On new inequalities of Hermite–Hadamard–Fejer type for harmonically convex functions via fractional integrals

In this paper, firstly, new Hermite–Hadamard type inequalities for harmonically convex functions in fractional integral forms are given. Secondly, Hermite–Hadamard–Fejer inequalities for harmonically convex functions in fractional integral forms are built. Finally, an integral identity and some Hermite–Hadamard–Fejer type integral inequalities for harmonically convex functions in fractional integral forms are obtained. Some results presented here provide extensions of others given in earlier works.

We recall the following inequality and special functions which are known as Beta and hypergeometric function respectively: Lemma 1 (Prudnikov et al. 1981;Wang et al. 2013) For 0 < α ≤ 1 and 0 ≤ a < b we have The following definitions and mathematical preliminaries of fractional calculus theory are used further in this paper.
Definition 1 (Kilbas et al. 2006 . Because of the wide application of Hermite-Hadamard type inequalities and fractional integrals, many researchers extend their studies to Hermite-Hadamard type inequalities involving fractional integrals not limited to integer integrals. Recently, more and more Hermite-Hadamard inequalities involving fractional integrals have been obtained for different classes of functions; see Dahmani (2010), İşcan (2013b, 2014a), İşcan and Wu (2014), Mihai and Ion (2014), Sarıkaya et al. (2013), Wang et al. (2012), Wang et al. (2013). İşcan (2014b) can defined the so-called harmonically convex functions and established following Hermite-Hadamard type inequality for them as follows: Definition 2 Let I ⊂ R\{0} be a real interval. A function f : I → R is said to be harmonically convex, if Kilbas et al. 2006).
for all x, y ∈ I and t ∈ [0, 1]. If the inequality in (3) is reversed, then f is said to be harmonically concave.
Theorem 2 (İşcan 2014b) Let f : I ⊂ R\{0} → R be a harmonically convex function and a, b ∈ I. If f ∈ L[a, b] then the following inequalities holds: Latif et al. (2015) gave the following definition: Chen and Wu (2014) presented a Hermite-Hadamard-Fejer type inequality for harmonically convex functions as follows: Theorem 3 Let f : I ⊆ R\{0} → R be a harmonically convex function and a, b ∈ I. If f ∈ L[a, b] and g : [a, b] ⊆ R\{0} → R is nonnegative, integrable and harmonically symmetric with respect to 2ab a+b , then In this paper, we give new Hermite-Hadamard type inequalities for harmonically convex functions in fractional integral forms. We establish new Hermite-Hadamard-Fejer inequalities for harmonically convex functions in fractional integral forms. We obtain an integral identity and some Hermite-Hadamard-Fejer type integral inequalities for harmonically convex functions in fractional integral forms.

Main results
Throughout this section, we write g ∞ = sup t∈ [a,b] g(t) , for the continuous function g : [a, b]→ R.
Lemma 2 If g : [a, b] ⊆ R\{0}→ R is integrable and harmonically symmetric with respect to 2ab a+b , then Proof Since g is harmonically symmetric with respect to 2ab a+b , using Definition 3 we have g 1 Hence, in the following integral setting This completes the proof.
Theorem 4 Let f : If f is a harmonically convex function on [a, b], then the following inequalities for fractional integrals holds: Proof Since f is a harmonically convex function on [a, b], we have for all t ∈ [0, 1] Multiplying both sides of (7) by 2t α−1 , then integrating the resulting inequality with respect to t over 0, 1 2 , we obtain Setting x = tb+(1−t)a ab and dx = b−a ab dt gives and the first inequality is proved. For the proof of the second inequality in (6), we first note that, if f is a harmonically convex function, then, for all t ∈ [0, 1], it yields Then multiplying both sides of (8) by t α−1 and integrating the resulting inequality with respect to t over 0, 1 2 , we obtain i.e.
The proof is completed.
]→ R is nonnegative, integrable and harmonically symmetric with respect to 2ab a+b , then the following inequalities for fractional integrals holds: , multiplying both sides of (7) by 2t α−1 g ab tb+(1−t)a , then integrating the resulting inequality with respect to t over 0, 1 2 , we obtain Since g is harmonically symmetric with respect to 2ab a+b , using Definition 3 we have Therefore, by Lemma 2 we have and the first inequality is proved. For the proof of the second inequality in (9) we first note that if f is a harmonically convex function, then, multiplying both sides of (8) by t α−1 g ab tb+(1−t)a and integrating the resulting inequality with respect to t over 0, 1 2 , we obtain The proof is completed.
]→ R is integrable and harmonically symmetric with respect to 2ab a+b , then the following equality for fractional integrals holds: Proof It suffices to note that By integration by parts and Lemma 2 we get and similarly Thus, we can write Multiplying both sides by (Γ (α)) −1 we obtain (10). This completes the proof.
Proof From Lemma 3 we have and similarly we get If we use (15) and (16) in (14) , we have (11). This completes the proof.

Corollary 1 In Theorem 6:
(1) If we take α = 1 we have the following Hermite-Hadamard-Fejer inequality for harmonically convex functions which is related to the left-hand side of (5): (2) If we take g(x) = 1 we have following Hermite-Hadamard type inequality for harmonically convex functions in fractional integral forms which is related to the left-hand side of (6): (3) If we take α = 1 and g(x) = 1 we have the following Hermite-Hadamard type inequality for harmonically convex functions which is related to the left-hand side of (4): Proof Using (12) , power mean inequality and the harmonically convexity of f ′ q , it follows that For the appearing integrals, we have    (1 − u) α (ub + (1 − u)a) 2 udu = C 6 (α) − C 7 (α) = C 8 (α).