Journal of Inequalities in Pure and
Applied Mathematics
http://jipam.vu.edu.au/
Volume 2, Issue 2, Article 21, 2001
REFINEMENTS OF CARLEMAN’S INEQUALITY
BAO-QUAN YUAN
D EPARTMENT OF M ATHEMATICS ,
J IAOZUO I NSTITUTE OF T ECHNOLOGY,
J IAOZUO C ITY, H ENAN P ROVINCE 454000
P EOPLE ’ S R EPUBLIC OF C HINA
baoquanyuan@chinaren.com
Received 18 August, 2000; accepted 2 March, 2001.
Communicated by J. Pečarić
A BSTRACT. In this paper, we obtain a class of refined Carleman’s Inequalities with the arithmeticgeometric mean inequality by decreasing their weight coefficient.
Key words and phrases: Carleman’s inequality, arithmetic-geometric mean inequality, weight coefficient.
2000 Mathematics Subject Classification. 26D15.
1. I NTRODUCTION
P+∞
Let {an }+∞
be
a
non-negative
sequence
such
that
0
≤
n=1
n=1 an < +∞, then, we have
(1.1)
+∞
X
n=1
1/n
(a1 a2 . . . an )
≤e
+∞
X
an .
n=1
The equality in (1.1) holds if and only if an = 0, n = 1, 2, . . . . the coefficient e is optimal.
Inequality (1.1) is called Carleman’s inequality. For details please refer to [1, 2]. The Carleman’s inequality has found many applications in mathematics, and the study of the Carleman’s
inequality has a rich literature, for details, please refer to [3, 4]. Though the coefficient e is optimal, we can refine its weight coefficient. In this article we give a class of improved Carleman’s
inequalities by decreasing the weight coefficient with the arithmetic-geometric mean inequality.
2. T WO S PECIAL C ASES
In this section, we give two special cases of refined Carleman’s inequality. First we prove
two lemmas.
ISSN (electronic): 1443-5756
c 2001 Victoria University. All rights reserved.
The author is indebted to the referee for many helpful and valuable comments and suggestions.
029-00
2
BAO -Q UAN Y UAN
Lemma 2.1. For m = 1, 2, . . . , the inequality
m
1
1 − 2/e
(2.1)
1+
≤e 1−
m
m
holds, where the constant 1 −
2
e
≈ 0.2642411 is best possible.
Proof. Inequality
m
1
1+
m
m
1 + m1 .
(2.2)
is equivalent to β ≤ m −
m
e
β
≤e 1−
m
1
1
Let f (x) = x1 − ex
(1 + x) x , x ∈ (0, 1].
It is obvious that the function f is decreasing on the interval (0, 1]. Consequently, β = f (1) =
1 − 2e is the optimal value satisfying inequality (2.2), so (2.1) holds. The proof of Lemma 2.1
follows.
Lemma 2.2. For m = 1, 2, . . . , the inequality
m
1
e
(2.3)
1+
≤
1 −1
m
1 + m1 ln 2
holds, where the constant
1
ln 2
− 1 ≈ 0.442695 is the best possible.
Proof. Inequality
(2.4)
1
1+
m
m
e
≤
1+
1
m
α
is equivalent to
α≤
1
ln 1 +
1
m
− m.
Let
1
1
−
x ∈ (0, 1] .
ln(1 + x) x
Since the function f is decreasing on the interval (0, 1], α = f (1) = ln12 − 1 is the optimal value
satisfying inequality (2.4), and thus (2.3) holds. The proof of Lemma 2.2 follows.
P
+∞
Theorem 2.3. Let {an }+∞
n=1 be a non-negative sequence such that 0 ≤
n=1 an < +∞. Then
the following inequalities hold:
+∞
+∞ X
X
1 − 2/e
1/n
(2.5)
(a1 a2 . . . an ) ≤ e
1−
am ,
m
n=1
m=1
f (x) =
and
(2.6)
+∞
X
1/n
(a1 a2 . . . an )
n=1
≤e
+∞
X
m=1
1+
am
1 .
1 ln 2 −1
m
Proof. Let ci > 0 (i = 1, 2, . . . ). According to the arithmetic-geometric mean inequality, we
have
n
1X
1/n
(c1 a1 c2 a2 · · · cn an ) ≤
c m am .
n m=1
J. Inequal. Pure and Appl. Math., 2(2) Art. 21, 2001
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R EFINEMENTS OF C ARLEMAN ’ S I NEQUALITY
3
Consequently,
+∞
X
1/n
(a1 a2 · · · an )
=
n=1
=
≤
1/n
+∞ X
c 1 a1 c 2 a2 · · · c n an
n=1
+∞
X
n=1
+∞
X
c1 c2 · · · cn
(c1 c2 · · · cn )−1/n (c1 a1 c2 a2 · · · cn an )1/n
−1/n 1
(c1 c2 · · · cn )
n
X
n m=1
n=1
+∞
X
c m am
+∞
X
1
=
c m am
(c1 c2 · · · cn )−1/n .
n
m=1
n=m
Let cm =
(m+1)m
mm−1
(m = 1, 2, . . . ). Then c1 c2 · · · cn = (n + 1)n , and
+∞
+∞
X
X
1
1
1
(c1 c2 · · · cn )−1/n =
= .
n
n(n + 1)
m
n=m
n=m
Therefore
+∞
X
m
+∞
+∞ X
X
cm
1
(2.7)
(a1 a2 · · · an ) ≤
am =
1+
am .
m
m
n=1
m=1
m=1
m
According to Lemmas 2.1 and 2.2, and substituting for 1 + m1
of inequality (2.7), so (2.5)
and (2.6) follow from Lemmas 2.1 and 2.2.
The proof is complete.
1/n
3. A C LASS OF R EFINED C ARLEMAN ’ S I NEQUALITIES
In this section we give a class of refined Carleman’s inequalities. First we have the following
inequality
Lemma 3.1. For m = 1, 2, . . . , the inequality
m
β
e 1− m
1
(3.1)
1+
≤
α ,
m
1+ 1
m
holds, where 0 ≤ α ≤
1
ln 2
− 1, 0 ≤ β ≤ 1 −
2
,
e
and eβ + 21+α = e.
Proof. Inequality (3.1) is equivalent to
m
β ≤m−
e
(3.2)
1
1+
m
m+α
.
If
1
1
1
1
−
(1 + x) x +α , x ∈ (0, 1], 0 ≤ α ≤
− 1,
x ex
ln 2
then f is decreasing on interval (0, 1]. Consequently, β = f (1) = 1 − 1e 21+α is the optimal
value satisfying inequality (3.2). Moreover, 0 ≤ β ≤ 1 − 2e , and eβ + 21+α = e. So (3.1) holds,
The proof is complete.
f (x) =
Remark 3.2. If α = 0, then β = 1 − 2e , and we obtain Lemma 2.1; if β = 0, then α =
and we obtain Lemma 2.2.
Similar to Theorem 2.3, according to Lemma 3.1, we have
J. Inequal. Pure and Appl. Math., 2(2) Art. 21, 2001
1
ln 2
− 1,
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4
BAO -Q UAN Y UAN
Theorem 3.3. Let an ≥ 0 (n = 1, 2, . . . ), 0 ≤
+∞
X
(a1 a2 · · · an )1/n
n=1
P+∞
an < +∞, then
β
1− m
≤e
α am ,
1
m=1 1 + m
n=1
+∞
X
where α, β satisfy 0 ≤ α ≤ ln12 − 1, 0 ≤ β ≤ 1 − 2e , and eβ + 21+α = e.
Remark 3.4. Theorem 2.3 gives two special cases of Theorem 3.3. If α = 0, β = 1 − 2e , and
α = ln12 − 1, β = 0, we can obtain (2.5) and (2.6) in Theorem 2.3 respectively.
R EFERENCES
[1] G.H. HARDY, J.E. LITTLEWOOD AND G. POLYA, Inequalities, Cambridge Univ. Press, London,
1952.
[2] JI-CHANG KUANG, Applied Inequalities, Hunan Education Press (second edition), Changsha,
China, 1993.(Chinese)
[3] PING YAN AND GUOZHENG SUN, A strengthened Carleman’s inequality, J. Math. Anal. Appl.,
240 (1999), 290–293.
[4] BICHENG YANG AND L. DEBNATH, Some inequalities involving the constant e, and an application to Carleman’s inequality, J. Math. Anal. Appl., 223 (1998), 347–353.
J. Inequal. Pure and Appl. Math., 2(2) Art. 21, 2001
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