Journal of Inequalities in Pure and
Applied Mathematics
http://jipam.vu.edu.au/
Volume 6, Issue 4, Article 128, 2005
ON OSTROWSKI-GRÜSS-ČEBYŠEV TYPE INEQUALITIES FOR FUNCTIONS
WHOSE MODULUS OF DERIVATIVES ARE CONVEX
B.G. PACHPATTE
57 S HRI N IKETAN C OLONY
N EAR A BHINAY TALKIES
AURANGABAD 431 001
(M AHARASHTRA ) I NDIA
bgpachpatte@hotmail.com
Received 17 August, 2005; accepted 30 August, 2005
Communicated by I. Gavrea
A BSTRACT. The aim of the present paper is to establish some new Ostrowski-Grüss-Čebyšev
type inequalities involving functions whose modulus of the derivatives are convex.
Key words and phrases: Ostrowski-Grüss-Čebyšev type inequalities, Modulus of derivatives, Convex, Log-convex, Integral
identities.
2000 Mathematics Subject Classification. 26D15, 26D20.
1. I NTRODUCTION
In 1938, A.M. Ostrowski [5] proved the following classial inequality.
Let f : [a, b] → R be 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 (x)| ≤ M < ∞. Then

!2 
Z b
a+b
x− 2
1
f (x) − 1
 (b − a) M,
(1.1)
f (t) dt ≤  +
b−a a
4
b−a
for all x ∈ [a, b] , where M is a constant.
For two absolutely continuous functions f, g : [a, b] → R, consider the functional
Z b
Z b
Z b
1
1
1
(1.2) T (f, g) =
f (x) g (x) dx −
f (x) dx
g (x) dx ,
b−a a
b−a a
b−a a
provided the involved integrals exist. In 1882, P.L.Čebyšev [6] proved that, if f 0 , g 0 ∈ L∞ [a, b],
then
1
(1.3)
|T (f, g)| ≤
(b − a)2 kf 0 k∞ kg 0 k∞ .
12
ISSN (electronic): 1443-5756
c 2005 Victoria University. All rights reserved.
245-05
2
B.G. PACHPATTE
In 1934, G. Grüss [6] showed that
1
(M − m) (N − n) ,
4
provided m, M, n, N are real numbers satisfying the condition −∞ < m ≤ f (x) ≤ M < ∞,
−∞ < n ≤ g (x) ≤ N < ∞, for all x ∈ [a, b].
During the past few years many researchers have given considerable attention to the above
inequalities and various generalizations, extensions and variants of these inequalities have appeared in the literature, see [1] – [10] and the references cited therein. Motivated by the recent
results given in [1] – [3], in the present paper, we establish some inequalities similar to those
given by Ostrowski, Grüss and Čebyšev, involving functions whose modulus of derivatives are
convex. The analysis used in the proofs is elementary and based on the use of integral identities
proved in [1] and [2].
|T (f, g)| ≤
(1.4)
2. S TATEMENT OF R ESULTS
Let I be a suitable interval of the real line R. A function f : I → R is called convex if
f (λx + (1 − λ) y) ≤ λf (x) + (1 − λ) f (y) ,
for all x, y ∈ I and λ ∈ [0, 1] . A function f : I → (0, ∞) is said to be log-convex, if
f (tx + (1 − t) y) ≤ [f (x)]t [f (y)]1−t ,
for all x, y ∈ I and t ∈ [0, 1] (see [10]). We need the following identities proved in [1] and [2]
respectively:
Z 1
Z b
Z b
1
1
0
f (x) =
f (t) dt +
(x − t)
f [(1 − λ) x + λt] dt dt,
b−a a
b−a a
0
1
f (x) =
b−a
Z
a
b
1
f (t) dt + (x − a)
b−a
2
Z
1
λf 0 [(1 − λ) a + λx] dλ
0
Z 1
1
2
− (b − x)
λf 0 [λx + (1 − λ) b] dλ,
b−a 0
for x ∈ [a, b] where f : [a, b] → R is an absolutely continuous function on [a, b] and λ ∈ [0, 1].
We use the following notation to simplify the details of presentation:
Z b
Z b
1
S (f, g) = f (x) g (x) −
g (x)
f (t) dt + f (x)
g (t) dt ,
2 (b − a)
a
a
and define k·k∞ as the usual Lebesgue norm on L∞ [a, b] i.e., khk∞ := ess supt∈[a,b] |h (t)| for
h ∈ L∞ [a, b] .
The following theorems deal with Ostrowski type inequalities involving two functions.
Theorem 2.1. Let f, g : [a, b] → R be absolutely continuous functions on [a, b].
(a1 ) If |f 0 | , |g 0 | are convex on [a, b], then

!2 
a+b
x− 2
1 1
 (b − a)
(2.1) |S (f, g)| ≤  +
4 4
b−a
× {|g (x)| [|f 0 (x)| + kf 0 k∞ ] + |f (x)| [|g 0 (x)| + kg 0 k∞ ]} ,
for x ∈ [a, b].
J. Inequal. Pure and Appl. Math., 6(4) Art. 128, 2005
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O N O STROWSKI -G RÜSS - Č EBYŠEV TYPE INEQUALITIES
3
(a2 ) If |f 0 | , |g 0 | are log-convex on [a, b], then
Z b
A−1
1
0
|g (x)| |f (x)|
dt
(2.2) |S (f, g)| ≤
|x − t|
2 (b − a)
log A
a
Z b
B−1
0
+ |f (x)| |g (x)|
|x − t|
dt ,
log B
a
for x ∈ [a, b], where
A=
(2.3)
|f 0 (t)|
|g 0 (t)|
,
B
=
.
|f 0 (x)|
|g 0 (x)|
Theorem 2.2. Let f, g : [a, b] → R be absolutely continuous functions on [a, b].
(b1 ) If |f 0 | , |g 0 | are convex on [a, x] and [x, b], then
(2.4)
|S (f, g)| ≤
1
{|g (x)| F (x) + |f (x)| G (x)} ,
2
for x ∈ [a, b], where
2
2
x−a
b−x
1h 0
0
(2.5) F (x) = |f (a)|
+ |f (b)|
6
b−a
b−a


+ 1+4

1h
(2.6) G (x) = |g 0 (a)|
6
x−a
b−a
2
2
b−x
+ |g (b)|
b−a


+ 1+4

0
a+b
2
x−
b−a
!2 


|f 0 (x)| (b − a) ,

a+b
2
x−
b−a
!2 


|g 0 (x)| (b − a) ,

for x ∈ [a, b].
(b2 ) If |f 0 | , |g 0 | are log-convex on [a, x] and [x, b], then
(2.7)
|S (f, g)| ≤ |g (x)| H (x) + |f (x)| L (x) ,
for x ∈ [a, b], where
"
2
1
x − a A1 log A1 + 1 − A1
0
(2.8) H (x) = (b − a) |f (a)|
2
b−a
(log A1 )2
#
2
b
−
x
B
log
B
+
1
−
B
1
1
1
+ |f 0 (b)|
,
2
b−a
(log B1 )
"
2
1
x − a A2 log A2 + 1 − A2
0
(2.9) L (x) = (b − a) |g (a)|
2
b−a
(log A2 )2
#
2
b
−
x
B
log
B
+
1
−
B
2
2
2
+ |g 0 (b)|
,
b−a
(log B2 )2
J. Inequal. Pure and Appl. Math., 6(4) Art. 128, 2005
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4
B.G. PACHPATTE
and
|f 0 (x)|
,
|f 0 (a)|
|g 0 (x)|
A2 = 0
,
|g (a)|
A1 =
(2.10)
(2.11)
|f 0 (x)|
,
|f 0 (b)|
|g 0 (x)|
B2 = 0
,
|g (b)|
B1 =
for x ∈ [a, b].
The Grüss type inequalities are embodied in the following theorems.
Theorem 2.3. Let f, g : [a, b] → R be absolutely continuous functions on [a, b].
(c1 ) If |f 0 | , |g 0 | are convex on [a, b], then
(2.12)
|T (f, g)|
1
≤
4 (b − a)2
Z
b
[|g (x)| [|f 0 (x)| + kf 0 k∞ ] + |f (x)| [|g 0 (x)| + kg 0 k∞ ]]E (x) dx,
a
where
(x − a)2 + (b − x)2
,
E (x) =
2
(2.13)
for x ∈ [a, b].
(c2 ) If |f 0 | , |g 0 | are log-convex on [a, b], then
(2.14)
1
|T (f, g)| ≤
2 (b − a)2
Z b
Z b
A−1
0
|g (x)|
|x − t| |f (x)|
dt
log A
a
a
Z b
B−1
0
+ |f (x)|
|x − t| |g (x)|
dt dx,
log B
a
where A, B are defined by (2.3).
Theorem 2.4. Let f, g : [a, b] → R be absolutely continuous functions on [a, b].
(d1 ) If |f 0 | , |g 0 | are convex on [a, b], then
(2.15)
1
|T (f, g)| ≤
2
Z b "
a
2 1 0
1 0
|g (x)|
|f (a)| + |f (x)|
6
3
1 0
1 0
+ |f (x)|
|g (a)| + |g (x)|
6
3
2 b−x
1 0
1 0
+
|g (x)|
|f (x)| + |f (b)|
b−a
3
6
1 0
1 0
+ |f (x)|
|g (x)| + |g (b)|
dx,
3
6
x−a
b−a
J. Inequal. Pure and Appl. Math., 6(4) Art. 128, 2005
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O N O STROWSKI -G RÜSS - Č EBYŠEV TYPE INEQUALITIES
5
(d2 ) If |f 0 | , |g 0 | are log-convex on [a, x] and [x, b], then
2
Z "
A1 log A1 + 1 − A1
x−a
1 b
{|g (x)| |f 0 (a)|
(2.16) |T (f, g)| ≤
2 a
b−a
(log A1 )2
A2 log A2 + 1 − A2
0
+ |f (x)| |g (a)|
(log A2 )2
2 B1 log B1 + 1 − B1
b−x
|g (x)| |f 0 (b)|
+
b−a
(log B1 )2
B2 log B2 + 1 − B2
0
dx,
+ |f (x)| |g (b)|
(log B2 )2
where A1 , B1 and A2 , B2 are defined by (2.10) and (2.11).
The next theorem contains Čebyšev type inequalities.
Theorem 2.5. Let f, g : [a, b] → R be absolutely continuous functions on [a, b].
(e1 ) If |f 0 | , |g 0 | are convex on [a, b], then
Z b
1
[|f 0 (x)| + kf 0 k∞ ] [|g 0 (x)| + kg 0 k∞ ] E 2 (x) dx,
(2.17)
|T (f, g)| ≤
3
4 (b − a) a
where E(x) is given by (2.13).
(e2 ) If |f 0 | , |g 0 | are log-convex on [a, b], then
Z b Z b
1
A−1
0
dt
(2.18) |T (f, g)| ≤
|x − t| |f (x)|
log A
(b − a)3 a
a
Z b
B−1
0
×
|x − t| |g (x)|
dt dx,
log B
a
where A, B are defined by (2.3).
3. P ROOFS OF T HEOREMS 2.1
AND
2.2
Proof of Theorem 2.1. From the hypotheses of Theorem 2.1, the following identities hold:
Z 1
Z b
Z b
1
1
0
(3.1)
f (x) =
f (t) dt +
(x − t)
f [(1 − λ) x + λt]dλ dt,
b−a a
b−a a
0
(3.2)
1
g (x) =
b−a
Z
a
b
1
g (t) dt +
b−a
Z
b
Z
(x − t)
a
1
g [(1 − λ) x + λt]dλ dt,
0
0
for x ∈ [a, b] . Multiplying both sides of (3.1) and (3.2) by g(x) and f (x) respectively, adding
the resulting identities and rewriting we have
Z 1
Z b
1
0
(3.3) S (f, g) =
g (x)
(x − t)
f [(1 − λ) x + λt]dλ dt
2 (b − a)
a
0
Z 1
Z b
0
+f (x)
(x − t)
g [(1 − λ) x + λt]dλ dt .
a
J. Inequal. Pure and Appl. Math., 6(4) Art. 128, 2005
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6
B.G. PACHPATTE
(a1 ) Since |f 0 | , |g 0 | are convex on [a, b] , from (3.3) we observe that
Z 1
Z b
1
0
|S (f, g)| ≤
|g (x)|
|x − t|
|f [(1 − λ) x + λt]| dλ dt
2 (b − a)
a
0
Z 1
Z b
0
+ |f (x)|
|x − t|
|g [(1 − λ) x + λt]| dλ dt
a
0
Z 1
Z b
1
0
0
≤
|g (x)|
|x − t|
{(1 − λ) |f (x)| + λ |f (t)|} dλ dt
2 (b − a)
a
0
Z 1
Z b
0
0
+ |f (x)|
|x − t|
{(1 − λ) |g (x)| + λ |g (t)|} dλ dt
a
0
Z b
Z 1
Z 1
1
0
0
=
|g (x)|
|x − t| |f (x)|
(1 − λ) dλ + |f (t)|
λdλ dt
2 (b − a)
a
0
0
Z b
Z 1
Z 1
0
0
+ |f (x)|
|x − t| |g (x)|
(1 − λ) dλ + |g (t)|
λdλ dt
a
0
0
Z b
1
1
=
|g (x)|
|x − t| [|f 0 (x)| + |f 0 (t)|] dt
2 (b − a)
2
a
Z b
1 0
0
+ |f (x)|
|x − t| [|g (x)| + |g (t)|] dt
2
a
Z b
1
ess. sup
0
0
≤
|g (x)|
[|f (x)| + |f (t)|]
|x − t| dt
t ∈ [a, b]
4 (b − a)
a
Z b
ess. sup
0
0
+ |f (x)|
[|g (x)| + |g (t)|]
|x − t| dt
t ∈ [a, b]
a
1
{|g (x)| [|f 0 (x)| + kf 0 k∞ ]
=
4 (b − a)
Z
0
b
0
+ |f (x)| |g (x)| + kg k∞
|x − t| dt
a
"
#
1 (x − a)2 + (b − x)2
=
4
2 (b − a)
× {|g (x)| [|f 0 (x)| + kf 0 k∞ ] + |f (x)| |g 0 (x)| + kg 0 k∞

!2 
a+b
x
−
1 1
2

=  +
4 4
b−a
× (b − a) {|g (x)| [|f 0 (x)| + kf 0 k∞ ] + |f (x)| |g 0 (x)| + kg 0 k∞ .
This is the required inequality in (2.1).
(a2 ) Since |f 0 | , |g 0 | are log-convex on [a, b] , from (3.3) we observe that
Z 1
Z b
1
0
|S (f, g)| ≤
|g (x)|
|x − t|
|f [(1 − λ) x + λt]| dλ dt
2 (b − a)
a
0
Z 1
Z b
0
+ |f (x)|
|x − t|
|g [(1 − λ) x + λt]| dλ dt
a
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O N O STROWSKI -G RÜSS - Č EBYŠEV TYPE INEQUALITIES
7
Z 1
Z b
1−λ
λ
0
0
|g (x)|
|x − t|
[|f (x)|]
[|f (t)|] dλ dt
a
0
Z 1
Z b
1−λ
λ
0
0
+ |f (x)|
|x − t|
[|g (x)|]
[|g (t)|] dλ dt
a
0
(
"
λ #
Z b
Z 1 0
|f
(t)|
1
=
|g (x)|
|x − t| |f 0 (x)|
dλ dt
2 (b − a)
|f 0 (x)|
a
0
"
λ # )
Z b
Z 1 0
|g
(t)|
+ |f (x)|
|x − t| |g 0 (x)|
dλ dt
|g 0 (x)|
a
0
Z b
1
A−1
0
|g (x)| |f (x)|
|x − t|
dt
=
2 (b − a)
logA
a
Z b
B−1
0
dt .
+ |f (x)| |g (x)|
|x − t|
log B
a
1
≤
2 (b − a)
This completes the proof of the inequality (2.2).
Proof of Theorem 2.2. From the hypotheses of Theorem 2.2, the following identities hold:
1
(3.4) f (x) =
b−a
Z
1
(3.5) g (x) =
b−a
Z
b
a
a
1
f (t) dt + (x − a)
b−a
Z
1
1
g (t) dt + (x − a)
b−a
Z
1
2
b
2
λf 0 [(1 − λ) a + λx] dλ
0
Z 1
1
2
λf 0 [λx + (1 − λ) b] dλ,
− (b − x)
b−a 0
λg 0 [(1 − λ) a + λx] dλ
0
Z 1
1
2
− (b − x)
λg 0 [λx + (1 − λ) b] dλ.
b−a 0
Multiplying both sides of (3.4) and (3.5) by g(x) and f (x) respectively, adding the resulting
identities and rewriting we have
1
(3.6) S (f, g) =
2
Z 1
1
λf 0 [(1 − λ) a + λx] dλ
b−a 0
Z 1
1
2
0
− (b − x)
λf [λx + (1 − λ) b] dλ
b−a 0
Z 1
1
2
+ f (x) (x − a)
λg 0 [(1 − λ) a + λx] dλ
b−a 0
Z 1
1
2
0
− (b − x)
λg [λx + (1 − λ) b] dλ .
b−a 0
g (x) (x − a)2
(b1 ) Since |f 0 | , |g 0 | are convex on [a, x] and [x, b], from (3.6) we observe that
(3.7)
|S (f, g)| ≤
J. Inequal. Pure and Appl. Math., 6(4) Art. 128, 2005
1
{|g (x)| M (x) + |f (x)| N (x)} ,
2
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B.G. PACHPATTE
where
(3.8)
(3.9)
1
M (x) = (x − a)
b−a
2
(x − a)2
N (x) =
b−a
Z
Z
0
1
λ |f 0 [(1 − λ) a + λx]| dλ
Z 1
1
2
+ (b − x)
λ |f 0 [λx + (1 − λ) b]| dλ,
b−a 0
1
λ |g 0 [(1 − λ) a + λx]| dλ
0
(b − x)2
+
b−a
Next, we observe that
Z 1
Z
0
0
(3.10)
λ |f [(1 − λ) a + λx]| dλ ≤ |f (a)|
0
Z
1
λ |g 0 [λx + (1 − λ) b]| dλ.
0
1
Z
0
λ (1 − λ) dλ + |f (x)|
0
1
λ2 dλ
0
1
1
= |f 0 (a)| + |f 0 (x)|
6
3
and
Z
1
0
0
Z
λ |f [λx + (1 − λ) b]| dλ ≤ |f (x)|
(3.11)
0
1
2
0
1
λ dλ + |f (b)|
0
=
Z
λ (1 − λ) dλ
0
1 0
1
|f (x)| + |f 0 (b)| .
3
6
Similarly we have
1
Z
(3.12)
λ |g 0 [(1 − λ) a + λx]| dλ ≤
1 0
1
|g (a)| + |g 0 (x)| ,
6
3
λ |g 0 [λx + (1 − λ) b]| dλ ≤
1 0
1
|g (x)| + |g 0 (b)| .
3
6
0
Z
(3.13)
1
0
From (3.8), (3.10) and (3.11) we observe that
"
2 Z 1
x−a
(3.14)
M (x) =
λ |f 0 [(1 − λ) a + λx]| dλ
b−a
0
#
2 Z 1
b−x
+
λ |f 0 [λx + (1 − λ) b]| dλ (b − a)
b−a
0
"
2 x−a
1 0
1 0
≤
|f (a)| + |f (x)|
b−a
6
3
2 #
b−x
1 0
1 0
+
|f (x)| + |f (b)| (b − a)
b−a
3
6
J. Inequal. Pure and Appl. Math., 6(4) Art. 128, 2005
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O N O STROWSKI -G RÜSS - Č EBYŠEV TYPE INEQUALITIES
9
#
2
2
x−a
b
−
x
|f 0 (a)| +
|f 0 (b)| (b − a)
b−a
b−a
"
2 2 #
1
x−a
b−x
+
+
|f 0 (x)| (b − a)
3
b−a
b−a
"
2
2
1
x−a
b−x
0
=
|f (a)| +
|f 0 (b)|
6
b−a
b−a
#
"
2 2 #
x−a
b−x
+2
+
|f 0 (x)| (b − a) .
b−a
b−a
1
=
6
"
It is easy to observe that
"
(3.15)
2
x−a
b−a
2
+
b−x
b−a
2 #
"
#
4
(x − a)2 + (b − x)2
=
b−a
2 (b − a)

!2 
a+b
x− 2
4 1
 (b − a)
+
=
b−a 4
b−a

!2 
a+b
x− 2
.
= 1 + 4
b−a
Using (3.15) in (3.14) we get
M (x) ≤ F (x) .
(3.16)
Similarly, from (3.9), (3.12), (3.13) we get
N (x) ≤ G (x) .
(3.17)
Using (3.16), (3.17) in (3.7) we get the required inequality in (2.4).
(b2 ) Since |f 0 | , |g 0 | are log-convex on [a, x] and [x, b] , from (3.6) we observe that
(3.18)
(
"
2 Z 1
1
x−a
|S (f, g)| ≤ (b − a) |g (x)|
λ |f 0 [(1 − λ) a + λx]|dλ
2
b−a
0
#
2 Z 1
b−x
+
λ |f 0 [λx + (1 − λ) b]|dλ
b−a
0
"
2 Z 1
x−a
+ |f (x)|
λ |g 0 [(1 − λ) a + λx]|dλ
b−a
0
#)
2 Z 1
b−x
+
λ |g 0 [λx + (1 − λ) b]|dλ
b−a
0
J. Inequal. Pure and Appl. Math., 6(4) Art. 128, 2005
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10
B.G. PACHPATTE
(
"
2 Z 1
1
x−a
1−λ
λ
≤ (b − a) |g (x)|
λ [|f 0 (a)|]
[|f 0 (x)|] dλ
2
b−a
0
#
2 Z 1
b−x
λ
1−λ
+
λ [|f 0 (x)|] [|f 0 (b)|]
b−a
0
"
2 Z 1
x−a
1−λ
λ
λ [|g 0 (a)|]
[|g 0 (x)|] dλ
+ |f (x)|
b−a
0
#)
2 Z 1
b−x
λ
1−λ
+
λ [|g 0 (x)|] [|g 0 (b)|] dλ
b−a
0
(
"
2
Z 1
1
x−a
0
= (b − a) |g (x)|
|f (a)|
λAλ1 dλ
2
b−a
0
#
2
Z 1
b−x
+
|f 0 (b)|
λB1λ dλ
b−a
0
"
2
Z 1
x−a
0
+ |f (x)|
|g (a)|
λAλ2 dλ
b−a
0
#)
2
Z 1
b−x
+
|g 0 (b)|
λB2λ dλ
.
b−a
0
A simple calculation shows that for any C > 0 we have (see [2])
Z
(3.19)
1
λC λ dλ =
0
C log C + 1 − C
.
(log C)2
Using this fact in (3.18) we get the required inequality in (2.7).
4. P ROOFS OF T HEOREMS 2.3
AND
2.4
Proof of Theorem 2.3. From the hypotheses of Theorem 2.3 the identities (3.1) – (3.3) hold.
Integrating both sides of (3.3) with respect to x from a to b and rewriting we have
1
(4.1) T (f, g) =
2 (b − a)2
Z
Z b
Z b
g (x)
(x − t)
a
a
1
f [(1 − λ) x + λt] dλ dt
0
0
Z
b
Z
(x − t)
+f (x)
a
1
g [(1 − λ) x + λt] dλ dt dx.
0
0
(c1 ) Since |f 0 | , |g 0 | are convex on [a, b], from (4.1) we observe that
1
|T (f, g)| ≤
2 (b − a)2
Z
Z b
Z b
|g (x)|
|x − t|
a
Z
+ |f (x)|
a
b
|x − t|
J. Inequal. Pure and Appl. Math., 6(4) Art. 128, 2005
[(1 − λ) |f (x)| + λ |f (t)|] dλ dt
0
0
0
Z
a
1
1
[(1 − λ) |g (x)| + λ |g (t)|] dλ dt dx
0
0
0
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O N O STROWSKI -G RÜSS - Č EBYŠEV TYPE INEQUALITIES
11
0
Z b
Z b
|f (x)| + |f 0 (t)|
dt
|g (x)|
|x − t|
2
a
a
0
Z b
|g (x)| + |g 0 (t)|
+ |f (x)|
|x − t|
dt dx
2
a
Z b
Z b
1
ess sup |f 0 (x)| + |f 0 (t)|
≤
|g (x)|
|x − t|
dt
t ∈ [a, b]
2
2 (b − a)2 a
a
Z b
ess sup |g 0 (x)| + |g 0 (t)|
+ |f (x)|
|x − t|
dt dx
t ∈ [a, b]
2
a
Z b
1
=
[|g (x)| [|f 0 (x)| + kf 0 k∞ ]dt
2
4 (b − a) a
Z b
0
0
+ |f (x)| [|g (x)| + kg k∞ ]]
|x − t|dt dx
1
=
2 (b − a)2
a
Z
=
b
1
[|g (x)| [|f 0 (x)| + kf 0 k∞ ]dt
4 (b − a)2 a
+ |f (x)| [|g 0 (x)| + kg 0 k∞ ]] E (x) dx.
This completes the proof of the inequality (2.14).
(c2 ) Since |f 0 | , |g 0 | are log-convex on [a, b] from (4.1) we observe that
Z 1
Z b
Z b
1
1−λ
λ
0
0
|T (f, g)| ≤
|g (x)|
|x − t|
[|f (x)|]
[|f (t)|] dλ dt
2 (b − a)2 a
a
0
Z 1
Z b
1−λ
λ
0
0
+ |f (x)|
|x − t|
[|g (x)|]
[|g (t)|] dλ dt dx
a
0
"
λ #
Z b"
Z b
Z 1 0
1
|f
(t)|
=
|g (x)|
|x − t| |f 0 (x)|
dλ dt
2
|f 0 (x)|
2 (b − a) a
a
0
"
λ # #
Z b
Z 1 0
|g
(t)|
+ |f (x)|
|x − t| |g 0 (x)|
dλ dt dx
|g 0 (x)|
a
0
Z b
Z b
A−1
1
0
|g (x)|
|x − t| |f (x)|
=
dt
log A
2 (b − a)2 a
a
Z b
B−1
0
+ |f (x)|
|x − t| |g (x)|
dt dx,
log B
a
where A, B are defined by (2.3). This is the required inequality in (2.14).
Proof of Theorem 2.4. From the hypotheses of Theorem 2.4 the identities (3.4) – (3.6) hold.
Integrating both sides of (3.6) with respect to x from a to b and rewriting we have
2 Z "
Z 1
1 b
x−a
(4.2) T (f, g) =
g (x)
λf 0 [(1 − λ) a + λx] dλ
2 a
b−a
0
Z 1
0
+f (x)
λg [(1 − λ) a + λx]dλ
0
J. Inequal. Pure and Appl. Math., 6(4) Art. 128, 2005
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12
B.G. PACHPATTE
−
b−x
b−a
2 Z
g (x)
0
1
λf 0 [λx + (1 − λ) b] dλ
Z 1
0
+ f (x)
λg [λx + (1 − λ) b]dλ dx.
0
(d1 ) Since |f 0 | , |g 0 | are convex on [a, x] and [x, b] from (4.2) we observe that
1
|T (f, g)| ≤
2
Z b "
2 Z 1
x−a
|g (x)|
λ |f 0 [(1 − λ) a + λx]| dλ
b
−
a
a
0
Z 1
0
+ |f (x)|
λ |g [(1 − λ) a + λx]|dλ
0
2 Z 1
b−x
+
|g (x)|
λ |f 0 [λx + (1 − λ) b]| dλ
b−a
0
Z 1
0
+ |f (x)|
λ |g [λx + (1 − λ) b]|dλ dx
0
2 Z "
Z 1
x−a
1 b
≤
|g (x)|
λ {(1 − λ) |f 0 (a)| + λ |f 0 (x)|} dλ
2 a
b−a
0
Z 1
0
0
+ |f (x)|
λ {(1 − λ) |g (a)| + λ |g (x)|}dλ
0
2 Z 1
b−x
+
|g (x)|
λ {λ |f 0 (x)| + (1 − λ) |f 0 (b)|} dλ
b−a
0
Z 1
0
0
+ |f (x)|
λ {λ |g (x)| + (1 − λ) |g (b)|}dλ dx
0
2 Z "
1 0
1 0
1 b
x−a
|g (x)|
|f (a)| + |f (x)|
=
2 a
b−a
6
3
1 0
1
+ |f (x)|
|g (a)| + |g 0 (x)|
6
3
2 b−x
1 0
1 0
+
|g (x)|
|f (x)| + |f (b)|
b−a
3
6
1 0
1
+ |f (x)|
|g (x)| + |g 0 (b)|
dx.
3
6
This proves the inequality in (2.15).
(d2 ) Since |f 0 | , |g 0 | are log-convex on [a, x] and [x, b], from (4.2) and the fact (3.19) we observe
that
2 Z "
Z 1
1 b
x−a
1−λ
λ
|T (f, g)| ≤
|g (x)|
λ [|f 0 (a)|]
[|f 0 (x)|] dλ
2 a
b−a
0
Z 1
1−λ
λ
0
0
+ |f (x)|
λ [|g (a)|]
[|g (x)|] dλ
0
J. Inequal. Pure and Appl. Math., 6(4) Art. 128, 2005
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O N O STROWSKI -G RÜSS - Č EBYŠEV TYPE INEQUALITIES
13
2 Z 1
b−x
λ
1−λ
+
|g (x)|
λ [|f 0 (x)|] [|f 0 (b)|]
dλ
b−a
0
Z 1
λ
1−λ
0
0
+ |f (x)|
λ [|g (x)|] [|g (b)|]
dλ dx
0
2 Z "
Z 1
Z 1
x−a
1 b
0
λ
0
λ
=
|g (x)| |f (a)|
λA1 dλ + |f (x)| |g (a)|
λB1 dλ
2 a
b−a
0
0
2 #
Z 1
Z 1
b−x
+
|g (x)| |f 0 (b)|
λAλ2 dλ+ |f (x)| |g 0 (b)|
λB2λ dλ dx
b−a
0
0
2 Z b "
1
x−a
A1 log A1 + 1 − A1
=
|g (x)| |f 0 (a)|
2 a
b−a
(log A1 )2
B1 log B1 + 1 − B1
0
+ |f (x)| |g (a)|
(log B1 )2
2 A2 log A2 + 1 − A2
b−x
+
|g (x)| |f 0 (b)|
b−a
(log A2 )2
B2 log B2 + 1 − B2
0
+ |f (x)| |g (b)|
dx.
(log B2 )2
This is the desired inequality in (2.16).
5. P ROOF OF T HEOREM 2.5
From the hypotheses, the identities (3.1) and (3.2) hold. From (3.1) and (3.2) we observe that
f (x) −
1
b−a
Z b
1
f (t) dt
g (x) −
g (t) dt
b−a a
a
Z 1
Z b
1
0
=
(x − t)
f [(1 − λ) x + λt] dλ dt
b−a a
0
Z 1
Z b
1
0
×
(x − t)
g [(1 − λ) x + λt] dλ dt
b−a a
0
Z
b
that is,
Z b
Z b
1
1
(5.1) f (x) g (x) − f (x)
g (t) dt − g (x)
f (t) dt
b−a a
b−a a
Z b
Z b
1
1
+
f (t) dt
g (t) dt
b−a a
b−a a
Z 1
Z b
1
0
=
(x − t)
f [(1 − λ) x + λt] dλ dt
b−a a
0
Z 1
Z b
1
0
×
(x − t)
g [(1 − λ) x + λt] dλ dt .
b−a a
0
J. Inequal. Pure and Appl. Math., 6(4) Art. 128, 2005
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14
B.G. PACHPATTE
Integrating both sides of (5.1) with respect to x from a to b and rewriting we have
Z 1
Z b
Z b
1
1
0
(5.2) T (f, g) =
(x − t)
f [(1 − λ) x + λt] dλ dt
b−a a b−a a
0
Z 1
Z b
1
0
×
(x − t)
g [(1 − λ) x + λt] dλ dt dx.
b−a a
0
(e1 ) Since |f 0 | , |g 0 | are convex on [a, b] , from (5.2) we observe that
Z 1
Z b
Z b
1
1
0
|x − t|
|f [(1 − λ) x + λt]| dλ dt
|T (f, g)| ≤
b−a a b−a a
0
Z 1
Z b
1
0
×
|x − t|
|g [(1 − λ) x + λt]| dλ dt dx
b−a a
0
Z 1
Z b Z b
1
0
0
≤
|x − t|
[(1 − λ) |f (x)| + λ |f (t)|] dλ dt
(b − a)3 a
a
0
Z b
Z 1
0
0
×
|x − t|
[(1 − λ) |g (x)| + λ |g (t)|] dλ dt dx
a
0
0
Z b Z b
1
|f (x)| + |f 0 (t)|
dt
=
|x − t|
2
(b − a)3 a
a
Z b
0
|g (x)| + |g 0 (t)|
×
|x − t|
dt dx.
2
a
The rest of the proof of inequality (2.17) can be completed by closely looking at the proof of
Theorem 2.3, part (c1 ).
(e2 ) The proof follows by closely looking at the proof of (e1 ) given above and the proof of
Theorem 2.3, part (c2 ) . We omit the details.
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inequalities for functions whose modulus of derivatives are convex and applications, RGMIA Res.
Rep. Coll., 5(2) (2002), 219–231. [ONLINE: http://rgmia.vu.edu.au/v5n2.html]
[2] P.CERONE AND S.S.DRAGOMIR, Ostrowski type inequalities for functions whose derivatives
satisfy certain convexity assumptions, Demonstratio Math., 37(2) (2004), 299–308.
[3] S.S.DRAGOMIR AND A.SOFO, Ostrowski type inequalities for functions whose derivatives are
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[8] B.G. PACHPATTE, A note on Hadamard type integral inequalities involving several log-convex
functions, Tamkang J. Math., 36(1) (2005), 43–47.
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