ON A GENERALIZATION OF THE
HERMITE-HADAMARD INEQUALITY II
Hermite-Hadamard Inequality
MATLOOB ANWAR
J. PEČARIĆ
1- Abdus Salam School of Mathematical Sciences
GC University, Lahore,
Pakistan
EMail: matloob_t@yahoo.com
University Of Zagreb
Faculty Of Textile Technology
Croatia
EMail: pecaric@mahazu.hazu.hr
M. Anwar and J. Pečarić
vol. 9, iss. 4, art. 105, 2008
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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.
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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 function f : [a, b] → R (see [1, pp. 137]):
Z b
a+b
1
f (a) + f (b)
(HH)
f
≤
f (x) dx ≤
.
2
b−a a
2
In [2] the first author with Sabir Hussain proved the following two theorems
Theorem 1. Assume that f is Lebesgue integrable and convex on (a, b). Then
1
b−a
Z
a
b
a+b
f (y)dy + f+0 (x) x −
− f (x)
2
Z b
1
(x − a)2 + (b − x)2
|f (y) − f (x)| dy − f+0 (x)
≥
b−a a
2(b − a)
Hermite-Hadamard Inequality
M. Anwar and J. Pečarić
vol. 9, iss. 4, art. 105, 2008
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Contents
for all x ∈ (a, b).
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Theorem 2. Assume that f : [a, b] → R is a convex function. Then
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Z b
1
f (b)(b − x) + f (a)(x − a)
1
f (x) +
−
f (y)dy
2
b−a
b−a a
Z b
Z b
1
1
1
≥
|f (x) − f (y)| dy −
|x − y| |f 0 (y)| dy
2 b−a a
b−a a
for all x ∈ (a, b).
Remark 1. For x = a+b
in Theorem 1 and x = a or x = b in Theorem 2, we obtain
2
improvements of inequality (HH).
In this paper we will prove further generalizations of these results.
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Theorem 3. Assume that f : [a, b] → R is a (2n − 1)-times differentiable and
(2n)−convex function. Then
Z b
2n−1
X (b − x)k+1 − (a − x)k+1
1
f (y)dy − (b − a)f (x) −
f (k) (x)
(b − a) a
(k + 1)!(b − a)
1
Z b
2n−2
X (y − x)k
1
≥
f (y) − f (x) −
f (k) (x) dy
(b − a) a
k!
1
(b − x)2n − (a − x)2n
− f (2n−1) (x)
(2n)!(b − a)
for all x ∈ (a, b).
M. Anwar and J. Pečarić
vol. 9, iss. 4, art. 105, 2008
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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 of f . By Taylor’s formula,
f (y) = f (x) + (y − x)f 0 (x) +
(y − x)2 00
f (x) + · · ·
2!
(y − x)2n−1 (2n−1)
(y − x)2n (2n)
+
f
(x) +
f (ξ),
2n − 1!
2n!
for x, y ∈ [a, b], ξ ∈ (a, b). Since f is (2n)−convex, we have f (2n) (x) ≥ 0.
So
(y − x)2n−1 (2n−1)
(y − x)2 00
0
f (x) + · · · +
f
(x)
f (y) ≥ f (x) + (y − x)f (x) +
2!
(2n − 1)!
and we can write
f (y) − f (x) − (y − x)f 0 (x) −
Hermite-Hadamard Inequality
(y − x)2n−1 (2n−1)
(y − x)2 00
f (x) − · · · −
f
(x) ≥ 0,
2!
2n − 1!
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i.e.,
(y − x)2 00
(y − x)2n−1 (2n−1)
f (x) − · · · −
f
(x)
2!
(2n − 1)!
(y − x)2 00
(y − x)2n−1 (2n−1)
f (x) − · · · −
f
(x) .
= f (y) − f (x) − (y − x)f 0 (x) −
2!
(2n − 1)!
f (y) − f (x) − (y − x)f 0 (x) −
Hermite-Hadamard Inequality
Now by using the triangle inequality
M. Anwar and J. Pečarić
2
2n−1
(y − x) 00
(y − x)
f (x) − · · · −
f (2n−1) (x)
2!
2n − 1!
2n−2
(y
−
x)
≥ f (y) − f (x) − (y − x)f 0 (x) − · · · −
f (2n−2) (x)
2n − 2!
(y − x)2n−1 (2n−1)
−
f
(x) .
2n − 1!
(1) f (y) − f (x) − (y − x)f 0 (x) −
Now integrating the last inequality with respect to y and using the triangle inequality
for integrals, we get
Z
b
f (y)dy − (b − a)f (x) −
a
2n−1
X
1
Z
≥
(b − x)k+1 − (a − x)k+1 (k)
f (x)
(k + 1)!
b
f (y) − f (x) −
a
2n−2
X
1
(y − x)k (k)
f (x) dy
k!
− f (2n−1) (x)
(b − x)2n − (a − x)2n
.
(2n)!
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
Z b
2n−1
X 2n − k
2n
[(x − b)k f (k−1) (b) − (x − a)k f (k−1) (a)]
f (x) −
f (y)dy −
(b − a) a
k!(b
−
a)
1
Z b
2n−2
X (x − y)k
1
f (x) − f (y) −
f (k) (y) dy
≥
b−a a
k!
1
Z b
1
(x − y)2n−1 (2n−1)
−
f
(y) dy .
b − a a (2n − 1)!
Proof. Integrating (1) with respect to x and by using the triangle inequality for integrals, we get
Z b
Z b 2n−1
X (y − x)k
(2) (b − a)f (y) −
f (x)dx −
f (k) (x)dx
k!
a
a
1
Z b
2n−2
X (y − x)k
≥
f (y) − f (x) −
f (k) (x) dx
k!
a
1
Z b
(y − x)2n−1 (2n−1)
−
f
(x) dx .
(2n − 1)!
a
By replacing x and y we obtain the required result.
Corollary 5. Suppose that f : [a, b] → R is a (2n − 1)−times differentiable and
(2n)−convex function. Then
2n−1
Z b
X b−a k+1 − a−b k+1
1
a+b
a+b
(k)
2
2
f (y)dy−(b−a)f
−
f
(b − a) a
2
(k + 1)!(b − a)
2
1
Hermite-Hadamard Inequality
M. Anwar and J. Pečarić
vol. 9, iss. 4, art. 105, 2008
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1
≥
(b − a)
Proof. Set x =
b
Z
f (y) − f
a
a+b
2
k
y − a+b
a+b
(k)
2
f
dy
−
k!
2
1
a + b (b − a)2n − (a − b)2n
(2n−1)
− f
.
2
(2n)!(b − a)22n
a+b
2
2n−2
X
in Theorem 3.
Hermite-Hadamard Inequality
Corollary 6. Suppose that f : [a, b] → R is a (2n − 1)−times differentiable and
(2n)−convex function. Then
2n
f (a) −
(b − a)
Z
b
f (y)dy −
a
1
≥
b−a
2n−1
X
1
Z
a
b
2n − k
[(a − b)k f (k−1) (b)]
k!(b − a)
2n−2
X
(a − y)k (k)
f (a) − f (y) −
f (y) dy
k!
1
Z b
1
(a − y)2n−1 (2n−1)
−
f
(y) dy .
b − a a (2n − 1)!
Proof. Set x = a in Theorem 4.
M. Anwar and J. Pečarić
vol. 9, iss. 4, art. 105, 2008
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References
[1] J.E. PEČARIĆ, 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].
Hermite-Hadamard Inequality
M. Anwar and J. Pečarić
vol. 9, iss. 4, art. 105, 2008
Title Page
Contents
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