ON THE OSTROWSKI TYPE INTEGRAL INEQUALITY
M. Z. SARIKAYA
Abstract. In this note, we establish an inequality of Ostrowski-type involving functions of two independent variables newly by using certain integral inequalities.
1. Introduction
In [3], Ujević proved the following double integral inequality:
Theorem 1. Let f : [a, b]→ R be a twice differentiable mapping on (a, b) and suppose that
00
γ ≤ f (t) ≤ Γ for all t ∈ (a, b). Then we have the double inequality
3S − Γ
f (a) + f (b)
1
(b − a)2 ≤
−
24
2
b−a
(1.1)
Zb
f (t)dt ≤
3S − γ
(b − a)2
24
a
0
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where S =
0
f (b) − f (a)
.
b−a
In a recent paper [2], Liu et al. have proved the following two sharp inequalities of perturbed
Ostrowski-type
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Received March 9, 2009; revised August 11, 2009.
2000 Mathematics Subject Classification. Primary 26D07, 26D15.
Key words and phrases. Ostrowski’s inequality.
Theorem 2. Under the assumptations of Theorem 1, we have
2
a + b Γ[(x − a)3 − (b − x)3 ] 1 b − a +
+ x −
(S − Γ)
12(b − a)
8
2
2 Zb
1
(x − a)f (a) + (b − x)f (b)
1
≤
f (x) +
−
f (t)dt
2
b−a
b−a
(1.2)
a
2
3
3
γ[(x − a) − (b − x) ] 1 b − a a + b ≤
+
+ x −
(S − γ),
12(b − a)
8
2
2 for all x ∈ [a, b],
where S =
f 0 (b) − f 0 (a)
. If γ, Γ are given by
b−a
γ = min f 00 (t),
t∈[a,b]
Γ = max f 00 (t)
t∈[a,b]
then the inequality given by (2) is sharp in the usual sense.
In [1], Cheng has proved the following integral inequality
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Theorem 3. Let I ⊂ R be an open interval, a, b ∈ I, a < b. f : I→ R is a differentiable
function such that there exist constants γ, Γ ∈ R with γ ≤ f 0 (x) ≤ Γ, x ∈ [a, b]. Then we have
Zb
1
(x
−
b)f
(b)
−
(x
−
a)f
(a)
1
f (x) −
−
f (t)dt
2
2(b − a)
b−a
(1.3)
a
(x − a)2 + (b − x)2
≤
(Γ − γ),
8(b − a)
for all x ∈ [a, b].
The main purpose of this paper is to establish new inequality similar to the inequalities (1)–(3)
involving functions of two independent variables.
2. Main Result
Theorem 4. Let f : [a, b] × [c, d]→ R be an absolutely continuous fuction such that the partial
2
f (t,s)
derivative of order 2 exists and supposes that there exist constants γ, Γ ∈ R with γ ≤ ∂ ∂t∂s
≤Γ
for all (t, s) ∈ [a, b] × [c, d]. Then, we have
Zb
Zd
1
1
1
f (x, y) + 1 H(x, y) −
f (t, y)dt −
f (x, s)ds
4
4
2(b − a)
2(d − c)
a
−
1
2(b − a)(d − c)
c
Zb
[(y − c)f (t, c) + (d − y)f (t, d)]dt
a
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(2.1)
−
1
2(b − a)(d − c)
[(x − a)f (a, s) + (b − x)f (b, s)]ds
c
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Zd
+
1
2(b − a)(d − c)
Zb Zd
a
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f (t, s)ds dt
c
[(x − a)2 + (b − x)2 ][(y − c)2 + (d − y)2 ]
≤
(Γ − γ),
32(b − a)(d − c)
for all (x, y) ∈ [a, b] × [c, d] where
H(x, y)
(x − a)[(y − c)f (a, c)+(d − y)f (a, d)]+(b − x)[(y − c)f (b, c)+(d − y)f (b, d)]
(b − a)(d − c)
(x − a)f (a, y) + (b − x)f (b, y) (y − c)f (x, c) + (d − y)f (x, d)
+
.
+
b−a
d−c
=
Proof. We define the functions: p : [a, b] × [a, b] → R, q : [c, d] × [c, d] → R given by
p(x, t)
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=

a+x


 t− 2 ,
t ∈ [a, x]


 t − b + x,
2
t ∈ (x, b]

c+y


 s− 2 ,
s ∈ [c, y]


 s − d + y,
2
s ∈ (y, d].
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and
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q(y, s)
=
By definitions of p(x, t) and q(y, s), we have
Zb Zd
a
∂ 2 f (t, s)
dsdt =
p(x, t)q(y, s)
∂t∂s
Zx Zy c + y ∂ 2 f (t, s)
a+x
s−
ds dt
t−
2
2
∂t∂s
a
c
c
Zx Zd a+x
d + y ∂ 2 f (t, s)
+
t−
s−
ds dt
2
2
∂t∂s
a
(2.2)
y
Zb Zy c + y ∂ 2 f (t, s)
b+x
s−
ds dt
t−
+
2
2
∂t∂s
x
c
Zb
Zd b+x
d + y ∂ 2 f (t, s)
t−
s−
ds dt.
2
2
∂t∂s
x
y
+
Integrating by parts twice, we can state:
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Zx Zy t−
a
a+x
2
c + y ∂ 2 f (t, s)
s−
ds dt
2
∂t∂s
c
(2.3) =
(x − a)(y − c)
[f (x, y) + f (a, y) + f (x, c) + f (a, c)]
4
Zx
Zy
Zx Zy
x−a
y−c
−
[f (t, y) + f (t, c)]dt −
[f (x, s) + f (a, s)]ds +
f (t, s)dsdt.
2
2
a
c
a
c
Zx Zd t−
a
a+x
2
d + y ∂ 2 f (t, s)
s−
ds dt
2
∂t∂s
y
(2.4) =
(x − a)(d − y)
[f (x, y) + f (x, d) + f (a, y) + f (a, d)]
4
Zx
Zd
Zx Zd
d−y
x−a
−
[f (t, d) + f (t, y)]dt −
[f (x, s) + f (a, s)]ds +
f (t, s)ds dt.
2
2
a
y
a
y
Zb Zy c + y ∂ 2 f (t, s)
b+x
s−
ds dt
t−
2
2
∂t∂s
x
c
(2.5) =
(b − x)(y − c)
[f (x, y) + f (b, y) + f (x, c) + f (b, c)]
4
Zb
Zy
Zb Zy
b−x
y−c
[f (t, c) + f (t, y)]dt −
[f (x, s) + f (b, s)]ds +
f (t, s)ds dt.
−
2
2
x
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Zb Zd
(t −
x
c
x
c
d + y ∂ 2 f (t, s)
b+x
)(s −
)
ds dt
2
2
∂t∂s
y
(2.6) =
(b − x)(d − y)
[f (x, y) + f (x, d) + f (b, y) + f (b, d)]
4
Zb
Zd
Zb Zd
d−y
b−x
−
[f (t, d) + f (t, y)]dt −
[f (x, s) + f (b, s)]ds +
f (t, s)ds dt.
2
2
x
y
x
y
Adding (6)–(9) and rewriting, we easily deduce
Zb Zd
p(x, t)q(y, s)
a
1
∂ 2 f (t, s)
ds dt = {(b − a)(d − c)f (x, y)
∂t∂s
4
c
+ [(x − a)f (a, y) + (b − x)f (b, y)](d − c)
+ [(y − c)f (x, c) + (d − y)f (x, d)](b − a)
+ [(y − c)f (a, c) + (d − y)f (a, d)](x − a)
+ [(y − c)f (b, c) + (d − y)f (b, d)](b − x)}
(2.7)
d−c
−
2
Zb
b−a
f (t, y)dt −
2
a
−
1
2
Zd
f (x, s)ds
c
Zb
[(y − c)f (t, c) + (d − y)f (t, d)]dt
a
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1
−
2
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Zb Zd
[(x − a)f (a, s) + (b − x)f (b, s)]ds +
c
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Zd
We also have
Zb Zd
(2.8)
p(x, t)q(y, s)ds dt = 0.
a
c
f (t, s)ds dt.
a
c
Let M =
Γ+γ
2 .
From (10) and (11), it follows that
Zb Zd
a
(2.9)
∂ 2 f (t, s)
− M ds dt
p(x, t)q(y, s)
∂t∂s
c
Zb Zd
=
p(x, t)q(y, s)
a
∂ 2 f (t, s)
ds dt.
∂t∂s
c
On the other hand, we get
Zb Zd
2
∂ f (t, s)
− M dsdt
p(x, t)q(y, s)
∂t∂s
(2.10)
a
c
2
Zb Zd
∂ f (t, s)
≤
max
− M |p(x, t)q(y, s)| ds dt.
∂t∂s
(t,s)∈[a,b]×[c,d]
a
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c
We also have
2
∂ f (t, s)
Γ−γ
− M ≤
max
2
(t,s)∈[a,b]×[c,d] ∂t∂s
(2.11)
and
Zb Zd
|p(x, t)q(y, s)| ds dt =
(2.12)
a
c
[(x − a)2 + (b − x)2 ][(y − c)2 + (d − y)2 ]
.
16
From (13) to (15), we easily get
Zb Zd
2
∂ f (t, s)
− M ds dt
p(x, t)q(y, s)
∂t∂s
(2.13)
a c
[(x − a)2 + (b − x)2 ][(y − c)2 + (d − y)2 ]
(Γ − γ).
32
From (12) and (16), we see that (4) holds.
≤
1. Cheng X. L., Improvement of some Ostrowski-Grüss type inequalities, Computers Math. Applic., 42 (2001),
109–114.
2. Liu W-J., Xue Q-L. and Wang S-F., Several new perturbed Ostrowski-like type inequalities, J. Inequal. Pure and
App.Math.(JIPAM), 8(4) (2007), Article:110.
3. Ujević N., Some double integral inequalities and applications, Appl. math. E-Notes, 7 (2007), 93–101.
M. Z. Sarikaya, Department of Mathematics, Faculty of Science and Arts, Afyon Kocatepe University, Afyon-Turkey,
e-mail: sarikaya@aku.edu.tr
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