We use basic properties of superquadratic functions to obtain some new Hermite-Hadamard type inequalities via Riemann-Liouville fractional integrals. For superquadratic functions which are also convex, we get refinements of existing results.

1. Introduction

Let f:I⊆R→R be a convex function and a,b∈I with a<b; then
(1)f(a+b2)≤1b-a∫abf(t)dt≤f(a)+f(b)2
is known as the Hermite-Hadamard inequality.

This remarkable result is well known in the literature as the Hermite-Hadamard inequality. Recently, the generalizations, refinements, and improvements of the classical Hermite-Hadamard inequality have been the subject of intensive research.

Definition 1 (see [<xref ref-type="bibr" rid="B1">1</xref>]).

A function f:[0,∞)→R is superquadratic provided that for all x≥0 there exists a constant Cx∈R such that
(2)f(y)≥f(x)+Cx(y-x)+f(|y-x|),
for all y≥0.

Theorem 2 (see [<xref ref-type="bibr" rid="B1">1</xref>]).

The inequality
(3)f(∫gdμ)≤∫[f(g(s))-f(|g(s)-∫gdμ|)]dμ(s)
holds for all probability measures μ and all nonnegative, μ-integrable functions g, if and only if f is superquadratic.

The discrete version of the above theorem is also used in the sequel.

Lemma 3 (see [<xref ref-type="bibr" rid="B3">2</xref>]).

Suppose that f is superquadratic. Let xi≥0, 1≤i≤n, and let x¯=∑i=1nλixi, where λi≥0 and ∑i=1nλi=1. Then
(4)∑i=1nλif(xi)≥f(x¯)+∑i=1nλif(|xi-x¯|).

Nonnegative superquadratic functions are much better behaved as we see next.

Lemma 4 (see [<xref ref-type="bibr" rid="B1">1</xref>]).

Let f be a superquadratic function with Cx∈R as in Definition 1. Then one gets the following:

f(0)≤0;

if f(0)=f′(0)=0, then Cx=f′(x) whenever f is differentiable at x>0;

if f≥0, then f is convex and f(0)=f′(0)=0.

The Hermite-Hadamard inequalities for superquadratic functions are established by Banic et al. in [3].

Theorem 5 (see [<xref ref-type="bibr" rid="B5">3</xref>]).

Let f:[0,∞)→R be a superquadratic function and 0≤a<b; then
(5)f(a+b2)+1b-a∫abf(|x-a+b2|)dx≤1b-a∫abf(x)dx≤f(a)+f(b)2-1(b-a)2∫ab[(b-x)f(x-a)+(x-a)f(b-x)]dx.

It is remarkable that Sarikaya et al. [4] proved the following interesting inequalities of Hermite-Hadamard type involving Riemann-Liouville fractional integrals.

Theorem 6 (see [<xref ref-type="bibr" rid="B11">4</xref>]).

Let f:[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:
(6)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 symbols Ja+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
(7)Ja+αf(x)=1Γ(α)∫ax(x-t)α-1f(t)dt,x>a,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.

Fractional integral operators are widely used to solve differential equations and integral equations. So a lot of work has been obtained on the theory and applications of fractional integral operators.

For more results concerning the fractional integral operators, we refer the reader to [5–10] and references cited therein.

In this paper, we establish some new Hermite-Hadamard type inequalities for superquadratic functions via Riemann-Liouville fractional integrals which refine the inequalities of (6) for superquadratic functions which are also convex.

2. Main Results

From Lemma 3 for n=2, we get for 0≤m≤z≤M, m≤M, for the superquadratic function f that
(8)f(z)≤M-zM-mf(m)+z-mM-mf(M)-M-zM-mf(z-m)-z-mM-mf(M-z),f(M+m-z)≤z-mM-mf(m)+M-zM-mf(M)-M-zM-mf(z-m)-z-mM-mf(M-z)
hold, and therefore
(9)f(z)+f(M+m-z)≤f(m)+f(M)-2M-zM-mf(z-m)-2z-mM-mf(M-z).
Let 0≤a≤x≤(a+b)/2; we get that
(10)a≤x≤a+b2≤a+b-x≤b.
Therefore, by replacing in (9)
(11)z=a+b2,M=a+b-x,m=x,
we get that
(12)2f(a+b2)≤f(x)+f(a+b-x)-2f(a+b2-x).

Theorem 7.

Let f be a superquadratic integrable function on [0,b] with 0≤a<b. Then
(13)f(a+b2)≤Γ(α+1)2(b-a)α[Ja+αf(b)+Jb-αf(a)]-α2(b-a)α∫abf|a+b2-x|×((b-x)α-1+(x-a)α-1)dx
with α>0.

Proof.

By inequality (12), we get that
(14)f(a+b2)=f(a+b2)α2(b-a)α×∫ab((b-x)α-1+(x-a)α-1)dx=f(a+b2)α(b-a)α×∫a(a+b)/2((b-x)α-1+(x-a)α-1)dx≤α2(b-a)α∫a(a+b)/2[f(x)+f(a+b-x)]×((b-x)α-1+(x-a)α-1)dx-α(b-a)α∫a(a+b)/2f(a+b2-x)×((b-x)α-1+(x-a)α-1)dx.
By the change of variable x→a+b-x, we get
(15)α2(b-a)α∫a(a+b)/2f(a+b-x)×((b-x)α-1+(x-a)α-1)dx=α(b-a)α∫(a+b)/2bf(x)×((b-x)α-1+(x-a)α-1)dx,α(b-a)α∫a(a+b)/2f(a+b2-x)×((b-x)α-1+(x-a)α-1)dx=α(b-a)α∫(a+b)/2bf(x-a+b2)×((b-x)α-1+(x-a)α-1)dx.
Therefore
(16)f(a+b2)=f(a+b2)α2(b-a)α×∫ab((b-x)α-1+(x-a)α-1)dx=f(a+b2)α(b-a)α×∫a(a+b)/2((b-x)α-1+(x-a)α-1)dx≤α2(b-a)α∫abf(x)×((b-x)α-1+(x-a)α-1)dx-α2(b-a)α∫abf|a+b2-x|×((b-x)α-1+(x-a)α-1)dx=Γ(α+1)2(b-a)α[Ja+αf(b)+Jb-αf(a)]-α2(b-a)α∫abf|a+b2-x|×((b-x)α-1+(x-a)α-1)dx.
We have completed the proof.

Corollary 8.

Putting α=1 in Theorem 7 gives
(17)f(a+b2)≤1b-a∫abf(x)dx-1b-a∫abf|a+b2-x|dx.

Let 0≤a≤x≤(a+b)/2; we get that
(18)a≤x≤a+b-x≤b.
Therefore, by replacing in (9)
(19)z=x,M=b,m=a,
we get that
(20)f(x)+f(a+b-x)≤f(a)+f(b)-2b-xb-af(x-a)-2x-ab-af(b-x).

Theorem 9.

Let f be a superquadratic integrable function on [0,b] with 0≤a<b. Then
(21)Γ(α+1)2(b-a)α[Ja+αf(b)+Jb-αf(a)]≤f(a)+f(b)2-α2(b-a)α∫ab[b-xb-af(x-a)+x-ab-af(b-x)]×((b-x)α-1+(x-a)α-1)dx
with α>0.

Proof.

By inequality (20), we get that
(22)f(a)+f(b)2=f(a)+f(b)2α2(b-a)α×∫ab((b-x)α-1+(x-a)α-1)dx=[f(a)+f(b)]α2(b-a)α×∫a(a+b)/2((b-x)α-1+(x-a)α-1)dx,Γ(α+1)2(b-a)α[Ja+αf(b)+Jb-αf(a)]=α2(b-a)α∫abf(x)×((b-x)α-1+(x-a)α-1)dx=α2(b-a)α∫a(a+b)/2[f(x)+f(a+b-x)]×((b-x)α-1+(x-a)α-1)dx≤α2(b-a)α∫a(a+b)/2[f(a)+f(b)]×((b-x)α-1+(x-a)α-1)dx-α(b-a)α∫a(a+b)/2b-xb-af(x-a)×((b-x)α-1+(x-a)α-1)dx-α(b-a)α∫a(a+b)/2x-ab-af(b-x)×((b-x)α-1+(x-a)α-1)dx=f(a)+f(b)2-α2(b-a)α∫ab[b-xb-af(x-a)+x-ab-af(b-x)]×((b-x)α-1+(x-a)α-1)dx.
The proof is completed.

Corollary 10.

Choosing α=1 in Theorem 9, one has
(23)1b-a∫abf(x)dx≤f(a)+f(b)2-1(b-a)2∫ab[(b-x)f(x-a)+(x-a)f(b-x)]dx.

Corollary 11.

Let f be defined as in Theorem 7; one gets
(24)f(a+b2)+α2(b-a)α×∫abf|a+b2-x|((b-x)α-1+(x-a)α-1)dx≤Γ(α+1)2(b-a)α[Ja+αf(b)+Jb-αf(a)]≤f(a)+f(b)2-α2(b-a)α×∫ab[b-xb-af(x-a)+x-ab-af(b-x)]×((b-x)α-1+(x-a)α-1)dx.

Corollary 12.

Taking α=1 in Corollary 11, one obtains
(25)f(a+b2)+1b-a∫abf|a+b2-x|dx≤∫abf(x)dx≤f(a)+f(b)2-1(b-a)2×∫ab[(b-x)f(x-a)+(x-a)f(b-x)]dx.

3. Conclusion

In this note, we obtain some new Hermite-Hadamard type inequalities for superquadratic functions via Riemann-Liouville fractional integrals. For superquadratic functions which are also convex, we get refinements of known results. The concept of superquadratic functions in several variables is introduced in [11]. An interesting topic is whether we can use the methods in this paper to establish the Hermite-Hadamard inequalities for superquadratic functions in several variables via Riemann-Liouville integrals.

Conflict of Interests

The authors declare that there is no conflict of interests regarding the publication of this paper.

Acknowledgment

This work is supported by the Youth Project of Chongqing Three Gorges University of China (no. 13QN11).

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