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Riemann-Liouville Fractional integral operators with respect to increasing functions and strongly (α; m)-convex functions

In this paper Hadamard type inequalities for strongly (α,m)-convex functions via generalized Riemann-Liouville fractional integrals are studied. These inequalities provide generalizations as well as refinements of several well known inequalities. The established results are further connected with fractional integral inequalities for Riemann-Liouville fractional integrals of convex, strongly convex and strongly m-convex functions. By using two fractional integral identities some more Hadamard type inequalities are proved.


Introduction
Fractional calculus deals with the equations which involve integrals and derivatives of fractional orders. The history of fractional calculus begins from the history of calculus. The role of fractional integral operators is very vital in the applications of this subject in other fields. Several well known phenomenas and their solutions are presented in fractional calculus which can not be studied in ordinary calculus. Inequalities are useful tools in mathematical modelling of real world problems, they also appear as constraints to initial/boundary value problems. Fractional integral/derivative inequalities are of great importance in the study of fractional differential models and fractional dynamical systems. In recent years study of fractional integral/derivative inequalities accelerate very fastly. Many well known classical inequalities have been generalized by using classical and newly defined integral operators in fractional calculus. For some recent work on fractional integral inequalities we refer the readers to [1][2][3][4][5][6] and references therein.
Our goal in this paper is to apply generalize Riemann-Liouville fractional integrals using a monotonically increasing function. The Hadamard inequalities are proved for these integral operators using strongly (α, m)-convex functions. Also error bounds of well known Hadamard inequalities are obtained by using two fractional integral identities. In connection with the results of this paper, we give generalizations and refinements of some well known results added recently in the literature of mathematical inequalities.
Next, we like to give some definitions and established results which are necessary and directly associated with the findings of this paper.
The well-known Hadamard inequality is a very nice geometrical interpretation of convex functions defined on the real line, it is stated as follows: Theorem 1. The following inequality holds: for convex function f : I → R, where I is an interval and x, y ∈ I, x < y.
The definition of Riemann-Liouville fractional integrals is given as follows: Definition 2. Let f ∈ L 1 [a, b]. Then left-sided and right-sided Riemann-Liouville fractional integrals of a function f of order µ where (µ) > 0 are defined by (1.3) and (1.4) The following theorems provide two Riemann-Liouville fractional versions of the Hadamard inequality for convex functions.
]. If f is a convex function on [a, b], then the following fractional integral inequality holds: with µ > 0.
, then the following fractional integral inequality holds: The k-analogue of Riemann-Liouville fractional integrals is defined as follows: (1.8) and The k-fractional versions of Hadamard type inequalities (1.5)-(1.7) are given in the following theorems. Theorem 5. [12] Let f : [a, b] → R be a positive function with 0 ≤ a < b. If f is a convex function on [a, b], then the following inequalities for k-fractional integrals hold: (1.10) Theorem 6. [13] Under the assumption of Theorem 5, the following fractional integral inequality holds: (1.11) , then the following inequality for k-fractional integrals holds: (1.12) In the following, we give the definition of generalized Riemann-Liouville fractional integrals by a monotonically increasing function. Definition 4. [14] Let f ∈ L 1 [a, b]. Also let ψ be an increasing and positive monotone function on (a, b], having a continuous derivative ψ on (a, b). The left-sided and right-sided fractional integrals of a function f with respect to another function ψ on [a, b] of order µ where (µ) > 0 are defined by (1.13) and (1.14) The k-analogue of generalized Riemann-Liouville fractional integrals is defined as follows: . Also let ψ be an increasing and positive monotone function on (a, b], having a continuous derivative ψ on (a, b). The left-sided and right-sided fractional integrals of a function f with respect to another function ψ on [a, b] of order µ where (µ) > 0, k > 0, are defined by (1.15) and (1.16) For more details of above defined fractional integrals, we refer the readers to see [15,16].
Rest of the paper is organized as follows: In Section 2, we find Hadamard type inequalities for generalized Riemann-Liouville fractional integrals with the help of strongly (α, m)-convex functions. The consequences of these inequalities are listed in remarks. Also some new fractional integral inequalities for convex functions, strongly convex functions and strongly m-convex functions are deduced in the form of corollaries. In Section 3, the error bounds of Hadamard type fractional inequalities are established via two fractional integral identities.

Main results
Theorem 8. Let f : [a, b] → R be a positive function with 0 ≤ a < mb and f ∈ L 1 [a, b]. Also suppose that f is strongly (α, m)-convex function on [a, b] with modulus c ≥ 0, ψ is positive strictly increasing function having continuous derivative ψ on (a, b). If [a, b] ⊂ Range(ψ), k > 0 and (α, m) ∈ (0, 1] 2 , then the following k-fractional integral inequality holds: Proof. Since f is strongly (α, m)-convex function, for x, y ∈ [a, b] we have By setting x = at + m(1 − t)b, y = a m (1 − t) + bt and integrating the resulting inequality after multiplying with t µ k −1 , we get 3), then multiplying µ k after applying Definition 5, we get the following inequality: Hence by rearranging the terms, the first inequality is established. On the other hand, f is strongly (α, m)-convex function, for t ∈ [0, 1], we have the following inequality: (2.5) Multiplying inequality (2.5) with t µ k −1 on both sides and then integrating over the interval [0, 1], we get in (2.6), then by applying Definition 5, the second inequality can be obtained.
Corollary 2. Under the assumption of Theorem 8 with k = 1 in (2.1), the following fractional integral inequality holds: Corollary 3. Under the assumption of Theorem 8 with ψ = I in (2.1), the following fractional integral inequality holds: Theorem 9. Under the assumption of Theorem 8, the following k-fractional integral inequality holds: with µ > 0.
2) and integrating the resulting inequality over [0, 1] after multiplying with t µ k −1 , we get (2.8), then by applying Definition 5, we get Hence by rearranging terms, the first inequality is established. Since f is strongly (α, m)-convex function with modulus c ≥ 0, for t ∈ [0, 1], we have following inequality  (2.11), then by applying Definition 5, the second inequality can be obtained.   .7), the following fractional integral inequality holds: Corollary 5. Under the assumption of Theorem 9 with k = 1 in (2.7), the following fractional integral inequality holds: Corollary 6. Under the assumption of Theorem 9 with ψ = I in (2.7), the following fractional integral inequality holds:

Error estimations of Hadamard type fractional inequalities for strongly (α, m)-convex function
In this section, we find the error estimations of Hadamard type fractional inequalities for strongly (α, m)-convex functions by using (1.15) and (1.16) that gives the refinements of already proved estimations. The following lemma is useful to prove the next results. Let a < b and f : [a, b] → R be a differentiable mapping on (a, b). Also, suppose that f ∈ L[a, b], ψ is positive strictly increasing function, having a continuous derivative ψ on (a, b). If [a, b] ⊂ Range(ψ), k > 0, then the following identity holds for generalized fractional integral operators: Proof. We cosider the right hand side of (3.1) as follows: Integrating by parts we get We have v ∈ [a, b] such that ψ(v) = ta + (1 − t)b, with this substitution one can have Similarly one can get after a little computation Theorem 10. Let f : [a, b] → R be a differentiable mapping on (a, b) with 0 ≤ a < b. Also suppose that | f | is strongly (α, m)-convex with modulus c ≥ 0, ψ is positive strictly increasing function having continuous derivative ψ on (a, b). If [a, b] ⊂ Range(ψ), k > 0 and (α, m) ∈ (0, 1] 2 , then the following k-fractional integral inequality holds: with µ > 0 and 2F1 1 + 2α, − µ k , 2(1 + α); 1 2 is regularized hypergeometric function. Proof. By Lemma 1, it follows that Since | f | is strongly (α, m)-convex function on [a, b] and t ∈ [0, 1], we have (3.7) Therefore (3.6) implies the following inequality In the following, we compute integrals appearing on the right side of the above inequality Corollary 8. Under the assumption of Theorem 10 with k = m = 1 and c = 0 in (3.5), the following inequality holds:

Corollary 9.
Under the assumption of Theorem 10 with ψ = I in (3.5), the following inequality holds: For next two results, we need the following lemma.
Proof. Applying Lemma 2 and strongly (α, m)-convexity of | f |, (for q = 1), we have Now for q > 1, we proceed as follows: From Lemma 2 and using power mean inequality, we get This completes the proof.
Corollary 13. Under the assumption of Theorem 12 with c = 0 in 3.14, the following inequality holds: