IMPROVED HERMITE HADAMARD TYPE INEQUALITIES FOR HARMONICALLY CONVEX FUNCTIONS VIA KATUGAMPOLA FRACTIONAL INTEGRALS

In this paper, we prove three new Katugampola fractional HermiteHadamard type inequalities for harmonically convex functions by using the left and the right fractional integrals independently. One of our Katugampola fractional Hermite-Hadamard type inequalities is better than given in [17]. Also, we give two new Katugampola fractional identities for differentiable functions. By using these identities, we obtain some new trapezoidal type inequalities for harmonically convex functions. Our results generalize many results from earlier papers.


Introduction
Let f : I R !R be a convex function de…ned on the interval I of real numbers and a; b 2 I with a < b.The inequality is well known in the literature as Hermite-Hadamard's inequality.There are so many generalizations and extensions of inequalities (1) for various classes of functions.One of this classes of functions is harmonically convex functions de…ned by • Işcan.
In [7], • Işcan gave the de…nition of harmonically convex functions as follows: De…nition 1. [7] Let I Rn f0g be a real interval.A function f : I !R is said to be harmonically convex, if for all x; y 2 I and t 2 [0; 1].If the inequality in (2) is reversed, then f is said to be harmonically concave.
In [9], • Işcan and Wu presented Hermite-Hadamard type inequalities for harmonically convex functions in fractional integral forms as follows: with > 0 and h(x) = 1=x.
In [20], Şanl¬ et al. proved the following three Riemann-Liouville fractional Hermite-Hadamard type inequalities for harmonically convex functions by using the left and the right fractional integrals seperately as follows: where h (x) = 1 x and > 0. Theorem 6.Let f : I (0; 1) !R be a harmonically convex function and a; b 2 I with a < b.If f 2 L [a; b], then the following inequality for the right Riemann-Liouville fractional integral holds: where h (x) = 1 x and > 0. Theorem 7. Let f : I (0; 1) !R be a harmonically convex function and a; b 2 I with a < b.If f 2 L [a; b], then the following inequality for the Riemann-Liouville fractional integral holds: where h (x) = 1 x and > 0. The following de…nitions of Katugampola fractional integrals could be found in [4,11].
R be a …nite interval.Then the left and right-side Katugampola fractional integrals of order > 0 of f 2 X p c (a; b) are de…ned by and with a < x < b and > 0, respectively.
(See [10], for the de…nition of the set X p c (a; b)) It is easily seen that if one takes ! 1 in the De…nition 8, one has the De…nition 3.
In [17, For the Theorem 10 , the correct inequality should be expressed as follows: In (8), if one takes ! 1, one obtaines the inequality (4) in the Theorem 4.

De…nition 11. [19, page 12]
A function f de…ned on I has a support at x 0 2 I if there exists an a¢ ne functions A (x) = f (x 0 ) + m (x x 0 ) such that A (x) f (x) for all x 2 I.The graph of the support function A is called a line of support for f at x 0 .
Theorem 12. [19, page 12] f : (a; b) !R is a convex function if and only if there is at least one line of support for f at each x 0 2 (a; b).
In literature, there are so many studies for Hermite-Hadamard type inequalities by using the left and right fractional integrals (such as Riemann-Liouville fractional integrals, Hadamard fractional integrals, Katugampola fractional integrals etc.).In all of them, the left and right fractional integrals are used together.As much as we know, the studies [20] are the …rst two works by using only the right fractional integrals or the left fractional integrals.
In this paper, our aim is to obtain new Katugampola fractional Hermite-Hadamard type inequalities by using only the right or the left fractional integrals separately for harmonically convex functions.

Katugampola Fractional Hermite Hadamard Type Inequalities for
Harmonically Convex Functions Theorem 14.Let f : then the following inequality for the left katugampola fractional integral holds: where > 0, > 0 and h then by using Remark 13 the function g (x) = f 1 x is convex on 1 b ; 1 a .Hence using Theorem 12, there is at least one line of support .From (10) and harmonically convexity of f , we have for all t 2 [0; 1].Multiplying all sides of (11) with t 1 and integrating over [0; 1] respect to t, we have This completes the proof.
(2) if one takes ! 1, and after that if one takes = 1, one has the inequality (3).
where > 0, > 0 and h (x) = 1 x .Proof.Let > 0. Since f is harmonically convex on [a ; b ], then by using Remark 13 the function g (x) = f 1 x is convex on 1 b ; 1 a .Hence using Theorem 12, there is at least one line of support for all x 2 1 b ; 1 a and m 2 . From (10) and harmonically convexity of f , we have for all t 2 [0; 1].Multiplying all sides of (14) with t 1 and integrating over [0; 1] respect to t, we have This completes the proof.
(2) If one takes ! 1, and after that if one takes = 1, one has the inequality (3).
Theorem 18.Let f : I (0; 1) !R be a function such that f 2 X p c (a ; b ), where a ; b 2 I with a < b.If f is a harmonically convex function on [a ; b ] , then the following inequality for katugampola fractional integrals hold: where > 0, > 0 and h (x) = 1 x Proof.Adding the inequalities (9) and (12) side by side, then multiplying the resulting inequalities by 1 2 , we have the inequalities (15).
(2) if one takes ! 1, and after that if one takes = 1, one has the inequality Corollary 20.The left hand side of (15) is better than the left hand side of (8).
Proof.Since f is harmonically convex on [a ; b ], it is clear from

Lemmas
In this section we will prove two new identities used in forward results.
Lemma 21.Let f : I R !R be a di¤ erentiable function on I , a ; b 2 I with a < b.If the fractional integrals exist and f 0 2 L [a ; b ], then the following equality for the left katugampola fractional integral holds: where > 0 and > 0.
Proof.It could be prove directly by applying the partial integration to the right hand side of the equation ( 16) as follows: This completes the proof.
Proof.It could be prove directly by applying the partial integration to the right hand side of the equation (17) as follows: This completes the proof.

Some new conformable fractional trapezoid type inequalities for harmonically convex functions
In this section, we will prove some new conformable fractional trapezoid type inequalities for harmonically convex functions by using Lemma 21 and Lemma 23. q is harmonically convex on [a ; b ] for q 1, then the following inequality for the left katugampola fractional integral holds: where Proof.By using Lemma 21, power mean inequality and harmonically convexity of f 0 q , we have (19) Calculating the appearing integrals in (19) we have,    q is harmonically convex on [a ; b ] for q > 1 and 1 q + 1 p = 1, then the following inequality for the left katugampola fractional integral holds: jf 0 (a )j q Z 5 (a; b; ; ) + jf 0 (b )j q Z6 ( ; ; p) with > 0 and > 0.
Proof.By using Lemma 21, Hölder inequality and harmonically convexity of f 0 q , we have Calculating the appearing integrals in (20), we have  jf 0 (a )j q Z 8 (a; b; ; ) + jf 0 (b )j q Z 9 (a; b; ; )  Proof.Similarly the proof of the Theorem 25, by using Lemma 23, power mean inequality and harmonically convexity of f 0 q , we have (28).
Proof.Similarly the proof of the Theorem 27, by using Lemma 23, Hölder inequality and harmonically convexity of f 0 q , we have (29).

Theorem 2 . [ 7 ]
Let f : I Rn f0g !R be a harmonically convex function and a; b 2 I with a < b.If f 2 L [a; b] ; then the following inequalities hold: de…nitions of the left and right side Riemann-Liouville fractional integrals are well known in the literature.De…nition 3. Let a; b 2 R with a < b and f 2 L [a; b].The left and right Riemann-Liouville fractional integrals J a+ f and J b f of order > 0 are de…ned by

Theorem 4 .
Let f : I (0; 1) !R be a function such that f 2 L [a; b], where a; b 2 I with a < b.If f is a harmonically convex function on [a; b], then the following inequalities for fractional integrals hold:

Theorem 5 .
Let f : I (0; 1) !R be a harmonically convex function and a; b 2 I with a < b.If f 2 L [a; b], then the following inequality for the left Riemann-Liouville fractional integral holds: Theorem 2.1], Mumcu et al. presented Hermite-Hadamard type inequalities for harmoncally convex functions in Katugampola fractional integral forms as follows: Theorem 10.Let > 0 and > 0. Let f : I (0; 1) !R be a function such that f 2 X p c (a ; b ), where a ; b 2 I with a < b.If f is a harmonically convex function on [a; b], then the following inequalities hold:

Theorem 16 .
Let f : I (0; 1) !R be a function such that f 2 X p c (a ; b ), where a ; b 2 I with a < b.If f is a harmonically convex function on [a ; b ] , then the following inequality for the right katugampola fractional integral holds:

( 1 )
if one takes ! 1, one has the inequality [20, Lemma 3].(2) if one takes ! 1, and after that if one takes = 1, one has the inequality [7, 2.5.Lemma].Lemma 23.Let f : I R !R be a di¤ erentiable function on I , a ; b 2 I with a < b .If the fractional integrals exist and f 0 2 L [a ; b ], then the following equality for the right katugampola fractional integral holds:

Theorem 25 .
Let f : I R !R be a di¤ erentiable function on I , a ; b 2 I with a < b.If f 0 2 L [a ; b ] and jf 0 j

Theorem 29 .
Let f : I R !R be a di¤ erentiable function on I , a ; b 2 I with a < b .If f 0 2 L [a ; b ] and jf 0 jq is harmonically convex on [a ; b ] for q 1, then the following inequality for the right katugampola fractional integral holds:f (a ) + f (b )