# On Hermite-Hadamard Type Inequalities for Riemann-Liouville Fractional Integrals via Two Kinds of Convexity

Authors: Feixiang Chen
Journal: Chinese Journal of Mathematics
 (2014)

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## Abstract

We obtain some Hermite-Hadamard type inequalities for products of twom-convex functions via Riemann-Liouville integrals. The analogous results for (α,m)-convex functions are also established.


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## Body

## 1. Introduction

Iff:I→R is a convex function on the interval I, then for any a,b∈I with a≠b we have the following double inequality:
(1)f(a+b2)≤1b-a∫ab‍f(t)dt≤f(a)+f(b)2.
This remarkable result is well known in the literature as the Hermite-Hadamard inequality.Since then, some refinements of the Hermite-Hadamard inequality for convex functions have been extensively obtained by a number of authors (e.g., [1–7]).In [8], Toader defined the concept of m-convexity as follows.Definition 1 (see [8]).
The functionf:[0,b]→R is said to be m-convex, where m∈(0,1], if for every x,y∈[0,b] and t∈(0,1) one has
(2)f(tx+m(1-t)y)≤tf(x)+m(1-t)f(y).In [3], Dragomir and Toader proved the following inequality of Hermite-Hadamard type for m-convex functions.Theorem 2 (see [3]).
Letf:[0,∞)→R be a m-convex function with m∈(0,1]; if 0≤a<b<∞ and f∈L1[a,b], then one has the following inequality:
(3)1b-a∫ab‍f(x)dx≤min⁡{f(a)+mf(b/m)2,f(b)+mf(a/m)2}.The notion ofm-convexity has been further generalized in [9] as it is stated in the following definition.Definition 3 (see [9]).
The functionf:[0,b]→R is said to be (α,m)-convex, where (α,m)∈[0,1]2, if for every x,y∈[0,b] and t∈[0,1] one has
(4)f(tx+m(1-t)y)≤tαf(x)+m(1-tα)f(y).In [10], Pachpatte established two new Hermite-Hadamard type inequalities for products of convex functions as follows.Theorem 4 (see [10]).
Letf and g be real-valued, nonnegative, and convex functions on [a,b]. Then
(5)1b-a∫ab‍f(x)g(x)dx≤13M(a,b)+16N(a,b),2f(a+b2)g(a+b2)≤1b-a∫ab‍f(x)g(x)dx+16M(a,b)+13N(a,b),
where M(a,b)=f(a)g(a)+f(b)g(b) and N(a,b)=f(a)g(b)+f(b)g(a).Some Hermite-Hadamard type inequalities for products of twom-convex and (α,m)-convex functions are established in [11].Theorem 5 (see [11]).
Letf,g:[0,∞)→[0,∞) be functions such that fg∈L1[a,b], where 0≤a<b<∞. If f is m1-convex and g is m2-convex on [a,b] for some fixed m1,m2∈(0,1], then
(6)1b-a∫ab‍f(x)g(x)dx≤min⁡{M1,M2},
where
(7)M1=13[f(a)g(a)+m1m2f(bm1)g(bm2)]+16[m2f(a)g(bm2)+m1f(bm1)g(a)],M2=13[f(b)g(b)+m1m2f(am1)g(am2)]+16[m2f(b)g(am2)+m1f(am1)g(b)].Theorem 6 (see [11]).
Letf,g:[0,∞)→[0,∞) be functions such that fg∈L1[a,b], where 0≤a<b<∞. If f is (α1,m1) convex and g is (α2,m2)-convex on [a,b] for some fixed α1,m1,α2,m2∈(0,1], then
(8)1b-a∫ab‍f(x)g(x)dx≤min⁡{N1,N2},
where
(9)N1=f(a)g(a)α1+α2+1+m2[11+α1-11+α1+α2]f(a)g(bm2)+m1[11+α2-11+α1+α2]f(bm1)g(a)+m1m2(1-11+α1-11+α2+11+α1+α2)×f(bm1)g(bm2),N2=f(b)g(b)α1+α2+1+m2[11+α1-11+α1+α2]f(b)g(am2)+m1[11+α2-11+α1+α2]f(am1)g(b)+m1m2(1-11+α1-11+α2+11+α1+α2)×f(am1)g(am2).Some new integral inequalities involving two nonnegative and integrable functions that are related to the Hermite-Hadamard type are also proposed by many authors. In [12], Pachpatte established some Hermite-Hadamard type inequalities involving two log-convex functions. An analogous result for s-convex functions is obtained by Kirmaci et al. in [13]. In [14], Sarikaya et al. presented some integral inequalities for two h-convex functions.It is remarkable that Sarikaya et al. [15] proved the following interesting inequalities of Hermite-Hadamard type involving Riemann-Liouville fractional integrals.Theorem 7 (see [15]).
Letf:[a,b]→R be a positive function with a<b and f∈L1[a,b]. If f is a convex function on [a,b], then the following inequalities for fractional integrals hold:
(10)f(a+b2)≤Γ(α+1)2(b-a)α[Ja+αf(b)+Jb-αf(a)]≤f(a)+f(b)2,
with α>0.We remark that the symbolsJa+α and Jb-αf denote the left-sided and right-sided Riemann-Liouville fractional integrals of the order α≥0 with a≥0 which are defined by
(11)Ja+αf(x)=1Γ(α)∫ax‍(x-t)α-1f(t)dt,x>a,(12)Jb-αf(x)=1Γ(α)∫xb‍(t-x)α-1f(t)dt,x<b,
respectively. Here, Γ(α) is the Gamma function defined by Γ(α)=∫0∞‍e-ttα-1dt.In this paper, we obtain some new Hermite-Hadamard type inequalities for products of twom-convex functions via Riemann-Liouville integrals. The analogous results for (α,m)-convex functions are also given.


## 2. Inequalities for Products of Two Functions for Riemann-Liouville Fractional Integrals

Theorem 8.
Letf,g:[0,∞)→[0,∞), 0≤a<b, be functions such that fg∈L1[a,b]. If f is m1-convex and g is m2-convex on [a,b] with m1,m2∈(0,1], then one has
(13)Γ(α)(b-a)αJa+αf(b)g(b)≤f(a)g(a)α+2+m2(α+1)(α+2)f(a)g(bm2)+m1(α+1)(α+2)g(a)f(bm1)+2m1m2α(α+1)(α+2)f(bm1)g(bm2),Γ(α)(b-a)αJb-αf(a)g(a)≤f(b)g(b)α+2+m2(α+1)(α+2)f(b)g(am2)+m1(α+1)(α+2)g(b)f(am1)+2m1m2α(α+1)(α+2)f(am1)g(am2).Proof.
Sincef is m1-convex and g is m2-convex on [a,b], then for t∈[0,1] we get
(14)f(ta+(1-t)b)≤tf(a)+m1(1-t)f(bm1),g(ta+(1-t)b)≤tg(a)+m2(1-t)g(bm2).
From (14), we get
(15)f(ta+(1-t)b)g(ta+(1-t)b)≤t2f(a)g(a)+m2f(a)g(bm2)t(1-t)+m1g(a)f(bm1)t(1-t)+m1m2f(bm1)g(bm2)(1-t)2.
Multiplying both sides of the above inequality by tα-1 and then integrating the resulting inequality with respect to t over [0,1], we obtain
(16)∫01‍tα-1f(ta+(1-t)b)g(ta+(1-t)b)dt=∫ba‍(b-ub-a)α-1f(u)g(u)dua-b=Γ(α)(b-a)αJa+αf(b)g(b)≤f(a)g(a)∫01‍tα+1dt+m2f(a)g(bm2)∫01‍tα(1-t)dt+m1g(a)f(bm1)∫01‍tα(1-t)dt+m1m2f(bm1)g(bm2)∫01‍tα-1(1-t)2dt=f(a)g(a)α+2+m2(α+1)(α+2)f(a)g(bm2)+m1(α+1)(α+2)g(a)f(bm1)+2m1m2α(α+1)(α+2)f(bm1)g(bm2).
Analogously, we obtain
(17)f((1-t)a+tb)g((1-t)a+tb)≤t2f(b)g(b)+m2f(b)g(am2)t(1-t)+m1g(b)f(am1)t(1-t)+m1m2f(am1)g(am2)(1-t)2.
Multiplying both sides of above inequality by tα-1 and then integrating the resulting inequality with respect to t over [0,1], we obtain
(18)∫01‍tα-1f((1-t)a+tb)g((1-t)a+tb)dt=∫ab‍(v-ab-a)α-1f(v)g(v)dvb-a=Γ(α)(b-a)αJb-αf(a)g(a)≤f(b)g(b)∫01‍tα+1dt+m2f(b)g(am2)∫01‍tα(1-t)dt+m1g(b)f(am1)∫01‍tα(1-t)dt+m1m2f(am1)g(am2)∫01‍tα-1(1-t)2dt=f(b)g(b)α+2+m2(α+1)(α+2)f(b)g(am2)+m1(α+1)(α+2)g(b)f(am1)+2m1m2α(α+1)(α+2)f(am1)g(am2),
which completes the proof.Corollary 9.
With assumptions in Theorem8, if α=1, one gets
(19)1b-a∫ab‍f(x)g(x)dx≤13[f(a)g(a)+m1m2f(bm1)g(bm2)]+16[m2f(a)g(bm2)+m1f(bm1)g(a)],1b-a∫ab‍f(x)g(x)dx≤13[f(b)g(b)+m1m2f(am1)g(am2)]+16[m2f(b)g(am2)+m1f(am1)g(b)],
which is just the result in Theorem 5.Corollary 10.
With assumptions in Theorem8, one gets
(20)Γ(α)2(b-a)α[Ja+αf(b)g(b)+Jb-αf(a)g(a)]≤12(α+2)[f(a)g(a)+f(b)g(b)]+m22(α+1)(α+2)[f(b)g(am2)+f(a)g(bm2)]+m12(α+1)(α+2)[g(a)f(bm1)+g(b)f(am1)]+m1m2α(α+1)(α+2)×[f(bm1)g(bm2)+f(am1)g(am2)].Corollary 11.
With assumptions in Theorem8, if one chooses g:[a,b]→R as g(x)=1 and m2=1 for all x∈[a,b], one has
(21)Γ(α)(b-a)αJa+αf(b)≤f(a)α+1+m1α(α+1)f(bm1),Γ(α)(b-a)αJb-αf(a)≤f(b)α+1+m1α(α+1)f(am1),Γ(α)(b-a)α[Ja+αf(b)+Jb-αf(a)]≤f(b)+f(a)α+1+m1α(α+1)[f(bm1)+f(am1)].Theorem 12.
Letf,g:[0,∞)→[0,∞),  0≤a<b, be functions such that fg∈L1[a,b]. If f is (α1,m1)-convex and g is (α2,m2)-convex on [a,b] with (α1,m1),(α2,m2)∈(0,1]2, respectively, then one has
(22)Γ(α)(b-a)αJa+αf(b)g(b)≤1α1+α2+αf(a)g(a)+α2(α+α1)(α+α1+α2)m2f(a)g(bm2)+α1(α+α2)(α+α1+α2)m1g(a)f(bm1)+(1α-1α+α1-1α+α2+1α+α1+α2)×m1m2f(bm1)g(bm2),Γ(α)(b-a)αJb-αf(a)g(a)≤1α1+α2+αf(b)g(b)+α2(α+α1)(α+α1+α2)m2f(b)g(am2)+α1(α+α2)(α+α1+α2)m1g(b)f(am1)+(1α-1α+α1-1α+α2+1α+α1+α2)×m1m2f(am1)g(am2).Proof.
Sincef is (α1,m1)-convex and g is (α2,m2)-convex on [a,b], then for t∈[0,1] we get
(23)f(ta+(1-t)b)≤tα1f(a)+m1(1-tα1)f(bm1),g(ta+(1-t)b)≤tα2g(a)+m2(1-tα2)g(bm2).
From (23), we get
(24)f(ta+(1-t)b)g(ta+(1-t)b)≤tα1+α2f(a)g(a)+m2f(a)g(bm2)tα1(1-tα2)+m1g(a)f(bm1)tα2(1-tα1)+m1m2f(bm1)g(bm2)(1-tα1)(1-tα2).
Multiplying both sides of above inequality by tα-1 and then integrating the resulting inequality with respect to t over [0,1], we obtain
(25)∫01‍tα-1f(ta+(1-t)b)g(ta+(1-t)b)dt=∫ba‍(b-ub-a)α-1f(u)g(u)dua-b=Γ(α)(b-a)αJa+αf(b)g(b)≤f(a)g(a)∫01‍tα1+α2+α-1dt+m2f(a)g(bm2)∫01‍tα-1tα1(1-tα2)dt+m1g(a)f(bm1)∫01‍tα-1tα2(1-tα1)dt+m1m2f(bm1)g(bm2)∫01‍tα-1(1-tα1)(1-tα2)dt=1α1+α2+αf(a)g(a)+α2m2(α+α1)(α+α1+α2)f(a)g(bm2)+α1m1(α+α2)(α+α1+α2)g(a)f(bm1)+(1α-1α+α1-1α+α2+1α+α1+α2)×m1m2f(bm1)g(bm2).
Similarly, we have
(26)f((1-t)a+tb)g((1-t)a+tb)≤tα1+α2f(b)g(b)+m2f(b)g(am2)tα1(1-tα2)+m1g(b)f(am1)tα2(1-tα1)+m1m2f(am1)g(am2)(1-tα1)(1-tα2),(27)∫01‍tα-1f((1-t)a+tb)g((1-t)a+tb)dt=∫ab‍(v-ab-a)α-1f(v)g(v)dvb-a=Γ(α)(b-a)αJb-αf(a)g(a)≤f(b)g(b)∫01‍tα1+α2+α-1dt+m2f(b)g(am2)∫01‍tα-1tα1(1-tα2)dt+m1g(b)f(am1)∫01‍tα-1tα2(1-tα1)dt+m1m2f(am1)g(am2)∫01‍tα-1(1-tα1)(1-tα2)dt=1α1+α2+αf(b)g(b)+α2(α+α1)(α+α1+α2)m2f(b)g(am2)+α1(α+α2)(α+α1+α2)m1g(b)f(am1)+(1α-1α+α1-1α+α2+1α+α1+α2)×m1m2f(am1)g(am2).
We get the desired result.Corollary 13.
With assumptions in Theorem12, if α=1, then
(28)1b-a∫ab‍f(x)g(x)dx≤f(a)g(a)α1+α2+1+m2[11+α1-11+α1+α2]×f(a)g(bm2)+m1[11+α2-11+α1+α2]f(bm1)g(a)+m1m2(1-11+α1-11+α2+11+α1+α2)×f(bm1)g(bm2),1b-a∫ab‍f(x)g(x)dx≤f(b)g(b)α1+α2+1+m2[11+α1-11+α1+α2]×f(b)g(am2)+m1[11+α2-11+α1+α2]f(am1)g(b)+m1m2(1-11+α1-11+α2+11+α1+α2)×f(am1)g(am2),
which is just the result in Theorem 6.Corollary 14.
With assumptions in Theorem12, one gets
(29)Γ(α)(b-a)α[Ja+αf(b)g(b)+Jb-αf(a)g(a)]≤1α1+α2+α[f(a)g(a)+f(b)g(b)]+α2m2(α+α1)(α+α1+α2)×[f(b)g(am2)+f(a)g(bm2)]+α1m1(α+α2)(α+α1+α2)×[g(b)f(am1)+g(a)f(bm1)]+(1α-1α+α1-1α+α2+1α+α1+α2)×m1m2[f(am1)g(am2)+f(bm1)g(bm2)].


## 3. Conclusion

In this paper, we obtain some new Hermite-Hadamard type inequalities for products of twom-convex functions via Riemann-Liouville integrals. The analogous results for (α,m)-convex functions are also established. An interesting topic is whether we can use the methods in this paper to establish the Hermite-Hadamard inequalities for products of two convex functions on the coordinates via Riemann-Liouville integrals.



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