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volume 6, issue 5, article 140, 2005.

Received 06 April, 2005;

accepted 20 September, 2005.

Communicated by:A. Lupa¸s

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Journal of Inequalities in Pure and Applied Mathematics

REFINEMENTS OF THE HERMITE-HADAMARD INEQUALITY FOR CONVEX FUNCTIONS

S.S. DRAGOMIR AND A. McANDREW

School of Computer Science and Mathematics Victoria University

PO Box 14428, MCMC 8001 VIC, Australia.

EMail:sever@csm.vu.edu.au URL:http://rgmia.vu.edu.au/dragomir/

EMail:Alasdair.Mcandrew@vu.edu.au URL:http://sci.vu.edu.au/˜amca/

c

2000Victoria University ISSN (electronic): 1443-5756 106-05

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Refinements of the Hermite-Hadamard Inequality

for Convex Functions S.S. Dragomir and A. McAndrew

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J. Ineq. Pure and Appl. Math. 6(5) Art. 140, 2005

http://jipam.vu.edu.au

Abstract

New refinements for the celebrated Hermite-Hadamard inequality for convex functions are obtained. Applications for special means are pointed out as well.

2000 Mathematics Subject Classification:Primary 26D15; Secondary 26D10.

Key words: Hermite-Hadamard Inequality.

This paper is based on the talk given by the second author within the “International Conference of Mathematical Inequalities and their Applications, I”, December 06- 08, 2004, Victoria University, Melbourne, Australia [http://rgmia.vu.edu.au/

conference]

1. Introduction

The following result is well known in the literature as the Hermite-Hadamard integral inequality:

(1.1) f

a+b 2

≤ 1 b−a

Z b a

f(t)dt≤ f(a) +f(b)

2 ,

provided thatf : [a, b]→Ris a convex function on[a, b].

The following refinements of theH·−H·inequality were obtained in [2]

(1.2) 1 b−a

Z b a

f(x)dx−f

a+b 2

≥

1 b−a

Z b a

f(x) +f(a+b−x) 2

dx− f

a+b 2

≥0.

and

(1.3) f(a) +f(b)

2 − 1

b−a Z b

a

f(t)dt

≥











|f(a)| − _b−a¹ Rb

a |f(x)|dx

if f(a) = f(b)

1 f(b)−f(a)

Rf(b)

f(a) |x|dx−_b−a¹ Rb

a|f(x)|dx

if f(a)6=f(b) for the general case of convex functionsf : [a, b]→R.

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If one would assume differentiability of f on (a, b), then the following bounds in terms of its derivative holds (see [3, pp. 30-31])

(1.4) f(a) +f(b)

2 − 1

b−a Z b

a

f(t)dt ≥max{|A|,|B|,|C|} ≥0 where

A := 1 b−a

Z b a

x− a+b 2

|f⁰(x)|dx−1 4

Z b a

|f⁰(x)|dx, B := f(b)−f(a)

4 − 1

b−a

"

Z ^a+b₂

a

f(x)dx− Z b

a+b 2

f(x)dx

#

and

C := 1 b−a

Z b a

x−a+b 2

|f⁰(x)|dx.

A different approach considered in [1] led to the following lower bounds (1.5) f(a) +f(b)

2 − 1

b−a Z b

a

f(t)dt ≥max{|D|,|E|,|F|} ≥0, where

D:= 1 b−a

Z b a

|xf⁰(x)|dx− 1 b−a

Z b a

|f⁰(x)|dx· 1 b−a

Z b a

|x|dx, E := 1

b−a Z b

a

x|f⁰(x)|dx−a+b 2 · 1

b−a Z b

a

|f⁰(x)|xdx

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and

F := 1 b−a

Z b a

|x|f⁰(x)dx− f(b)−f(a) b−a · 1

b−a Z b

a

|x|dx.

For other results connected to theH·−H· inequality see the recent mono- graph on line [3].

In the present paper, we use a different method to obtain other refinements of theH_·−H_·inequality. Applications for special means are pointed out as well.

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2. The Results

The following refinement of the Hermite-Hadamard inequality for differentiable convex functions holds.

Theorem 2.1. Assume that f : [a, b] → R is differentiable convex on (a, b). Then one has the inequality:

(2.1) 1 b−a

Z b a

f(t)dt−f

a+b 2

≥

1 b−a

Z b a

f(x)−f

a+b 2

dx− b−a 4 ·

f⁰

a+b 2

≥0.

Proof. Sincef is differentiable convex on(a, b),then for eachx, y ∈(a, b)one has the inequality

(2.2) f(x)−f(y)≥(x−y)f⁰(y). Using the properties of modulus, we have

f(x)−f(y)−(x−y)f⁰(y) = |f(x)−f(y)−(x−y)f⁰(y)|

(2.3)

≥ ||f(x)−f(y)| − |x−y| |f⁰(y)||

for eachx, y ∈(a, b).

If we choosey= ^a+b₂ in (2.3) we get (2.4) f(x)−f

a+b 2

−

x−a+b 2

f⁰

a+b 2

≥

f(x)−f

a+b 2

−

x−a+b 2

f⁰

a+b 2

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for anyx∈(a, b).

Integrating (2.4) on [a, b], dividing by (b−a) and using the properties of modulus, we have

1 b−a

Z b a

f(x)dx−f

a+b 2

−f⁰

a+b 2

· 1 b−a

Z b a

x− a+b 2

dx

≥ 1 b−a

Z b a

f(x)−f

a+b 2

−

x−a+b 2

f⁰

a+b 2

dx

≥

1 b−a

Z b a

f(x)−f

a+b 2

dx

−

f⁰

a+b 2

1 b−a

Z b a

x− a+b 2

dx and since

(2.5)

Z b a

x− a+b 2

dx= 0, Z b

a

x−a+b 2

dx= (b−a)² 4 , we deduce by (2.5) the desired result (2.1).

The second result is embodied in the following theorem.

Theorem 2.2. Assume that f : [a, b] → R is differentiable convex on (a, b).

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Then one has the inequality

(2.6) 1 2

f(a) +f(b)

2 +f

a+b 2

− 1 b−a

Z b a

f(x)dx

≥ 1 2

1 b−a

Z b a

f(x)−f

a+b 2

dx

− 1 b−a

Z b a

x−a+b 2

|f⁰(x)|dx

≥0.

Proof. We choosex= ^a+b₂ in (2.3) to get (2.7) f

a+b 2

−f(y)−

a+b 2 −y

f⁰(y)

≥ f

a+b 2

−f(y)

−

a+b 2 −y

|f⁰(y)|

. Integrating (2.7) overy, dividing by(b−a)and using the modulus properties, we get

(2.8) f

a+b 2

− 1 b−a

Z b a

f(y)dy− Z b

a

a+b 2 −y

f⁰(y)dy

≥

1 b−a

Z b a

f

a+b 2

−f(y)

dy

− 1 b−a

Z b a

a+b 2 −y

|f⁰(y)|dy .

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Since Z b

a

y− a+b 2

f⁰(y)dy= f(a) +f(b)

2 (b−a)− Z b

a

f(t)dt, then by (2.8) we deduce

f

a+b 2

+ f(a) +f(b)

2 − 2

b−a Z b

a

f(y)dy

≥

1 b−a

Z b a

f(y)−f

a+b 2

dy

− 1 b−a

Z b a

y− a+b 2

|f⁰(y)|dy which is clearly equivalent to (2.6).

The following result holding for the subclass of monotonic and convex functions is whort to mention.

Theorem 2.3. Assume that f : [a, b] → Ris monotonic and convex on(a, b). Then we have:

(2.9) 1 2

f(a) +f(b)

2 +f

a+b 2

− 1 b−a

Z b a

f(x)dx

≥ 1

4[f(b)−f(a)] + 1 b−a

Z b a

sgn

a+b 2 −x

f(x)dx .

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Proof. Since the class of differentiable convex functions in (a, b) is dense in uniform topology in the class of all convex functions defined on(a, b),we may assume, without loss of generality, thatfis differentiable convex and monotonic on(a, b).

Firstly, assume thatf is monotonic nondecreasing on[a, b].Then Z b

a

f(x)−f

a+b 2

dx= Z ^a+b₂

a

f

a+b 2

−f(x)

dx +

Z b

a+b 2

f(x)−f

a+b 2

dx

= Z b

a+b 2

f(x)dx− Z ^a+b₂

a

f(x)dx,

Z b a

x− a+b 2

|f⁰(x)|dx

= Z ^a+b₂

a

a+b 2 −x

f⁰(x)dx+ Z b

a+b 2

x− a+b 2

f⁰(x)dx

=

a+b 2 −x

f(x)

a+b 2

a

+ Z ^a+b₂

a

f(x)dx +

x− a+b 2

f(x)

b

a+b 2

− Z b

a+b 2

f(x)dx

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=−b−a

2 f(a) + Z ^a+b₂

a

f(x)dx+ b−a

2 f(b)− Z b

a+b 2

f(x)dx.

Using (2.6) we have 1

2

f(a) +f(b)

2 +f

a+b 2

− 1 b−a

Z b a

f(x)dx

≥ 1

2 (b−a)

Z b

a+b 2

f(x)dx− Z ^a+b₂

a

f(x)dx

−

"

b−a

2 f(b)− b−a

2 f(a) + Z ^a+b₂

a

f(x)dx− Z b

a+b 2

f(x)dx

#

= 1

2 (b−a) 2

Z b

a+b 2

f(x)dx−2 Z ^a+b₂

a

f(x)dx−b−a

2 [f(b)−f(a)]

, which is clearly equivalent to (2.9).

A similar argument may be done if f is monotonic nonincreasing and we omit the details.

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3. Applications for Special Means

Let us recall the following means:

a) The arithmetic mean

A(a, b) := a+b

2 , a, b >0, b) The geometric mean

G(a, b) := √

ab; a, b≥0, c) The harmonic mean

H(a, b) := 2

1

a +¹_b; a, b >0, d) The identric mean

I(a, b) :=









 1 e

b^b a^a

_b−a¹

if b 6=a

a if b =a

; a, b >0

e) The logarithmic mean

L(a, b) :=







b−a

lnb−lna if b 6=a

a if b =a

; a, b > 0

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f) Thep−logarithmic mean

Lp(a, b) :=







b^p+1−a^p+1 (p+1)(b−a)

¹_p

if b6=a, p∈R\ {−1,0}

a if b=a

; a, b >0.

It is well known that, if, on denotingL−1(a, b) := L(a, b)andL₀(a, b) :=

I(a, b),then the function R3 p → L_p(a, b) is strictly monotonic increasing and, in particular, the following classical inequalities are valid

(3.1) min{a, b} ≤H(a, b)≤G(a, b)

≤L(a, b)≤I(a, b)≤A(a, b)≤max{a, b}

for anya, b >0.

The following proposition holds:

Proposition 3.1. Let 0< a < b < ∞.Then we have the following refinement for the inequalityA ≥L:

(3.2) A−L≥ AL

b−a

"

G A

2

−ln G

A 2

−1

#

≥0.

The proof follows by Theorem2.1on choosingf : [a, b]→(0,∞), f(t) = 1/tand we omit the details.

The following proposition contains a refinementof the following well known inequality

1

2 A⁻¹ +H⁻¹

≥L⁻¹.

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Proposition 3.2. With the above assumption foraandbwe have (3.3) 1

2 A⁻¹+H⁻¹

−L⁻¹ ≥ 1 b−a

"

A G

2

−ln A

G 2

−1

#

≥0.

The proof follows by Theorem2.3for the same funcionf : [a, b]→(0,∞), f(t) = 1/t,which is monotonic and convex on[a, b],and the details are omit- ted.

One may state other similar results that improve classical inequalities for means by choosing appropriate convex functions f.However, they will not be stated below.

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References

[1] S.S. DRAGOMIR ANDS. MABZELA, Some error estimates in the trape- zoidal quadrature rule, Tamsui Oxford J. of Math. Sci., 16(2) (2000), 259–

272.

[2] S.S. DRAGOMIR, Refinements of the Hermite-Hadamard inequality for convex functions, Tamsui Oxford J. of Math. Sci., 17(2) (2001), 131–137.

[3] S.S. DRAGOMIR AND C.E.M. PEARCE, Selected Topics on Hermite- Hadamard Type Inequalities and Applications, RGMIA, Monographs, 2000. [ONLINE: http://rgmia.vu.edu.au/monographs.

html].