Volume 9 (2008), Issue 4, Article 105, 4 pp.
ON A GENERALIZATION OF THE HERMITE-HADAMARD INEQUALITY II
MATLOOB ANWAR AND J. PEČARIĆ
1- A BDUS S ALAM S CHOOL OF M ATHEMATICAL S CIENCES
GC U NIVERSITY, L AHORE , PAKISTAN
matloob_t@yahoo.com
U NIVERSITY O F Z AGREB
FACULTY O F T EXTILE T ECHNOLOGY
C ROATIA
pecaric@mahazu.hazu.hr
Received 01 November, 2007; accepted 12 November, 2007
Communicated by W.S. Cheung
A BSTRACT. Generalized form of Hermite-Hadamard inequality for (2n)-convex Lebesgue integrable functions are obtained through generalization of Taylor’s Formula.
Key words and phrases: Convex function, Hermite-Hadamard inequality, Taylor’s Formula.
2000 Mathematics Subject Classification. Primary 26A51; Secondary 26A46, 26A48.
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
b
f (y)dy +
a
for all x ∈ (a, b).
336-07
f+0 (x)
a+b
x−
− f (x)
2
Z b
1
0
(x − a)2 + (b − x)2 ≥
|f (y) − f (x)| dy − f+ (x)
b−a a
2(b − a)
2
M ATLOOB A NWAR AND J. P E ČARI Ć
Theorem 2. Assume that f : [a, b] → R is a convex function. Then
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
0
≥ |f (x) − f (y)| dy −
|x − y| |f (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 improve2
ments of inequality (HH).
In this paper we will prove further generalizations of these results.
Theorem 3. Assume that f : [a, b] → R is a (2n − 1)-times differentiable and (2n)−convex
function. Then
1
(b − a)
Z
b
(b − x)k+1 − (a − x)k+1 (k)
f (x)
(k + 1)!(b − a)
1
2n−2
1 Z b
X (y − x)k
(k)
≥
f (x) dy
f (y) − f (x) −
(b − a) a k!
1
(2n−1) (b − x)2n − (a − x)2n − f
(x)
(2n)!(b − a)
f (y)dy − (b − a)f (x) −
a
2n−1
X
for all x ∈ (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 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)2 00
(y − x)2n−1 (2n−1)
f (y) ≥ f (x) + (y − x)f 0 (x) +
f (x) + · · · +
f
(x)
2!
(2n − 1)!
and we can write
f (y) − f (x) − (y − x)f 0 (x) −
(y − x)2 00
(y − x)2n−1 (2n−1)
f (x) − · · · −
f
(x) ≥ 0,
2!
2n − 1!
i.e.,
(y − x)2 00
(y − x)2n−1 (2n−1)
f (y) − f (x) − (y − x)f (x) −
f (x) − · · · −
f
(x)
2!
(2n − 1)!
(y − x)2 00
(y − x)2n−1 (2n−1) 0
= f (y) − f (x) − (y − x)f (x) −
f (x) − · · · −
f
(x) .
2!
(2n − 1)!
0
J. Inequal. Pure and Appl. Math., 9(4) (2008), Art. 105, 4 pp.
http://jipam.vu.edu.au/
H ERMITE -H ADAMARD I NEQUALITY
3
Now by using the triangle inequality
(y − x)2 00
(y − x)2n−1 (2n−1)
(1) f (y) − f (x) − (y − x)f (x) −
f (x) − · · · −
f
(x)
2!
2n − 1!
(y − x)2n−2 (2n−2) ≥ f (y) − f (x) − (y − x)f 0 (x) − · · · −
f
(x)
2n − 2!
(y − x)2n−1 (2n−1) − f
(x) .
2n − 1!
0
Now integrating the last inequality with respect to y and using the triangle inequality for integrals, we get
b
2n−1
X
(b − x)k+1 − (a − x)k+1 (k)
f (x)
(k + 1)!
a
1
Z 2n−2
b
2n
2n X (y − x)k
(b
−
x)
−
(a
−
x)
.
≥
f (k) (x) dy − f (2n−1) (x)
f (y) − f (x) −
a k!
(2n)!
Z
f (y)dy − (b − a)f (x) −
1
Theorem 4. Assume that f : [a, b] → R is a (2n − 1)−times differentiable and (2n)−convex
function. Then
2n
f (x) −
(b − a)
b
2n−1
X
2n − k
[(x − b)k f (k−1) (b) − (x − a)k f (k−1) (a)]
k!(b
−
a)
a
1
2n−2
1 Z b
k
X
(x − y) (k) f (y) dy
≥
f (x) − f (y) −
b − a a k!
1
Z b
(x − y)2n−1 (2n−1) 1
−
f
(y) dy .
b − a a (2n − 1)!
Z
f (y)dy −
Proof. Integrating (1) with respect to x and by using the triangle inequality for integrals, we get
b
b 2n−1
X
(y − x)k (k)
f (x)dx
k!
a
a
1
Z Z b
2n−2
b
k
2n−1
X
(y − x) (k) (2n−1)
(y − x)
dx .
≥
f (x) dx −
f
(x)
f (y) − f (x) −
(2n − 1)!
a k!
a
1
Z
(2) (b − a)f (y) −
Z
f (x)dx −
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
1
(b − a)
Z
b
f (y)dy − (b − a)f
a
a+b
2
J. Inequal. Pure and Appl. Math., 9(4) (2008), Art. 105, 4 pp.
−
2n−1
X
1
b−a k+1
2
k+1
− a−b
a+b
(k)
2
f
(k + 1)!(b − a)
2
http://jipam.vu.edu.au/
4
M ATLOOB A NWAR AND J. P E ČARI Ć
2n−2
1 Z b
X y − a+b k
a+b
a
+
b
2
≥
−
f (k)
f (y) − f
dy
(b − a) a 2
k!
2
1
(2n−1) a + b (b − a)2n − (a − b)2n .
− f
2
(2n)!(b − a)22n Proof. Set x =
a+b
2
in Theorem 3.
Corollary 6. Suppose that f : [a, b] → R is a (2n − 1)−times differentiable and (2n)−convex
function. Then
Z b
2n−1
X 2n − k
2n
f (a) −
f (y)dy −
[(a − b)k f (k−1) (b)]
(b − a) a
k!(b − a)
1
2n−2
1 Z b
X (a − y)k
(k)
≥
f (y) dy
f (a) − f (y) −
b − a a k!
1
Z b
(a − y)2n−1 (2n−1) 1
f
(y) dy .
−
b − a a (2n − 1)!
Proof. Set x = a in Theorem 4.
R EFERENCES
[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].
J. Inequal. Pure and Appl. Math., 9(4) (2008), Art. 105, 4 pp.
http://jipam.vu.edu.au/