Journal of Inequalities in Pure and
Applied Mathematics
http://jipam.vu.edu.au/
Volume 2, Issue 3, Article 31, 2001
IMPROVEMENT OF AN OSTROWSKI TYPE INEQUALITY FOR MONOTONIC
MAPPINGS AND ITS APPLICATION FOR SOME SPECIAL MEANS
S.S. DRAGOMIR AND M.L. FANG
S CHOOL OF C OMMUNICATIONS AND I NFORMATICS
V ICTORIA U NIVERSITY OF T ECHNOLOGY
PO B OX 14428
M ELBOURNE C ITY MC
V ICTORIA 8001, AUSTRALIA .
sever@matilda.vu.edu.au
URL: http://rgmia.vu.edu.au/SSDragomirWeb.html
D EPARTMENT OF M ATHEMATICS ,
NANJING N ORMAL U NIVERSITY,
NANJING 210097, P. R. C HINA
mlfang@pine.njnu.edu.cn
Received 10 January, 2001; accepted 23 April, 2001
Communicated by B. Mond
A BSTRACT. We first improve two Ostrowski type inequalities for monotonic functions, then
provide its application for special means.
Key words and phrases: Ostrowski’s inequality, Trapezoid inequality, Special means.
2000 Mathematics Subject Classification. 26D15, 26D10.
1. I NTRODUCTION
In [1], Dragomir established the following Ostrowski’s inequality for monotonic mappings.
Theorem 1.1. Let f : [a, b] → R be a monotonic nondecreasing mapping on [a, b]. Then for all
x ∈ [a, b], we have the following inequality
Z b
Z b
1
1
f (x) −
f (t)dt ≤
[2x − (a + b)]f (x) +
sgn(t − x)f (t)dt
b−a
b−a
a
a
1
[(x − a)(f (x) − f (a)) + (b − x)(f (b) − f (x))]
"b − a #
1 x − a+b
2
≤
+
(f (b) − f (a)).
2
b−a
≤
(1.1)
ISSN (electronic): 1443-5756
c 2001 Victoria University. All rights reserved.
Project supported by the National Natural Science Foundation of China (Grant No.10071038).
006-01
2
S.S. D RAGOMIR AND M.L. FANG
The constant 21 is the best possible one.
In [2], Dragomir, Pečarić and Wang generalized Theorem 1.1 and proved
Theorem 1.2. Let f : [a, b] → R be a monotonic nondecreasing mapping on [a, b] and t1 , t2 ,
t3 ∈ (a, b) be such that t1 ≤ t2 ≤ t3 . Then
Z b
f (x)dx − [(t1 − a)f (a) + (t3 − t1 )f (t2 ) + (b − t3 )f (b)]
a
Z
≤ (b − t3 )f (b) + (2t2 − t1 − t3 )f (t2 ) − (t1 − a)f (a) +
b
T (x)f (x)dx
a
≤ (b − t3 )(f (b) − f (t3 )) + (t3 − t2 )(f (t3 ) − f (t2 ))
+ (t2 − t1 )(f (t2 ) − f (t1 )) + (t1 − a)(f (t1 ) − f (a))
≤ max{t1 − a, t2 − t1 , t3 − t2 , b − t3 }(f (b) − f (a)),
(1.2)
where T (x) = sgn(t1 − x), for x ∈ [a, t2 ], and T (x) = sgn(t3 − x), for x ∈ [t2 , b].
In the present paper, we firstly improve the above results, and then provide its application for
some special means.
2. M AIN R ESULT
We shall start with the following result.
Theorem 2.1. Let f : [a, b] → R be a monotonic nondecreasing mapping on [a, b] and let t1 ,
t2 , t3 ∈ [a, b] be such that t1 ≤ t2 ≤ t3 . Then
Z b
f (x)dx − [(t1 − a)f (a) + (t3 − t1 )f (t2 ) + (b − t3 )f (b)]
a
≤ max{(b − t3 )(f (b) − f (t3 )) + (t2 − t1 )(f (t2 ) − f (t1 )),
(t3 − t2 )(f (t3 ) − f (t2 )) + (t1 − a)(f (t1 ) − f (a))}
(2.1)
≤ max{t1 − a, t2 − t1 , t3 − t2 , b − t3 }(f (b) − f (a)).
(2.2)
Proof. Since f (x) is a monotonic nondecreasing mapping on [a, b], we have
Z b
f
(x)dx
−
[(t
−
a)f
(a)
+
(t
−
t
)f
(t
)
+
(b
−
t
)f
(b)]
1
3
1
2
3
a
Z t1
Z t3
Z b
= (f (x) − f (a))dx +
(f (x) − f (t2 ))dx +
(f (x) − f (b))dx
a
t1
t
Z t1
3
Z t3
(f (x) − f (t2 ))dx
= (f (x) − f (a))dx +
a
t2
Z
t2
−
Z
b
(f (t2 ) − f (x))dx +
t1
t3
(f (b) − f (x))dx ≤ max{(b − t3 )(f (b) − f (t3 )) + (t2 − t1 )(f (t2 ) − f (t1 )),
(t3 − t2 )(f (t3 ) − f (t2 )) + (t1 − a)(f (t1 ) − f (a))}
≤ max{t1 − a, t2 − t1 , t3 − t2 , b − t3 }(f (b) − f (a)).
Thus (2.1) and (2.2) is proved.
For t1 = t2 = t3 = x, Theorem 2.1 becomes the following corollary.
J. Inequal. Pure and Appl. Math., 2(3) Art. 31, 2001
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M ONOTONIC M APPINGS AND A PPLICATIONS
3
Corollary 2.2. Let f be defined as in Theorem 2.1. Then
Z b
f
(x)dx
−
[(x
−
a)f
(a)
+
(b
−
x)f
(b)]
a
≤ max{(b − x)(f (b) − f (x)), (x − a)(f (x) − f (a))}
≤ max{x − a, b − x} · max{(f (x) − f (a)), (f (b) − f (x))}
a + b 1
≤ (b − a) + x −
(f (b) − f (a)).
2
2 For x =
a+b
,
2
we get trapezoid inequality.
Corollary 2.3. Let f be defined as in Theorem 2.1. Then
Z b
f
(a)
+
f
(b)
f
(x)dx
−
(b
−
a)
2
a
b−a
a+b
a+b
(2.3)
max
f
− f (a) , f (b) − f
≤
2
2
2
1
≤ (b − a)(f (b) − f (a)).
2
For t1 = a, t2 = x, t3 = b, we get Theorem 1.1.
3. A PPLICATION
FOR
S PECIAL M EANS
In this section, we shall give application of Corollary 2.3. Let us recall the following means.
(1) The arithmetic mean:
A = A(a, b) :=
a+b
,
2
a, b ≥ 0.
(2) The geometric mean:
√
G = G(a, b) :=
ab,
a, b ≥ 0.
(3) The harmonic mean:
H = H(a, b) :=
2
,
1/a + 1/b
a, b ≥ 0.
(4) The logarithmic mean:
L = L(a, b) :=
b−a
,
ln b − ln a
a, b ≥ 0, a 6= b; If a = b, then L(a, b) = a.
(5) The identric mean:
1
I = I(a, b) :=
e
bb
aa
1
b−a
,
a, b ≥ 0, a 6= b; If a = b, then I(a, b) = a.
(6) The p-logarithmic mean:
p+1
1
b
− ap+1 p
Lp = Lp (a, b) :=
,
(p + 1)(b − a)
a 6= b; If a = b, then Lp (a, b) = a,
where p 6= −1, 0 and a, b > 0.
J. Inequal. Pure and Appl. Math., 2(3) Art. 31, 2001
http://jipam.vu.edu.au/
4
S.S. D RAGOMIR AND M.L. FANG
The following simple relationships are known in the literature
H ≤ G ≤ L ≤ I ≤ A.
We are going to use inequality (2.3) in the following equivalent version:
Z b
1
f
(a)
+
f
(b)
f (t)dt −
b − a
2
a
1
a+b
a+b
(3.1)
≤ max
f
− f (a) , f (b) − f
2
2
2
1
≤ (f (b) − f (a)),
2
where f : [a, b] → R is monotonic nondecreasing on [a, b].
3.1. Mapping f (x) = xp . Consider the mapping f : [a, b] ⊂ (0, ∞) → R, f (x) = xp , p > 0.
Then
Z b
1
f (t)dt = Lpp (a, b),
b−a a
f (a) + f (b)
= A(ap , bp ),
2
f (b) − f (a) = p(b − a)Lp−1
p−1 .
Then by (3.1), we get
(3.2)
p
p p
1
a
+
b
a
+
b
p
p
p
p
Lp (a, b) − A(a , b ) ≤ max
− a ,b −
2
2
2
p a+b
1 p
b −
=
2
2
p
1 p
1
a+b
p
p
= (b − a ) −
−a
2
2
2
1
p(b − a)ap−1
≤ p(b − a)Lp−1
−
.
p−1
2
4
Remark 3.1. The following result was proved in [2].
p
Lp (a, b) − A(ap , bp ) ≤ 1 p(b − a)Lp−1
p−1 .
2
3.2. Mapping f (x) = −1/x. Consider the mapping f : [a, b] ⊂ (0, ∞) → R, f (x) = − x1 .
Then
Z b
1
f (t)dt = −L−1 (a, b),
b−a a
f (a) + f (b)
A(a, b)
=− 2
,
2
G (a, b)
f (b) − f (a) =
J. Inequal. Pure and Appl. Math., 2(3) Art. 31, 2001
b−a
.
G2 (a, b)
http://jipam.vu.edu.au/
M ONOTONIC M APPINGS AND A PPLICATIONS
5
Then by (3.1), we get
A(a, b)
1
1
2
2
1
−1
G2 (a, b) − L (a, b) ≤ 2 max a − a + b , a + b − b
1
b−a
1 b−a 1
b−a
= ·
− ·
= ·
2 a(a + b)
2
ab
2 b(a + b)
1 b−a
1
b−a
≤ · 2
− ·
.
2 G (a, b) 2 b(a + b)
Thus we get
1 b
(b − a)L.
2a+b
Remark 3.2. The following result was proved in [2].
1
0 ≤ AG − G2 ≤ (b − a)L.
2
0 ≤ AL − G2 ≤
(3.3)
3.3. Mapping f (x) = ln x. Consider the mapping f : [a, b] ⊂ (0, ∞) → R, f (x) = ln x.
Then
Z b
1
f (t)dt = ln I(a, b),
b−a a
f (a) + f (b)
= ln G(a, b),
2
b−a
f (b) − f (a) =
.
L(a, b)
Then by (3.1), we get
1
a+b
a+b
|ln I(a, b) − ln G(a, b)| ≤ max ln
− ln a, ln b − ln
2
2
2
1 a+b
1 b−a
1
2b
= ln
=
− ln
.
2
2a
2 L(a, b) 2 a + b
Thus we get
r
b−a
I
a + b 12 · L(a,b)
(3.4)
1≤
≤
e
.
G
2b
Remark 3.3. The following result was proved in [2].
1≤
1 b−a
I
≤ e 2 · L(a,b) .
G
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J. Inequal. Pure and Appl. Math., 2(3) Art. 31, 2001
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