Journal of Inequalities in Pure and
Applied Mathematics
ON AN INEQUALITY OF OSTROWSKI TYPE
JOSIP PEČARIĆ AND SIME UNGAR
volume 7, issue 4, article 151,
2006.
Received 22 August, 2006;
accepted 28 August, 2006.
Faculty of Textile Technology
University of Zagreb
Pierottijeva 6
10000 Zagreb, Croatia
EMail: pecaric@hazu.hr
Communicated by: W.-S. Cheung
Department of Mathematics
University of Zagreb
Pierottijeva 6
10000 Zagreb, Croatia
EMail: ungar@math.hr
Abstract
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Abstract
We prove an inequality of Ostrowski type for p-norm, generalizing a result of
Dragomir [1].
2000 Mathematics Subject Classification: 26D15, 26D10.
Key words: Ostrowski’s inequality, p-norm.
Contents
1
Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3
2
The Main Result . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4
3
The Weighted Case . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 10
References
On an Inequality of Ostrowski
type
Josip Pečarić and Sime Ungar
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1.
Introduction
For a differentiable function f : [a, b] → R, a · b > 0, Dragomir has in [1]
proved, using Pompeiu’s mean value theorem [3], the following Ostrowski type
inequality:
Z b
1
a + b f (x)
·
−
f (t) dt ≤ D(x) · kf − ι f 0 k∞ ,
(1.1)
2
x
b−a a
On an Inequality of Ostrowski
type
where ι(t) = t, t ∈ [a, b], and

(1.2)
b − a 1
D(x) =
+
|x|
4
a+b
2
x−
b−a
!2 
Josip Pečarić and Sime Ungar
.
Title Page
We are going to prove a general estimate with the p-norm, 1 ≤ p ≤ ∞,
which will for p = ∞ give the Dragomir result.
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2.
The Main Result
Theorem 2.1. Let the function f : [a, b] → R be continuous on [a, b] and differentiable on (a, b) with 0 < a < b. Then for p1 + 1q = 1, with 1 ≤ p, q ≤ ∞, and
all x ∈ [a, b], the following inequality holds:
Z b
a + b f (x)
1
(2.1)
·
−
f (t) dt ≤ P U (x, p) · kf − ι f 0 kp
2
x
b−a a
where ι(t) = t, t ∈ [a, b], and
On an Inequality of Ostrowski
type
1
x2−q − a1+q x1−2q q
a2−q − x2−q
·
+
(2.2) P U (x, p) = (b − a)
(1 − 2q)(2 − q)
(1 − 2q)(1 + q)
2−q
1 #
b
− x2−q
x2−q − b1+q x1−2q q
+
+
.
(1 − 2q)(2 − q)
(1 − 2q)(1 + q)
1
−1
p
"
Note that in cases (p, q) = (1, ∞), (∞, 1), and (2, 2) the constant P U (x, p)
has to be taken as the limit as p → 1, ∞, and 2, respectively.
Proof. Define F : [1/b, 1/a] → R by F (t) := t f ( 1t ). The function F is continuous and is differentiable on (1/b, 1/a), and for all x1 , x2 ∈ [1/b, 1/a] we
have
Z x1
F (x1 ) − F (x2 ) =
F 0 (t) dt
x
Z 2x1 1
1 0 1
1
(2.3)
=
f
− f
dt
u :=
t
t
t
t
x2
Josip Pečarić and Sime Ungar
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Z
(2.4)
1/x1
=−
(f (u) − uf 0 (u))
1/x2
1
du .
u2
Denote x1 =: 1/x and x2 =: 1/t. Then for all x, t ∈ [a, b] from (2.4) we get
Z t
1
1
1
f (x) − f (t) =
(f (u) − uf 0 (u)) 2 du,
x
t
u
x
i.e.,
Z
(2.5)
tf (x) − xf (t) = x t
x
t
1
(f (u) − uf (u)) 2 du .
u
0
Integrating on t and dividing by x, we obtain
Z b
Z b Z t
b2 − a2 f (x)
1
0
·
−
f (t) dt =
t
(f (u) − uf (u)) 2 du dt ,
2
x
u
a
a
x
and therefore
b
b2 − a2 f (x)
·
−
f (t) dt
2
x
a
Z b Z t
t
≤
(f (u) − uf 0 (u)) 2 du dt
u
Za x Zx x
t
=
(f (u) − uf 0 (u)) 2 du dt
u
a
t
Z b Z t
t
+
(f (u) − uf 0 (u)) 2 du dt .
u
x
x
Z
(2.6)
On an Inequality of Ostrowski
type
Josip Pečarić and Sime Ungar
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First, consider the case 1 < p, q < ∞. Applying Hölder’s inequality, the sum
in the last line of (2.5) is
Z x Z x
p1 Z x Z x q
1q
t
p
0
≤
|f (u) − uf (u)| du dt ·
du dt
(2.7)
2q
a
t
a
t u
Z b Z t
p1 Z b Z t q 1q
t
p
0
+
|f (u) − uf (u)| du dt ·
du dt
2q
x
x
x
x u
Z b Z b
p1
p
≤
|f (u) − uf 0 (u)| du dt
a
a
" Z Z
1q #
1q Z b Z t q
x q
x
t
t
.
+
du dt
×
du dt
2q
2q
x u
a
t u
x
The first factor in (2.7) equals
Z b Z b
p1
1
p
(2.8)
|f (u) − uf 0 (u)| du dt
= (b − a) p kf − ι f 0 kp ,
a
a
and for the second factor, for p, q 6= 2, we get
Z x Z x q
1q Z b Z t q
1q
t
t
(2.9)
du dt
+
du dt
2q
2q
a
t u
x
x u
1
2−q
a
− x2−q
x2−q − a1+q x1−2q q
=
+
(1 − 2q)(2 − q)
(1 − 2q)(1 + q)
1
2−q
x2−q − b1+q x1−2q q
b
− x2−q
+
+
.
(1 − 2q)(2 − q)
(1 − 2q)(1 + q)
On an Inequality of Ostrowski
type
Josip Pečarić and Sime Ungar
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Putting (2.8) and (2.9) into (2.7) and dividing (2.6) and (2.7) by (b − a)
gives the required inequality (2.1) in the case 1 < p, q < ∞, p, q 6= 2.
For p = q = 2, instead of (2.9) we obtain
Z
x
12 Z b Z t 2
12
t2
t
du dt
+
du dt
4
4
t u
x
x u
21 12 !
x 3 a3
x 3 b3
1
=
ln
+ 3 −1
+ ln
+ 3 −1
3
a
x
b
x
Z
(2.10)
a
x
which is easily shown to be equal to the limit of the right hand side in (2.9) for
q → 2, i. e.
(2.11)
p→2
12 12 !
x 3 a3
x 3 b3
ln
+ 3 −1
+ ln
+ 3 −1
.
a
x
b
x
1
1
3(b − a) 2
Now consider the case p = ∞, q = 1. The last line in (2.6) is
(2.12)
Josip Pečarić and Sime Ungar
Title Page
lim P U (x, p)
=
On an Inequality of Ostrowski
type
≤ sup |f (u) − u f 0 (u)|
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a≤u≤b
Z b Z t
t
t
×
du dt +
du dt
2
2
a
t u
x
x u
2
a + b2
0
= kf − ι f k∞ ·
+x−a−b .
2x
Z
x
Z
x
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Putting (2.12) into (2.6) and dividing by (b − a) gives
1
a + b f (x)
·
−
2
x
b−a
b
Z
f (t) dt
a
1
≤
b−a
a2 + b 2
+ x − a − b · kf − ι f 0 k∞
2x
which reproves the Dragomir’s result [1].
It is easy to see that
2
1
a + b2
+ x − a − b = lim P U (x, p)
p→∞
b−a
2x
proving (2.1) in the case p = ∞, q = 1.
Finally consider the case p = 1, q = ∞. In this case the last line of (2.6) is
x
Z
(2.13)
Z
x
|f (u) − uf 0 (u)| max
≤
a
t≤u≤x
a≤t≤x
t
Z
b
Z
t
du dt
u2
t
t
du dt
x≤u≤t u2
x
x
x≤t≤b
Z bZ b
1
b
0
≤
|f (u) − u f (u)| du dt ·
+
a x2
a
a
1
b
= (b − a)
+
· kf − ι f 0 k1 .
a x2
+
|f (u) − uf 0 (u)| max
On an Inequality of Ostrowski
type
Josip Pečarić and Sime Ungar
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Appending (2.13) to (2.6) and dividing by (b − a) gives
Z b
1
a + b f (x)
1
b
(2.14)
·
−
f (t) dt ≤
+ 2 · kf − ι f 0 k1 .
2
x
b−a a
a x
It is not too difficult to show that
(2.15)
lim P U (x, p) =
p→1
1
b
+ 2,
a x
so (2.14) proves formula (2.1) for p = 1, q = ∞, proving the theorem.
Remark 1. We have considered only the positive case, 0 < a < b. In the case
a < b < 0, a similar but more cumbersome formula holds, where most a’s, b’s
and x’s have to be replaced by their absolute values.
On an Inequality of Ostrowski
type
Josip Pečarić and Sime Ungar
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3.
The Weighted Case
In the weighted case we have the following result:
Theorem 3.1. Let the function f : [a, b] → R be continuous on [a, b] and differentiable on (a, b) with 0 < a < b, and let w : [a, b] → R be a nonnegative
integrable function. Then for p1 + 1q = 1, with 1 ≤ p, q ≤ ∞, and all x ∈ [a, b],
the following inequality holds:
f (x)
x
Z
b
Z
b
t w(t) dt −
f (t) w(t) dt
a
a
"
1
1q
Z x
Z x
p
(b
−
a)
q
q
1−2q
q
1−q
0
x
t (w(t)) dt −
t
(w(t)) dt
≤ kf − ιf kp ·
1 ·
a
a
(1 − 2q) q
Z b
1q #
Z b
.
+
t1−q (w(t))q dt − x1−2q
tq (w(t))q dt
(3.1)
x
x
Proof. Multiplying (2.5) by w(t)/x and integrating on t, we get
f (x)
x
Z
b
Z
t w(t) dt −
a
b
f (t) w(t) dt
Z t
Z b
1
0
=
t w(t)
(f (u) − u f (u)) 2 du dt ,
u
a
x
a
On an Inequality of Ostrowski
type
Josip Pečarić and Sime Ungar
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and as in the proof of Theorem 2.1 we have
Z
Z b
f (x) b
t w(t) dt −
f (t) w(t) dt
x a
a
Z x Z x
t w(t)
du dt
≤
(f (u) − uf 0 (u))
u2
a
t
Z b Z t
t w(t)
+
(f (u) − uf 0 (u))
du dt
u2
x
x
Z x Z x
p1 Z
p
0
≤
|f (u) − u f (u)| du dt
·
1q
t (w(t))q
du dt
u2q
a
t
a
t
1q
Z b Z t
p1 Z b Z t q
q
t
(w(t))
p
du dt
+
|f (u) − u f 0 (u)| du dt
·
u2q
x
x
x
x
Z b Z b
p1
p
0
≤
|f (u) − u f (u)| du dt
a
a
" Z Z
1q Z b Z t q
1q #
x
x q
t (w(t))q
t (w(t))q
·
du dt
+
du dt
u2q
u2q
a
t
x
x
which gives (3.1).
x
Z
x q
On an Inequality of Ostrowski
type
Josip Pečarić and Sime Ungar
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References
[1] S.S. DRAGOMIR, An inequality of Ostrowski type via Pompeiu’s mean
value theorem, J. of Inequal. in Pure and Appl. Math., 6(3) (2005),
Art. 83. [ONLINE: http://jipam.vu.edu.au/article.php?
sid=532].
[2] A. OSTROWSKI, Über die Absolutabweichung einer differentienbaren
Funktionen von ihren Integralmittelwert, Comment. Math. Hel., 10 (1938),
226–227.
[3] D. POMPEIU, Sur une proposition analogue au théorème des accroissements finis, Mathematica (Cluj, Romania), 22 (1946), 143–146.
On an Inequality of Ostrowski
type
Josip Pečarić and Sime Ungar
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