Some Inequalities of Čebyšev Type for Conformable k-Fractional Integral Operators

In the article, the authors present several inequalities of the Čebyšev type for conformable k-fractional integral operators.


Introduction
The Čebyšev inequality [1] reads that where f and g are two integrable and synchronous functions on [a, b] and two functions f and g are called synchronous on [a, b] if The inequality (1) has many applications in diverse research subjects such as numerical quadrature, transform theory, probability, existence of solutions of differential equations, and statistical problems (see ( [2], Chapter IX) and the paper [3]).Many authors have investigated, generalized, and applied the Čebyšev inequality (1).For detailed information, please refer to [4,5] and closely related references.
In this paper, we introduce the conformable k-fractional integral operator When k = 1, the Equations ( 7) to ( 8) reduces to the Equations ( 4) to ( 6), respectively.

Main Results
In this section, we present several Čebyšev type inequalities for conformable k-fractional integral operators defined in the Equation (8).
Proof.Since f and g are synchronous on [0, ∞), we have Multiplying both sides of the Equation ( 9) by Further integrating both sides with respect to u over (0, x) gives Consequently, it follows that and where Multiplying both sides of the Equation ( 10) by Therefore, we have The proof of Theorem 5 is complete.
Corollary 1.Let f and g be two integrable functions which are synchronous on [0, ∞).Then, Proof.This follows from taking α = 1 in Theorem 5.
Proof.Multiplying both sides of the equality (10) by .
Further integrating both sides with respect to v over (0, x) leads to Therefore, we have .
Further integrating with respect to v over (0, x), as did in the proof of Theorem 5, concludes Theorem 6.
Proof.This follows from taking α = 1 in Theorem 6. Theorem 7. Let f i for 1 ≤ i ≤ n be positive and increasing functions on [a, b].For α, β > 0, we have Proof.We prove this theorem by induction on n ∈ N. Obviously, the case n = 1 of (11) holds.
For n = 2, since f 1 and f 2 are increasing, we have Now, the left proof of the inequality (11) for n = 2 is the same as that of Theorem 5. Assume that the inequality ( 11) is true for some n ≥ 3. We observe that, since f i is increasing, f = ∏ n i=1 f i is increasing.Let g = f n+1 .Then, applying the case n = 2 to the functions f and g yields where the induction hypothesis for n is used in the deduction of the second inequality.The proof of Theorem 7 is complete.
Corollary 3. Let f i for 1 ≤ i ≤ n be positive and increasing functions on [a, b].For α, β > 0, we have Proof.This follows from taking α = 1 in Theorem 7.
Theorem 8. Let α, β > 0 and the functions f , g : [0, ∞) → R be such that f is increasing, g is differentiable, and g has a lower bound m = inf t∈[0,∞) g (t).Then, where i(x) is the identity function.
Corollary 4.Under conditions of Theorem 8, we have where i(x) is the identity function.
Proof.This follows from taking α = 1 in Theorem 8.

Conclusions
In this paper, we established several Čebyšev type inequalities for conformable k-fractional integral operators.We observed that, if allowing k = 1, inequalities obtained in this paper will reduce to those inequalities in [21].Similarly, if letting α = k = 1, inequalities obtained in this paper will reduce to those inequalities in [11].