IJRRAS 5 (2) ● November 2010 Amrahov ● A Note On Hadamard Inequalities
168
A NOTE ON HADAMARD INEQUALITIES FOR THE PRODUCT
OF THE CONVEX FUNCTIONS
Şahin Emrah Amrahov
The Computer Engineering Department of Engineering Faculty, Ankara University, Keçiören, Ankara, Turkey.
ABSTRACT
The main aim of the present note is to prove new Hadamard like integral inequalities for the product of the convex functions.
Keywords: integral inequality, convex functions, Hadamard inequality
Mathematics subject classification (2000):26D15, 26D99
1. INTRODUCTION
Let
f
be a real valued convex function defined on the interval[
a
,
b
]
. Then2
)
(
)
(
)
(
1
2
b
f
a
f
dx
x
f
a
b
b
a
f
b
a
(1)is known as Hadamard’s inequality for the convex function. [1] If
f
uv
andu
,
v
are convex functions then we have2
)
(
)
(
)
(
)
(
)
(
)
(
1
2
2
b
v
b
u
a
v
a
u
dx
x
v
x
u
a
b
b
a
v
b
a
u
b
a
(2)by Hadamard inequality.
Therefore by Cauchy-Scwartz inequality
2
)
(
)
(
)
(
)
(
2
)
(
)
(
)
(
)
(
a
v
a
u
b
v
b
u
2a
u
2b
v
2a
v
2b
u
(3)
the inequality
2
)
(
)
(
)
(
)
(
)
(
)
(
1
u
2a
u
2b
v
2a
v
2b
dx
x
v
x
u
a
b
b
a
(4)holds.
Note that if
u
,
v
are convex functions, then the functionf
uv
may not be convex function. Example1.u
(
x
)
x
2,
v
(
x
)
(
1
x
)
2 are convex functions defined on[
0
,
1
]
,but the function2 2 2
2
(
1
)
(
)
)
(
)
(
)
(
x
u
x
v
x
x
x
x
x
f
is not convex function on this interval.In fact, the first and second derivatives
f
(
x
)
,f
(
x
)
off
(
x
)
can be calculated by the formulas.)
1
2
)(
(
2
)
(
2
x
x
x
x
f
)
(
4
)
1
2
(
2
)
(
x
x
2x
2x
f
Therefore we have
1
2
1
f
.It means that the function
f
uv
is not convex function on the interval[
0
,
1
]
.Example2. For the convex functions
u
(
x
)
x
2,
v
(
x
)
(
1
x
)
2 defined on[
0
,
1
]
we have30
1
)
(
)
(
)
(
1
10
2
2
a
u
x
v
x
dx
x
x
dx
b
b
a
IJRRAS 5 (2) ● November 2010 Amrahov ● A Note On Hadamard Inequalities
169
.
0
)
1
(
)
1
(
,
0
)
0
(
)
0
(
v
u
v
u
Hence the Hadamard inequality
0
2
)
1
(
)
1
(
)
0
(
)
0
(
30
1
)
(
)
(
)
(
1
10
2
2
v
u
v
u
dx
x
x
dx
x
v
x
u
a
b
b
a
for the non-convex function
f
uv
does not hold.. On the other hand, since0
)
1
(
,
1
)
0
(
,
1
)
1
(
,
0
)
0
(
2 2 2 2
v
v
u
u
the inequality
2
1
2
)
1
(
)
0
(
)
1
(
)
0
(
30
1
)
(
)
(
)
(
1
1 2 2 2 20
2
2
v
v
u
u
dx
x
x
dx
x
v
x
u
a
b
b
a
holds. It means that although the function
f
uv
is no convex function we have that the inequality (4) is true for these functions. Our aim is to investigate the inequality (4) whenf
uv
is non-convex function.2. MAIN RESULTS
Our main result is the following theorem.
Theorem. Let
u
andv
are nonnegative convex functions defined on the interval[
a
,
b
]
. Then the inequality (4) holds.To prove of the theorem we need the following lemma.
Lemma. Let
u
is a nonnegative convex function defined on the interval[
a
,
b
]
. Then the functionu
2 is also convex function on the interval[
a
,
b
]
Proof. For arbitrary
x
,
y
[
a
,
b
]
andk
[
0
,
1
]
we have0
)
(
)
(
)
(
2
)
(
0
))
(
)
(
(
2 2
2
y
u
y
u
x
u
x
u
y
u
x
u
)
(
)
(
)
(
)
(
2
u
x
u
y
u
2x
u
2y
(5) Multiplying both sides of the inequality (5) byk
(
1
k
)
we get)
(
)
1
(
)
(
)
1
(
)
(
)
(
)
1
(
2
k
k
u
x
u
y
k
k
u
2x
k
k
u
2y
Therefore
)
(
]
)
1
(
)
1
[(
)
(
)
(
)
(
)
(
)
1
(
2
k
k
u
x
u
y
k
k
2u
2x
k
k
2u
2y
So)
(
)
1
(
)
(
)
1
(
)
(
)
(
)
(
)
(
)
1
(
2
k
k
u
x
u
y
ku
2x
k
2u
2x
k
u
2y
k
2u
2y
Hence
)
(
)
1
(
)
(
)
(
)
1
(
)
(
)
(
)
1
(
2
)
(
2 2 2 22 2
y
u
k
x
ku
y
u
k
y
u
x
u
k
k
x
u
k
Therefore
)
(
)
1
(
)
(
)]
(
)
1
(
)
(
[
ku
x
k
u
y
2
ku
2x
k
u
2y
(6) Sinceu
(
x
)
is a nonnegative convex function we have)
(
)
1
(
)
(
)
)
1
(
(
kx
k
y
ku
x
k
u
y
u
2 2
(
kx
(
1
k
)
y
)
[
ku
(
x
)
(
1
k
)
u
(
y
)]
u
(7)IJRRAS 5 (2) ● November 2010 Amrahov ● A Note On Hadamard Inequalities
170
)
(
)
1
(
)
(
)
)
1
(
(
2 22
kx
k
y
ku
x
k
u
y
u
(8)The inequality (8) proves that the function
u
2 is a convex function. Proof of the theorem.By the lemma the functions
u
2 andv
2 are convex functions. By the Hadamard inequality for these functions wehave
2
)
(
)
(
)
(
1
2u
2a
u
2b
dx
x
u
a
b
b
a
(9)2
)
(
)
(
)
(
1
v
2x
dx
v
2a
v
2b
a
b
b
a
(10)Multiplying the inequalities (9) and (10) we get
2
)
(
)
(
2
)
(
)
(
)
(
)
(
)
(
1
2 2 2 2 2 22
b
v
a
v
b
u
a
u
dx
x
v
dx
x
u
a
b
b
a
b
a
(11)By Cauchy-Scwartz inequality
ba
b
a b
a
dx
x
v
dx
x
u
dx
x
v
x
u
(
)
(
)
2(
)
2(
)
2
(12)
Hence by (11) and (12) we get
2
)
(
)
(
2
)
(
)
(
)
(
)
(
)
(
1
2 2 2 22
2
b
v
a
v
b
u
a
u
dx
x
v
x
u
a
b
b
a
The last inequality means that the inequality (4) holds.
Example3. Prove the inequality
sin
8
2
10
2
dx
x
x
Solution. Since the functions
u
(
x
)
sin
x
8
andx
x
v
(
)
1
are nonnegative convexfunctions on
,
2
we have by inequality (4)2
4
1
1
)
8
2
(sin
)
8
(sin
8
sin
1
2 22 2
2
dx
x
x
Hence
4
5
8
8
8
sin
1
2
2
2
dx
x
x
Therefore
10
2
8
sin
2
dx
x
x
Note that
x
x
x
f
(
)
sin
8
is not a convex function on
,
2
. 3. REFERENCES[1] J.E.Pecaric, F.Proschan and Y.L.Tong, “Convex Functions, Partial Orderings and Statistical
