Volume 9 (2008), Issue 2, Article 37, 5 pp.
NOTE ON AN OPEN PROBLEM
GHOLAMREZA ZABANDAN
D EPARTMENT OF M ATHEMATICS
FACULTY OF M ATHEMATICAL S CIENCE AND C OMPUTER E NGINEERING
T EACHER T RAINING U NIVERSITY
599 TALEGHANI AVENUE
T EHRAN 15618, IRAN
Zabandan@saba.tmu.ac.ir
Received 23 August, 2007; accepted 13 March, 2008
Communicated by F. Qi
A BSTRACT. 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].
Key words and phrases: Integral inequality.
2000 Mathematics Subject Classification. 26D15.
1. I NTRODUCTION
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
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).
I am grateful to the referee for his comments, especially for Theorem 2.3.
275-07
2
G HOLAMREZA Z ABANDAN
Lemma 1.1. If (1.1) holds, then for each x ∈ [0, 1] we have
Z 1
1 − xk+2
tk f (t)dt ≥
(k ∈ N).
k+2
x
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
1 k 1
1
k−1
y
f (t)dt dy = y
f (t)dt +
y k f (y)dy
k
k
x
y
y
x
x
Z 1
Z 1
1
1
= − xk
f (t)dt +
y k f (y)dy.
k
k
x
x
Thus
Z
Z
1 k 1
1 1 k
1
1
1
f (t)dt +
y f (y)dy ≥
− xk +
xk+2
− x
k
k
k(k
+
2)
2k
2(k
+
2)
x
x
Z 1
Z 1
1
1
k
k
k
=⇒
y f (y)dy ≥ x
f (t)dt +
− xk +
xk+2
k
+
2
2
2(k
+
2)
x
x
1 1 2
1
1
k
k
≥x
− x +
− xk +
xk+2
2 2
k+2 2
2(k + 2)
1 − xk+2
.
=
k+2
=
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
2
Lemma 1.2. Let f be a non-negative continuous function on [0, 1] such that x f (t)dt ≥ 1−x
2
(0 ≤ x ≤ 1). Then for each x ∈ [0, 1] and k ∈ N, we have
Z 1
1 − xk+1
f k (t)dt ≥
.
k+1
x
Proof. Since
Z
1
(f (t) − t)(f k (t) − tk )dt
x
Z 1
Z 1
Z
k+1
k
=
f (t)dt −
t f (t)dt −
0≤
x
x
J. Inequal. Pure and Appl. Math., 9(2) (2008), Art. 37, 5 pp.
0
1
k
Z
tf (t)dt +
1
tk+1 dt
x
http://jipam.vu.edu.au/
N OTE O N A N O PEN P ROBLEM
3
it follows that
1
Z
k+1
f
1
Z
(t)dt ≥
1
Z
k
x
x
1
(1 − xk+2 ).
k+2
tf k (t)dt −
t f (t)dt +
x
By using Lemma 1.1, we get
1
Z
f
(1.2)
k+1
1
Z
tf k (t)dt.
(t)dt ≥
x
x
We
the proof by mathematical
induction. The assertion is obvious for k = 1. Let
R 1 k+1
R 1 continue
k+2
1−xk+1
k
f (t)dt ≥ k+1 , we show that x f (t)dt ≥ 1−x
. We have
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
Z 1 Z 1
Z 1
k
k
f (t)dt dy = y
f (t)dt +
x
y
y
x
Z
1
f k (t)dt +
= −x
x
Z
1
yf k (y)dy
x
1
yf k (y)dy.
x
Thus
Z
−x
1
1
Z
k
x
1
1
1
−
x+
xk+2
k+2 k+1
(k + 1)(k + 2)
yf k (y)dy ≥
f (t)dt +
x
and hence
Z
1
1
Z
k
x
1
1
1
−
x+
xk+2
k+2 k+1
(k + 1)(k + 2)
1
1
1
+
−
x+
xk+2
k+2 k+1
(k + 1)(k + 2)
f k (t)dt +
yf (y)dy ≥ x
x
1 − xk+1
k+1
1 − xk+2
=
.
k+2
≥x
So by (1.2) we get
Z
1
f
k+1
Z
(t)dt ≥
x
1
tf k (t)dt ≥
x
1 − xk+2
,
k+2
which completes the proof.
2. M AIN R ESULTS
Theorem 2.1. Let f be a non-negative and continuous function on [0, 1]. If
(0 ≤ x ≤ 1), then for each m, n ∈ N,
Z 1
Z 1
m+n
f
(x)dx ≥
xm f n (x)dx.
0
J. Inequal. Pure and Appl. Math., 9(2) (2008), Art. 37, 5 pp.
R1
x
f (t)dt ≥
1−x2
2
0
http://jipam.vu.edu.au/
4
G HOLAMREZA Z ABANDAN
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
which implies
n
m+n
Hence
Z
1
Z
1
f
m+n
0
m
(x)dx +
m+n
1
Z
x
m+n
0
Z
dx ≥
1
xm f n (x)dx.
0
Z 1
m
m
f
(x)dx ≥
x f (x)dx +
f m+n (x)dx −
m+n 0
(m + n)(m + n + 1)
0
0
Z 1
Z 1
1
m
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.
m+n
Z
1
m n
0
0
Theorem 2.2. Let f be a continuous function such that f (x) ≥ 1 (0 ≤ x ≤ 1). If
1−x2
, then for each α, β > 0,
2
Z 1
Z 1
α+β
(2.1)
f
(x)dx ≥
xα f β (x)dx.
0
R1
x
f (t)dt ≥
0
Proof.
to that used in the proof of Theorem
2.1 the inequality (2.1) holds
R 1 By a similar method
R1
1
1
if 0 f α+β (x)dx ≥ α+β+1
. So it is enough to prove that 0 f γ (x)dx ≥ γ+1
(γ > 0). Since
f (x) ≥ 1 (0 ≤ x ≤ 1) and [γ] ≤ γ < [γ] + 1, we have
Z 1
Z 1
γ
f (x)dx >
f [γ] (x)dx.
0
By Lemma 1.2 we obtain
Z 1
γ
0
Z
f (x)dx ≥
0
0
1
f [γ] (x)dx ≥
1
1
≥
.
[γ] + 1
γ+1
R1
Remark 2. The condition f (x) ≥ 1 (0 ≤ x ≤ 1) in Theorem 2.2 is necessary for 0 f γ (x)dx ≥
1
(γ > 0). For example, let
γ+1

0
0 ≤ x ≤ 12
f (x) =
2(2x − 1) 1 < x ≤ 1
2
√
R
R1 1
2
1
and γ = 21 , then f is continuous on [0, 1] and x f (t)dt ≥ 1−x
, but 0 f 2 (x)dx = 32 < 23 .
2
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.
J. Inequal. Pure and Appl. Math., 9(2) (2008), Art. 37, 5 pp.
http://jipam.vu.edu.au/
N OTE O N A N O PEN P ROBLEM
5
Theorem 2.3. Let f be a non-negative continuous function on [0, 1]. If
(0 ≤ x ≤ 1), then for each α, β > 0 such that α + β ≥ 1, we have
Z 1
1
f α+β (x)dx ≥
.
α+β+1
0
R1
x
f (t)dt ≥
1−x2
2
Proof. By using Theorem A of [5] for g(t) = t, α = 1, a = 0, and b = 1, the assertion is
obvious.
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J. Inequal. Pure and Appl. Math., 9(2) (2008), Art. 37, 5 pp.
http://jipam.vu.edu.au/