ON A GENERALIZATION OF THE HERMITE-HADAMARD INEQUALITY II

Hermite-Hadamard Inequality M. Anwar and J. Peˇcari´c vol. 9, iss. 4, art. 105, 2008

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ON A GENERALIZATION OF THE HERMITE-HADAMARD INEQUALITY II

MATLOOB ANWAR J. PE ˇCARI ´C

1- Abdus Salam School of Mathematical Sciences University Of Zagreb GC University, Lahore, Faculty Of Textile Technology

Pakistan Croatia

EMail:matloob_t@yahoo.com EMail:pecaric@mahazu.hazu.hr

Received: 01 November, 2007

Accepted: 12 November, 2007

Communicated by: W.S. Cheung

2000 AMS Sub. Class.: Primary 26A51; Secondary 26A46, 26A48.

Key words: Convex function, Hermite-Hadamard inequality, Taylor’s Formula.

Abstract: Generalized form of Hermite-Hadamard inequality for(2n)-convex Lebesgue integrable functions are obtained through generalization of Taylor’s Formula.

Hermite-Hadamard Inequality M. Anwar and J. Peˇcari´c vol. 9, iss. 4, art. 105, 2008

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The classical Hermite-Hadamard inequality gives us an estimate, from below and from above, of the mean value of a convex functionf : [a, b]→R(see [1, pp. 137]):

(HH) f

a+b 2

≤ 1 b−a

Z b

a

f(x)dx≤ f(a) +f(b)

2 .

In [2] the first author with Sabir Hussain proved the following two theorems Theorem 1. Assume thatf is Lebesgue integrable and convex on(a, b).Then

1 b−a

Z b

a

f(y)dy+f₊⁰ (x)

x− a+b 2

−f(x)

≥

1 b−a

Z b

a

|f(y)−f(x)|dy− f₊⁰ (x)

(x−a)²+ (b−x)² 2(b−a)

for allx∈(a, b).

Theorem 2. Assume thatf : [a, b]→Ris a convex function. Then 1

2

f(x) + f(b)(b−x) +f(a)(x−a) b−a

− 1 b−a

Z b

a

f(y)dy

≥ 1 2

1 b−a

Z b

a

|f(x)−f(y)|dy− 1 b−a

Z b

a

|x−y| |f⁰(y)|dy

for allx∈(a, b).

Remark 1. Forx = ^a+b₂ in Theorem1andx=aorx= bin Theorem2, we obtain improvements of inequality (HH).

In this paper we will prove further generalizations of these results.

Hermite-Hadamard Inequality M. Anwar and J. Peˇcari´c vol. 9, iss. 4, art. 105, 2008

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Theorem 3. Assume that f : [a, b] → R is a (2n − 1)-times differentiable and (2n)−convex function. Then

1 (b−a)

Z b

a

f(y)dy−(b−a)f(x)−

2n−1

X

1

(b−x)^k+1−(a−x)^k+1

(k+ 1)!(b−a) f^(k)(x)

≥

1 (b−a)

Z b

a

f(y)−f(x)−

2n−2

X

1

(y−x)^k

k! f^(k)(x)

dy

−

f⁽²ⁿ⁻¹⁾(x)(b−x)²ⁿ−(a−x)²ⁿ (2n)!(b−a)

for allx∈(a, b).

Proof. It is well known that a continuous(2n)−convex function can be uniformly approximated by a (2n)−convex polynomial. So we can suppose that we have (2n)−derivatives off. By Taylor’s formula,

f(y) = f(x) + (y−x)f⁰(x) + (y−x)²

2! f⁰⁰(x) +· · · + (y−x)²ⁿ⁻¹

2n−1! f⁽²ⁿ⁻¹⁾(x) + (y−x)²ⁿ

2n! f⁽²ⁿ⁾(ξ), forx, y ∈[a, b],ξ ∈(a, b). Sincef is(2n)−convex, we havef⁽²ⁿ⁾(x)≥0.

So

f(y)≥f(x) + (y−x)f⁰(x) + (y−x)²

2! f⁰⁰(x) +· · ·+(y−x)²ⁿ⁻¹

(2n−1)! f⁽²ⁿ⁻¹⁾(x) and we can write

f(y)−f(x)−(y−x)f⁰(x)−(y−x)²

2! f⁰⁰(x)− · · · −(y−x)²ⁿ⁻¹

2n−1! f⁽²ⁿ⁻¹⁾(x)≥0,

Hermite-Hadamard Inequality M. Anwar and J. Peˇcari´c vol. 9, iss. 4, art. 105, 2008

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i.e.,

f(y)−f(x)−(y−x)f⁰(x)− (y−x)²

2! f⁰⁰(x)− · · · −(y−x)²ⁿ⁻¹

(2n−1)! f⁽²ⁿ⁻¹⁾(x)

=

f(y)−f(x)−(y−x)f⁰(x)− (y−x)²

2! f⁰⁰(x)− · · · − (y−x)²ⁿ⁻¹

(2n−1)! f⁽²ⁿ⁻¹⁾(x) .

Now by using the triangle inequality

(1) f(y)−f(x)−(y−x)f⁰(x)− (y−x)²

2! f⁰⁰(x)− · · · − (y−x)²ⁿ⁻¹

2n−1! f⁽²ⁿ⁻¹⁾(x)

≥

f(y)−f(x)−(y−x)f⁰(x)− · · · − (y−x)²ⁿ⁻²

2n−2! f⁽²ⁿ⁻²⁾(x)

−

(y−x)²ⁿ⁻¹

2n−1! f⁽²ⁿ⁻¹⁾(x) . Now integrating the last inequality with respect toyand using the triangle inequality for integrals, we get

Z b

a

f(y)dy−(b−a)f(x)−

2n−1

X

1

(b−x)^k+1−(a−x)^k+1

(k+ 1)! f^(k)(x)

≥

Z b

a

f(y)−f(x)−

2n−2

X

1

(y−x)^k

k! f^(k)(x)

dy

−

f⁽²ⁿ⁻¹⁾(x)(b−x)²ⁿ−(a−x)²ⁿ (2n)!

.

Hermite-Hadamard Inequality M. Anwar and J. Peˇcari´c vol. 9, iss. 4, art. 105, 2008

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Theorem 4. Assume that f : [a, b] → R is a (2n −1)−times differentiable and (2n)−convex function. Then

f(x)− 2n (b−a)

Z b

a

f(y)dy−

2n−1

X

1

2n−k

k!(b−a)[(x−b)^kf^(k−1)(b)−(x−a)^kf^(k−1)(a)]

≥

1 b−a

Z b

a

f(x)−f(y)−

2n−2

X

1

(x−y)^k

k! f^(k)(y)

dy

− 1 b−a

Z b

a

(x−y)²ⁿ⁻¹

(2n−1)! f⁽²ⁿ⁻¹⁾(y)

dy . Proof. Integrating (1) with respect toxand by using the triangle inequality for inte- grals, we get

(2) (b−a)f(y)− Z b

a

f(x)dx− Z b

a 2n−1

X

1

(y−x)^k

k! f^(k)(x)dx

≥

Z b

a

f(y)−f(x)−

2n−2

X

1

(y−x)^k

k! f^(k)(x)

dx

− Z b

a

(y−x)²ⁿ⁻¹

(2n−1)! f⁽²ⁿ⁻¹⁾(x)

dx . By replacingxandywe obtain the required result.

Corollary 5. Suppose that f : [a, b] → R is a(2n−1)−times differentiable and (2n)−convex function. Then

1 (b−a)

Z b

a

f(y)dy−(b−a)f

a+b 2

−

2n−1

X

1 b−a

2

k+1

− ^a−b₂ k+1

(k+ 1)!(b−a) f^(k)

a+b 2

Hermite-Hadamard Inequality M. Anwar and J. Peˇcari´c vol. 9, iss. 4, art. 105, 2008

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≥

1 (b−a)

Z b

a

f(y)−f

a+b 2

−

2n−2

X

1

y− ^a+b₂ k

k! f^(k)

a+b 2

dy

−

f⁽²ⁿ⁻¹⁾

a+b 2

(b−a)²ⁿ−(a−b)²ⁿ (2n)!(b−a)2²ⁿ

.

Proof. Setx= ^a+b₂ in Theorem3.

Corollary 6. Suppose that f : [a, b] → R is a(2n−1)−times differentiable and (2n)−convex function. Then

f(a)− 2n (b−a)

Z b

a

f(y)dy−

2n−1

X

1

2n−k

k!(b−a)[(a−b)^kf^(k−1)(b)]

≥

1 b−a

Z b

a

f(a)−f(y)−

2n−2

X

1

(a−y)^k

k! f^(k)(y)

dy

− 1 b−a

Z b

a

(a−y)²ⁿ⁻¹

(2n−1)! f⁽²ⁿ⁻¹⁾(y)

dy . Proof. Setx=ain Theorem4.

Hermite-Hadamard Inequality M. Anwar and J. Peˇcari´c vol. 9, iss. 4, art. 105, 2008

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References

[1] J.E. PE ˇCARI ´C, F. PROSCHAN AND Y.C. TONG, Convex Functions, Partial Orderings and Statistical Applications, Academic Press, New York, 1992.

[2] S. HUSSAIN AND M. ANWAR, On generalization of the Hermite-Hadamard inequality, J. Inequal. Pure and Appl. Math., 8(2) (2007), Art. 60. [ONLINE:

http://jipam.vu.edu.au/article.php?sid=873].