Volume 9 (2008), Issue 2, Article 42, 5 pp.
NOTES ON AN INEQUALITY
N. S. HOANG
D EPARTMENT OF M ATHEMATICS , K ANSAS S TATE U NIVERSITY
M ANHATTAN , KS 66506-2602, USA
nguyenhs@math.ksu.edu
URL: http://www.math.ksu.edu/~nguyenhs
Received 13 November, 2007; accepted 26 April, 2008
Communicated by S.S. Dragomir
A BSTRACT. In this note we prove a generalized version of an inequality which was first introduced by A. Q. Ngo, et al. and later generalized and proved by W. J. Liu, et al. in the paper: "On
an open problem concerning an integral inequality", J. Inequal. Pure & Appl. Math., 8(3) 2007.
Key words and phrases: Integral inequality, Young’s inequality.
2000 Mathematics Subject Classification. 26D15.
1. I NTRODUCTION
In [2] the following result was proved: If f ≥ 0 is a continuous function on [0, 1] such that
Z 1
Z 1
(1.1)
f (t)dt ≥
tdt, ∀x ∈ [0, 1],
x
then
x
1
Z
f
α+1
1
Z
xα f (x)dx,
(x)dx ≥
0
∀α > 0.
0
The following question was raised in [2]: If f satisfies the above assumptions, under what
additional assumptions can one claim that:
Z 1
Z 1
α+β
f
(x)dx ≥
xα f β (x)dx, ∀α, β > 0?
0
0
It was proved in [1] that if f ≥ 0 is a continuous function on [0, 1] satisfying
Z b
Z b
α
f (t)dt ≥
tα dt, α, b > 0, ∀x ∈ [0, b],
x
x
then
Z
b
f
0
α+β
Z
(x)dx ≥
b
xα f β (x)dx,
∀β > 0.
0
In this paper, we prove more general results, namely, Theorems 2.4 and 2.5 below.
The author wishes to express his thanks to Prof. A.G. Ramm for helpful comments during the preparation of the paper.
339-07
2
N. S. H OANG
2. R ESULTS AND P ROOFS
Let us recall the following result:
Lemma 2.1 (Young’s inequality). Let α and β be positive real numbers satisfying α + β = 1.
Then for all positive real numbers x and y, we have:
αx + βy ≥ xα y β .
Throughout the paper, [a, b] denotes a bounded interval and all functions are real-valued. Let
us prove the following lemma:
Lemma 2.2. Let f ∈ L1 [a, b], g ∈ C 1 [a, b]. Suppose f ≥ 0, g > 0 is nondecreasing. If
Z b
Z b
f (t)dt ≥
g(t)dt, ∀x ∈ [a, b],
x
x
then ∀α > 0 the following inqualities hold
Z b
Z b
α
(2.1)
g (x)f (x)dx ≥
g α+1 (x)dx,
a
a
Z b
Z b
(2.2)
f α+1 (x)dx ≥
f α (x)g(x)dx,
a
a
Z b
Z b
(2.3)
f α+1 (x)dx ≥
f (x)g α (x)dx.
a
a
∗
Proof. First, let us prove (2.1). Let A, A denote
Z x
Z b
∗
Af (x) :=
f (t)dt, A f (x) :=
f (t)dt,
a
x ∈ [a, b], f ∈ L1 [a, b].
x
Note that these are continuous functions. From the assumption one has
A∗ f (x) ≥ A∗ g(x),
∀x ∈ [a, b].
This means
(A∗ f − A∗ g)(x) ≥ 0, ∀x ∈ [a, b].
Then ∀h ∈ L1 [a, b], h ≥ 0, one obtains
Z b
∗
∗
(2.4)
hA f − A g, hi :=
(A∗ f − A∗ g)(x)h(x)dx ≥ 0.
a
Note that the left-hand side of (2.4) is finite since A∗ f, A∗ g are bounded and h ∈ L1 [a, b]. Thus,
by Fubini’s Theorem, one has
(2.5)
hf − g, Ahi = hA∗ f − A∗ g, hi ≥ 0,
∀h ≥ 0, h ∈ L1 [a, b].
Denote h(x) = αg(x)α−1 g 0 (x). One has
Z x
Ah(x) =
h(t)dt = g α (x) − g α (a),
∀x ∈ [a, b].
a
By the assumption,
(2.6)
α
α
Z
hf − g, g (a)i = g (a)
b
(f (x) − g(x))dx ≥ 0.
a
Since h ≥ 0, from (2.5) and (2.6) one gets
(2.7)
hf − g, g α i = hf − g, Ahi + hf − g, g α (a)i ≥ 0,
∀α ≥ 0.
Hence, (2.1) is obtained.
J. Inequal. Pure and Appl. Math., 9(2) (2008), Art. 42, 5 pp.
http://jipam.vu.edu.au/
N OTES ON AN I NEQUALITY
3
Since
(f (x) − g(x))(f α (x) − g α (x)) ≥ 0,
∀x ∈ [a, b],
∀α ≥ 0,
one gets
hf − g, f α − g α i ≥ 0,
(2.8)
∀α ≥ 0.
Inequalities (2.7) and (2.8) imply
hf − g, f α i = hf − g, f α − g α i + hf − g, g α i ≥ 0,
∀α > 0.
Thus, (2.2) holds.
By Lemma 2.1,
1
α α+1
f α+1 (x) +
g (x) ≥ g α (x)f (x),
α+1
α+1
∀x ∈ [a, b].
Thus,
(2.9)
1
α+1
Z
b
f
α+1
a
α
(x)dx +
α+1
Z
b
g
α+1
Z
(x)dx ≥
a
g α (x)f (x)dx,
∀α > 0.
a
From (2.1) and (2.9) one obtains
Z b
Z b
α+1
f
(x)dx ≥
g α (x)f (x)dx,
a
b
∀α ≥ 0.
a
The proof is complete.
In particular, one has the following result
Corollary 2.3. Suppose f ∈ L1 [a, b], g ∈ C 1 [a, b] f, g ≥ 0, g is nondecreasing. If
Z b
Z b
f (t)dt ≥
g(t)dt, ∀x ∈ [a, b]
x
x
then the following inequality holds
Z b
Z b
β
(2.10)
f (x)dx ≥
g β (x)dx,
a
∀β ≥ 1.
a
Proof. Denote f := f + , g := g + where > 0. It is clear that g > 0 and
Z b
Z b
f (t)dt ≥
g (t)dt, ∀x ∈ [a, b].
x
x
By (2.1) and (2.3) in Lemma 2.2 one has
Z b
Z b
β
(2.11)
f (x)dx ≥
gβ (x)dx,
a
∀β ≥ 1.
a
Inequality (2.10) is obtained from (2.11) by letting → 0.
Theorem 2.4. Suppose f ∈ L1 [a, b], g ∈ C 1 [a, b], f, g ≥ 0, g is nondecreasing. If
Z b
Z b
f (t)dt ≥
g(t)dt, ∀x ∈ [a, b],
x
x
then ∀α, β ≥ 0, α + β ≥ 1, the following inequality holds
Z b
Z b
α+β
(2.12)
f
(x)dx ≥
f α (x)g β (x)dx.
a
J. Inequal. Pure and Appl. Math., 9(2) (2008), Art. 42, 5 pp.
a
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N. S. H OANG
Proof. Lemma 2.1 shows that
α
β
f (x)α+β +
g(x)α+β ≥ f α (x)g β (x), ∀x ∈ [a, b], ∀α, β > 0.
α+β
α+β
Therefore, ∀α, β > 0 one has
Z b
Z b
Z b
α
β
α+β
α+β
(2.13)
f (x) dx +
g(x) dx ≥
f α (x)g β (x)dx.
α+β a
α+β a
a
Corollary 2.3 implies
Z b
Z b
α+β
(2.14)
f (x) dx ≥
g(x)α+β dx, ∀α, β ≥ 0, α + β ≥ 1.
a
a
Inequality (2.12) is obtained from (2.13) and (2.14).
Remark 1. Theorem 2.4 is not true if we drop the assumption α + β ≥ 1. Indeed, take g ≡ 1,
[a, b] = [0, 1], and define
f (x) = c(1 − x)c−1 , 0 ≤ x ≤ 1,
where c ∈ (0, 1). One has
Z 1
Z 1
c
(1 − x) =
f (t)dt ≥
g(t)dt = (1 − x), ∀x ∈ [0, 1], c ∈ (0, 1),
x
but
x
√
Z 1p
Z 1p
2 c
=
f (t)dt <
g(t)dt = 1,
c+1
0
0
∀c ∈ (0, 1).
Assuming that the condition g ∈ C 1 [a, b] can be dropped and replaced by g ∈ L1 [a, b], we
have the following result:
Theorem 2.5. Suppose f, g ∈ L1 [a, b], f, g ≥ 0, g is nondecreasing. If
Z b
Z b
(2.15)
f (t)dt ≥
g(t)dt, ∀x ∈ [a, b],
x
x
then
Z
b
f
(2.16)
α+β
Z
(x)dx ≥
a
b
f α (x)g β (x)dx,
∀α, β ≥ 0, α + β ≥ 1.
a
1
Proof. Since C 1 [a, b] is dense in L1 , there exists a sequence (gn )∞
n=1 ∈ C [a, b] such that gn is
nondecreasing, gn % g a.e. Since gn % g a.e.,
Z b
Z b
(2.17)
g(t)dt ≥
gn (t)dt, ∀x ∈ [a, b], ∀n.
x
x
Inequalities (2.15), (2.17) and Theorem 2.4 imply
Z b
Z b
α+β
(2.18)
f
(x)dx ≥
f α (x)gnβ (x)dx,
a
∀n, ∀α, β ≥ 0, α + β ≥ 1.
a
Since f α gnβ % f α g β a.e., f α gnβ ≥ 0 is measurable satisfying (2.18), by the Monotone convergence theorem (see [3, 4]) kf α gnβ → f α g β kL1 → 0 as n → ∞. Hence,
Z b
Z b
α+β
f
(x)dx ≥
f α (x)g β (x)dx, ∀α, β ≥ 0, α + β ≥ 1.
a
a
The proof is complete.
J. Inequal. Pure and Appl. Math., 9(2) (2008), Art. 42, 5 pp.
http://jipam.vu.edu.au/
N OTES ON AN I NEQUALITY
5
Remark 2. One may wish to extend Theorem 2.5 to the case where [a, b] is unbounded. Note
that the case b = ∞ is not meaningful. It is because if g 6= 0 a.e., then both sides of (2.15) are
infinite. If b < ∞ and a = −∞ and inequality (2.15) holds for a = ∞, then it holds as well
for all finite a < 0. Hence, inequality (2.16) holds for all a < 0. Thus, by letting a → −∞ in
Theorem 2.5, one gets the result of Theorem 2.5 in the case a = −∞.
R EFERENCES
[1] W.J. LIU, C.C. LI AND J.W. DONG, On an open problem concerning an integral inequality, J.
Inequal. Pure & Appl. Math., 8(3) (2007), Art. 74. [ONLINE: http://jipam.vu.edu.au/
article.php?sid=882].
[2] Q.A. NGO, D.D. THANG, T.T. DAT AND D.A. TUAN, Notes on an integral inequality, J. Inequal. Pure & Appl. Math., 7(4) (2006), Art. 120. [ONLINE: http://jipam.vu.edu.au/
article.php?sid=737].
[3] M. REED AND B. SIMON, Methods of Modern Mathematicals Physics, Functional Analysis I,
Academic Press, Revised and enlarged edition, (1980).
[4] W. RUDIN, Real and Complex Analysis, McGraw-Hill Series in Higher Mathematics, Second edition, (1974).
J. Inequal. Pure and Appl. Math., 9(2) (2008), Art. 42, 5 pp.
http://jipam.vu.edu.au/