volume 2, issue 2, article 21, 2001.
Received 18 August, 2000;
accepted 2 March, 2001.
Communicated by:J. Peˇcari´c
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Journal of Inequalities in Pure and Applied Mathematics
REFINEMENTS OF CARLEMAN’S INEQUALITY
BAO-QUAN YUAN
Department of Mathematics Jiaozuo Institute of Technology Jiaozuo City, Henan 454000
THE PEOPLE’S REPUBLIC OF CHINA EMail:baoquanyuan@chinaren.com
2000c Victoria University ISSN (electronic): 1443-5756 029-00
Refinements of Carleman’s Inequality Bao-Quan Yuan
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J. Ineq. Pure and Appl. Math. 2(2) Art. 21, 2001
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Abstract
In this paper, we obtain a class of refined Carleman’s Inequalities with the arithmetic-geometric mean inequality by decreasing their weight coefficient.
2000 Mathematics Subject Classification:26D15.
Key words: Carleman’s inequality, arithmetic-geometric mean inequality, weight co- efficient.
The author is indebted to the referee for many helpful and valuable comments and suggestions.
Contents
1 Introduction. . . 3 2 Two Special Cases. . . 4 3 A Class of Refined Carleman’s Inequalities. . . 8
References
Refinements of Carleman’s Inequality Bao-Quan Yuan
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1. Introduction
Let{an}+∞n=1be a non-negative sequence such that0 ≤P+∞
n=1an < +∞, then, we have
(1)
+∞
X
n=1
(a1a2. . . an)1/n ≤e
+∞
X
n=1
an.
The equality in (1) holds if and only ifan = 0, n = 1,2, . . .. the coefficient eis optimal.
Inequality (1) is called Carleman’s inequality. For details please refer to [1, 2]. The Carleman’s inequality has found many applications in mathemat- ics, 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 in- equalities by decreasing the weight coefficient with the arithmetic-geometric mean inequality.
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2. Two Special Cases
In this section, we give two special cases of refined Carleman’s inequality. First we prove two lemmas.
Lemma 2.1. Form= 1,2, . . . ,the inequality
(2)
1 + 1
m m
≤e
1−1−2/e m
holds, where the constant1− 2e ≈0.2642411is best possible.
Proof. Inequality
(3)
1 + 1
m m
≤e
1− β m
is equivalent toβ≤m−me 1 + m1m
. Letf(x) = 1x −ex1 (1 +x)x1,x∈(0,1].
It is obvious that the functionf is decreasing on the interval(0,1]. Conse- quently,β =f(1) = 1− 2e is the optimal value satisfying inequality (3), so (2) holds. The proof of Lemma2.1follows.
Lemma 2.2. Form= 1,2, . . . ,the inequality
(4)
1 + 1
m m
≤ e
1 + m1ln 21 −1
holds, where the constant ln 21 −1≈0.442695is the best possible.
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Proof. Inequality
(5)
1 + 1
m m
≤ e
1 + m1α is equivalent to
α≤ 1
ln 1 + m1 −m.
Let
f(x) = 1
ln(1 +x) − 1
x x∈(0,1].
Since the function f is decreasing on the interval(0,1], α = f(1) = ln 21 −1 is the optimal value satisfying inequality (5), and thus (4) holds. The proof of Lemma2.2follows.
Theorem 2.3. Let{an}+∞n=1be a non-negative sequence such that0≤P+∞
n=1an <
+∞. Then the following inequalities hold:
(6)
+∞
X
n=1
(a1a2. . . an)1/n ≤e
+∞
X
m=1
1−1−2/e m
am,
and
(7)
+∞
X
n=1
(a1a2. . . an)1/n≤e
+∞
X
m=1
am 1 + m1ln 21 −1
.
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Proof. Letci >0 (i = 1,2, . . .). According to the arithmetic-geometric mean inequality, we have
(c1a1c2a2· · ·cnan)1/n ≤ 1 n
n
X
m=1
cmam. Consequently,
+∞
X
n=1
(a1a2· · ·an)1/n =
+∞
X
n=1
c1a1c2a2· · ·cnan c1c2· · ·cn
1/n
=
+∞
X
n=1
(c1c2· · ·cn)−1/n(c1a1c2a2· · ·cnan)1/n
≤
+∞
X
n=1
(c1c2· · ·cn)−1/n1 n
n
X
m=1
cmam
=
+∞
X
m=1
cmam
+∞
X
n=m
1
n(c1c2· · ·cn)−1/n. Letcm = (m+1)mm−1m (m= 1,2, . . .). Thenc1c2· · ·cn= (n+ 1)n, and
+∞
X
n=m
1
n(c1c2· · ·cn)−1/n =
+∞
X
n=m
1
n(n+ 1) = 1 m. Therefore
(8)
+∞
X
n=1
(a1a2· · ·an)1/n ≤
+∞
X
m=1
cm mam =
+∞
X
m=1
1 + 1
m m
am.
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According to Lemmas2.1and2.2, and substituting for 1 + m1m
of inequality (8), so (6) and (7) follow from Lemmas2.1and2.2.
The proof is complete.
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3. A Class of Refined Carleman’s Inequalities
In this section we give a class of refined Carleman’s inequalities. First we have the following inequality
Lemma 3.1. Form= 1,2, . . . ,the inequality
(9)
1 + 1
m m
≤ e 1− mβ 1 + m1α,
holds, where0≤α≤ ln 21 −1,0≤β ≤1− 2e, andeβ+ 21+α =e.
Proof. Inequality (9) is equivalent to
(10) β ≤m− m
e
1 + 1 m
m+α
. If
f(x) = 1 x − 1
ex(1 +x)1x+α, x ∈(0,1], 0≤α≤ 1 ln 2 −1,
then f is decreasing on interval(0,1]. Consequently,β = f(1) = 1− 1e21+α is the optimal value satisfying inequality (10). Moreover,0 ≤ β ≤ 1− 2e, and eβ+ 21+α =e.So (9) holds, The proof is complete.
Remark 3.1. Ifα = 0, thenβ = 1− 2e, and we obtain Lemma 2.1; ifβ = 0, thenα= ln 21 −1, and we obtain Lemma2.2.
Similar to Theorem2.3, according to Lemma3.1, we have
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Theorem 3.2. Letan ≥0 (n = 1,2, . . .), 0≤P+∞
n=1an <+∞, then
+∞
X
n=1
(a1a2· · ·an)1/n ≤e
+∞
X
m=1
1−mβ 1 + m1αam,
whereα,β satisfy0≤α≤ ln 21 −1,0≤β ≤1− 2e, andeβ+ 21+α =e.
Remark 3.2. Theorem 2.3 gives two special cases of Theorem 3.2. If α = 0, β = 1− 2e, andα= ln 21 −1,β = 0, we can obtain (6) and (7) in Theorem2.3 respectively.
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References
[1] G.H. HARDY, J.E. LITTLEWOOD AND G. POLYA, Inequalities, Cam- bridge 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 inequal- ity, 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.