Volume 10 (2009), Issue 2, Article 45, 7 pp.
A NEW REFINEMENT OF THE HERMITE-HADAMARD INEQUALITY FOR
CONVEX FUNCTIONS
G. ZABANDAN
zabandan@tmu.ac.ir
D EPARTMENT OF M ATHEMATICS
FACULTY OF M ATHEMATICAL S CIENCE AND C OMPUTER E NGINEERING
T EACHER T RAINING U NIVERSITY, 599 TALEGHANI AVENUE
T EHRAN 15618 IRAN.
Received 13 August, 2008; accepted 12 February, 2009
Communicated by S.S. Dragomir
A BSTRACT. In this paper we establish a new refinement of the Hermite-Hadamard inequality
for convex functions.
Key words and phrases: Hermite-Hadamard inequality.
2000 Mathematics Subject Classification. 26D15, 26D10.
1. I NTRODUCTION
Let f : [a, b] → R be a convex function, then the following inequality:
Z b
a+b
1
f (a) + f (b)
≤
f (x)dx ≤
(1.1)
f
2
b−a a
2
is known as the Hermite-Hadamard inequality [5].
In recent years there have been many extensions, generalizations and similar results 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. If f : [a, b] → R is a convex function, and H is defined on [0, 1] by
Z b 1
a+b
H(t) =
f tx + (1 − t)
dx,
b−a a
2
then H is convex, increasing on [0, 1], and for all t ∈ [0, 1], we have
Z b
a+b
1
f
= H(0) ≤ H(t) ≤ H(1) =
f (x)dx.
2
b−a a
In [6] Yang and Hong established the following theorem which is a refinement of the right
side of inequality (1.1).
226-08
2
G. Z ABANDAN
Theorem 1.2. If f : [a, b] → R is a convex function, and F is defined by
Z b 1+t
1−t
1+t
1−t
1
F (t) =
f
a+
x +f
b+
x dx,
2(b − a) a
2
2
2
2
then F is convex, increasing on [0, 1], and for all t ∈ [0, 1], we have
Z b
1
f (a) + f (b)
f (x)dx = F (0) ≤ F (t) ≤ F (1) =
.
b−a a
2
In this paper we establish a refinement of the both sides of inequality (1.1). For this we first
define two sequences {xn } and {yn } by
n
(1.2)
(1.3)
2
1 X
b−a b−a
xn = n
f a + i n − n+1
2 i=1
2
2
2n
1 X
1 b−a
= n
f a+ i−
,
2 i=1
2
2n
2n i
i
i−1
i−1
1 X
yn = n+1
f
1 − n a + nb + f
1− n
a+ n b
2
2
2
2
2
i=1
"
#
n
2X
−1 i
i
1
f
1 − n a + nb
= n+1 f (a) + f (b) + 2
2
2
2
i=1
and we prove the following
a+b
1
3a + b
a + 3b
f
= x0 ≤
f
+f
2
2
4
4
= x1 ≤ · · · ≤ xn ≤ · · ·
Z b
1
≤
f (x)dx ≤ · · · ≤ yn ≤ · · · ≤ y1
b−a a
1
a+b
f (a) + 2f
+ f (b)
=
4
2
f (a) + f (b)
,
≤ y0 =
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].
2. A R EFINEMENT R ESULT
In this section, using the terminologies of the introduction, we refine the Hermite-Hadamard
inequality via the sequences {xn } and {yn }.
Theorem 2.1. Let f be a convex function on [a, b]. Then we have
Z b
a+b
1
f (a) + f (b)
f
≤ xn ≤
f (x)dx ≤ yn ≤
.
2
b−a a
2
J. Inequal. Pure and Appl. Math., 10(2) (2009), Art. 45, 7 pp.
http://jipam.vu.edu.au/
H ERMITE -H ADAMARD I NEQUALITY
3
Proof. By the right side of Hermite-Hadamard inequality (1.1) we have
Z b
1
f (x)dx
b−a a
b−a
2n Z
1 X a+i 2n
f (x)dx
=
b − a i=1 a+(i−1) b−a
n
2
n 2
+ f a + (i − 1) b−a
b−a
b − a f a + i b−a
1 X
2n
2n
≤
a + i n − a − (i − 1) n
2
b − a i=1
2
2
" 2n #
X
1
i
i
i−1
i−1
= n+1
f
1 − n a + nb + f
1− n a + n b
2
2
2
2
2
i=1
= yn .
By the convexity of f we obtain
2n 1 X
i
i
i−1
i−1
yn ≤ n+1
1 − n f (a) + n f (b) + 1 − n
f (a) + n f (b)
2
2
2
2
2
i=1
"
#
2n 2n X
X
1
i
1
i
1
= n+1 f (a)
2 − n−1 + n + f (b)
−
2
2
2
2n−1 2n
i=1
i=1
1
1 2n (2n + 1) 2n
1
2n (2n + 1) 2n
n+1
= n+1 f (a) 2
− n−1
+ n + f (b) n−1 ·
− n
2
2
2
2
2
2
2
1
f (a) + f (b)
,
= n+1 [f (a)(2n+1 − 2n ) + f (b)(2n )] =
2
2
so
Z b
f (a) + f (b)
1
f (x)dx ≤ yn ≤
.
b−a a
2
On the other hand, by the left side of inequality (1.1) we have
b−a
Z b
2n Z
2n
1
1 X a+i 2n
1 Xb−a
f (x)dx =
f (x)dx ≥
,
b−a a
b − a i=1 a+(i−1) b−a
b − a i=1 2n
2n
!
2n
b−a
X
a + i b−a
+
a
+
(i
−
1)
1
b−a b−a
n
n
2
2
f
= n
f a + i n − n+1 = xn .
2
2 i=1
2
2
By the convexity of f and Jensen’s inequality we obtain
2n
1 X
b−a b−a
xn = n
f a + i n − n+1
2 i=1
2
2
"
#
2n 1 X
b−a b−a
a + i n − n+1
≥f n
2 i=1
2
2
1
b − a 2n (2n + 1) b − a n
n
=f n 2 a+ n ·
− n+1 2
2
2
2
2
b−a
a+b
=f a+
=f
.
2
2
J. Inequal. Pure and Appl. Math., 10(2) (2009), Art. 45, 7 pp.
http://jipam.vu.edu.au/
4
G. Z ABANDAN
Theorem 2.2. Let f be a convex function on [a, b], then {xn } is increasing, {yn } is decreasing
and
Z b
1
lim xn = lim yn =
f (x)dx.
n→∞
n→∞
b−a a
Proof. We have
2n
1 X
b−a b−a
xn = n
f a + i n − n+1
2 i=1
2
2
n+1
2n
1 X
(2
− 2i + 1)a + (2i − 1)b
= n
f
2 i=1
2n+1
2n
1 X
1 (2n+3 − 8i + 4)a + (8i − 4)b
= n
f
·
2 i=1
2
2n+2
2n
1 X
1 (2n+2 + 3 − 4i)a + (4i − 3)b + (2n+2 + 1 − 4i)a + (4i − 1)b
= n
f
·
2 i=1
2
2n+2
n+2
2n
1 X
(2
+ 3 − 4i)a + (4i − 3)b
≤ n+1
f
2
2n+2
i=1
n+2
2n
(2
+ 1 − 4i)a + (4i − 1)b
1 X
f
+ n+1
2
2n+2
i=1
set A = {1, 3, . . . , 2n+1 − 1} and B = {2, 4, . . . , 2n+1 }, thus we obtain
n+2
X n+2
2n
X
(2
+ 3 − 4i)a + (4i − 3)b
(2
+ 1 − 2i)a + (2i − 1)b
f
=
f
2n+2
2n+2
i=1
A
n+2
X n+2
2n
X
(2
+ 1 − 4i)a + (4i − 1)b
(2
+ 1 − 2i)a + (2i − 1)b
f
=
f
,
n+2
n+2
2
2
i=1
B
which implies that
xn ≤
1
"
2n+1
#
X (2n+2 + 1 − 2i)a + (2i − 1)b f
= xn+1 ,
n+2
2
A∪B
so {xn } is increasing. On the other hand we have
yn+1 =
1
2n+2
"
f (a) + f (b) + 2
2n+1
X−1
f
1−
i
a+
i
2n+1
i=1
"
#
2n+1
X−1 (2n+1 − i)a + ib 1
= n+2 f (a) + f (b) + 2
f
.
2
2n+1
i=1
J. Inequal. Pure and Appl. Math., 10(2) (2009), Art. 45, 7 pp.
2n+1
#
b
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H ERMITE -H ADAMARD I NEQUALITY
5
Setting C = {2, 4, 6, . . . , 2n+1 − 2}, we obtain
yn+1 =
=
=
≤
=
=
=
#
X (2n+1 − i)a + ib X (2n+1 − i)a + ib f (a) + f (b) + 2
f
+2
f
n+1
2n+2
2
2n+1
i∈C
i∈A
"
n −1
n+1
2X
(2
− 2i)a + 2ib
1
f (a) + f (b) + 2
f
2n+2
2n+1
i=1
n+1
#
2n
X
(2
− 2i + 1)a + (2i − 1)b
+2
f
2n+1
i=1
"
n −1
n
2X
1
(2 − i)a + ib
f (a) + f (b) + 2
f
2n+2
2n
i=1
#
2n
n
n
X
1 (2 − i)a + ib + (2 − i + 1)a + (i − 1)b
+2
f
·
2
2n
i=1
"
n −1
n
2X
1
(2 − i)a + ib
f (a) + f (b) + 2
f
2n+2
2n
i=1
n
X
n
#
2n
2n
X
(2 − i)a + ib
(2 − i + 1)a + (i − 1)b
+
f
+
f
n
2
2n
i=1
i=1
"
n −1
n −1
n
2X
n
2X
1
(2 − i)a + ib
(2 − i)a + ib
f (a) + f (b) + 2
f
+
f
n
2n+2
2
2n
i=1
i=1
#
n
2n
X
(2 − i + 1)a + (i − 1)b
+f (b) + f (a) +
f
2n
i=2
"
n −1
n −1
n
2X
n
#
2X
1
(2 − i)a + ib
(2 − i)a + ib
2f (a) + 2f (b) + 3
f
+
f
2n+2
2n
2n
i=1
i=1
"
n −1
n
#
2X
1
(2 − i)a + ib
f (a) + f (b) + 2
f
= yn ,
n
2n+1
2
i=1
1
"
so {yn } is decreasing.
For the proof of the last assertions, since f is continuous on [a, b], we use the following well
known equality:
Z b
n
b−a
b−aX
lim
f a+i
=
f (x)dx.
n→∞
n i=1
n
a
So we obtain
1
lim xn = lim yn =
n→∞
n→∞
b−a
Z
b
f (x)dx.
a
J. Inequal. Pure and Appl. Math., 10(2) (2009), Art. 45, 7 pp.
http://jipam.vu.edu.au/
6
G. Z ABANDAN
Remark 1. Let f be a convex function on [a, b]. In conclusion, we can state that
a+b
3a + b
a + 3b
1
f
= x0 ≤ f
+f
2
2
4
4
= x1 ≤ · · · ≤ xn ≤ · · ·
Z b
1
≤
f (x)dx ≤ · · · ≤ yn ≤ · · · ≤ y1
b−a a
a+b
1
=
f (a) + 2f
+ f (b)
4
2
f (a) + f (b)
≤ y0 =
.
2
3. A PPLICATIONS FOR S PECIAL M EANS
Recall the following means
a) The arithmetic mean
A(a, b) =
b) The geometric neam
a+b
2
(a, b > 0);
√
ab
G(a, b) =
(a, b > 0);
c) The harmonic mean
H(a, b) =
1
a
2
+
1
b
(a, b > 0);
d) The logarithmic mean
(
L(a, b) =
b−a
ln b−ln a
a
b 6= a;
b = 0;
(a, b > 0).
We define the two new means by the following:
e) The n−harmonic mean
"
n −1
2X
1
n+1 1
Hn (a, b) = 2
+2
i
a
1 − 2n a +
i=1
(n = 0, 1, 2, . . . ,
i
b
2n
+
1
b
#−1
a, b > 0)
f) The n−arithmetic mean
"
An (a, b) = 2n
#−1
n
2
X
i=1
1−
i
2n
1
1
+ 2n+1 a +
i
2n
−
1
2n+1
b
(n = 0, 1, 2, . . . ; a, b > 0).
It is clear that H0 (a, b) = H(a, b) and A0 (a, b) = A(a, b). By the above terminology we have
the following simple proposition:
Proposition 3.1. Let 0 < a < b < ∞. Then we have
H(a, b) ≤ Hn (a, b) ≤ L(a, b) ≤ An (a, b) ≤ A(a, b),
lim Hn (a, b) = lim An (a, b) = L(a, b).
n→∞
Proof. Let f : [a, b] → (0, ∞), f (x) =
n→∞
1
x
J. Inequal. Pure and Appl. Math., 10(2) (2009), Art. 45, 7 pp.
and use Remark 1. We omit the details.
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H ERMITE -H ADAMARD I NEQUALITY
7
R EFERENCES
[1] S.S. DRAGOMIR, Some remarks on Hadamard’s inequalities for convex functions, 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. DRAGOMIR AND A. 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. CHO AND S.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
consid́erée 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 Hermite Hadamard Inequality and
Applications, RGMIA Monographs, Victoria University, 2000. [ONLINE: http://www.staff.
vu.edu.au/RGMIA/monographs/hermite_hadamard.html].
J. Inequal. Pure and Appl. Math., 10(2) (2009), Art. 45, 7 pp.
http://jipam.vu.edu.au/