NOTE ON AN OPEN PROBLEM
LAZHAR BOUGOFFA
Al-imam Muhammad Ibn Saud Islamic University
Faculty of Science, Department of Mathematics
P.O.Box 84880, Riyadh 11681, Saudi Arabia
EMail: bougoffa@hotmail.com
Note on an Open Problem
Lazhar Bougoffa
vol. 8, iss. 2, art. 58, 2007
Title Page
Received:
17 December, 2006
Accepted:
1 April, 2007
Communicated by:
P.S. Bullen
2000 AMS Sub. Class.:
26D15.
Key words:
Integral inequality.
Abstract:
The aim of this short note is to establish an integral inequality and its reverse
which give an affirmative answer to an open problem posed by QUÔC ANH
NGÔ, DU DUC THANG, TRANT TAT DAT, and DANG ANH TUAN, in the
paper [Notes on an integral inequality, J. Ineq. Pure and Appl. Math., 7(4)(2006),
Art. 120.]
Contents
JJ
II
J
I
Page 1 of 10
Go Back
Full Screen
Close
Contents
1
Introduction
3
2
Main Results
5
Note on an Open Problem
Lazhar Bougoffa
vol. 8, iss. 2, art. 58, 2007
Title Page
Contents
JJ
II
J
I
Page 2 of 10
Go Back
Full Screen
Close
1.
Introduction
Very recently, in the paper [1] the authors studied some integral inequalities and
proposed the following open problem:
Problem 1.1. Letf be a continuous function on [0, 1] satisfying
Z 1
Z 1
(1.1)
f (t)dt ≥
tdt, ∀x ∈ [0, 1].
x
Lazhar Bougoffa
x
Under what conditions does the inequality
Z 1
Z
α+β
(1.2)
f
(x)dx ≥
0
Note on an Open Problem
vol. 8, iss. 2, art. 58, 2007
1
xα f β (x)dx
Title Page
0
Contents
hold for α and β?
This type of integral inequality is a complement, variant and continuation of Qi’s
inequality [2]. Before giving an affirmative answer to Problem 1.1 and its reverse,
we establish the following essential lemma:
Lemma 1.2. Let f (x) be nonnegative function, continuous on [a, b] and differentiable on (a, b).
Rb
Rb
If x f (t)dt ≤ x (t − a)dt, ∀x ∈ [a, b], then
(1.3)
f (x) ≤ x − a.
Rb
Rb
If x f (t)dt ≥ x (t − a)dt, ∀x ∈ [a, b], then
(1.4)
f (x) ≥ x − a.
JJ
II
J
I
Page 3 of 10
Go Back
Full Screen
Close
Proof. In order to prove (1.3), set
Z b
G(x) =
[f (t) − (t − a)]dt ≤ 0,
∀x ∈ [a, b],
x
we have
0
G (x) = x − a − f (x),
∀x ∈ [a, b].
0
We shall give an indirect proof, we suppose f (x) ≥ x − a, then G (x) ≤ 0, G(x)
decreases, and G(x) ≥ 0, because of G(b) = 0. This contradiction establishes (1.3).
The proof of (1.4) is the same as the proof of (1.3).
Now, our results can be stated as follows:
Note on an Open Problem
Lazhar Bougoffa
vol. 8, iss. 2, art. 58, 2007
Title Page
Contents
JJ
II
J
I
Page 4 of 10
Go Back
Full Screen
Close
2.
Main Results
Theorem 2.1. Let f (x) be a nonnegative function, continuous on [a, b] and differentiable on (a, b), and let α and β be positive numbers.
Rb
Rb
If x f (t)dt ≤ x (t − a)dt, ∀x ∈ [a, b], then
Z b
Z b
α+β
(2.1)
f
(x)dx ≤
(x − a)α f β (x)dx.
a
If
Rb
x
f (t)dt ≥
Rb
(2.2)
x
a
Lazhar Bougoffa
(t − a)dt, ∀x ∈ [a, b], then
Z b
Z b
α+β
f
(x)dx ≥
(x − a)α f β (x)dx.
a
Note on an Open Problem
vol. 8, iss. 2, art. 58, 2007
Title Page
a
Contents
Proof. Set
x
Z
α+β
f
(t) − (t − a)α f β (t)dt ,
F (x) =
∀x ∈ [a, b].
a
We can see that
JJ
II
J
I
Page 5 of 10
0
F (x) = f α+β (x) − (x − a)α f β (x),
Go Back
so that
0
F (x) = [f α (x) − (x − a)α ] f β (x).
If
Z
b
Z
f (t)dt ≤
x
Full Screen
Close
b
(t − a)dt,
∀x ∈ [a, b],
x
and from (1.3) of Lemma 1.2, we have, f (x) ≤ (x − a), so that f α (x) ≤ (x − a)α .
0
Thus F (x) ≤ 0, and F (x) is decreasing on [a, b]. Since F (a) = 0, we have F (x) ≤
0, ∀x ∈ [a, b], which gives the inequality (2.1).
When
b
Z
b
Z
f (t)dt ≥
x
α
(t − a)dt,
∀x ∈ [a, b],
x
we have from (1.4) f (x) ≥ (x − a)α , and the rest of the proof is the same as that
of (2.1).
Remark 1. If f (x) = 0 or f (x) = x − a, the equality in (2.1) holds.
Now we establish new integral inequalities similar to (2.1) and (2.2) involving n
functions: fi (x), i = 1, . . . , n.
Theorem 2.2. Let fi (x), i = 1, 2, . . . , n be nonnegative functions, continuous on
[a, b] and differentiable on (a, b), and let αi , βi , i = 1, . . . , n be positive numbers.
If
Z
Z
fi (t)dt ≤
x
(t − a)dt,
x
fiαi +βi (x)dx ≤
Z
b
Pn
(x − a)
a i=1
i=1
αi
a
n
Y
fiβi (x)dx.
i=1
b
Z
fi (t)dt ≥
x
(2.4)
a i=1
fiαi +βi (x)dx
JJ
II
J
I
Page 6 of 10
Full Screen
b
(t − a)dt,
∀x ∈ [a, b],
x
then
Z bY
n
Contents
Go Back
If
Z
Title Page
∀x ∈ [a, b],
then
(2.3)
Lazhar Bougoffa
vol. 8, iss. 2, art. 58, 2007
b
b
Z bY
n
Note on an Open Problem
Z
≥
b
Pn
(x − a)
a
i=1
αi
n
Y
i=1
fiβi (x)dx.
Close
Proof. As in proof of (2.1) and (2.2), we let
#
Z x "Y
n
n
Pn
Y
F (x) =
fiαi +βi (t) − (t − a) i=1 αi
fiβi (t) dt,
a
thus
i=1
i=1
"
0
F (x) =
n
Y
Pn
fiαi (x) − (x − a)
i=1
#
αi
i=1
when
Z b
Z b
fi (t)dt ≤
(t − a)dt
x
n
Y
fiβi (x),
Note on an Open Problem
Lazhar Bougoffa
i=1
vol. 8, iss. 2, art. 58, 2007
Z
Z
fi (t)dt ≥
resp.
x
b
x
b
(t − a)dt ,
∀x ∈ [a, b],
x
Title Page
Contents
using Lemma 1.2, we have
fi (x) ≤ x − a
(resp. fi (x) ≥ x − a) ,
i = 1, 2, . . . , n,
and
fiαi (x) ≤ (x − a)αi
(resp. fiαi (x) ≥ (x − a)αi ) ,
JJ
II
J
I
Page 7 of 10
i = 1, 2, . . . , n,
Go Back
thus
Full Screen
n
Y
i=1
Pn
fiαi (x) ≤ (x − a)
i=1
αi
resp.
n
Y
Pn
fiαi (x) ≥ (x − a)
i=1
!
αi
.
i=1
The rest of the proof is the same as in Theorem 2.1, and we omit (2.3) and (2.4).
In order to illustrate a possible practical use of this result, we shall give in the
following a simple example in which we can apply Theorem 2.1.
Close
Example 2.1. For α = β = 1 :
i) Let f (t) = cos t + t on [0, π2 ], we see that all the conditions of Theorem 2.1 are
fulfilled. Indeed,
Z π
2
1 π2
2
(cos t + t) dt = 1 − sin x +
−x
2 4
x
Z π
h πi
2
1 π2
2
≥
tdt =
− x , ∀x ∈ 0,
,
2 4
2
x
and straightforward computation yields
Z π
2
5π π 3
(cos x + x)2 dx =
+
−2
4
24
0
Z π
2
x (cos x + x) dx
>
0
π π3
= +
− 1.
2 24
That is,
Z
0
π
2
f 2 (x)dx >
Z
Note on an Open Problem
Lazhar Bougoffa
vol. 8, iss. 2, art. 58, 2007
Title Page
Contents
JJ
II
J
I
Page 8 of 10
Go Back
π
2
xf (x)dx.
Full Screen
0
ii) Let f (t) = t− π2 +cos t on [ π2 , π], all the conditions of Theorem 2.1 be satisfied.
We see that
Z π
Z π
hπ i
π
π
t − + cos t dt ≤
t−
dt, ∀x ∈
,π ,
2
2
2
x
x
Close
which is equivalent to
Z π
cos tdt = − sin x ≤ 0,
x∈
x
hπ
2
i
,π ,
and direct computation yields
Z π
2
π π3
π
−2
x − + cos x dx = +
π
2
4
24
2
Z π
π
x−
f (x)dx
<
π
2
2
=
That is
Z
π
f 2 (x)dx <
π
2
Z
π
π
2
π3
− 1.
24
x−
π
f (x)dx.
2
Note on an Open Problem
Lazhar Bougoffa
vol. 8, iss. 2, art. 58, 2007
Title Page
Contents
JJ
II
J
I
Page 9 of 10
Go Back
Full Screen
Close
References
[1] QUÔC ANH NGÔ, DU DUC THANG, TRAN TAT DAT, AND DANG ANH
TUAN, Notes on an integral inequality, J. Ineq. Pure and Appl. Math.,
7(4) (2006), Art. 120. [ONLINE: http://jipam.vu.edu.au/article.
php?sid=737].
[2] FENG QI, Several integral inequalities, J. Ineq. Pure and Appl. Math., 1(2)
(2000), Art. 19. [ONLINE: http://jipam.vu.edu.au/article.php?
sid=113].
Note on an Open Problem
Lazhar Bougoffa
vol. 8, iss. 2, art. 58, 2007
Title Page
Contents
JJ
II
J
I
Page 10 of 10
Go Back
Full Screen
Close