NOTE ON AN OPEN PROBLEM
GHOLAMREZA ZABANDAN
Department of Mathematics
Faculty of Mathematical Science and Computer Engineering
Teacher Training University
599 Taleghani Avenue
Tehran 15618, IRAN
Note On An Open Problem
Gholamreza Zabandan
vol. 9, iss. 2, art. 37, 2008
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EMail: Zabandan@saba.tmu.ac.ir
Contents
Received:
23 August, 2007
Accepted:
13 March, 2008
Communicated by:
F. Qi
2000 AMS Sub. Class.:
26D15.
Key words:
Integral inequality.
Abstract:
In this paper we give an affirmative answer to an open problem proposed by Quôc
Anh Ngô, Du Duc Thang, Tran Tat Dat, and Dang Anh Tuan [6].
Acknowledgement:
I am grateful to the referee for his comments, especially for Theorem 2.3.
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Contents
1
Introduction
3
2
Main Results
8
Note On An Open Problem
Gholamreza Zabandan
vol. 9, iss. 2, art. 37, 2008
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1.
Introduction
In [6] the authors proved some integral inequalities and proposed the following question:
Let f be a continuous function on [0, 1] satisfying
Z 1
1 − x2
(1.1)
f (t)dt ≥
,
(0 ≤ x ≤ 1).
2
x
Under what conditions does the inequality
Z 1
Z
α+β
f
(x)dx ≥
0
Note On An Open Problem
Gholamreza Zabandan
vol. 9, iss. 2, art. 37, 2008
1
xα f β (x)dx
0
hold for α, β?
In [1] the author has given an answer to this open problem, but there is a clear
gap in the proof of Lemma 1.1, so that the other results of the paper break down too.
In this paper we give an affirmative answer to this problem by presenting stronger
results. First we prove the following two essential lemmas.
Throughout this paper, we always assume that f is a non-negative continuous
function on [0, 1], satisfying (1.1).
Lemma 1.1. If (1.1) holds, then for each x ∈ [0, 1] we have
Z 1
1 − xk+2
(k ∈ N).
tk f (t)dt ≥
k+2
x
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Proof. By our assumptions, we have
Z 1
Z 1
Z
k−1
y
f (t)dt dy ≥
x
y
1
y k−1
x
1
=
2
=
Z
1 − y2
dy
2
1
(y k−1 − y k+1 )dy
x
1
1
1
− xk +
xk+2 .
k(k + 2) 2k
2(k + 2)
On the other hand, integrating by parts, we also obtain
1
Z 1
Z 1
Z
Z
1 k 1
1 1 k
k−1
y
f (t)dt dy = y
f (t)dt +
y f (y)dy
k
k x
x
y
y
x
Z
Z
1 k 1
1 1 k
=− x
f (t)dt +
y f (y)dy.
k
k x
x
Thus
1
− xk
k
Z
1
f (t)dt +
x
1
k
Z
1
y k f (y)dy ≥
x
1
1
1
− xk +
xk+2
k(k + 2) 2k
2(k + 2)
Note On An Open Problem
Gholamreza Zabandan
vol. 9, iss. 2, art. 37, 2008
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Z
=⇒
x
1
Z
1
1
1
k
f (t)dt +
− xk +
xk+2
k+2 2
2(k + 2)
x
1 1 2
1
1
k
≥ xk
− x +
− xk +
xk+2
2 2
k+2 2
2(k + 2)
k+2
1−x
.
=
k+2
y k f (y)dy ≥ xk
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Remark 1. By a similar argument, we can show that Lemma 1.1 also holds when k
is a real number in [1, ∞). That is
Z 1
1 − xα+2
tα f (t)dt ≥
(∀α ≥ 1).
α+2
x
It is also interesting to note that the result of [5, Lemma 1.3] holds if we take x = 0
in Lemma 1.1.
R1
Lemma 1.2. Let f be a non-negative continuous function on [0, 1] such that x f (t)dt ≥
1−x2
(0 ≤ x ≤ 1). Then for each x ∈ [0, 1] and k ∈ N, we have
2
Z 1
1 − xk+1
.
f k (t)dt ≥
k+1
x
Proof. Since
Z
Gholamreza Zabandan
vol. 9, iss. 2, art. 37, 2008
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1
(f (t) − t)(f k (t) − tk )dt
Zx 1
Z 1
Z
k+1
k
=
f (t)dt −
t f (t)dt −
0≤
x
x
it follows that
Z 1
Z
k+1
f (t)dt ≥
x
Note On An Open Problem
1
k
Z
tf (t)dt +
0
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1
tk+1 dt
x
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1
tk f (t)dt +
x
Z
1
tf k (t)dt −
x
By using Lemma 1.1, we get
Z 1
Z
k+1
(1.2)
f (t)dt ≥
x
x
1
tf k (t)dt.
1
(1 − xk+2 ).
k+2
Close
We continue
the proof by mathematical induction.
is obvious for k =
R1
R 1 k+1 The assertion
k+1
1−xk+2
1. Let x f k (t)dt ≥ 1−x
,
we
show
that
f
(t)dt
≥
.
We have
k+1
k+2
x
Z 1 Z 1
Z 1
1 − y k+1
k
f (t)dt dy ≥
dy
k+1
x
y
x
1
1
1 k+2 =
y−
y
k+1
k+2
x
1
1
1
−
x+
xk+2 .
k+2 k+1
(k + 1)(k + 2)
On the other hand, integrating by parts, we also obtain
1 Z 1
Z 1 Z 1
Z 1
k
k
f (t)dt dy = y
f (t)dt +
yf k (y)dy
=
x
y
y
x
Z
= −x
x
1
f k (t)dt +
Z
x
1
yf k (y)dy.
x
Thus
Z
−x
1
k
Z
1
f (t)dt +
x
x
and hence
Z 1
Z
k
yf (y)dy ≥ x
x
yf k (y)dy ≥
1
1
1
−
x+
xk+2
k+2 k+1
(k + 1)(k + 2)
Note On An Open Problem
Gholamreza Zabandan
vol. 9, iss. 2, art. 37, 2008
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1
1
1
1
f (t)dt +
−
x+
xk+2
k+2 k+1
(k + 1)(k + 2)
x
k+1
1−x
1
1
1
≥x
+
−
x+
xk+2
k+1
k+2 k+1
(k + 1)(k + 2)
1 − xk+2
=
.
k+2
k
Close
So by (1.2) we get
Z
1
f
k+1
x
Z
(t)dt ≥
x
1
1 − xk+2
tf (t)dt ≥
,
k+2
k
which completes the proof.
Note On An Open Problem
Gholamreza Zabandan
vol. 9, iss. 2, art. 37, 2008
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2.
Main Results
Theorem 2.1. Let f be a non-negative and continuous function on [0, 1]. If
1−x2
(0 ≤ x ≤ 1), then for each m, n ∈ N,
2
Z 1
Z 1
m+n
f
(x)dx ≥
xm f n (x)dx.
0
R1
x
f (t)dt ≥
0
Note On An Open Problem
Proof. By using the general Cauchy inequality [5, Theorem 3.1], we have
n
m
f m+n (x) +
xm+n ≥ xm f n (x),
m+n
m+n
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which implies
n
m+n
Gholamreza Zabandan
vol. 9, iss. 2, art. 37, 2008
Z
1
f
0
m+n
m
(x)dx +
m+n
1
Z
x
0
m+n
Z
dx ≥
Contents
1
xm f n (x)dx.
0
Hence
Z 1
Z 1
Z 1
m
m
m+n
m n
f m+n (x)dx −
f
(x)dx ≥
x f (x)dx +
m+n 0
(m + n)(m + n + 1)
0
0
Z 1
Z 1
m
1
m n
m+n
=
x f (x)dx +
f
(x)dx −
.
m+n
m+n+1
0
0
R1
1
By Lemma 1.2, we have 0 f m+n (x)dx ≥ m+n+1
. Therefore
Z 1
Z 1
m+n
f
(x)dx ≥
xm f n (x)dx.
0
0
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2.2. Let f be a continuous function such that f (x) ≥ 1 (0 ≤ x ≤ 1). If
RTheorem
2
1
f (t)dt ≥ 1−x
, then for each α, β > 0,
2
x
Z 1
Z 1
α+β
(2.1)
f
(x)dx ≥
xα f β (x)dx.
0
0
Proof. By a similar
method to that used in the proof of Theorem 2.1Rthe inequality
R 1 α+β
1
1
(2.1) holds if 0 f
(x)dx ≥ α+β+1
. So it is enough to prove that 0 f γ (x)dx ≥
1
(γ > 0). Since f (x) ≥ 1 (0 ≤ x ≤ 1) and [γ] ≤ γ < [γ] + 1, we have
γ+1
Z
1
Z
γ
0
f [γ] (x)dx.
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0
By Lemma 1.2 we obtain
Z 1
Z
γ
f (x)dx ≥
Gholamreza Zabandan
vol. 9, iss. 2, art. 37, 2008
1
f (x)dx >
0
Note On An Open Problem
Contents
1
f [γ] (x)dx ≥
0
1
1
≥
.
[γ] + 1
γ+1
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Remark
f (x) ≥ 1 (0 ≤ x ≤ 1) in Theorem 2.2 is necessary for
R 1 γ 2. The condition
1
f (x)dx ≥ γ+1 (γ > 0). For example, let
0

0
0 ≤ x ≤ 12
f (x) =
2(2x − 1) 1 < x ≤ 1
2
and
γ = 12 , then f is continuous on [0, 1] and
√
2
< 23 .
3
R1
x
f (t)dt ≥
1−x2
,
2
but
R1
0
1
f 2 (x)dx =
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In the following theorem, we show that the condition f (x) ≥ 1 (0 ≤ x ≤ 1) in
Theorem 2.2 can be removed if we assume that α + β ≥ 1.
R1
Theorem 2.3. Let f be a non-negative continuous function on [0, 1]. If x f (t)dt ≥
1−x2
(0 ≤ x ≤ 1), then for each α, β > 0 such that α + β ≥ 1, we have
2
Z 1
1
f α+β (x)dx ≥
.
α+β+1
0
Note On An Open Problem
Gholamreza Zabandan
Proof. By using Theorem A of [5] for g(t) = t, α = 1, a = 0, and b = 1, the
assertion is obvious.
vol. 9, iss. 2, art. 37, 2008
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References
[1] L. BOUGOFFA, Note on an open problem, J. Ineq. Pure and Appl. Math., 8(2)
(2007), Art. 58. [ONLINE: http://jipam.vu.edu.au/article.php?
sid=871].
[2] K. BOUKERRIOUA AND A. GUEZANE-LAKOUD, On an open question regarding an integral inequality, J. Ineq. Pure and Appl. Math., 8(3) (2007), Art.
77. [ONLINE: http://jipam.vu.edu.au/article.php?sid=885].
[3] W.J. LIU, C.C. LI AND J.W. DONG, On an open problem concerning an integral inequality, J. Ineq. Pure and Appl. Math., 8(3) (2007), Art. 74. [ONLINE:
http://jipam.vu.edu.au/article.php?sid=882].
[4] F. 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].
[5] QUÔC ANH NGÔ, On the inverse of an integral inequality, RGMIA Research
Report Collection, 10(4) (2007), Art. 10. [ONLINE: http://www.staff.
vu.edu.au/RGMIA/v10n4.asp].
Note On An Open Problem
Gholamreza Zabandan
vol. 9, iss. 2, art. 37, 2008
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[6] 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].
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