Journal of Inequalities in Pure and Applied Mathematics

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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 DEPARTMENT OFMATHEMATICS, JIAOZUOINSTITUTE OFTECHNOLOGY, JIAOZUOCITY, HENANPROVINCE454000

PEOPLE’SREPUBLIC OFCHINA

[email protected]

Received 18 August, 2000; accepted 2 March, 2001.

Communicated by J. Peˇcari´c

ABSTRACT. In this paper, we obtain a class of refined Carleman’s Inequalities with the arithmetic- geometric 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. INTRODUCTION

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

n=1a_n <+∞, then, we have (1.1)

+∞

X

n=1

(a1a2. . . an)^1/n ≤e

+∞

X

n=1

an.

The equality in (1.1) holds if and only ifan= 0, n= 1,2, . . .. the coefficienteis optimal.

Inequality (1.1) is called Carleman’s inequality. For details please refer to [1, 2]. The Carle- man’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 coefficienteis 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. TWO SPECIALCASES

In this section, we give two special cases of refined Carleman’s inequality. First we prove two lemmas.

ISSN (electronic): 1443-5756

The author is indebted to the referee for many helpful and valuable comments and suggestions.

029-00

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2 BAO-QUANYUAN

Lemma 2.1. Form= 1,2, . . . ,the inequality (2.1)

1 + 1

m m

≤e

1− 1−2/e m

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

Proof. Inequality (2.2)

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 functionfis decreasing on the interval(0,1]. Consequently,β =f(1) = 1− ²_e is the optimal value satisfying inequality (2.2), so (2.1) holds. The proof of Lemma 2.1

follows.

1 + 1

m m

≤ e

1 + _m¹_{ln 2}¹ ⁻¹ holds, where the constant _{ln 2}¹ −1≈0.442695is the best possible.

Proof. Inequality (2.4)

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 functionf is decreasing on the interval(0,1],α =f(1) = _{ln 2}¹ −1is the optimal value satisfying inequality (2.4), and thus (2.3) holds. The proof of Lemma 2.2 follows.

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

n=1a_n <+∞. Then the following inequalities hold:

(2.5)

+∞

X

n=1

(a1a2. . . an)^1/n≤e

+∞

X

m=1

1− 1−2/e m

am, and

(2.6)

+∞

X

n=1

(a₁a₂. . . a_n)^1/n ≤e

+∞

X

m=1

a_m 1 + _m¹_{ln 2}¹ −1.

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.

J. Inequal. Pure and Appl. Math., 2(2) Art. 21, 2001 http://jipam.vu.edu.au/

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REFINEMENTS OFCARLEMAN’SINEQUALITY 3

Consequently,

+∞

X

n=1

(a₁a₂· · ·a_n)^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

cmam +∞

X

n=m

1

n(c1c2· · ·cn)^−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

(2.7)

+∞

X

n=1

(a₁a₂· · ·a_n)^1/n ≤

+∞

X

m=1

c_m ma_m =

+∞

X

m=1

1 + 1

m m

a_m. According to Lemmas 2.1 and 2.2, and substituting for 1 + _m¹m

of inequality (2.7), so (2.5) and (2.6) follow from Lemmas 2.1 and 2.2.

The proof is complete.

3. A CLASS OFREFINEDCARLEMAN’SINEQUALITIES

In this section we give a class of refined Carleman’s inequalities. First we have the following inequality

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 (3.1) is equivalent to

(3.2) 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 (3.2). Moreover,0≤β ≤1−²_e, andeβ+ 2^1+α =e.So (3.1) holds,

The proof is complete.

Remark 3.2. Ifα= 0, thenβ = 1−²_e, and we obtain Lemma 2.1; ifβ= 0, thenα= _{ln 2}¹ −1, and we obtain Lemma 2.2.

Similar to Theorem 2.3, according to Lemma 3.1, we have

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4 BAO-QUANYUAN

Theorem 3.3. 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,βsatisfy0≤α ≤ _{ln 2}¹ −1,0≤β ≤1− ²_e, andeβ+ 2^1+α =e.

Remark 3.4. Theorem 2.3 gives two special cases of Theorem 3.3. Ifα = 0, β = 1− ²_e, and α= _{ln 2}¹ −1,β = 0, we can obtain (2.5) and (2.6) in Theorem 2.3 respectively.

REFERENCES

[1] G.H. HARDY, J.E. LITTLEWOODANDG. POLYA, Inequalities, Cambridge Univ. Press, London, 1952.

[2] JI-CHANG KUANG, Applied Inequalities, Hunan Education Press (second edition), Changsha, China, 1993.(Chinese)

[3] PING YANANDGUOZHENG SUN, A strengthened Carleman’s inequality, J. Math. Anal. Appl., 240 (1999), 290–293.

[4] BICHENG YANGANDL. DEBNATH, Some inequalities involving the constante,and an applica- tion to Carleman’s inequality, J. Math. Anal. Appl., 223 (1998), 347–353.