1. Introduction

A function is said to be convex function if the inequality

(1.1)

holds for all x,y∈[a,b] and t∈[0,1].

One of the most famous inequality for convex functions is so called Hermite-Hadamard’s inequality as follows: Let be a convex function defined on the interval I of real numbers and a.b∈I with a < b. Then:

(1.2)

Fejér ^[22] gave a generalization of the inequalities (1.2) as the following:

If is a convex function, and is nonnegative, integrable and symmetric to then

(1.3)

For some results which generalize, improve, and extend the inequalities (1.3), see ([16-21]^[16]).

In ^[23], Hudzik and Maligrada considered among others the class of functions which are s-convex in the second sense.

Definition 1. A function is said to be s-convex in the second sense if

(1.4)

for all x,y∈[0,∞), ∈[0,1] and for some fixed s ∈(0,1].

It can be easily seen that s = 1, s-convexity reduces to ordinary convexity of functions defined on [0,∞).

In ^[24], Dragomir and Fitzpatrick proved Hermite-Hadamard’s inequality which holds for s-convex functions in the second sense.

Theorem 1. Suppose that f: [0,∞)→[0,∞) is an s-convex functions in the second sense, where s ∈ (0,1), and let a, b∈[0,∞), a < b. If f∈L[a,b], then the following inequalities hold:

(1.5)

We give some necessary definitions and mathematical preliminaries of fractional calculus theory which are used throught this paper.

Definition 2. Let f∈[a,b]. The Riemann-Liouville integrals and of order α > 0 with a ≥ 0 are defined by

(1.6)

and

respectively where . Here is .

In the case of α= 1, the fractional integral reduces to the classical integral.

Let us consider the following special functions:

(1) The Beta function:

(2) The incomplete Beta function:

In ^[13], Sarikaya et. al. represented Hermite-Hadamard’s inequality in fractional integral forms as follows.

Theorem 2. Let be positive function with 0 ≤ a < b and f ∈ L[a,b].

If f is a convex function on [a,b], then the following inequalities for fractional integrals hold

(1.7)

with α > 0.

In ^[13], Sarikaya et. al. proved the following lemma.

Lemma 1. Let be a differentiable mapping on (a,b) with a < b. If f’ ∈L[a,b] then the following equality for fractional integrals holds:

The following Hermite-Hadamard type inequality was proved using the above lemma.

Theorem 3. ^[13] Let be a differentiable mapping on (a,b) with a < b. If |f’| is convex on [a,b] then the following inequality for fractional integrals holds:

Properties concerning this operator and more and more Hermite-Hadamard type inequalities involving fractional integrals for different classes of functions can be found ([1-15]^[1]).

Now, let us give the following lemma which we will use the proof.

Lemma 2. For 0 < α ≤ 1 and 0 ≤ a < b, we have

Işcan ^[17] established following lemmas and theorems connected with the right-hand side of Hermite-Hadamard-Fejer type integral inequality for the fractional integrals.

Lemma 3. If is integrable and symmetric to with a < b, then

with α > 0.

Theorem 4. Let be convex function with a < b and f∈L[a,b]. If is nonnegative, integrable and symmetric to (a + b)/2, then the following inequalities for fractional integrals hold

with α > 0.

Lemma 4. Let be a differentiable mapping on (a,b) with a < b and f’∈L[a,b]. If is integrable and symmetric to (a + b)/2 then the following equality for fractional integral holds

with α > 0.

Theorem 5. Let be a differentiable mapping on and f’∈L[a,b] with a < b. If |f’| is convex on [a,b] and is continuous and symmetric to (a + b)/2, then the following inequality for fractional integral holds

(1.8)

with α > 0.

Theorem 6. Let be a differentiable mapping on and f’ ∈L[a,b] with a < b: If |f’|^q, q > 1, is convex on [a,b] and is continuous and symmetric to (a + b)/2, then the following inequality for fractional integral holds

(1.9)

where α > 0 and .

Theorem 7. Let be a differentiable mapping on and f’ ∈L[a,b] with a < b. If |f’|^q, q > 1, is convex on [a,b] and is continuous and symmetric to (a + b)/2, then the following inequality for fractional integral holds

(i)

(1.10)

with α> 0.

(ii)

(1.11)

for 0 < α ≤ 1 where .

The main of this paper is to establish some new inequalities related to the left-hand side of the Hermete-Hadamard-Fejer type inequalities for s-convex functions in the second sense via Riemann-Liouville fractional integrals.

2. Main Results

Theorem 8. Let be a differentiable mapping on and f’∈L[a,b] with a < b. If |f’| is s-convex on [a,b] for some fixed s ∈(0,1], and is continious and symmetric to (a + b)/2, then the following inequality for fractional integral holds

(2.1)

with α > 0.

Proof. From Lemma 4 we have

(2.2)

Since |f’| is s-convex on [a,b] for some fixed s∈(0,1], we know that for t ∈[a,b]

(2.3)

and since is symmetric to (a + b)/2 we write

then we have

(2.4)

By virtue of (2.2), (2.3) and (2.4), we get

where

(2.5)

and

(2.6)

Remark 1. In Theorem 8, if we take s = 1, then the inequality (2.1) becomes inequality (1.8) of Theorem 5.

Theorem 9. Let be a differentiable mapping on Io and f’ ∈L[a,b] with a < b. If |f’|^q, q > 1, is s-convex on [a,b] for some fixed s∈(0,1], and is continious and symmetric to (a + b)/2, then the following inequality for fractional integral holds

(2.7)

where α > 0.

Proof. Using Lemma 4, Hölder’s inequality, (2.4) and the s-convexity of |f’|^q, it follows that

(2.8)

where it is easily seen that

Hence, if we use (2.5) and (2.6) in (2.8), we have

Remark 2. In Theorem 9, if we take s = 1, then the inequality (2.7) becomes inequality (1.9) of Theorem 6.

Theorem 10. Let be a differentiable mapping on and f’ ∈L[a,b] with a < b. If |f’|^q, q > 1, is s-convex on [a,b] for some fixed s∈(0,1], and is continious and symmetric to (a + b)/2, then the following inequality for fractional integral holds

(i)

(2.9)

with α>0.

(ii)

(2.10)

for 0 < α < 1, where .

Proof. (i) Using Lemma 4, Hölder’s inequality, the inequality (2.4) and the s-convexity of |f’|^q, it follows that

Here we use

for t ∈[0,1/2] and

for t ∈[1/2,1], which follows from

for any A ≥ B ≥ 0 and q ≥ 1. Hence the inequality (2.9) is proved.

(ii) The inequality (2.10) is easily proved using (2.11) and Lemma 2.

Remark 3. In Theorem 10, if we take s = 1, then inequalities (2.9) and (2.10) becomes inequalities (1.10) and (1.11) of Theorem 7.

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