# Fejér and Hermite-Hadamard Type Inequalities for Harmonically Convex Functions

Authors: Feixiang Chen; Shanhe Wu
Journal: Journal of Applied Mathematics
 (2014)

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## Abstract

We establish a Fejér type inequality for harmonically convex functions. Our results are the generalizations of some known results. Moreover, some properties of the mappings in connection with Hermite-Hadamard and Fejér type
inequalities for harmonically convex functions are also considered.


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## Body

## 1. Introduction

Letf: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.Inequality (1) is known in the literature as the Hermite-Hadamard inequality. Fejér [1] established the following weighted generalization of inequality (1).Theorem 1.
Iff:[a,b]→R is a convex function, then the following inequality holds:
(2)f(a+b2)∫ab‍p(x)dx≤∫ab‍f(x)p(x)dx≤f(a)+f(b)2∫ab‍p(x)dx,
where p:[a,b]→R is positive, integrable, and symmetric with respect to x=(a+b)/2.Some generalizations, refinements, variations, and improvements of inequalities (1) and (2) were investigated by Wu [2], Chen and Liu [3], Sarikaya and Ogunmez [4], and Xiao et al. [5], respectively.In [6], Dragomir proposed an interesting Hermite-Hadamard type inequality which refines the left hand side of inequality of (1) as follows.Theorem 2 (see [6]).
Letf be a convex function defined on [a,b]. Then H is convex, increasing on [0,1], and for all t∈[0,1], one has
(3)f(a+b2)=H(0)≤H(t)≤H(1)=1b-a∫ab‍f(x)dx,
where
(4)H(t)=1b-a∫ab‍f(tx+(1-t)a+b2)dx.An analogous result for convex functions which refines the right hand side of inequality (1) was obtained by Yang and Hong in [7] as follows.Theorem 3 (see [7]).
Letf be a convex function defined on [a,b]. Then F is convex, increasing on [0,1], and for all t∈[0,1], one has
(5)1b-a∫ab‍f(x)dx=F(0)≤F(t)≤F(1)=f(a)+f(b)2,
where
(6)F(t)=12(b-a)∫ab‍[f((1+t2)a+(1-t2)x)+f((1+t2)b+(1-t2)x)]dx.Yang and Tseng in [8] established the following Fejér type inequalities, which is the generalization of inequalities (3) and (5) as well as the refinement of the Fejér inequality (2).Theorem 4 (see [8]).
Iff is convex on [a,b], p:[a,b]→R is positive, integrable, and symmetric about x=(a+b)/2. Then P and Q are convex, increasing on [0,1], and for all t∈[0,1], one has
(7)f(a+b2)∫ab‍p(x)dx=P(0)≤P(t)≤P(1)=∫ab‍f(x)p(x)dx=Q(0)≤Q(t)≤Q(1)=f(a)+f(b)2∫ab‍p(x)dx,
where
(8)P(t)=∫ab‍f(tx+(1-t)a+b2)p(x)dx,(9)Q(t)=12∫ab‍[f((1+t2)a+(1-t2)x)p(x+a2)+f((1+t2)b+(1-t2)x)p(x+b2)]dx.In [9, 10], İşcan and Wu gave the definition of harmonic convexity as follows.Definition 5.
LetI⊆R∖{0} be a real interval. A function f:I→R is said to be harmonically convex if
(10)f(xytx+(1-t)y)≤tf(y)+(1-t)f(x),
for all x,y∈I and t∈[0,1]. If the inequality in (10) is reversed, then f is said to be harmonically concave.The following Hermite-Hadamard inequality for harmonically convex functions holds true.Theorem 6 (see [9]).
Letf:I⊆R∖{0}→R be a harmonically convex function and a,b∈I with a<b. If f∈L(a,b), then one has
(11)f(2aba+b)≤abb-a∫ab‍f(x)x2dx≤f(a)+f(b)2.In [10], İşcan and Wu established the following Hermite-Hadamard inequalities for harmonically convex functions via the Riemann-Liouville fractional integral.Theorem 7 (see [10]).
Letf:I⊆(0,∞)→R be a function such that f∈L(a,b), where a,b∈I with a<b. If f is a harmonically convex function on [a,b], then the following inequalities for fractional integrals hold:
(12)f(2aba+b)≤Γ(α+1)2(abb-a)α[J1/a-α(f∘g)(1b)+J1/b+α(f∘g)(1a)]≤f(a)+f(b)2,
where α>0 and g(x)=1/x.The Riemann-Liouville fractional integralsJa+αf and Jb-αf of order α>0 with a≥0 are defined by
(13)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,
where Γ(α) is the Gamma function defined by Γ(α)=∫0∞‍e-ttα-1dt.In this paper, we establish a Fejér type inequality for harmonically convex functions; our main result includes, as special cases, the inequalities given by Theorems6 and 7. Moreover, we investigate some properties of the mappings in connection to Hermite-Hadamard and Fejér type inequalities for harmonically convex functions.


## 2. Fejér Type Inequality for Harmonically Convex Functions

The following Fejér inequality for harmonically convex functions holds true.Theorem 8.
Letf:I⊆R∖{0}→R be a harmonically convex function and a,b∈I with a<b. If f∈L(a,b), then one has
(14)f(2aba+b)∫ab‍p(x)x2dx≤∫ab‍f(x)x2p(x)dx≤f(a)+f(b)2∫ab‍p(x)x2dx,
where p:[a,b]→R is nonnegative and integrable and satisfies
(15)p(abx)=p(aba+b-x).Proof.
Sincef is a harmonically convex function on [a,b], we have, for all x,y∈[a,b],
(16)f(2xyx+y)≤f(y)+f(x)2.
Choosing x=ab/(tb+(1-t)a) and y=ab/(ta+(1-t)b), we have
(17)f(2aba+b)≤f(ab/(tb+(1-t)a))+f(ab/(ta+(1-t)b))2≤f(a)+f(b)2.
Since p is nonnegative and satisfies the condition of (15), we obtain
(18)f(2aba+b)p(abtb+(1-t)a)≤(f(abtb+(1-t)a)p(abtb+(1-t)a)+f(abta+(1-t)b)p(abta+(1-t)b))×2-1≤f(a)+f(b)2p(abtb+(1-t)a).
Integrating both sides of the above inequalities with respect to t over [0,1], we obtain
(19)f(2aba+b)∫01‍p(abtb+(1-t)a)dt≤∫01((f(abtb+(1-t)a)p(abtb+(1-t)a)+f(abta+(1-t)b)p(abta+(1-t)b))×2-1)dt≤f(a)+f(b)2∫01‍p(abtb+(1-t)a)dt.
The proof of Theorem 8 is completed.Remark 9.
Puttingp(x)≡1 in Theorem 8, we obtain inequality (11).Remark 10.
Choosing(20)p(x)=α2(abb-a)α{(1x-1b)α-1+(1a-1x)α-1},(α>0,0<a<b),
in Theorem 8, it is easy to observe that p(ab/x)=p(ab/(a+b-x)).
Since(21)∫ab‍p(x)x2dx=∫ab‍p(x)x2dx=α2(abb-a)α∫ab‍1x2{(1x-1b)α-1+(1a-1x)α-1}dx=α2(abb-a)α∫ab‍1x2{(1x-1b)α-1+(1a-1x)α-1}dx=α2(abb-a)α∫1/b1/a‍{(u-1b)α-1+(1a-u)α-1}du=1,∫ab‍f(x)x2p(x)dx=∫ab‍f(x)p(x)x2dx=α2(abb-a)α×∫ab‍f(x)x2{(1x-1b)α-1+(1a-1x)α-1}dx=α2(abb-a)α×∫1/b1/a‍f(1u){(u-1b)α-1+(1a-u)α-1}du=α2(abb-a)α{∫1/b1/a‍f(1u)(u-1b)α-1du+∫1/b1/a‍f(1u)(1a-u)α-1du}=α2(abb-a)α{∫1/b1/a‍f∘g(u)(u-1b)α-1du+∫1/b1/a‍f∘g(u)(1a-u)α-1du}=α2(abb-a)α×{Γ(α)[J1/a-α(f∘g)(1b)+J1/b+α(f∘g)(1a)]}=Γ(α+1)2(abb-a)α×[J1/a-α(f∘g)(1b)+J1/b+α(f∘g)(1a)],
where g(x)=1/x, which implies that inequality (14) can be transformed to inequality (12) under an appropriate selection of p(x).Remark 11.
In Theorem8, taking p(ab/x)=ω(x), where 0<a<b, ω(x) is nonnegative, integrable, and symmetric with respect to x=(a+b)/2. Then inequality (14) becomes
(22)f(2aba+b)∫ab‍ω(x)dx≤∫ab‍f(abx)ω(x)dx≤f(a)+f(b)2∫ab‍ω(x)dx.


## 3. Some Mappings in connection with Hermite-Hadamard and Fejér Inequalities for Harmonically Convex Functions

Lemma 12.
Letf:I⊆R∖{0}→R be a harmonically convex function and a,b∈I with a<b, and let
(23)h(t)=12f(2aba+b-t)+12f(2aba+b+t),t∈[0,b-a]. Then h is convex, increasing on [0,b-a], and for all t∈[0,b-a],
(24)f(2aba+b)≤h(t)≤f(a)+f(b)2.Proof.
Firstly, forx,y∈[0,b-a], we have
(25)h(tx+(1-t)y)=12f(2aba+b-[tx+(1-t)y])+12f(2aba+b+[tx+(1-t)y])=12f(2abt(a+b-x)+(1-t)(a+b-y))+12f(2abt(a+b+x)+(1-t)(a+b+y))≤t2f(2aba+b-x)+1-t2f(2aba+b-y)+t2f(2aba+b+x)+1-t2f(2aba+b+y)=th(x)+(1-t)h(y),
and hence h is convex on [0,b-a].
Next, ift∈[0,b-a], it follows from the harmonic convexity of f that
(26)h(t)=12f(2aba+b-t)+12f(2aba+b+t)≥f(2ab(1/2)(a+b-t)+(1/2)(a+b+t))=f(2aba+b).
It is easy to observe that(27)h(t)=12f(2aba+b-t)+12f(2aba+b+t)=12f(2ab×(b-a+tb-aa+b-a-tb-ab)-1)+12f(2ab×(b-a-tb-aa+b-a+tb-ab)-1)≤12(b-a)+t2(b-a)f(b)+12(b-a)-t2(b-a)f(a)+12(b-a)+t2(b-a)f(a)+12(b-a)-t2(b-a)f(b)=f(a)+f(b)2.
Thus inequality (24) holds.
Finally, for0<t1<t2≤b-a, since h is convex, it follows from (24) that
(28)h(t2)-h(t1)t2-t1≥h(t1)-h(0)t1-0=h(t1)-f(2ab/(a+b))t1≥0,
and hence, h(t2)≥h(t1), which means that h is increasing on [0,b-a]. This completes the proof of Lemma 12.Theorem 13.
Letf:I⊆R∖{0}→R be a harmonically convex function and a,b∈I with a<b. If f∈L(a,b) and H is defined by
(29)H(t)=12(b-a)∫0b-a‍f(2aba+b-tx)dx+12(b-a)∫0b-a‍f(2aba+b+tx)dx=1b-a∫ab‍f(ab(1-t)((a+b)/2)+tx)dx,
then H is convex and increasing on [0,1], and
(30)f(2aba+b)=H(0)≤H(t)≤H(1)=abb-a∫ab‍f(x)x2dx.Proof.
It follows from Lemma12 that
(31)h(t)=12f(2aba+b-t)+12f(2aba+b+t)
is convex and increasing on [0,b-a]. Hence H(t) is convex and increasing on [0,1]. Further, inequality (30) can be deduced from (24). Theorem 13 is proved.Theorem 14.
Letf:I⊆R∖{0}→R be a harmonically convex function and a,b∈I with a<b. If f∈L(a,b) and G is defined by
(32)G(t)=12(b-a)∫0b-a‍f(2ab2a+(1-t)x)dx+12(b-a)∫0b-a‍f(2ab2b-(1-t)x)dx=12(b-a)∫ab‍f(2ab(1+t)a+(1-t)x)dx+12(b-a)∫ab‍f(2ab(1+t)b+(1-t)x)dx,
then G is convex and increasing on [0,1], and
(33)abb-a∫ab‍f(x)x2dx=G(0)≤G(t)≤G(1)=f(a)+f(b)2.Proof.
We note that iff is convex and g is linear, then the composition f∘g is convex. It follows from Lemma 12 that
(34)h(t)=12f(2aba+b-t)+12f(2aba+b+t),
and k(t)=b-a-(1-t)x are increasing on [0,b-a] and [0,1], respectively. Hence,
(35)h(k(t))=f(2ab2a+(1-t)x)+f(2ab2b-(1-t)x)
is convex and increasing on [0,1]. We infer that G is convex and increasing on [0,1]. Furthermore, inequality (33) follows directly from (24). The proof of Theorem 14 is completed.Theorem 15.
Letf:I⊆R∖{0}→R be a harmonically convex function and a,b∈I with a<b. If f∈L(a,b) and P is defined by
(36)P(t)=12ab∫0b-a‍f(2aba+b-tx)p(2abb+a-x)dx+12ab∫0b-a‍f(2aba+b+tx)p(2abb+a+x)dx=1ab∫ab‍f(ab((a+b)/2)(1-t)+tx)p(abx)dx,
where p:[a,b]→R is nonnegative and integrable and satisfies the condition of (15), then P is convex and increasing on [0,1], and
(37)f(2aba+b)∫ab‍p(x)x2dx=P(0)≤P(t)≤P(1)=∫ab‍f(x)x2p(x)dx.Proof.
From Lemma12 we obtain that
(38)h(t)=12f(2aba+b-t)+12f(2aba+b+t)
is convex and increasing on [0,b-a]. Since p(2ab/(a+b+x)) is nonnegative and satisfies p(2ab/(a+b+x))=p(2ab/(a+b-x)), it follows that P(t) is convex and increasing on [0,1], while inequality (37) can be deduced from (24). Theorem 15 is proved.Theorem 16.
Letf:I⊆R∖{0}→R be a harmonically convex function and a,b∈I with a<b. If f∈L(a,b) and Q is defined by
(39)Q(t)=12ab∫0b-a‍f(2ab2a+(1-t)x)p(2ab2a+x)dx+12ab∫0b-a‍f(2ab2b-(1-t)x)p(2ab2b-x)dx=12ab∫ab‍f(2ab(1+t)a+(1-t)x)p(2abx+a)dx+12ab∫ab‍f(2ab(1+t)b+(1-t)x)p(2abx+b)dx,
where p:[a,b]→R is nonnegative and integrable and satisfies the condition of (15), then Q is convex and increasing on [0,1], and
(40)∫ab‍f(x)x2p(x)dx=Q(0)≤Q(t)≤Q(1)=f(a)+f(b)2∫ab‍p(x)x2dx.Proof.
By using the same method as in the proof of Theorem14, we obtain from Lemma 12 that
(41)h(k(t))=f(2ab2a+(1-t)x)+f(2ab2b-(1-t)x)
is convex and increasing on [0,1]. Since p(2ab/(2a+x)) is nonnegative and satisfies p(2ab/(2a+x))=p(2ab/(2b-x)), we deduce that Q(t) is convex and increasing on [0,1]. Inequality (40) follows from (24) and the identity
(42)12ab∫0b-a‍p(2ab2a+x)dx=12ab∫0b-a‍p(2ab2b-x)dx=12{12ab∫0b-a‍p(2ab2a+x)dx+12ab∫0b-a‍p(2ab2b-x)dx}=12∫ab‍p(x)x2dx.
This completes the proof of Theorem16.Remark 17.
If we put(43)p(x)=α2(abb-a)α{(1x-1b)α-1+(1a-1x)α-1},
in inequalities (37) and (40), respectively, we obtain the refined versions of inequality (12).



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