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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:[email protected]

2000c Victoria University ISSN (electronic): 1443-5756 029-00

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Refinements of Carleman’s Inequality Bao-Quan Yuan

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J. Ineq. Pure and Appl. Math. 2(2) Art. 21, 2001

http://jipam.vu.edu.au

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.

1. Introduction

Let{a_n}^+∞_n=1be a non-negative sequence such that0 ≤P+∞

n=1a_n < +∞, then, we have

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+∞

X

n=1

(a₁a₂. . . a_n)^1/n ≤e

+∞

X

n=1

a_n.

The equality in (1) holds if and only ifa_n = 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 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.

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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

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1 + 1

m m

≤e

1−1−2/e m

holds, where the constant1− ²_e ≈0.2642411is best possible.

Proof. Inequality

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1 + 1

m m

≤e

1− β m

is equivalent toβ≤m−^m_e 1 + _m¹m

. Letf(x) = ¹_x −_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− ²_e is the optimal value satisfying inequality (3), so (2) holds. The proof of Lemma2.1follows.

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1 + 1

m m

≤ e

1 + _m¹_{ln 2}¹ −1

holds, where the constant _{ln 2}¹ −1≈0.442695is the best possible.

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Proof. Inequality

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1 + 1

m m

≤ e

1 + _m¹^α 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 2}¹ −1 is the optimal value satisfying inequality (5), and thus (4) holds. The proof of Lemma2.2follows.

Theorem 2.3. Let{a_n}^+∞_n=1be a non-negative sequence such that0≤P+∞

n=1a_n <

+∞. Then the following inequalities hold:

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+∞

X

n=1

(a₁a₂. . . a_n)^1/n ≤e

+∞

X

m=1

1−1−2/e m

a_m,

and

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+∞

X

n=1

(a₁a₂. . . a_n)^1/n≤e

+∞

X

m=1

a_m 1 + _m¹_{ln 2}¹ ⁻¹

.

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Proof. Letc_i >0 (i = 1,2, . . .). According to the arithmetic-geometric mean inequality, we have

(c₁a₁c₂a₂· · ·c_na_n)^1/n ≤ 1 n

n

X

m=1

c_ma_m. Consequently,

+∞

X

n=1

(a1a2· · ·an)^1/n =

+∞

X

n=1

c₁a₁c₂a₂· · ·c_na_n c₁c₂· · ·c_n

1/n

=

+∞

X

n=1

(c₁c₂· · ·c_n)^−1/n(c₁a₁c₂a₂· · ·c_na_n)^1/n

≤

+∞

X

n=1

(c₁c₂· · ·c_n)^−1/n1 n

n

X

m=1

c_ma_m

=

+∞

X

m=1

c_ma_m

+∞

X

n=m

1

n(c₁c₂· · ·c_n)^−1/n. Letc_m = ^(m+1)_mm−1^m (m= 1,2, . . .). Thenc₁c₂· · ·c_n= (n+ 1)ⁿ, and

+∞

X

n=m

1

n(c1c2· · ·cn)^−1/n =

+∞

X

n=m

1

n(n+ 1) = 1 m. Therefore

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+∞

X

n=1

(a1a2· · ·an)^1/n ≤

+∞

X

m=1

c_m 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

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1 + 1

m m

≤ e 1− _m^β 1 + _m¹^α,

holds, where0≤α≤ _{ln 2}¹ −1,0≤β ≤1− ²_e, andeβ+ 2^1+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)¹^x^+α, x ∈(0,1], 0≤α≤ 1 ln 2 −1,

then f is decreasing on interval(0,1]. Consequently,β = f(1) = 1− ¹_e2^1+α is the optimal value satisfying inequality (10). Moreover,0 ≤ β ≤ 1− ²_e, and eβ+ 2^1+α =e.So (9) holds, The proof is complete.

Remark 3.1. Ifα = 0, thenβ = 1− ²_e, and we obtain Lemma 2.1; ifβ = 0, thenα= _{ln 2}¹ −1, and we obtain Lemma2.2.

Similar to Theorem2.3, according to Lemma3.1, we have

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Theorem 3.2. Leta_n ≥0 (n = 1,2, . . .), 0≤P+∞

n=1a_n <+∞, then

+∞

X

n=1

(a₁a₂· · ·a_n)^1/n ≤e

+∞

X

m=1

1−_m^β 1 + _m¹^αa_m,

whereα,β satisfy0≤α≤ _{ln 2}¹ −1,0≤β ≤1− ²_e, andeβ+ 2^1+α =e.

Remark 3.2. Theorem 2.3 gives two special cases of Theorem 3.2. If α = 0, β = 1− ²_e, andα= _{ln 2}¹ −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.

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