Journal of Inequalities in Pure and
Applied Mathematics
http://jipam.vu.edu.au/
Volume 7, Issue 2, Article 58, 2006
ON ČEBYŠEV TYPE INEQUALITIES INVOLVING FUNCTIONS WHOSE
DERIVATIVES BELONG TO Lp SPACES
B.G. PACHPATTE
57 S HRI N IKETAN C OLONY
N EAR A BHINAY TALKIES
AURANGABAD 431 001 (M AHARASHTRA )
I NDIA
bgpachpatte@gmail.com
Received 10 November, 2005; accepted 16 November, 2005
Communicated by I. Gavrea
A BSTRACT. In this note we establish new Čebyšev type integral inequalities involving functions
whose derivatives belong to Lp spaces via certain integral identities.
Key words and phrases: Čebyšev type inequalities, Lp spaces, Integral identities, Absolutely continuous functions, Hölder’s
integral inequality.
2000 Mathematics Subject Classification. 26D15, 26D20.
1. I NTRODUCTION
One of the classical and important inequalities discovered by P.L.Čebyšev [1] is the following
integral inequality (see also [10, p. 207]):
1
(1.1)
|T (f, g)| ≤
(b − a)2 kf 0 k∞ kg 0 k∞ ,
12
where f, g : [a, b] → R are absolutely continuous functions whose derivatives f 0 , g 0 ∈ L∞ [a, b]
and
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
which is called the Čebyšev functional, provided the integrals in (1.2) exist.
Because of fundamental importance of (1.1) in analysis and applications, many researchers
have given considerable attention to it and a number of extensions, generalizations and variants
have appeared in the literature, see [5], [6], [8] – [10] and the references given therein. The
main purpose of the present note is to establish new inequalities similar to the inequality (1.1)
involving functions whose derivatives belong to Lp spaces. The analysis used in the proofs is
based on the integral identities proved in [3] and [2] and our results provide new estimates on
these types of inequalities.
ISSN (electronic): 1443-5756
c 2006 Victoria University. All rights reserved.
340-05
2
B.G. PACHPATTE
2. S TATEMENT OF R ESULTS
In what follows R and 0 denote the set of real numbers and the derivative of a function. Let
[a, b] ⊂ R, a < b; and as usual for any function h ∈ Lp [a, b] , p > 1 we define khkp =
R
p1
b
p
|h (t)| dt . We use the following notations to simplify the details of presentation. For
a
suitable functions f, g : [a, b] → R we set
1 f (a) + f (b)
a+b
F =
+ 2f
,
3
2
2
1 g (a) + g (b)
a+b
G=
+ 2g
,
3
2
2
Z b
Z b
1
S (f, g) = F G −
F
g (x) dx + G
f (x) dx
b−a
a
a
Z b
Z b
1
1
+
f (x) dx
g (x) dx ,
b−a a
b−a a
1
H (f, g) =
b−a
Z
a
b
[F g (x) + Gf (x)] dx
Z b
Z b
1
1
−2
f (x) dx
g (x) dx .
b−a a
b−a a
Now, we state our main results as follows.
Theorem 2.1. Let f, g : [a, b] → R be absolutely continuous functions whose derivatives
f 0 , g 0 ∈ Lp [a, b], p > 1. Then we have the inequalities
Z b
2
1
0
0
(2.1)
|T (f, g)| ≤
(B (x)) q dx,
3 kf kp kg kp
(b − a)
a
(2.2)
1
|T (f, g)| ≤
2 (b − a)2
Z bh
i
1
|g (x)| kf 0 kp + |f (x)| kg 0 kp (B (x)) q dx,
a
where
B (x) =
(2.3)
for x ∈ [a, b] and
1
p
+
1
q
(x − a)q+1 + (b − x)q+1
,
q+1
= 1.
The following variants of the inequalities in (2.1) and (2.2) hold.
Theorem 2.2. Let f, g : [a, b] → R be absolutely continuous functions whose derivatives
f 0 , g 0 ∈ Lp [a, b], p > 1. Then we have the inequalities
(2.4)
(2.5)
|S (f, g)| ≤
1
1
q
|H (f, g)| ≤
M
2
(b − a)
J. Inequal. Pure and Appl. Math., 7(2) Art. 58, 2006
2
1
q
M
kf 0 kp kg 0 kp ,
2
(b − a)
Z bh
0
0
i
|g (x)| kf kp + |f (x)| kg kp dx,
a
http://jipam.vu.edu.au/
Č EBYŠEV TYPE INEQUALITIES
3
where
M=
(2.6)
and
1
p
+
1
q
(2q+1 + 1) (b − a)q+1
,
3 (q + 1) 6q
= 1.
3. P ROOF OF T HEOREM 2.1
From the hypotheses we have the following identities (see [3, 7]):
Z b
Z b
1
1
(3.1)
f (x) −
f (t) dt =
k (x, t) f 0 (t) dt,
b−a a
b−a a
(3.2)
1
g (x) −
b−a
Z
a
b
1
g (t) dt =
b−a
Z
b
k (x, t) g 0 (t) dt,
a
for x ∈ [a, b] , where
(
(3.3)
k (x, t) =
t − a if t ∈ [a, x]
t − b if t ∈ (x, b]
.
Multiplying the left sides and right sides of (3.1) and (3.2) we have
Z b
Z b
1
1
g (t) dt − g (x)
f (t) dt
(3.4) f (x) g (x) − f (x)
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 b
Z b
1
0
0
=
k (x, t) f (t) dt
k (x, t) g (t) dt .
(b − a)2
a
a
Integrating both sides of (3.4) with respect to x over [a, b] and dividing both sides of the resulting
identity by (b − a) we get
Z b
Z b Z b
1
0
0
k (x, t) f (t) dt
k (x, t) g (t) dt dx.
(3.5)
T (f, g) =
(b − a)3 a
a
a
From (3.5) and using the properties of modulus and Hölder’s integral inequality we have
Z b
Z b Z b
1
0
0
|T (f, g)| ≤
|k (x, t)| |f (t)| dt
|k (x, t)| |g (t)| dt dx
(b − a)3 a
a
a
1q Z b
p1 !
Z b Z b
1
p
|f 0 (t)| dt
≤
|k (x, t)|q
3
(b − a) a
a
a
1
Z b
q Z b
p1 !
p
|g 0 (t)| dt
dx
×
|k (x, t)|q dt
a
a
(3.6)
1
=
kf 0 kp kg 0 kp
(b − a)3
J. Inequal. Pure and Appl. Math., 7(2) Art. 58, 2006
Z
a
b
Z
b
|k (x, t)|q
1q !2
dx.
a
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4
B.G. PACHPATTE
A simple calculation shows that (see [4])
Z b
Z x
Z b
q
q
|k (x, t)| dt =
|t − a| dt +
|t − b|q dt
a
a
x
Z x
Z b
=
(t − a)q dt +
(b − t)q dt
a
x
q+1
(x − a)
=
(3.7)
+ (b − x)q+1
= B (x) .
q+1
Using (3.7) in (3.6) we get (2.1).
Multiplying both sides of (3.1) and (3.2) by g(x) and f (x) respectively and adding the resulting identities we get
Z b
Z b
1
1
(3.8) 2f (x) g (x) − g (x)
f (t) dt + f (x)
g (t) dt
b−a a
b−a a
Z b
Z b
1
1
0
0
= g (x)
k (x, t) f (t) dt + f (x)
k (x, t) g (t) dt .
b−a a
b−a a
Integrating both sides of (3.8) with respect to x over [a, b] and rewriting we obtain
Z b
Z b
Z b
1
0
0
(3.9) T (f, g) =
g (x)
k (x, t) f (t) dt+ f (x)
k (x, t) g (t) dt dx.
2 (b − a)2 a
a
a
From (3.9) and using the properties of modulus, Hölder’s integral inequality and (3.7) we have
Z b
Z b
Z b
1
0
0
|g (x)|
|k (x, t)| |f (t)| dt + |f (x)|
|k (x, t)| |g (t)| dt dx
|T (f, g)| ≤
2 (b − a)2 a
a
a
Z b
1q Z b
p1
Z b"
1
p
q
0
≤
|g (x)|
|k (x, t)| dt
|f (t)| dt
2 (b − a)2 a
a
a
p1 #
Z b
1q Z b
p
q
|g 0 (t)| dt
dx
+ |f (x)|
|k (x, t)| dt
a
a
1
=
2 (b − a)2
Z bh
1
2 (b − a)2
Z bh
=
0
0
|g (x)| kf kp + |f (x)| kg kp
i Z
a
b
1q
|k (x, t)| dt dx
q
a
i
1
|g (x)| kf 0 kp + |f (x)| kg 0 kp (B (x)) q dx.
a
This is the required inequality in (2.2). The proof is complete.
4. P ROOF OF T HEOREM 2.2
From the hypotheses we have the following identities (see [2]):
Z b
Z b
1
1
(4.1)
F−
f (x) dx =
m (x) f 0 (x) dx,
b−a a
b−a a
(4.2)
1
G−
b−a
Z
a
J. Inequal. Pure and Appl. Math., 7(2) Art. 58, 2006
b
1
g (x) dx =
b−a
Z
b
m (x) g 0 (x) dx,
a
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Č EBYŠEV TYPE INEQUALITIES
5
where
(
m (x) =
(4.3)
x−
5a+b
6
x−
a+5b
6
if x ∈ a, a+b
2
a+b .
if x ∈ 2 , b
Multiplying the left sides and right sides of (4.1) and (4.2) we get
Z b
Z b
1
0
0
(4.4)
S (f, g) =
m (x) f (x) dx
m (x) g (x) dx .
(b − a)2
a
a
From (4.4) and using the properties of modulus and Hölder’s integral inequality, we have
Z b
Z b
1
0
0
|S (f, g)| ≤
|m (x)| |f (x)| dx
|m (x)| |g (x)| dx
(b − a)2
a
a
Z b
1q Z b
p1 !
1
p
q
0
≤
|m (x)| dx
|f (x)| dx
(b − a)2
a
a
p1 !
Z b
1q Z b
p
×
|m (x)|q dx
|g 0 (x)| dx
a
(4.5)
1
=
(b − a)2
a
Z
b
|m (x)|q dx
1q !2
kf 0 kp kg 0 kp .
a
A simple computation gives (see [2])
q
Z b
Z a+b Z b a + 5b q
2 5a + b q
|m (x)| dx =
x − 6 dx + a+b x− 6 dx
a
a
2
q
q
Z 5a+b Z a+b 6
2
5a + b
5a + b
=
− x dx +
dx
x−
5a+b
6
6
a
6
q
q
Z b Z a+5b 6
a + 5b
a + 5b
− x dx +
x−
dx
+
a+b
a+5b
6
6
2
6
"
q+1 q+1
1
5a + b
a + b 5a + b
=
−a
+
−
q+1
6
2
6
q+1 q+1 #
a + 5b a + b
a + 5b
+
−
+ b−
6
2
6
(4.6)
=
(2q+1 + 1) (b − a)q+1
= M.
3 (q + 1) 6q
Using (4.6) in (4.5) we get the required inequality in (2.4).
Multiplying both sides of (4.1) and (4.2) by g(x) and f (x) respectively and adding the resulting identities we get
Z b
Z b
1
1
(4.7) F g (x) + Gf (x) − g (x)
f (x) dx + f (x)
g (x) dx
b−a a
b−a a
Z b
Z b
1
1
0
0
= g (x)
m (x) f (x) dx + f (x)
m (x) g (x) dx .
b−a a
b−a a
J. Inequal. Pure and Appl. Math., 7(2) Art. 58, 2006
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6
B.G. PACHPATTE
Integrating both sides of (4.7) with respect to x over [a, b] and dividing both sides of the resulting
identity by (b − a) we get
Z b
Z b
Z b
1
0
0
(4.8)
H (f, g) =
g (x)
m (x) f (x) dx + f (x)
m (x) g (x) dx dx.
(b − a)2 a
a
a
From (4.8) and using the properties of modulus, Hölder’s integral inequality and (4.6) we have
Z b
Z b
Z b
1
0
0
|g (x)|
|m (x)| |f (x)| dx + |f (x)|
|H (f, g)| ≤
|m (x)| |g (x)| dx dx
(b − a)2 a
a
a
Z b
1q Z b
p1
Z b"
1
p
q
≤
|g (x)|
|m (x)| dx
|f 0 (x)| dx
(b − a)2 a
a
a
p1 #
Z b
1q Z b
p
+ |f (x)|
|m (x)|q dx
|g 0 (x)| dx
dx
a
1
=
(b − a)2
Z bh
a
|g (x)| kf 0 kp + |f (x)| kg 0 kp
a
i Z
1q
b
q
|m (x)| dx
a
Z bh
i
1
1
0
0
q
=
M
|g (x)| kf kp + |f (x)| kg kp dx.
(b − a)2
a
This is the desired inequality in (2.5). The proof is complete.
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J. Inequal. Pure and Appl. Math., 7(2) Art. 58, 2006
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