Some New Estimates of Hermite-Hadamard Inequalities for Harmonically Convex Functions with Applications

ISSN 2291-8639

Volume 13, Number 1 (2017), 15-21

http://www.etamaths.com

SOME NEW ESTIMATES OF HERMITE-HADAMARD INEQUALITIES FOR HARMONICALLY CONVEX FUNCTIONS WITH APPLICATIONS

WEN WANG1,2,∗ AND JIBING QI1

Abstract. In this paper, we first establish an integral identity. Further, using this identity, some new estimates for Hermite-Hadamard inequalities for harmonically convex functions are established. Finally, some applications to special mean are showed.

1. _Introduction

In this article, let R= (−∞,∞),R++= (0,∞).

Theory of convex functions and theory of inequalities are closely related to each other. Therefore, some literature on inequalities can be found for convex functions.

One of the most extensively research on inequalities is Hermite-Hadamard type inequalities.

Definition 1.1 ( [1,2])A functionf :I⊂R→Ris convex function on I, if

f(tx+ (1−t)y)≤tf(x) + (1−t)f(y), ∀x, y∈I, t∈[0,1]. (1.1) f is concave function if−f is convex function.

Let f : I ⊂ R → R be a convex function. The following inequality is the well-known Hermite-Hadamard’s inequality

f _a+b

2

≤ 1

b−a Z b

a

f(x)dx≤ f(a) +f(b)

2 , a, b∈I with a < b. (1.2)

Estimates for Hermite-Hadamard inequality for convex functions are studied in a rich literature [8–16].

Theorem 1.2 ( [1])Let f :I0⊂R→Rbe a differentiable mapping on I0 anda, b∈I0 witha < b.

If|f0(x)|q _{is a convex function forq >1, then}

f(a) +f(b)

2 −

1 b−a

Z b

a

f(x)dx

≤(b−a)(|f

0_(a)|+|f0_(b)|)

8 . (1.3)

Recently, ˙Iscan [3] introduced the concept of harmonically convex functions and established Hermite-Hadamard type inequality for harmonically convex functions.

Definition 1.3( [3,4])A function f :I⊂_R\{0} →_Ris said to be harmonically convex function onI, if

f

₁

tx−1_{+ (1−t)y}−1

≤tf(x) + (1−t)f(y), ∀x, y∈I, t∈[0,1]. (1.4)

f is said to be harmonically concave function if−f is harmonically convex function.

Theorem 1.4 ( [3,4]) Let f : I ⊂ _R\{0} → _R be a harmonically convex function a, b ∈ I with

a < b. Iff ∈L[a, b] , then

f

2ab a+b

≤ ab

b−a Z b

a

f(x) x2 dx≤

f(a) +f(b)

2 . (1.5)

Received 1st_{July, 2016; accepted 18}th _{September, 2016; published 3}rd_{January, 2017.}

2010Mathematics Subject Classification. 26D15, 26A51.

Key words and phrases. Hermite-Hadamard’s inequality; harmonically convex function; estimate; mean; inequality.

c

2017 Authors retain the copyrights of their papers, and all open access articles are distributed under the terms of the Creative Commons Attribution License.

Theorem 1.5( [3,4])Letf :I⊂R\{0} →Rbe a harmonically convex functiona, b∈Iwitha < b

andf0∈L[a, b]. If |f0(x)|q _{is harmonically convex on [a, b]forq >1,} 1 p+

1

q = 1, then

f(a) +f(b)

2 −

ab b−a

Z b

a

f(x) x2 dx

≤ ab(b−a)

2

₁

p+ 1 1p

(µ1|f0(a)|q_+µ2|f0_(b)|q₎1_{q, (1.6)}

where

µ1=

a2−2q+b1−2q[(b−a)(1−2q)−a] 2(b−a)2_{(1−q)(1−2q)} ,

µ2=

a2−2q+b1−2q[(b−a)(1−2q)−b] 2(b−a)2_{(1−q)(1−2q)} .

For many recent results related to Hermite-Hadamard type inequalities for harmonically functions, see [3–7].

The aim of this paper is first to establish a integral identity. Then, using this identity, some new estimates for Hermite-Hadamard inequalities for harmonically convex functions are established by ˙I. ˙Iscan in [3] are derived.

2. Some lemmas

In order to prove our main results we need some lemmas.

Lemma 2.1 Let f :I⊂R\{0} →R be a differentiable function onI0 anda, b∈I0 witha < b. If f0 ∈L[0,1], then forλ∈[0,1], one has

(1−λ)f

_ab

(1−λ)a+λb

+λf

_ab

(1−λ)a+λb

− ab

b−a Z b

a

f(x) x2 dx

=b−a ab

" Z 1−λ

0

t

[tb−1_{+ (1−t)a}−1_]f 0

₁

tb−1_{+ (1−t)a}−1

dt

−

Z 1

1−λ

t−1

[tb−1_{+ (1−t)a}−1_]f 0

₁

tb−1_{+ (1−t)a}−1

dt

. (2.1)

Proof. Letx=_tb−1_+(1−t)a1 −1,t∈[0,1] andx∈[a, b], then

₁

a− 1 b

Z 1−λ

0

t

(tb−1_{+ (1−t)a}−1₎2f 0

₁

tb−1_{+ (1−t)a}−1

dt

=tf

₁

tb−1_{+ (1−t)a}−1

1−λ

0 −

Z u

a

f(x) x2

ab b−adx

= (1−λ)f

_ab

(1−λ)a+λb

− ab

b−a Z u

a

f(x)

x2 dx. (2.2)

and

₁

b − 1 a

Z 1

1−λ

t

(tb−1_{+ (1−t)a}−1₎2f

0 1

tb−1_{+ (1−t)a}−1

dt

= tf

₁

tb−1_{+ (1−t)a}−1

1

1−λ −

Z b

u

f(x) x2

ab b−adx

=λf

_ab

(1−λ)a+λb

− ab

b−a Z b

u

f(x)

x2 dx, (2.3)

Remark 2.2 From (7) we derive the following two identities.

f

2ab a+b

− 2ab

b−a Z b

a

f(x) x2 dx

= 2(b−a) ab

" Z 12

0

t

[tb−1_{+ (1−t)a}−1_]f

0 1

tb−1_{+ (1−t)a}−1

dt

−

Z 1

1 2

t−1

[tb−1_{+ (1−t)a}−1_]f

0 1

tb−1_{+ (1−t)a}−1

dt #

; (2.4)

f(a) +f(b)

2 −

ab b−a

Z b

a

f(x) x2 dx

= b−a 2ab

Z 1

0

t

[tb−1_{+ (1−t)a}−1_]f 0

₁

tb−1_{+ (1−t)a}−1

dt

−

Z 1

0

t−1

[tb−1_{+ (1−t)a}−1_]f

0 1

tb−1_{+ (1−t)a}−1

dt

. (2.5)

Proof. Takeλ= 1₂ in (7), we can derive (10).

We respectively takeλ= 0 andλ= 1 in (7) and add two inequalities, then (11) is obtained.

Lemma 2.3By integral calculation, then

C1(a, b, λ) =

Z 1−λ

0

t(1−t)[tb+ (1−t)a]2dt

= (1−λ)3 ₁

5(b−a)

2_(1−λ)2₊1

2a(b−a)(1−λ) + 1 3λ

2

; (2.6)

C2(a, b, λ) =

Z 1−λ

0

t2[tb+ (1−t)a]2dt

= (1−λ)2 ₁

20+ 1 5λ

(b−a)2(1−λ)2+ ₁

6+ 1 2λ

a(b−a)(1−λ) + ₁

6+ 1 3λ

λ2

;

(2.7)

C3(a, b, λ) =

Z 1

1−λ

t(1−t)[tb+ (1−t)a]2dt= Z λ

0

(1−u)u[(1−u)b+ua]2du

=−1

5(b−a)

2_λ5₊1

4(a

2_{+ 3b}2_−4ab)λ4₊1

3(2ab−3b

2_)λ3₊1

2b

2_λ2_{; (2.8)}

C4(a, b, λ) =

Z 1

1−λ

(1−t)2[tb+ (1−t)a]2dt= Z λ

0

u2[(1−u)b+ua]2du

=1 5(b−a)

2_λ5₊1

2b(a−b)λ

4₊1

3b

2_λ3_{, (2.9)}

whereu= 1−t.

3. _{Main results}

Theorem 3.1 Let f :I ⊂R\{0} →R be a differentiable function andf0 ∈L[a, b]. If |f0(x)|q is

harmonically convex on[a, b] with0≤a < bforq≥1, then

(1−λ)f

_ab

(1−λ)a+λb

+λf

_ab

(1−λ)a+λb

− ab

b−a Z b

a

f(x) x2 dx

≤ b−a

ab n

(C1(a, b, λ))1−

1

q_|f0_(b)|q_C

1(a, b, λ) +|f0(a)|qC2(a, b, λ)

1q

+ (C3(a, b, λ))1−

1

q_|f0_(b)|q_C

3(a, b, λ) +|f0(a)|qC4(a, b, λ)

1_qo

. (3.1)

Proof. From Lemma 2.1 and using H¨older inequality, further, since|f0(x)|q _{is harmonically convex on}

[a, b], we have

I=

(1−λ)f

_ab

(1−λ)a+λb

+λf

_ab

(1−λ)a+λb

− ab

b−a Z b

a

f(x) x2 dx

≤b−a

ab "

Z 1−λ

0

t

[tb−1_{+ (1−t)a}−1_]

f0 ₁

tb−1_{+ (1−t)a}−1

dt + Z 1 1−λ

1−t [tb−1_{+ (1−t)a}−1_]

f0 ₁

tb−1_{+ (1−t)a}−1

dt

≤b−a

ab   Z 1−λ 0 t

[tb−1_{+ (1−t)a}−1_]2dt

!1−1q _Z 1−λ

0

t

[tb−1_{+ (1−t)a}−1_]2

f0 ₁

tb−1_{+ (1−t)a}−1

q dt

!1q

+ Z 1

1−λ

1−t

[tb−1_{+ (1−t)a}−1_]2dt

1−1qZ 1

1−λ

1−t [tb−1_{+ (1−t)a}−1_]2

f0 ₁

tb−1_{+ (1−t)a}−1

q dt #

≤b−a

ab    Z 1−λ 0 t

[tb−1_{+ (1−t)a}−1_]2dt

!1−1_q" Z 1−λ

0

t

t|f0(b)|q_{+ (1−t)|f}0_(a)|q [tb−1_{+ (1−t)a}−1_]2 dt

#1_q

+ Z 1

1−λ

1−t

[tb−1_{+ (1−t)a}−1_]2dt

1−1q"Z 1

1−λ

(1−t)t|f0(b)|q_{+ (1−t)|f}0_(a)|q [tb−1_{+ (1−t)a}−1_]2 dt

#1q



 .

Noticing that [tb−1+ (1−t)a−1]−1≤ta+ (1−t)b, and using (12-15), then, from above inequality we have

I≤b−a

ab    Z 1−λ 0

t[tb+ (1−t)a]2dt !1−1q"

Z 1−λ

0

t[tb+ (1−t)a]2

t|f0(b)|q_{+ (1−t)|f}0_(a)|q_dt #q1

+ Z 1

1−λ

(1−t)[tb+ (1−t)a]2dt

1−1qZ 1

1−λ

(1−t)[tb+ (1−t)a]2t|f0(b)|q+ (1−t)|f0(a)|q

dt 1q)

= b−a ab

 



(C1(a, b, λ))1−

1 q " Z 1−λ 0 h

|f0(b)|qt2[tb+ (1−t)a]2+|f0(a)|qt(1−t)[tb+ (1−t)a]2idt #q1

+ (C3(a, b, λ)) 1−1

q

Z 1

1−λ

h

|f0(b)|q_{t(1−t)[tb+ (1−t)a]}2_+|f0_(a)|q_(1−t)2_{[tb+ (1−t)a]}2i_dt

1q)

= b−a ab

n

(C1(a, b, λ)) 1−1

q |_f0_(b)|q_C

1(a, b, λ) +|f0(a)|qC2(a, b, λ)

q1

+ (C3(a, b, λ)) 1−1

q _|f0_(b)|q_C

3(a, b, λ) +|f0(a)|qC4(a, b, λ)

1qo_.

Corollary 3.2 Assume that all the assumptions of Theorem 3.1 are satisfied. If we takeq= 1 and

M = max{|f0(a)|,|f0(b)|}, then

(1−λ)f

_ab

(1−λ)a+λb

+λf

_ab

(1−λ)a+λb

− ab

b−a Z b

a

f(x) x2 dx

≤ M(b−a)

ab h

C1(a, b, λ) +C2(a, b, λ) +C3(a, b, λ) +C4(a, b, λ)

i

. (3.2)

Corollary 3.3 Assume that all the assumptions of Theorem 3.1 are satisfied. If we takeλ=q= 1

andM = max{|f0(a)|,|f0(b)|}, then

f(a)− ab

b−a Z b

a

f(x) x2 dx

≤M(b−a)

ab ₁

4(b−a)

2₊1

6b(b−4a)

. (3.3)

Corollary 3.4 Assume that all the assumptions of Theorem 3.1 are satisfied. If we take λ = 0,

q= 1 andM = max{|f0(a)|,|f0(b)|}, then

f(b)− ab

b−a Z b

a

f(x) x2 dx

≤ M(b−a)

ab

1 4(b−a)

2₊2

3a(b−a)

. (3.4)

Corollary 3.5 Assume that all the assumptions of Theorem 3.1 are satisfied. If we take λ = 1₂,

q= 1 andM = max{|f0(a)|,|f0(b)|}, then

f

_2ab

a+b

− ab

b−a Z b

a

f(x) x2 dx

≤M(b−a)

ab

×

C1(a, b,

1 2)

C1(a, b,

1

2) +C2(a, b, 1 2)

+C3(a, b,

1 2)

C3(a, b,

1

2) +C4(a, b, 1 2)

. (3.5)

Remark 3.6 By (18) and (19) we obtain the following inequality

f(a) +f(b)

2 −

ab b−a

Z b

a

f(x) x2 dx

≤ M(b−a)

ab ₁

12(4b

2_−6ab−a2₎

. (3.6)

4. Some Applications for Special Means

Leta, bare two nonnegative number witha < b. Let us recall the following special means ofaand b.

(1) The arithmetic meanA=A(a, b) :=a+b

2 ; (2) The geometric meanG=G(a, b) := √

ab; (3) The harmonic mean H =H(a, b) :=_a+b2ab; (4) The logarithmic meanL=L(a, b) := _lnb−a_b−lna; (5) The -logarithmic mean

Lp=Lp(a, b) :=

_bp+1_−ap+1

(p+ 1)(b−a) 1p

, a6=b, p∈R, p6= 0,−1.

These means are often applied to numerical approximation and in other areas. However, the fol-lowing simple relationships are known in the literature:

H≤G≤L≤A. (4.1)

It is also known that Lp is monotonically increasing respecting to p ∈ R, denoting L0 = I and

Proposition 1. Let 0< a < b. Then we obtain the following inequalities:

H−G 2

L

≤ b−a

ab

×

C1(a, b,

1 2)

C1(a, b,

1

2) +C2(a, b, 1 2)

+C3(a, b,

1 2)

C3(a, b,

1

2) +|C4(a, b, 1 2)

;

A−G 2

L

≤ b−a

ab ₁

12(4b

2

−6ab−a2)

.

Proof. The assertion follows from inequalities (20) and (21), respectively, forf(x) =x,x∈R++.

Proposition 2. Let 0< a < b. Then we derive the following inequalities:

H2−G2≤

2(b−a) a

×

C1(a, b,

1 2)

C1(a, b,

1

2) +C2(a, b, 1 2)

+C3(a, b,

1 2)

C3(a, b,

1

2) +|C4(a, b, 1 2)

;

A(a2, b2)−G 2

L

≤ 2(b−a)

a ₁

12(4b

2_−6ab−a2₎

.

Proof. The assertion follows from inequalities (20) and (21), respectively, forf(x) =x2,x∈R++.

Proposition 3. Let 0< a < b. Then we have the following inequalities:

Hp+2−LppG 2

≤

2bp+2_(b−a)

ab

×

C1(a, b,

1 2)

C1(a, b,

1

2) +C2(a, b, 1 2)

+C3(a, b,

1 2)

C3(a, b,

1

2) +|C4(a, b, 1 2)

;

A(ap+2, bp+2)−Lp_pG2≤

2bn+2(b−a) ab

₁

12(4b

2_−6ab−a2₎

.

Proof. The assertion follows from inequalities (20) and (21), respectively, for f(x) = xp+2_{, x∈R} ++

andp∈(−1,∞)\{0}.

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1

School of Mathematics and Statistics, Hefei Normal University, Hefei 230601,P.R.China

2

School of Mathematical Science, University of Science and Technology of China, Hefei 230026, China ∗