INEQUALITIES USING MIXED CONFORMABLE FRACTIONAL INTEGRALS

. During the past two decades or so, fractional integral operators have been one of the most important tools in the development of inequalities theory. By this means, a lot generalized intergral inequalities involving various the fractional integral operators have been presented in the literature. Very recently, mixed conformable fractional integral operators has been introduced by T. Abdeljawad and with the help of these operators some new integral inequalities are obtained. The main aim of the paper is to establish some new Chebyshev type fractional integral inequalities by using mixed conformable fractional integral operators.


Introduction and Preliminaries
In the present paper, our work is based on a celebrated functional introduced by Chebyshev [4], which is de…ned by (1) where f and g are two integrable functions which are synchronous on [a; b], i.e.
(f (x) f (y)) (g(x) g(y)) 0 for any x; y 2 [a; b], then the Chebyshev inequality is given by T (f; g) 0. The Chebyshev functional (1) has many applications in numerical quadrature, transform theory, probability, study of existence for solutions of di¤erential equations, and in statistical problems. Moreover, under suitable assumptions (Chebyshev inequality, Grüss inequality, Minkowski inequality, Hermite-Hadamard inequality, Ostrowski inequality etc.), inequalities are playing a signi…cant role in the …eld of mathematical sciences, particularly, in the theory of approximations.
A remarkably large number inequalities of above type involving the special fractional integral (such as the Riemann-Liouville, conformable, Erdélyi-Kober, Katugampola, Hadamard and Weyl types) have been investigated by many researchers and received considerable attention to it (see [8-10, 14, 16]). Now, some fractional integral operators and Chebyshev type inequalities obtained with the help of these operators will be given in the following order: and where is the familiar Gamma function (see, e.g., [19,Section 1.1]). It is noted that J 1 a+ f (x) and J 1 b f (x) become the usual Riemann integrals. Theorem 2.
[13] Let p be a positive function and let f and g be two di¤ erentiable functions on for all t > 1, > 0 and > 0.
and (ii) The mixed right conformable fractional integral of f is de…ned by For recent results related to this operators, we refer the reader [1, 6, 18].

Main Results
We obtain in this section certain integral inequalities for the di¤erentiable functions involving the mixed conformable fractional integral operator.
Proof. Let f and g be two functions satisfying the conditions of Theorem 17 and let p be a positive function on [0; 1[. De…ne H(x; y) := (f (x) f (y)(g(x) g(y))); x; y 2 (0; t); t > 0: and integrating the resulting identity with respect to x from 0 to t, we can write Now, multiplying (7) by 1 (b y) 1 p(y) and integrating the resulting identity with respect to y from 0 to t, we can write On the other hand, we have Using Hölder inequality for double integral, we can write Since and then, we can estimate H as follows: On the other hand, we have Applying again Hölder inequality to right-hand side of (14), we can write Now, using the fact that Z y x jf 0 (u)j r du jjf 0 jj r r ; Z y x jg 0 (w)j s dw jjg 0 jj s s ; we obtain From (17), we get Since r 1 + s 1 = 1, then we have By the relations (6) and (19) and using the properties of the modulus, we get the …rst inequality in Theorem 17 . Now we shall prove the second inequality of Theorem 17, we have 0 x t; 0 y t: Hence 0 jx yj t: 1066 B. ÇEL · IK, E. SET Therefore, we have Theorem 17 is thus proved.
where H(x; y) are the same as given in (6).
Proof. Using the identity (7), we can write From the relation (13), we can obtain the following estimation Therefore, we have Applying Hölder inequality for double integral to the right-hand side of (24), yields (b y) 1 p(x)p(y)jH(x; y)jdxdy " 1 r ( ) Z y x jf 0 (u)j r du dxdy Z y x jg 0 (w)j s dw dxdy  (16) and (25), we get (b y) 1 p(x)p(y)jH(x; y)jdxdy jjf 0 jj r jjg 0 jj s ( ) ( ) Using (22) and (26) and the properties of modulus, we get the …rst inequality in (21).

Remarks
Now, let us brie ‡y consider some special cases of the main results. In Theorem 17 and Theorem 18, if we choose = 1 and make use of the relationship (3), then the main results are reduced to Theorem 2 and Theorem 3 obtained by Dahmani et al. [7].