightarrow\mathbb{R}$ is said to be $(s, m)$-convex function, where $(s, m)\in\left[0, 1\right]^2$ and $b > 0$, if for every $x, y\in[0, b]$ and $t\in[0, 1]$ we have

For some recent citations and utilizations of $(s, m)$-convex functions one can see ^{[5,10,18,19,23,31]} and references therein. Convex functions and related definitions have been widely used to develop the theory of inequalities and their applications. A huge amount of work by many authors had/has been dedicated to theory and applications of mathematical inequalities, see ^[22,24,27]. The aim of this paper is the study of boundedness, continuity of fractional integral operators containing Mittag-Leffler functions via $(s, m)$-convex functions.

The Mittag-Leffler function denoted by $E_{\alpha}(.)$ was introduced by Gosta Mittag-Leffler in 1903 ^[21]

where $t, \alpha\in\mathbb{C}, \, \Re{(\alpha)} > 0$ and $\Gamma(.)$ is the gamma function.

In the solution of fractional integral equations and fractional differential equations the Mittag-Leffler function arises naturally. The Mittag-Leffler function is a direct generalization of some special functions. It was consequently explored by Wiman, Pollard, Humbert, Agarwal and Feller, see ^[15]. It is further generalized and extended by various authors, for details see ^{[4,15,26,28,29]}. Andrić et al. introduced the following extended Mittag-Leffler function:

Definition 5. ^[3] Let $\mu, \alpha, l, \gamma, c\in \mathbb{C}$, $\Re(\mu), \Re(\alpha), \Re(l) > 0$, $\Re(c) > \Re(\gamma) > 0$ with $p\geq0$, $\delta > 0$ and $0 < k\leq\delta+\Re(\mu)$. Then the extended generalized Mittag-Leffler function $E_{\mu, \alpha, l}^{\gamma, \delta, k, c}(t; p)$ is defined by:

where $\beta_{p}$ is defined by

and $(c)_{nk} = \frac{\Gamma(c+nk)}{\Gamma(c)}$.

A derivative formula of the extended generalized Mittag-Leffler function is given in the following lemma.

Lemma 1. ^[3] If $m\in\mathbb{N}, \omega, \mu, \alpha, l, \gamma, c\in\mathbb{C}, \Re(\mu), \Re(\alpha), \Re(l) > 0, \Re(c) > \Re(\gamma) > 0$ with $p\geq0, \delta > 0$ and $0 < k < \delta+\Re(\mu)$, then

Remark 1. The extended Mittag-Leffler function (1.1) produces the related functions defined in ^{[25,26,28,29,30]}, see [32,Remark 1.3].

Next we give the definition of the fractional integral operator containing the extended generalized Mittag-Leffler function (1.1).

Definition 6. ^[3] Let $\omega, \mu, \alpha, l, \gamma, c\in \mathbb{C}$, $\Re(\mu), \Re(\alpha), \Re(l) > 0$, $\Re(c) > \Re(\gamma) > 0$ with $p\geq0$, $\delta > 0$ and $0 < k\leq\delta+\Re(\mu)$. Let $f\in L_{1}[a, b]$ and $x\in[a, b].$ Then the generalized fractional integral operators containing Mittag-Leffler function are defined by:

and

Remark 2. The operators (1.3) and (1.4) produce in particular several kinds of known fractional integral operators, see [32,Remark 1.4]

The classical Riemann-Liouville fractional integral operator is defined as follows:

Definition 7. ^[30] Let $f\in L_{1}[a, b]$. Then Riemann-Liouville fractional integral operators of order $\alpha\in\mathbb{C}\, \, (\mathcal{R}(\alpha) > 0)$ are defined as follows:

It can be noted that $\left(\epsilon_{\mu, \alpha, l, 0, a^+}^{\gamma, \delta, k, c}f\right)(x; 0) = I_{a^+}^{\alpha}f(x)$ and $\left(\epsilon_{\mu, \alpha, l, 0, b^-}^{\gamma, \delta, k, c}f\right)(x; 0) = I_{b^-}^{\alpha}f(x)$. From fractional integral operators (1.3) and (1.4), we have (see ^[13]):

Now a days integral operators have been proved very useful in the advancement of mathematical inequalities. Recently, several authors have established fractional integral inequalities by utilizing different types of integral operators, see ^{[1,6,7,8,9,11,12,13,14,17,20,32]} and references therein.

In the upcoming section upper bounds of generalized fractional integral operators are derived by using $(s, m)$-convexity, and some particular results are produced. By using these bounds continuity of these operators is established. Furthermore a modulus inequality is established for differentiable function $f$ such that $|f'|$ is $(s, m)$-convex. By imposing an additional condition Hadamard type inequality is obtained for $(s, m)$-convex functions. Also the results of this paper are connected with already known results.

2.   Main results

Theorem 1. Let $f:[a, b]\longrightarrow \mathbb{R}$ be a real valued function. If $f$ is positive and $(s, m)$-convex, then for $\alpha, \beta\geq1$, the following inequality holds for generalized fractional integral operators:

Proof. Let $x\in [a, b]$. Then for $t\in[a, x)$ and $\alpha\geq1$, one can has the following inequality:

The function $f$ is $(s, m)$-convex, therefore one can obtain

By multiplying (2.2) and (2.3) and then integrating over $[a, x]$, we get

that is, the left integral operator satisfies the following inequality:

Now on the other hand for $t\in (x, b]$ and $\beta\geq1$, one can has the following inequality:

Again from $(s, m)$-convexity of $f$, we have

By multiplying (2.5) and (2.6) and then integrating over $[x, b]$, we have

that is, the right integral operator satisfies the following inequality:

By Adding (2.4) and (2.7), the required inequality (2.1) is established.

Some particular results are stated in the following corollaries.

Corollary 1. If we set $\alpha = \beta$ in (2.1), then the following inequality is obtained:

Corollary 2. Along with assumptions of Theorem 1, if $f\in L_{\infty}[a, b]$, then the following inequality is obtained:

Corollary 3. For $\alpha = \beta$ in (2.9), we get the following result:

Corollary 4. For $s = 1$ in (2.9), we get the following result:

Theorem 2.2. With the assumptions of Theorem 1 if $f\in L_{\infty}[a, b]$, then operator defined in (1.3) and (1.4) are bounded and continuous.

Proof. If $f\in L_{\infty}[a, b]$, then from (2.4) we have

that is

where $M = \frac{2(b-a)J_{\alpha-1, a^+}(b; p)(1+ms)}{s+1}$. Therefore $\left(\epsilon_{\mu, \alpha, l, \omega, a^+}^{\gamma, \delta, k, c}f\right)(x; p)$ is bounded also it is easy to see that it is linear, hence this is continuous operator. Also on the other hand from (2.7) we can obtain:

where $K = \frac{2(b-a)J_{\beta-1, b^-}(a; p)(1+ms)}{s+1}$. Therefore $\left(\epsilon_{\mu, \beta, l, \omega, b^-}^{\gamma, \delta, k, c}f\right)(x; p)$ is bounded also it is linear, hence continuous.

Theorem 3. Let $f: [a, b]\longrightarrow \mathbb{R}$ be a real valued function. If $f$ is differentiable and $|f'|$ is $(s, m)$-convex, then for $\alpha, \beta\geq1$, the following fractional integral inequality for generalized integral operators (1.3) and (1.4) holds:

Proof. As $x\in [a, b]$ and $t\in [a, x)$, by using $(s, m)$-convexity of $|f'|$, we have

From (2.14), one can has

The product of (2.2) and (2.15), gives the following inequality:

After integrating above inequality over $[a, x]$, we get

The left hand side of (2.17) is calculated as follows:

put $x-t = z$ that is $t = x-z$, also using the derivative property (1.2) of Mittag-Leffler function, we have

now put $x-z = t$ in second term of the right hand side of the above equation and then using (1.3), we get

Therefore (2.17) takes the following form:

Also from (2.14), one can has

Following the same procedure as we did for (2.15), one can obtain:

From (2.19) and (2.21), we get

Now we let $x\in[a, b]$ and $t\in(x, b]$. Then by using $(s, m)$-convexity of $|f'|$ we have

on the same lines as we have done for (2.2), (2.15) and (2.20) one can get from (2.5) and (1.7), the following inequality:

From inequalities (2.22) and (2.24) via triangular inequality (2.13) is obtained.

Corollary 5. If we put $\alpha = \beta$ in (2.13), then the following inequality is obtained:

It is easy to prove the next lemma which will be helpful to produce Hadamard type estimations for the generalized fractional integral operators.

Lemma 2. Let $f: [a, b]\rightarrow\mathbb{R}$ be $(s, m)$-convex function. If $f$ is $f(\frac{a+mb-x}{m}) = f(x)$ and $(s, m)\in\mathbb[0, 1]^2$, then the following inequality holds:

Proof. For $t\in[0, 1]$ we have

As $f$ is $(s, m)$-convex function, we have

Let $x = a(1-t)+mtb$. Then we have $a+mb-x = ta+m(1-t)b$.

Hence by using $f(\frac{a+mb-x}{m}) = f(x)$, the inequality (2.26) can be obtained.

Theorem 4. Let $f: [a, b]\longrightarrow \mathbb{R}$, $a > b$, be a real valued function. If $f$ is positive, $(s, m)$-convex and $f(a+mb-x) = f(x)$, then for $\alpha, \beta > 0$, the following inequality holds for generalized fractional integral operators:

Proof. For $x\in[a, b]$, we have

As $f$ is $(s, m)$-convex so for $x\in[a, b]$, we have:

By multiplying (2.31) and (2.32) and then integrating over $[a, b]$, we get

From which we have

that is

Now on the other hand for $x\in[a, b]$, we have

By multiplying (2.32) and (2.35) and then integrating over $[a, b]$, we get

From which we have

that is

Adding (2.34) and (2.37), we get;

Multiplying (2.26) with $(x-a)^{\beta}E_{\mu, \beta, l}^{\gamma, \delta, k, c}(\omega(x-a)^{\mu}; p)$ and integrating over $[a, b]$, we get

By using (1.4) and (1.7), we get

By multiplying (2.26) with $(b-x)^{\alpha}E_{\mu, \alpha, l}^{\gamma, \delta, k, c}(\omega(b-x)^{\mu}; p)$ and integrating over $[a, b]$, also using (1.3) and (1.7), we get

By adding (2.40) and (2.41), we get;

By combining (2.38) and (2.42), inequality (2.30) can be obtained.

Corollary 6. If we put $\alpha = \beta$ in (2.30), then the following inequality is obtained:

3.   Conclusion

This work deals with the boundedness of generalized fractional integral operators given in (1.3) and (1.4), by using $(s, m)$-convex functions. The results of this paper provide the boundedness and continuity of several known integral operators defined in ^{[25,26,28,29,30]}. By applying $(s, m)$-convexity of functions $f$ and $|f'|$, variable bounds of sum of left and right definitions of these operators are obtained, while by imposing an additional condition a Hadamard inequality is proved. All the results hold for convex, $m$-convex and $s$-convex functions and for integral operators given in^{[25,26,28,29,30]}. The reader can obtain results for $s$-convex functions and for convex functions proved in ^[11]. The method adopted in this paper can be applied to derive bounds of other kinds of well known integral operators already exist in literature.

Acknowledgments

The research work of first author is supported by Higher Education Commission of Pakistan under NRPU 2016, Project No. 5421, the research work of fourth author is supported by the Ministry of Education and Science of the Russian Federation (the Agreement No. 02.a03.21.0008).

Conflict of interest

The authors declare that there is no conflicts of interest in this paper.