NOTES ON AN INEQUALITY
N. S. HOANG
Department of Mathematics
Kansas State University
Manhattan, KS 66506-2602, USA
EMail: nguyenhs@math.ksu.edu
URL: http://www.math.ksu.edu/∼nguyenhs
Notes on an Inequality
N. S. Hoang
vol. 9, iss. 2, art. 42, 2008
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Received:
13 November, 2007
Accepted:
26 April, 2008
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Communicated by:
S.S. Dragomir
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2000 AMS Sub. Class.:
26D15.
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Key words:
Integral inequality, Young’s inequality.
Abstract:
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), Art. 74.
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Acknowledgements:
The author wishes to express his thanks to Prof. A.G. Ramm for helpful comments during the preparation of the paper.
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Contents
1
Introduction
3
2
Results and Proofs
4
Notes on an Inequality
N. S. Hoang
vol. 9, iss. 2, art. 42, 2008
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1.
Introduction
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
x
Notes on an Inequality
then
1
Z
f α+1 (x)dx ≥
0
N. S. Hoang
1
Z
xα f (x)dx,
∀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
vol. 9, iss. 2, art. 42, 2008
α+β
Z
(x)dx ≥
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b
α β
x f (x)dx,
∀β > 0.
0
In this paper, we prove more general results, namely, Theorems 2.4 and 2.5 below.
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2.
Results and Proofs
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 .
Notes on an Inequality
Throughout the paper, [a, b] denotes a bounded interval and all functions are realvalued. 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
x ∈ [a, b], f ∈ L1 [a, b].
N. S. Hoang
vol. 9, iss. 2, art. 42, 2008
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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.
Notes on an Inequality
N. S. Hoang
vol. 9, iss. 2, art. 42, 2008
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,
∀x ∈ [a, b].
a
α
α
Z
hf − g, g (a)i = g (a)
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b
(f (x) − g(x))dx ≥ 0.
a
Since h ≥ 0, from (2.5) and (2.6) one gets
hf − g, g α i = hf − g, Ahi + hf − g, g α (a)i ≥ 0,
Hence, (2.1) is obtained.
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By the assumption,
(2.7)
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∀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),
(2.6)
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∀α ≥ 0.
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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.
Notes on an Inequality
Inequalities (2.7) and (2.8) imply
N. S. Hoang
α
α
α
α
hf − g, f i = hf − g, f − g i + hf − g, g i ≥ 0,
∀α > 0.
Thus, (2.2) holds.
By Lemma 2.1,
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α α+1
1
f α+1 (x) +
g (x) ≥ g α (x)f (x),
α+1
α+1
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∀x ∈ [a, b].
Thus,
(2.9)
1
α+1
Z
vol. 9, iss. 2, art. 42, 2008
b
f
α+1
a
α
(x)dx +
α+1
Z
a
b
g α+1 (x)dx
Z b
≥
g α (x)f (x)dx,
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∀α > 0.
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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
The proof is complete.
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∀α ≥ 0.
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
f (x)dx ≥
g β (x)dx,
(2.10)
a
Notes on an Inequality
∀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
N. S. Hoang
vol. 9, iss. 2, art. 42, 2008
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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
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
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Theorem 2.4. Suppose f ∈ L [a, b], g ∈ C [a, b], f, g ≥ 0, g is nondecreasing. If
Z b
Z b
f (t)dt ≥
g(t)dt, ∀x ∈ [a, b],
x
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Inequality (2.10) is obtained from (2.11) by letting → 0.
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Proof. Lemma 2.1 shows that
α
β
f (x)α+β +
g(x)α+β ≥ f α (x)g β (x),
α+β
α+β
∀x ∈ [a, b], ∀α, β > 0.
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,
a
Notes on an Inequality
N. S. Hoang
vol. 9, iss. 2, art. 42, 2008
∀α, β ≥ 0, α + β ≥ 1.
a
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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
c−1
f (x) = c(1 − x)
where c ∈ (0, 1). One has
Z 1
Z
c
(1 − x) =
f (t)dt ≥
x
but
,
0 ≤ x ≤ 1,
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g(t)dt = (1 − x),
∀x ∈ [0, 1], c ∈ (0, 1),
x
√
Z 1p
Z 1p
2 c
=
f (t)dt <
g(t)dt = 1,
c+1
0
0
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∀c ∈ (0, 1).
Assuming that the condition g ∈ C 1 [a, b] can be dropped and replaced by g ∈
L [a, b], we have the following result:
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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
b
Z
f
(2.16)
α+β
Z
b
f α (x)g β (x)dx,
(x)dx ≥
a
∀α, β ≥ 0, α + β ≥ 1.
Notes on an Inequality
a
1
N. S. Hoang
1
(gn )∞
n=1
Proof. Since C [a, b] is dense in L , there exists a sequence
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
1
∈ C [a, b] such
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
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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, ∀n, ∀α, β ≥ 0, α + β ≥ 1.
a
vol. 9, iss. 2, art. 42, 2008
a
The proof is complete.
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.,
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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 = −∞.
Notes on an Inequality
N. S. Hoang
vol. 9, iss. 2, art. 42, 2008
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References
[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).
Notes on an Inequality
N. S. Hoang
vol. 9, iss. 2, art. 42, 2008
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