A NEW REFINEMENT OF THE HERMITE-HADAMARD INEQUALITY FOR CONVEX FUNCTIONS

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Hermite-Hadamard Inequality G. Zabandan vol. 10, iss. 2, art. 45, 2009

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A NEW REFINEMENT OF THE

HERMITE-HADAMARD INEQUALITY FOR CONVEX FUNCTIONS

G. ZABANDAN

Department of Mathematics

Faculty of Mathematical Science and Computer Engineering Teacher Training University, 599 Taleghani Avenue Tehran 15618 IRAN.

EMail:zabandan@tmu.ac.ir

Received: 13 August, 2008

Accepted: 12 February, 2009

Communicated by: S.S. Dragomir 2000 AMS Sub. Class.: 26D15, 26D10.

Key words: Hermite-Hadamard inequality.

Abstract: In this paper we establish a new refinement of the Hermite-Hadamard inequality for convex functions.

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1. Introduction

Letf : [a, b]→Rbe a convex function, then the following inequality:

(1.1) f

a+b 2

≤ 1 b−a

Z b

a

f(x)dx≤ f(a) +f(b) 2 is known as the Hermite-Hadamard inequality [5].

In recent years there have been many extensions, generalizations and similar re- sults of the inequality (1.1).

In [2], Dragomir established the following theorem which is a refinement of the left side of (1.1).

Theorem 1.1. Iff : [a, b]→Ris a convex function, andHis defined on[0,1]by H(t) = 1

b−a Z b

a

f

tx+ (1−t)a+b 2

dx,

thenH is convex, increasing on[0,1], and for allt∈[0,1],we have f

a+b 2

=H(0)≤H(t)≤H(1) = 1 b−a

Z b

a

f(x)dx.

In [6] Yang and Hong established the following theorem which is a refinement of the right side of inequality (1.1).

Theorem 1.2. Iff : [a, b]→Ris a convex function, andF is defined by F(t) = 1

2(b−a) Z b

a

f

1 +t 2

a+

1−t 2

x

+f

1 +t 2

b+

1−t 2

x

dx,

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thenF is convex, increasing on[0,1], and for allt∈[0,1], we have 1

b−a Z b

a

f(x)dx=F(0) ≤F(t)≤F(1) = f(a) +f(b)

2 .

In this paper we establish a refinement of the both sides of inequality (1.1). For this we first define two sequences{x_n}and{y_n}by

x_n= 1 2ⁿ

2ⁿ

X

i=1

f

a+ib−a

2ⁿ − b−a 2ⁿ⁺¹

(1.2)

= 1 2ⁿ

2ⁿ

X

i=1

f

a+

i− 1 2

b−a 2ⁿ

,

y_n= 1 2ⁿ⁺¹

2ⁿ

X

i=1

f

1− i

2ⁿ

a+ i 2ⁿb

+f

1−i−1 2ⁿ

a+i−1 2ⁿ b

(1.3)

= 1 2ⁿ⁺¹

"

f(a)+f(b)+2

2ⁿ−1

X

i=1

f

1− i 2ⁿ

a+ i

2ⁿb #

and we prove the following f

a+b 2

=x0 ≤ 1 2

f

3a+b 4

+f

a+ 3b 4

=x₁ ≤ · · · ≤x_n ≤ · · ·

≤ 1 b−a

Z b

a

f(x)dx≤ · · · ≤y_n ≤ · · · ≤y₁

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= 1 4

f(a) + 2f

a+b 2

+f(b)

≤y₀ = f(a) +f(b)

2 ,

which is a new refinement of the Hermite-Hadamard inequality (1.1). For a similar discussion, see [1] or the monograph online [7, p. 19 – 22].

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2. A Refinement Result

In this section, using the terminologies of the introduction, we refine the Hermite- Hadamard inequality via the sequences{x_n}and{y_n}.

Theorem 2.1. Letf be a convex function on[a, b]. Then we have f

a+b 2

≤xn ≤ 1 b−a

Z b

a

f(x)dx≤yn≤ f(a) +f(b)

2 .

Proof. By the right side of Hermite-Hadamard inequality (1.1) we have 1

b−a Z b

a

f(x)dx

= 1

b−a

2ⁿ

X

i=1

Z a+i^b−a₂_n

a+(i−1)^b−a₂n

f(x)dx

≤ 1 b−a

2ⁿ

X

i=1

a+ib−a

2ⁿ −a−(i−1)b−a 2ⁿ

f a+i^b−a₂n

+f a+ (i−1)^b−a₂n

2

= 1 2ⁿ⁺¹

" ₂ⁿ X

i=1

f

1− i 2ⁿ

a+ i

2ⁿb

+f

1− i−1 2ⁿ a

+i−1 2ⁿ b

#

=y_n.

By the convexity off we obtain y_n ≤ 1

2ⁿ⁺¹

2ⁿ

X

i=1

1− i

2ⁿ

f(a) + i

2ⁿf(b) +

1− i−1 2ⁿ

f(a) + i−1 2ⁿ f(b)

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= 1 2ⁿ⁺¹

"

f(a)

2ⁿ

X

i=1

2− i

2ⁿ⁻¹ + 1 2ⁿ

+f(b)

2ⁿ

X

i=1

i

2ⁿ⁻¹ − 1 2ⁿ

#

= 1 2ⁿ⁺¹

f(a)

2ⁿ⁺¹− 1 2ⁿ⁻¹

2ⁿ(2ⁿ+ 1) 2 + 2ⁿ

2ⁿ

+f(b) 1

2ⁿ⁻¹ · 2ⁿ(2ⁿ+ 1) 2 − 2ⁿ

2ⁿ

= 1

2ⁿ⁺¹[f(a)(2ⁿ⁺¹−2ⁿ) +f(b)(2ⁿ)] = f(a) +f(b)

2 ,

so

1 b−a

Z b

a

f(x)dx≤y_n ≤ f(a) +f(b)

2 .

On the other hand, by the left side of inequality (1.1) we have 1

b−a Z b

a

f(x)dx= 1 b−a

2ⁿ

X

i=1

Z a+i^b−a₂n

a+(i−1)^b−a₂_n

f(x)dx≥ 1 b−a

2ⁿ

X

i=1

b−a 2ⁿ ,

f a+i^b−a₂n +a+ (i−1)^b−a₂n

2

!

= 1 2ⁿ

2ⁿ

X

i=1

f

a+ib−a

2ⁿ − b−a 2ⁿ⁺¹

=x_n. By the convexity off and Jensen’s inequality we obtain

x_n= 1 2ⁿ

2ⁿ

X

i=1

f

a+ib−a

2ⁿ − b−a 2ⁿ⁺¹

≥f

"

1 2ⁿ

2ⁿ

X

i=1

a+ib−a

2ⁿ − b−a 2ⁿ⁺¹

#

=f 1

2ⁿ

2ⁿa+b−a

2ⁿ · 2ⁿ(2ⁿ+ 1)

2 − b−a 2ⁿ⁺¹2ⁿ

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=f

a+b−a 2

=f

a+b 2

.

Theorem 2.2. Letf be a convex function on[a, b], then{x_n}is increasing,{y_n}is decreasing and

n→∞lim x_n= lim

n→∞y_n = 1 b−a

Z b

a

f(x)dx.

Proof. We have x_n= 1

2ⁿ

X

i=1

f

a+ib−a

2ⁿ − b−a 2ⁿ⁺¹

= 1 2ⁿ

2ⁿ

X

i=1

f

(2ⁿ⁺¹−2i+ 1)a+ (2i−1)b 2ⁿ⁺¹

= 1 2ⁿ

2ⁿ

X

i=1

f 1

2 ·(2ⁿ⁺³−8i+ 4)a+ (8i−4)b 2ⁿ⁺²

= 1 2ⁿ

2ⁿ

X

i=1

f 1

2 ·(2ⁿ⁺²+ 3−4i)a+ (4i−3)b+ (2ⁿ⁺²+ 1−4i)a+ (4i−1)b 2ⁿ⁺²

≤ 1 2ⁿ⁺¹

2ⁿ

X

i=1

f

(2ⁿ⁺²+ 3−4i)a+ (4i−3)b 2ⁿ⁺²

+ 1 2ⁿ⁺¹

2ⁿ

X

i=1

f

(2ⁿ⁺²+ 1−4i)a+ (4i−1)b 2ⁿ⁺²

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setA={1,3, . . . ,2ⁿ⁺¹−1}andB ={2,4, . . . ,2ⁿ⁺¹}, thus we obtain

2ⁿ

X

i=1

f

(2ⁿ⁺²+ 3−4i)a+ (4i−3)b 2ⁿ⁺²

=X

A

f

(2ⁿ⁺²+ 1−2i)a+ (2i−1)b 2ⁿ⁺²

2ⁿ

X

i=1

f

(2ⁿ⁺²+ 1−4i)a+ (4i−1)b 2ⁿ⁺²

=X

B

f

(2ⁿ⁺²+ 1−2i)a+ (2i−1)b 2ⁿ⁺²

,

which implies that x_n≤ 1

2ⁿ⁺¹

"

X

A∪B

f

(2ⁿ⁺²+ 1−2i)a+ (2i−1)b 2ⁿ⁺²

#

=x_n+1, so{x_n}is increasing. On the other hand we have

y_n+1 = 1 2ⁿ⁺²

"

f(a) +f(b) + 2

2ⁿ⁺¹−1

X

i=1

f

1− i 2ⁿ⁺¹

a+ i 2ⁿ⁺¹b

#

= 1 2ⁿ⁺²

"

f(a) +f(b) + 2

2ⁿ⁺¹−1

X

i=1

f

(2ⁿ⁺¹−i)a+ib 2ⁿ⁺¹

# .

SettingC ={2,4,6, . . . ,2ⁿ⁺¹−2}, we obtain y_n+1 = 1

2ⁿ⁺²

"

f(a) +f(b) + 2X

i∈C

f

(2ⁿ⁺¹−i)a+ib 2ⁿ⁺¹

+ 2X

i∈A

f

(2ⁿ⁺¹−i)a+ib 2ⁿ⁺¹

#

= 1 2ⁿ⁺²

"

f(a) +f(b) + 2

2ⁿ−1

X

i=1

f

(2ⁿ⁺¹−2i)a+ 2ib 2ⁿ⁺¹

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+ 2

2ⁿ

X

i=1

f

(2ⁿ⁺¹−2i+ 1)a+ (2i−1)b 2ⁿ⁺¹

#

= 1 2ⁿ⁺²

"

f(a) +f(b) + 2

2ⁿ−1

X

i=1

f

(2ⁿ−i)a+ib 2ⁿ

+ 2

2ⁿ

X

i=1

f 1

2 · (2ⁿ−i)a+ib+ (2ⁿ−i+ 1)a+ (i−1)b 2ⁿ

#

≤ 1 2ⁿ⁺²

"

f(a) +f(b) + 2

2ⁿ−1

X

i=1

f

(2ⁿ−i)a+ib 2ⁿ

+

2ⁿ

X

i=1

f

(2ⁿ−i)a+ib 2ⁿ

+

2ⁿ

X

i=1

f

(2ⁿ−i+ 1)a+ (i−1)b 2ⁿ

#

= 1 2ⁿ⁺²

"

f(a) +f(b) + 2

2ⁿ−1

X

i=1

f

(2ⁿ−i)a+ib 2ⁿ

+

2ⁿ−1

X

i=1

f

(2ⁿ−i)a+ib 2ⁿ

+f(b) +f(a) +

2ⁿ

X

i=2

f

(2ⁿ−i+ 1)a+ (i−1)b 2ⁿ

#

= 1 2ⁿ⁺²

"

2f(a) + 2f(b) + 3

2ⁿ−1

X

i=1

f

(2ⁿ−i)a+ib 2ⁿ

+

2ⁿ−1

X

i=1

f

(2ⁿ−i)a+ib 2ⁿ

#

= 1 2ⁿ⁺¹

"

f(a) +f(b) + 2

2ⁿ−1

X

i=1

f

(2ⁿ−i)a+ib 2ⁿ

#

=y_n, so{y_n}is decreasing.

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For the proof of the last assertions, since f is continuous on [a, b], we use the following well known equality:

n→∞lim b−a

n

X

i=1

f

a+ib−a n

= Z b

a

f(x)dx.

So we obtain

n→∞lim x_n= lim

n→∞y_n = 1 b−a

Z b

a

f(x)dx.

Remark 1. Letf be a convex function on[a, b]. In conclusion, we can state that f

a+b 2

=x₀ ≤ 1 2f

3a+b 4

+f

a+ 3b 4

=x₁ ≤ · · · ≤x_n ≤ · · ·

≤ 1 b−a

Z b

a

f(x)dx≤ · · · ≤y_n ≤ · · · ≤y₁

= 1 4

f(a) + 2f

a+b 2

+f(b)

≤y₀ = f(a) +f(b)

2 .

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3. Applications for Special Means

Recall the following means a) The arithmetic mean

A(a, b) = a+b

2 (a, b > 0);

b) The geometric neam

G(a, b) =√

ab (a, b >0);

c) The harmonic mean

H(a, b) = 2

1

a+ ¹_b (a, b >0);

d) The logarithmic mean L(a, b) =

( _b−a

lnb−lna b6=a;

a b= 0; (a, b >0).

We define the two new means by the following:

e) Then−harmonic mean H_n(a, b) = 2ⁿ⁺¹

"

1 a + 2

2ⁿ−1

X

i=1

1 1− ₂ⁱn

a+₂ⁱnb + 1 b

#⁻¹

(n= 0,1,2, . . . , a, b >0)

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f) Then−arithmetic mean A_n(a, b) = 2ⁿ

" ₂n

X

i=1

1 1− ₂ⁱn + ₂n+1¹

a+ ₂ⁱn − ₂n+1¹

b

#⁻¹

(n = 0,1,2, . . .;a, b >0).

It is clear thatH₀(a, b) = H(a, b)andA₀(a, b) = A(a, b). By the above termi- nology we have the following simple proposition:

Proposition 3.1. Let0< a < b <∞. Then we have

H(a, b)≤H_n(a, b)≤L(a, b)≤A_n(a, b)≤A(a, b),

n→∞lim H_n(a, b) = lim

n→∞A_n(a, b) =L(a, b).

Proof. Letf : [a, b] → (0,∞), f(x) = _x¹ and use Remark1. We omit the details.

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References

[1] S.S. DRAGOMIR, Some remarks on Hadamard’s inequalities for convex func- tions, Extracta Math., 9(2) (1994), 88–94.

[2] S.S. DRAGOMIR, Two mappings in connection to Hadamard’s inequalities, J.

Math. Anal. Appl., 167 (1992), 42–56.

[3] S.S. DRAGOMIRANDA. McANDREW, Refinement of the Hermite-Hadamard inequality for convex functions, J. Inequal. Pure and Appl. Math., 6(5) (2005), Art. 140. [ONLINE:http://jipam.vu.edu.au/article.php?sid=

614].

[4] S.S. DRAGOMIR, Y.J. CHOANDS.S. KIM. Inequalities of Hadamard’s type for Lipschitzian mappings and their applications, J. Math. Anal. Appl., 245 (2000), 489–501.

[5] J. HADAMARD, Étude sur les proprietes des fonctions entieres en particulier d’une fonction consi´derée par Riemann, J. Math. Pures Appl., 58 (1893), 171–

215.

[6] G.S. YANG AND M.C. HONG, A note on Hadamard’s inequality, Tamkang. J.

Math., 28(1) (1997), 33–37.

[7] S.S. DRAGOMIR AND C.E.M. PEARCE, Selected Topics on the Her- mite Hadamard Inequality and Applications, RGMIA Monographs, Victoria University, 2000. [ONLINE: http://www.staff.vu.edu.au/RGMIA/

monographs/hermite_hadamard.html].