Journal of Inequalities in Pure and
Applied Mathematics
http://jipam.vu.edu.au/
Volume 7, Issue 5, Article 182, 2006
A GENERALIZATION FOR OSTROWSKI’S INEQUALITY IN R2
BAOGUO JIA
S CHOOL OF M ATHEMATICS AND S CIENTIFIC C OMPUTER
Z HONGSHAN U NIVERSITY
G UANGZHOU , C HINA , 510275
mcsjbg@zsu.edu.cn
Received 27 February, 2006; accepted 28 April, 2006
Communicated by G.A. Anastassiou
A BSTRACT. We establish a new Ostrowski’s inequality in R2 by using an idea of B.G. Pachpatte.
Key words and phrases: Ostrowski’s inequality, Mean value theorem.
2000 Mathematics Subject Classification. 33B15, 33C05.
A.M. Ostrowski proved the following inequality (see [1, p. 226–227]):
#
"
Z b
a+b 2
x
−
1
1
2
f (x) −
(1)
f (t)dt ≤
+
(b − a)||f 0 ||∞ ,
b−a a
4
(b − a)2
for all x ∈ [a, b], where f : [a, b] ⊆ R → R is continuous on [a, b] and differentiable on (a, b),
whose derivative f 0 : (a, b) → R is bounded on (a, b), i.e., ||f 0 ||∞ = sup |f 0 (x)| < ∞.
x∈(a,b)
Recently, by using a fairly elementary analysis, B.G. Pachpatte [2] established the following
inequality of type (1) involving two functions and their derivatives.
Theorem 1. Let f, g : [a, b] ⊆ R → R be continuous functions on [a, b] and differentiable on
(a, b), whose derivative f 0 , g 0 : (a, b) → R are bounded on (a, b), i.e., ||f 0 ||∞ = sup |f 0 (x)| <
x∈(a,b)
∞, ||g 0 ||∞ = sup |g 0 (x)| < ∞. Then
x∈(a,b)
Z b
Z b
1
f (x)g(x) −
g(x)
f (y)dy + f (x)
g(y)dy 2(b − a)
a
a
#
a+b 2
x
−
1
1
2
≤ {|g(x)| ||f 0 ||∞ + |f (x)| ||g 0 ||∞ }
+
(b − a),
2
4
(b − a)2
"
ISSN (electronic): 1443-5756
c 2006 Victoria University. All rights reserved.
This project was supported in part by the Foundations of the Natural Science Committee and Zhongshan University Advanced Research
Centre, China.
056-06
2
B. J IA
for all x ∈ [a, b].
In this paper, by means of B.G. Pachpatte’s idea, we prove the following
Theorem 2. Let D = [a, b]×[a, b], intD = (a, b)×(a, b), f, g : D → R be continuous functions
on D and differentiable on intD, whose partial derivatives fx , fy , gx , gy : intD → R are
bounded on intD, i.e., ||fx ||∞ = sup |fx (x, y)| < ∞, ||fy ||∞ = sup |fy (x, y)| < ∞,
(x,y)∈intD
||gx ||∞ =
sup
|gx (x, y)| < ∞ , ||gy ||∞ =
(x,y)∈intD
(x,y)∈intD
sup
|gy (x, y)| < ∞. Then
(x,y)∈intD
ZZ
1
f (u1 , v1 )g(u1 , v1 ) −
g(u1 , v1 )
f (u2 , v2 )du2 dv2
2(b − a)2
D
ZZ
+ f (u1 , v1 )
g(u2 , v2 )du2 dv2 D
"
2 #
u1 − a+b
1
1
2
≤ [|g(u1 , v1 )| ||fx ||∞ + |f (u1 , v1 )| ||gx ||∞ ]
+
(b − a)
2
4
(b − a)2
"
2 #
v1 − a+b
1
1
2
+ [|g(u1 , v1 )| ||fy ||∞ + |f (u1 , v1 )| ||gy ||∞ ]
+
(b − a)
2
4
(b − a)2
for all (u1 , v1 ) ∈ D.
By taking g(x, y) = 1 in Theorem 2, we get the following Ostrowski like inequality in R2 ,
Corollary 3.
ZZ
1
f (u1 , v1 ) −
f
(u
,
v
)du
dv
2
2
2
2
(b − a)2
D
#
"
"
#
a+b 2
a+b 2
u
−
v
−
1
1
1
1
2
2
+
(b − a) + ||fy ||∞
+
(b − a).
≤ ||fx ||∞
2
4
(b − a)
4
(b − a)2
Proof of Theorem 2. By the mean value theorem, there exist ξ1 , η1 , ξ2 , η2 ∈ (a, b) such that
(2)
f (u1 , v1 ) − f (u2 , v2 ) = fx (ξ1 , v1 )(u1 − u2 ) + fy (u2 , η1 )(v1 − v2 ),
and
(3)
g(u1 , v1 ) − g(u2 , v2 ) = gx (ξ2 , v1 )(u1 − u2 ) + gy (u2 , η2 )(v1 − v2 ).
Multiplying both sides of (2) and (3) by g(u1 , v1 ) and f (u1 , v1 ) respectively and adding we get
2f (u1 , v1 )g(u1 , v1 ) − [f (u2 , v2 )g(u1 , v1 ) + f (u1 , v1 )g(u2 , v2 )]
= g(u1 , v1 )[fx (ξ1 , v1 )(u1 − u2 ) + fy (u2 , η1 )(v1 − v2 )]
+ f (u1 , v1 )[gx (ξ2 , v1 )(u1 − u2 ) + gy (u2 , η2 )(v1 − v2 )].
Integrate both sides with respect to u2 , v2 over D. Note that by the proof of the mean value
theorem, we know that fx (ξ1 , v1 ), fy (u2 , η1 ), gx (ξ2 , v1 ) and gy (u2 , η2 ) are Riemann-integrable
J. Inequal. Pure and Appl. Math., 7(5) Art. 182, 2006
http://jipam.vu.edu.au/
A G ENERALIZATION FOR O STROWSKI ’ S I NEQUALITY
IN
R2
3
for (u2 , v2 ) ∈ D. Rewriting we get
ZZ
1
g(u1 , v1 )
f (u2 , v2 )du2 dv2
f (u1 , v1 )g(u1 , v1 ) −
2(b − a)2
D
ZZ
+f (u1 , v1 )
g(u2 , v2 )du2 dv2
D
ZZ
1
=
g(u1 , v1 )
[fx (ξ1 , v1 )(u1 − u2 ) + fy (u2 , η1 )(v1 − v2 )]du2 dv2
2(b − a)2
D
ZZ
1
+
f (u1 , v1 )
[gx (ξ2 , v1 )(u1 − u2 ) + gy (u2 , η2 )(v1 − v2 )]du2 dv2 .
2(b − a)2
D
So
ZZ
1
f (u1 , v1 )g(u1 , v1 ) −
g(u1 , v1 )
f (u2 , v2 )du2 dv2
2(b − a)2
D
ZZ
+f (u1 , v1 )
g(u2 , v2 )du2 dv2 D
ZZ
ZZ
1
≤
|g(u1 , v1 )| ||fx ||∞
|u1 − u2 |du2 dv2 + ||fy ||∞
|v1 − v2 |du2 dv2
2(b − a)2
D
D
ZZ
ZZ
1
+
|f (u1 , v1 )| ||gx ||∞
|u1 − u2 |du2 dv2 + ||gy ||∞
|v1 − v2 |du2 dv2 .
2(b − a)2
D
D
Note that
(u1 − a)2 + (u1 − b)2
|u1 − u2 |du2 dv2 = (b − a)
2
D
"
#
2
u1 − a+b
1
2
+
(b − a)
=
4
(b − a)2
ZZ
and
2 #
v1 − a+b
1
2
|v1 − v2 |du2 dv2 =
+
(b − a).
2
4
(b
−
a)
D
ZZ
"
We obtain that
f (u1 , v1 )g(u1 , v1 ) −
ZZ
1
g(u1 , v1 )
f (u2 , v2 )du2 dv2
2(b − a)2
D
ZZ
+f (u1 , v1 )
g(u2 , v2 )du2 dv2 D
"
2 #
u1 − a+b
1
1
2
≤ [|g(u1 , v1 )| ||fx ||∞ + |f (u1 , v1 )| ||gx ||∞ ]
+
(b − a)
2
4
(b − a)2
"
2 #
v1 − a+b
1
1
2
+ [|g(u1 , v1 )| ||fy ||∞ + |f (u1 , v1 )| ||gy ||∞ ]
+
(b − a).
2
4
(b − a)2
Remark 4. Let f (x, y) = f (x), g(x, y) = g(x) in Theorem 2. Then fy = 0, gy = 0. We obtain
Theorem 1.
J. Inequal. Pure and Appl. Math., 7(5) Art. 182, 2006
http://jipam.vu.edu.au/
4
B. J IA
Remark 5. Let f (x, y) = f (x), g(x, y) = 1 in Theorem 2, we recapture the well known
Ostrowski inequality.
R EFERENCES
[1] A.M. OSTROWSKI, Uber die Absolutabweichung einer differentiebaren funktion von ihrem integralmitelwert, Comment. Math. Helv., 10 (1938), 226–227.
[2] B.G. PACHPATTE, A note on Ostrowski like inequalities, J. Ineq. Pure Appl. Math., 6(4) (2005),
Art. 114. [ONLINE: http://jipam.vu.edu.au/article.php?sid=588].
J. Inequal. Pure and Appl. Math., 7(5) Art. 182, 2006
http://jipam.vu.edu.au/