Journal of Inequalities and Applications
Volume 2009 (2009), Article ID 494257, 18 pages
doi:10.1155/2009/494257
Research Article

A Hilbert-Type Linear Operator with the Norm and Its Applications

Wuyi Zhong

Department of Mathematics, Guangdong Institute of Education, Guangzhou, Guangdong 510303, China

Received 9 February 2009; Accepted 9 March 2009

Academic Editor: Nikolaos Papageorgiou

Copyright © 2009 Wuyi Zhong. This is an open access article distributed under the Creative Commons Attribution License, which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited.

Abstract

A Hilbert-type linear operator 𝑇 ∶ ℓ 𝑝 𝜙 → ℓ 𝑝 𝜓 is defined. As for applications, a more precise operator inequality with the norm and its equivalent forms are deduced. Moreover, three equivalent reverses from them are given as well. The constant factors in these inequalities are proved to be the best possible.

1. Introduction

In 1925, Hardy [1] extended Hilbert inequality as follows.

If 𝑝 > 1 , 1 / 𝑝 + 1 / 𝑞 = 1 , 𝑎 𝑛 , 𝑏 𝑛 ∑ ≥ 0 , 0 < ∞ 𝑛 = 1 𝑎 𝑝 𝑛 < ∞ , and ∑ 0 < ∞ 𝑛 = 1 𝑏 𝑞 𝑛 < ∞ , then ∞ ∞ 𝑛 = 1 𝑚 = 1 𝑎 𝑚 𝑏 𝑛 < 𝜋 𝑚 + 𝑛 s i n ( 𝜋 / 𝑝 ) ∞ 𝑛 = 1 𝑎 𝑝 𝑛 1 / 𝑝 ∞ 𝑛 = 1 𝑏 𝑞 𝑛 1 / 𝑞 , ( 1 . 1 ) ∞ 𝑛 = 1 ∞ 𝑚 = 1 𝑎 𝑚 𝑚 + 𝑛 𝑝 < 𝜋 s i n ( 𝜋 / 𝑝 ) 𝑝 ∞ 𝑛 = 1 𝑎 𝑝 𝑛 , ( 1 . 2 ) where ( 𝑝 , 𝑞 ) is a pair of conjugate exponents. The constant factors 𝜋 / s i n ( 𝜋 / 𝑝 ) and [ 𝜋 / s i n ( 𝜋 / 𝑝 ) ] 𝑝 are the best possible. The expression (1.1) is the famous Hardy-Hilbert's inequality.

Under the same conditions, there are the classic inequalities [2]: ∞ ∞ 𝑛 = 1 𝑚 = 1 l n ( 𝑚 / 𝑛 ) 𝑎 𝑚 𝑏 𝑛 < 𝜋 𝑚 − 𝑛 s i n ( 𝜋 / 𝑝 ) 2 ∞ 𝑛 = 1 𝑎 𝑝 𝑛 1 / 𝑝 ∞ 𝑛 = 1 𝑏 𝑞 𝑛 1 / 𝑞 , ( 1 . 3 ) ∞ 𝑛 = 1 ∞ 𝑚 = 1 l n ( 𝑚 / 𝑛 ) 𝑎 𝑚 𝑚 − 𝑛 𝑝 < 𝜋 s i n ( 𝜋 / 𝑝 ) ∞ 2 𝑝 𝑛 = 1 𝑎 𝑝 𝑛 , ( 1 . 4 ) where the constant factors [ 𝜋 / s i n ( 𝜋 / 𝑝 ) ] 2 and [ 𝜋 / s i n ( 𝜋 / 𝑝 ) ] 2 𝑝 are also the best possible. The expression (1.3) is well known as a Hilbert-type inequality.

By setting a real space of sequences: ℓ 𝑝 ∶ = { 𝑎 ; 𝑎 = { 𝑎 𝑛 } ∞ 𝑛 = 0 , ‖ 𝑎 ‖ 𝑝 ∑ = { ∞ 𝑛 = 1 | 𝑎 𝑛 | 𝑝 } 1 / 𝑝 < ∞ } and defining a linear operator 𝑇 ∶ ℓ 𝑝 → ℓ 𝑝 , ( 𝑇 𝑎 ) ( 𝑛 ) = 𝐶 𝑛 = ∑ ∞ 𝑛 = 1 ( l n ( 𝑚 / 𝑛 ) 𝑎 𝑚 / ( 𝑚 − 𝑛 ) )   ( 𝑛 ∈ 𝑁 0 ) , the expressions (1.3) and (1.4) can be rewritten as ( 𝑇 𝑎 , 𝑏 ) < ‖ 𝑇 ‖ ‖ 𝑎 ‖ 𝑝 ‖ 𝑏 ‖ 𝑞 , ( 1 . 5 ) ‖ 𝑇 𝑎 ‖ 𝑝 < ‖ 𝑇 ‖ ‖ 𝑎 ‖ 𝑝 , ( 1 . 6 ) respectively, where ‖ 𝑇 ‖ = [ 𝜋 / s i n ( 𝜋 / 𝑝 ) ] 2 , 𝑏 ∈ ℓ 𝑞 . ( 𝑇 𝑎 , 𝑏 ) is the formal inner product of 𝑇 𝑎 and 𝑏 .

The inequalities (1.1)–(1.4) play important roles in theoretical analysis and applications [3]. These inequalities and their integral forms have been recently extended or strengthened in [4–8]. Zhao and Debnath [9] obtained a Hilbert-Pachpatte's reverse inequality. Zhong and Yang [10, 11] have given some reverses concerning some extensions of (1.1). Papers in [12–15] studied some multiple Hardy-Hilbert-type or Hilbert-type inequalities. Articles in [16, 17] got some Hilbert-type linear operator inequalities. In 2006, Yang [18] deduced a new Hilbert-type inequality as follows.

Set ( 𝑝 , 𝑞 ) as a pair of conjugate exponents, and 𝑝 > 1 , 1 / 2 ≤ 𝛼 ≤ 1 , 𝑎 𝑛 , 𝑏 𝑛 ≥ 0 , such that ∑ 0 < ∞ 𝑛 = 1 𝑎 𝑝 𝑛 < ∞ , ∑ 0 < ∞ 𝑛 = 1 𝑏 𝑞 𝑛 < ∞ , then one has ∞ ∞ 𝑛 = 0 𝑚 = 0 l n ( ( 𝑚 + 𝛼 ) / ( 𝑛 + 𝛼 ) ) 𝑎 𝑚 𝑏 𝑛 < 𝜋 𝑚 − 𝑛 s i n ( 𝜋 / 𝑝 ) 2 ∞ 𝑛 = 0 𝑎 𝑝 𝑛 1 / 𝑝 ∞ 𝑛 = 0 𝑏 𝑞 𝑛 1 / 𝑞 , ( 1 . 7 ) ∞ 𝑛 = 0 ∞ 𝑚 = 0 l n ( ( 𝑚 + 𝛼 ) / ( 𝑛 + 𝛼 ) ) 𝑎 𝑚 𝑚 − 𝑛 𝑝 < 𝜋 s i n ( 𝜋 / 𝑝 ) ∞ 2 𝑝 𝑛 = 0 𝑎 𝑝 𝑛 . ( 1 . 8 ) It has been proved that (1.7) and (1.8) are two equivalent inequalities and their constant factors [ 𝜋 / s i n ( 𝜋 / 𝑝 ) ] 2 and [ 𝜋 / s i n ( 𝜋 / 𝑝 ) ] 2 𝑝 are the best possible. When 𝛼 = 1 , the expressions (1.7) and (1.8) can be reduced to (1.3) and (1.4), respectively.

This paper reports the studies on a Hilbert-type linear operator 𝑇 ∶ ℓ 𝑝 𝜙 → ℓ 𝑝 𝜓 . As for the applications, a more precise linear operator's general form of Hilbert-type inequality (1.3) incorporating the norm and its equivalent form are deduced. Moreover, three equivalent reverses of the new general forms are deduced as well. The constant factors in these inequalities are all the best possible.

At first, two known results are introduced.

(1) If 𝑠 > 1 , ( 𝑟 , 𝑠 ) is a pair of conjugate exponents, then the Beta function is defined as follows (cf. [2, Theorem 342]), ∞ 0 l n 𝑢 𝑢 𝑢 − 1 1 / 𝑠 − 1 𝜋 𝑑 𝑢 = s i n ( 𝜋 / 𝑠 ) 2 = 𝐵 1 𝑠 , 1 𝑟 2 = 𝐵 1 𝑟 , 1 𝑠 2 . ( 1 . 9 )

(2) (Euler-Maclaurin's summation formula). Set 𝑓 ∈ 𝐶 3 [ 0 , ∞ ) , if ( − 1 ) 𝑖 𝑓 ( 𝑖 ) ( 𝑥 ) > 0 , 𝑓 ( 𝑖 ) ( ∞ ) = 0 ( 𝑖 = 0 , 1 , 2 , 3 ) , then (cf. [19, Lemma 1]) ∞ 𝑛 = 0 𝑓 ( 𝑛 ) < ∞ 0 1 𝑓 ( 𝑥 ) 𝑑 𝑥 + 2 1 𝑓 ( 0 ) − 𝑓 1 2 ( 0 ) , ( 1 . 1 0 ) ∞ 𝑛 = 0 𝑓 ( 𝑛 ) > ∞ 0 𝑓 1 ( 𝑥 ) 𝑑 𝑥 + 2 𝑓 ( 0 ) . ( 1 . 1 1 )

2. Lemmas

Lemma 2.1. Set ( 𝑟 , 𝑠 ) as a pair of conjugate exponents, 𝑠 > 1 , 𝛼 > 0 , 0 < 𝜆 ≤ 1 , and define 𝑔 ( 𝑢 ) ∶ = l n 𝑢 𝑓 𝑢 − 1 , 𝑢 ∈ ( 0 , 1 ) ∪ ( 1 , ∞ ) , 1 , 𝑢 = 1 , ( 2 . 1 ) 𝑠 ( 𝑥 ) ∶ = ℎ 𝑚 , 𝜆 1 𝑥 , 𝑠 ∶ = 𝑔 𝑥 + 𝛼 𝑚 + 𝛼 𝜆 𝑥 + 𝛼 𝑚 + 𝛼 𝜆 1 / 𝑠 − 1 / 𝜆 , 𝑥 ∈ ( − 𝛼 , ∞ ) , 𝑚 ∈ 𝑁 0 . ( 2 . 2 ) Then, one has the following: (1)the function 𝑓 𝑠 ( 𝑥 ) satisfies the conditions of (1.10) and (1.11). This means ( − 1 ) 𝑖 𝑓 𝑠 ( 𝑖 ) ( 𝑥 ) > 0 ( 𝑥 > − 𝛼 ) , 𝑓 𝑠 ( 𝑖 ) ( ∞ ) = 0 ( 𝑖 = 0 , 1 , 2 , 3 ) , ( 2 . 3 ) (2) 𝑘 𝜆 1 ( 𝑠 ) ∶ = 𝜆 ( 𝑚 + 𝛼 ) ∞ − 𝛼 𝑓 𝑠 ( 𝑥 ) 𝑑 𝑥 = 𝐵 ( 1 / 𝑠 , 1 / 𝑟 ) 𝜆 2 = 𝜋 𝜆 s i n ( 𝜋 / 𝑠 ) 2 . ( 2 . 4 )

Proof. (1) For 𝛼 > 0 , 𝑥 > − 𝛼 , 𝑚 ∈ 𝑁 0 , 0 < 𝜆 ≤ 1 and 𝑠 > 1 , set 𝑧 ( 𝑥 ) = 𝑔 ( ( ( 𝑥 + 𝛼 ) / ( 𝑚 + 𝛼 ) ) 𝜆 ) , 𝑡 ( 𝑥 ) = [ ( ( 𝑥 + 𝛼 ) / ( 𝑚 + 𝛼 ) ) 𝜆 ] 1 / 𝑠 − 1 / 𝜆 = ( ( 𝑥 + 𝛼 ) / ( 𝑚 + 𝛼 ) ) 𝜆 / 𝑠 − 1 and 𝑢 = ( ( 𝑥 + 𝛼 ) / ( 𝑚 + 𝛼 ) ) 𝜆 . These show that 𝑧 ( 𝑥 ) = 𝑔 ( 𝑢 ) and 𝑓 𝑠 ( 𝑥 ) = 𝑧 ( 𝑥 ) 𝑡 ( 𝑥 ) = 𝑔 ( 𝑢 ) 𝑡 ( 𝑥 ) when 𝑢 > 0 . With the settings, ( − 1 ) 𝑖 𝑔 ( 𝑖 ) ( 𝑢 ) > 0 , 𝑔 ( 𝑖 ) ( ∞ ) = 0 ( 𝑢 > 0 , 𝑖 = 0 , 1 , 2 , 3 ) (cf. [16, Lemma 2.2]), one has 𝑧 ( 𝑥 ) > 0 , 𝑡 ( 𝑥 ) > 0 , 𝑧 ( 𝑥 ) = 𝑔 𝜆 ( 𝑢 ) 𝑚 + 𝛼 𝑥 + 𝛼 𝑚 + 𝛼 𝜆 − 1 𝑧 < 0 , ( 𝑥 ) = 𝑔 𝜆 ( 𝑢 ) 𝑚 + 𝛼 𝑥 + 𝛼 𝑚 + 𝛼 𝜆 − 1 2 + 𝑔 ( 𝑢 ) 𝜆 ( 𝜆 − 1 ) ( 𝑚 + 𝛼 ) 2 𝑥 + 𝛼 𝑚 + 𝛼 𝜆 − 2 𝑧 > 0 , ( 𝑥 ) = 𝑔 𝜆 ( 𝑢 ) 𝑚 + 𝛼 𝑥 + 𝛼 𝑚 + 𝛼 𝜆 − 1 3 + 3 𝑔 𝜆 ( 𝑢 ) 2 ( 𝜆 − 1 ) ( 𝑚 + 𝛼 ) 3 𝑥 + 𝛼 𝑚 + 𝛼 2 𝜆 − 3 + 𝑔 ( 𝑢 ) 𝜆 ( 𝜆 − 1 ) ( 𝜆 − 2 ) ( 𝑚 + 𝛼 ) 3 𝑥 + 𝛼 𝑚 + 𝛼 𝜆 − 3 𝑡 < 0 , ( 𝑥 ) = 𝜆 / 𝑠 − 1 𝑚 + 𝛼 𝑥 + 𝛼 𝑚 + 𝛼 𝜆 / 𝑠 − 2 𝑡 < 0 , ( 𝑥 ) = ( 𝜆 / 𝑠 − 1 ) ( 𝜆 / 𝑠 − 2 ) ( 𝑚 + 𝛼 ) 2 𝑥 + 𝛼 𝑚 + 𝛼 𝜆 / 𝑠 − 3 𝑡 > 0 , ( 𝑥 ) = ( 𝜆 / 𝑠 − 1 ) ( 𝜆 / 𝑠 − 2 ) ( 𝜆 / 𝑠 − 3 ) ( 𝑚 + 𝛼 ) 3 𝑥 + 𝛼 𝑚 + 𝛼 𝜆 / 𝑠 − 4 < 0 . ( 2 . 5 ) These are followed by 𝑓 𝑠 ( 𝑥 ) = 𝑧 ( 𝑥 ) 𝑡 ( 𝑥 ) > 0 , 𝑓 𝑠 𝑓 ( ∞ ) = 0 , 𝑠 ( 𝑥 ) = 𝑧 ( 𝑥 ) 𝑡 ( 𝑥 ) + 𝑧 ( 𝑥 ) 𝑡 ( 𝑥 ) < 0 , 𝑓 𝑠 𝑓 ( ∞ ) = 0 , ( 2 . 6 ) 𝑠 ( 𝑥 ) = 𝑧 ( 𝑥 ) 𝑡 ( 𝑥 ) + 2 𝑧 ( 𝑥 ) 𝑡 ( 𝑥 ) + 𝑧 ( 𝑥 ) 𝑡 ( 𝑥 ) > 0 , 𝑓 𝑠 𝑓 ( ∞ ) = 0 , 𝑠 ( 𝑥 ) = 𝑧 ( 𝑥 ) 𝑡 ( 𝑥 ) + 3 𝑧 ′ ′ ( 𝑥 ) 𝑡 ( 𝑥 ) + 3 𝑧 ( 𝑥 ) 𝑡 ( 𝑥 ) + 𝑧 ( 𝑥 ) 𝑡 ( 𝑥 ) < 0 , 𝑓 𝑠 ( ∞ ) = 0 . ( 2 . 7 ) Then inequality (2.3) holds.
(2) For 𝑥 > − 𝛼 , 𝑚 ∈ 𝑁 0 and 𝜆 > 0 , 𝑠 > 1 , set 𝑢 = ( ( 𝑥 + 𝛼 ) / ( 𝑚 + 𝛼 ) ) 𝜆 , then one has 1 𝜆 ( 𝑚 + 𝛼 ) ∞ − 𝛼 𝑓 𝑠 1 ( 𝑥 ) 𝑑 𝑥 = 𝜆 ∞ − 𝛼 l n ( ( 𝑥 + 𝛼 ) / ( 𝑚 + 𝛼 ) ) 𝜆 ( ( 𝑥 + 𝛼 ) / ( 𝑚 + 𝛼 ) ) 𝜆 − 1 𝑥 + 𝛼 𝑚 + 𝛼 𝜆 1 / 𝑠 − 1 / 𝜆 𝑑 𝑥 + 𝛼 = 1 𝑚 + 𝛼 𝜆 2 ∞ 0 l n 𝑢 𝑢 𝑢 − 1 1 / 𝑠 − 1 𝑑 𝑢 . ( 2 . 8 ) By (1.9), then (2.4) holds. Lemma 2.1 is proved.

Lemma 2.2. Set ( 𝑟 , 𝑠 ) as a pair of conjugate exponents, 𝑠 > 1 , 𝛼 ≥ 1 / 2 ,  0 < 𝜆 ≤ 1 and define 𝜔 𝜆 ( 𝑚 , 𝑠 ) ∶ = ∞ 𝑛 = 0 l n ( ( 𝑛 + 𝛼 ) / ( 𝑚 + 𝛼 ) ) ( 𝑛 + 𝛼 ) 𝜆 − ( 𝑚 + 𝛼 ) 𝜆 ⋅ ( 𝑚 + 𝛼 ) 𝜆 / 𝑟 ( 𝑛 + 𝛼 ) 1 − 𝜆 / 𝑠 𝑚 ∈ 𝑁 0 , ( 2 . 9 ) then, one has 0 < 𝜔 𝜆 ( 𝑚 , 𝑠 ) < 𝑘 𝜆 ( 𝑠 ) , ( 2 . 1 0 ) 0 < 𝜔 𝜆 ( 𝑛 , 𝑟 ) < 𝑘 𝜆 ( 𝑟 ) = 𝑘 𝜆 ( 𝑠 ) 𝑛 ∈ 𝑁 0 , ( 2 . 1 1 ) where 𝑘 𝜆 ( 𝑠 ) is defined by (2.4).

Proof. By (2.9) and (2.2), it is evident that 0 < 𝜔 𝜆 ( 1 𝑚 , 𝑠 ) = 𝜆 ( 𝑚 + 𝛼 ) ∞ 𝑛 = 0 l n ( ( 𝑛 + 𝛼 ) / ( 𝑚 + 𝛼 ) ) 𝜆 ( ( 𝑛 + 𝛼 ) / ( 𝑚 + 𝛼 ) ) 𝜆 − 1 𝑛 + 𝛼 𝑚 + 𝛼 𝜆 1 / 𝑠 − 1 / 𝜆 = 1 𝜆 ( 𝑚 + 𝛼 ) ∞ 𝑛 = 0 ℎ 𝑚 , 𝜆 1 𝑛 , 𝑠 = 1 𝜆 ( 𝑚 + 𝛼 ) ∞ 𝑛 = 0 𝑓 𝑠 ( 𝑛 ) . ( 2 . 1 2 ) In view of (2.3), (1.10), and (2.4), one has 𝜔 𝜆 1 ( 𝑚 , 𝑠 ) < 𝜆 ( 𝑚 + 𝛼 ) ∞ 0 𝑓 𝑠 1 ( 𝑥 ) 𝑑 𝑥 + 2 𝑓 𝑠 1 ( 0 ) − 𝑓 1 2 𝑠 = 1 ( 0 ) 𝜆 ( 𝑚 + 𝛼 ) ∞ − 𝛼 𝑓 𝑠 ( 𝑥 ) 𝑑 𝑥 − 0 − 𝛼 𝑓 𝑠 1 ( 𝑥 ) 𝑑 𝑥 − 2 𝑓 𝑠 1 ( 0 ) + 𝑓 1 2 𝑠 ( 0 ) = 𝑘 𝜆 1 ( 𝑠 ) − 𝜆 ( 𝑚 + 𝛼 ) 𝑅 ( 𝑠 , 𝑚 ) , ( 2 . 1 3 ) where ∫ 𝑅 ( 𝑠 , 𝑚 ) ∶ = 0 − 𝛼 𝑓 𝑠 ( 𝑥 ) 𝑑 𝑥 − ( 1 / 2 ) 𝑓 𝑠 ( 0 ) + ( 1 / 1 2 ) 𝑓 𝑠 ( 0 ) ( 𝑚 ∈ 𝑁 0 ) . With (2.6), it follows that 𝑓 𝑠 𝛼 ( 0 ) = 𝑧 ( 0 ) 𝑡 ( 0 ) = 𝑔 𝑚 + 𝛼 𝜆 𝛼 𝑚 + 𝛼 𝜆 / 𝑠 − 1 𝑓 , ( 2 . 1 4 ) 𝑠 ( 0 ) = 𝑧 ( 0 ) 𝑡 ( 0 ) + 𝑧 ( 0 ) 𝑡 = ( 0 ) 𝜆 − 𝑠 𝑔 𝛼 𝑠 𝛼 𝑚 + 𝛼 𝜆 𝛼 𝑚 + 𝛼 𝜆 / 𝑠 − 1 + 𝜆 𝛼 𝑔 𝛼 𝑚 + 𝛼 𝜆 𝛼 𝑚 + 𝛼 𝜆 / 𝑠 + 𝜆 − 1 . ( 2 . 1 5 ) Set 𝑢 = ( ( 𝑥 + 𝛼 ) / ( 𝑚 + 𝛼 ) ) 𝜆 , with the partial integration, by the strictly monotonic increase of 𝑔 ( 𝑢 ) ( 𝑔 ( 𝑢 ) > 0 ) and 𝑠 = 𝑟 / ( 𝑟 − 1 ) , it gives 0 − 𝛼 𝑓 ( 𝑥 ) 𝑑 𝑥 = ( 𝑚 + 𝛼 ) 0 − 𝛼 l n ( ( 𝑥 + 𝛼 ) / ( 𝑚 + 𝛼 ) ) 𝜆 ( ( 𝑥 + 𝛼 ) / ( 𝑚 + 𝛼 ) ) 𝜆 − 1 𝑥 + 𝛼 𝑚 + 𝛼 𝜆 1 / 𝑠 − 1 / 𝜆 𝑑 𝑥 + 𝛼 = 𝑚 + 𝛼 𝑚 + 𝛼 𝜆 ( 𝛼 / ( 𝑚 + 𝛼 ) ) 𝜆 0 𝑔 ( 𝑢 ) 𝑢 1 / 𝑠 − 1 = 𝑑 𝑢 𝑠 ( 𝑚 + 𝛼 ) 𝜆 ( 𝛼 / ( 𝑚 + 𝛼 ) ) 𝜆 0 𝑔 ( 𝑢 ) 𝑑 𝑢 1 / 𝑠 = 𝑠 ( 𝑚 + 𝛼 ) 𝜆 𝑔 𝛼 𝑚 + 𝛼 𝜆 𝛼 𝑚 + 𝛼 𝜆 / 𝑠 − 𝑠 ( 𝑚 + 𝛼 ) 𝜆 ( 𝛼 / ( 𝑚 + 𝛼 ) ) 𝜆 0 𝑢 1 / 𝑠 𝑔 > ( 𝑢 ) 𝑑 𝑢 𝑠 𝛼 𝜆 𝑔 𝛼 𝑚 + 𝛼 𝜆 𝛼 𝑚 + 𝛼 𝜆 / 𝑠 − 1 − 𝑟 ( 𝑚 + 𝛼 ) 𝑔 𝜆 ( 𝑟 − 1 ) 𝛼 𝑚 + 𝛼 𝜆 ( 𝛼 / ( 𝑚 + 𝛼 ) ) 𝜆 0 𝑢 1 − 1 / 𝑟 = 𝑑 𝑢 𝑠 𝛼 𝜆 𝑔 𝛼 𝑚 + 𝛼 𝜆 𝛼 𝑚 + 𝛼 𝜆 / 𝑠 − 1 − 𝑟 2 ( 𝑚 + 𝛼 ) 𝑔 𝜆 ( 𝑟 − 1 ) ( 2 𝑟 − 1 ) 𝛼 𝑚 + 𝛼 𝜆 𝛼 𝑚 + 𝛼 𝜆 2 − 1 / 𝑟 = 𝑠 𝛼 𝜆 𝑔 𝛼 𝑚 + 𝛼 𝜆 𝛼 𝑚 + 𝛼 𝜆 / 𝑠 − 1 − 𝑟 2 𝛼 𝑔 𝜆 ( 𝑟 − 1 ) ( 2 𝑟 − 1 ) 𝛼 𝑚 + 𝛼 𝜆 𝛼 𝑚 + 𝛼 𝜆 / 𝑠 + 𝜆 − 1 . ( 2 . 1 6 ) In view of (2.13)–(2.16), one has 𝑅 ( 𝑠 , 𝑚 ) > 𝑠 𝛼 𝜆 − 1 2 + 𝜆 − 𝑠 𝑔 𝛼 1 2 𝑠 𝛼 𝑚 + 𝛼 𝜆 𝛼 𝑚 + 𝛼 𝜆 / 𝑠 − 1 − 𝑟 2 𝛼 − 𝜆 𝜆 ( 𝑟 − 1 ) ( 2 𝑟 − 1 ) 𝑔 1 2 𝛼 𝛼 𝑚 + 𝛼 𝜆 𝛼 𝑚 + 𝛼 𝜆 / 𝑠 + 𝜆 − 1 . ( 2 . 1 7 ) If 𝛼 ≥ 1 / 2 , 𝑠 > 1 ( 𝑟 > 1 ) , 0 < 𝜆 ≤ 1 , 𝑔 ( 𝑢 ) > 0 , − 𝑔 ( 𝑢 ) > 0 , one has 𝑠 𝛼 𝜆 − 1 2 + 𝜆 − 𝑠 = 1 2 𝑠 𝛼 1 2 𝑠 2 𝛼 2 − 6 𝑠 𝛼 𝜆 + 𝜆 ( 𝜆 − 𝑠 ) = 1 2 𝑠 𝛼 𝜆 6 𝑠 𝛼 ( 2 𝑠 𝛼 − 𝜆 ) − 𝜆 ( 𝑠 − 𝜆 ) ≥ 1 2 𝑠 𝛼 𝜆 6 𝑠 𝛼 ( 𝑠 − 𝜆 ) − 𝜆 ( 𝑠 − 𝜆 ) = 1 2 𝑠 𝛼 𝜆 ( 6 𝑠 𝛼 − 𝜆 ) ( 𝑠 − 𝜆 ) 𝑟 1 2 𝑠 𝛼 𝜆 > 0 , 2 𝛼 − 𝜆 𝜆 ( 𝑟 − 1 ) ( 2 𝑟 − 1 ) = 1 2 𝛼 1 2 𝑟 2 𝛼 2 − 𝜆 2 ( 𝑟 − 1 ) ( 2 𝑟 − 1 ) = 1 2 𝜆 𝛼 ( 𝑟 − 1 ) ( 2 𝑟 − 1 ) 2 𝑟 2 6 𝛼 2 − 𝜆 2 + 𝜆 2 ( 3 𝑟 − 1 ) 1 2 𝜆 𝛼 ( 𝑟 − 1 ) ( 2 𝑟 − 1 ) > 0 . ( 2 . 1 8 ) This means that 𝑅 ( 𝑠 , 𝑚 ) > 0 . By (2.13) and (2.4), the inequalities (2.10) and (2.11) hold. Lemma 2.2 is proved.

Lemma 2.3. Set ( 𝑟 , 𝑠 ) as a pair of conjugate exponents, 𝑠 > 1 , 𝛼 ≥ 1 / 2 , 0 < 𝜆 ≤ 1 , and 𝜔 𝜆 ( 𝑚 , 𝑠 ) , 𝑘 𝜆 ( 𝑠 ) are defined by (2.9), (2.4), respectively, then, ( 1 ) 𝜔 𝜆 ( 𝑚 , 𝑠 ) > 𝑘 𝜆 ( 𝑠 ) 1 − 𝜂 𝜆 ( 𝑚 ) , ( 2 . 1 9 ) ( 2 ) 0 < 𝜂 𝜆 ( 𝑚 ) < 𝜃 𝜆 𝜃 ( 𝑟 ) < 1 𝜆 1 ( 𝑟 ) ∶ = 𝑘 𝜆 ( 𝑠 ) 𝜆 2 1 0 l n 𝑢 𝑢 𝑢 − 1 − 1 / 𝑟 𝑑 𝑢 , ( 2 . 2 0 ) ( 3 ) 𝜂 𝜆 1 ( 𝑚 ) = 𝑂 𝑚 + 𝛼 𝜆 / 2 𝑠 ( 𝑚 ⟶ ∞ ) , ( 2 . 2 1 ) where 𝜂 𝜆 ( 𝑚 ) ∶ = ( 1 / 𝑘 𝜆 ∫ ( 𝑠 ) 𝜆 ( 𝑚 + 𝛼 ) ) [ 0 − 𝛼 𝑓 𝑠 ( 𝑥 ) 𝑑 𝑥 − ( 1 / 2 ) 𝑓 𝑠 ( 0 ) ] , 𝑓 𝑠 ( 𝑥 ) is defined by (2.2).

Proof. By (2.12), (1.11), and (2.4), 𝜔 𝜆 ( 1 𝑚 , 𝑠 ) = 𝜆 ( 𝑚 + 𝛼 ) ∞ 𝑛 = 0 𝑓 𝑠 ( 1 𝑛 ) > 𝜆 ( 𝑚 + 𝛼 ) ∞ 0 𝑓 𝑠 ( 1 𝑥 ) 𝑑 𝑥 + 2 𝑓 𝑠 ( = 1 0 ) 𝜆 ( 𝑚 + 𝛼 ) ∞ − 𝛼 𝑓 𝑠 ( 𝑥 ) 𝑑 𝑥 − 0 − 𝛼 𝑓 𝑠 ( 1 𝑥 ) 𝑑 𝑥 + 2 𝑓 𝑠 ( 0 ) = 𝑘 𝜆 1 ( 𝑠 ) 1 − 𝑘 𝜆 ( 𝑠 ) 𝜆 ( 𝑚 + 𝛼 ) 0 − 𝛼 𝑓 𝑠 1 ( 𝑥 ) 𝑑 𝑥 − 2 𝑓 𝑠 ( 0 ) = 𝑘 𝜆 ( 𝑠 ) 1 − 𝜂 𝜆 . ( 𝑚 ) ( 2 . 2 2 ) This implies that (2.19) holds.
From the monotonic decrease of the function 𝑓 𝑠 ( 𝑥 ) (see (2.3), 𝑓 𝑠 ( 0 ) > 0 and 𝛼 ≥ 1 / 2 , one has 𝜂 𝜆 ( 𝑚 ) > ( 1 / 𝑘 𝜆 ( 𝑠 ) 𝜆 ( 𝑚 + 𝛼 ) ) [ 𝛼 𝑓 𝑠 ( 0 ) − ( 1 / 2 ) 𝑓 𝑠 ( 0 ) ] ≥ 0 . On the other hand, if 𝑓 𝑠 ( 0 ) > 0 and by the computation as in (2.16), 𝜂 𝜆 1 ( 𝑚 ) = 𝑘 𝜆 ( 𝑠 ) 𝜆 ( 𝑚 + 𝛼 ) 0 − 𝛼 𝑓 𝑠 1 ( 𝑥 ) 𝑑 𝑥 − 2 𝑓 𝑠 < 1 ( 0 ) 𝑘 𝜆 ( 𝑠 ) 𝜆 ( 𝑚 + 𝛼 ) 0 − 𝛼 𝑓 𝑠 1 ( 𝑥 ) 𝑑 𝑥 = 𝑘 𝜆 ( 𝑠 ) 𝜆 2 ( 𝛼 / ( 𝑚 + 𝛼 ) ) 𝜆 0 l n 𝑢 𝑢 𝑢 − 1 1 / 𝑠 − 1 𝑑 𝑢 ≤ 𝜃 𝜆 ( 𝑟 ) < 1 . ( 2 . 2 3 ) Equation (2.20) is valid.
Since l i m 𝑢 → 0 + ( l n 𝑢 / ( 𝑢 − 1 ) ) 𝑢 1 / 2 𝑠 = 0 ( 𝑠 > 1 ) , there exists a constant 𝐿 > 0 , such that | ( l n 𝑢 / ( 𝑢 − 1 ) ) 𝑢 1 / 2 𝑠 | ≤ 𝐿 ( 𝑢 ∈ ( 0 , ( 𝛼 / ( 𝑚 + 𝛼 ) ) 𝜆 ) ) . Then, 0 < 𝜂 𝜆 ( 𝐿 𝑚 ) < 𝑘 𝜆 ( 𝑠 ) 𝜆 2 ( 𝛼 / ( 𝑚 + 𝛼 ) ) 𝜆 0 𝑢 1 / 2 𝑠 − 1 𝑑 𝑢 = 2 𝑠 𝐿 𝑘 𝜆 ( 𝑠 ) 𝜆 2 𝛼 𝑚 + 𝛼 𝜆 / 2 𝑠 . ( 2 . 2 4 ) This means that 𝜂 𝜆 ( 𝑚 ) = 𝑂 ( ( 1 / ( 𝑚 + 𝛼 ) ) 𝜆 / 2 𝑠 ) ( 𝑚 → ∞ ) , the proof is finished.

Lemma 2.4. Set ( 𝑝 , 𝑞 ) and ( 𝑟 , 𝑠 ) as two pairs of conjugate exponents, 𝑝 > 1 , 𝑟 > 1 , 𝛼 > 0 , 𝜆 > 0 , 0 < 𝜀 < 𝑝 𝜆 / 2 𝑟 , ̃ 𝑎 𝑚 ∶ = ( 𝑚 + 𝛼 ) 𝜆 / 𝑟 − 𝜀 / 𝑝 − 1 , ̃ 𝑏 𝑛 ∶ = ( 𝑛 + 𝛼 ) 𝜆 / 𝑠 − 𝜀 / 𝑞 − 1 , and 𝑘 𝜆 ( 𝑠 ) is defined by (2.4). Defining 𝐼 1 ∶ = 𝜀 ∞ 𝑚 = 0 ( 𝑚 + 𝛼 ) 𝑝 ( 1 − 𝜆 / 𝑟 ) − 1 ̃ 𝑎 𝑝 𝑚 1 / 𝑝 ∞ 𝑛 = 0 ( 𝑛 + 𝛼 ) 𝑞 ( 1 − 𝜆 / 𝑠 ) − 1 ̃ 𝑏 𝑞 𝑛 1 / 𝑞 , 𝐼 2 ∶ = 𝜀 ∞ 1 − 𝛼 ( 𝑥 + 𝛼 ) 𝜆 / 𝑟 − 𝜀 / 𝑝 − 1 ( 𝑦 + 𝛼 ) 𝜆 / 𝑠 − 𝜀 / 𝑞 − 1 l n ( ( 𝑥 + 𝛼 ) / ( 𝑦 + 𝛼 ) ) ( 𝑥 + 𝛼 ) 𝜆 − ( 𝑦 + 𝛼 ) 𝜆 𝑑 𝑥 𝑑 𝑦 , ( 2 . 2 5 ) then ( 1 ) 0 < 𝐼 1 < 𝜀 𝛼 1 + 𝜀 + 1 𝛼 𝜀 , ( 2 . 2 6 ) ( 2 ) 𝐼 2 ≥ 𝑘 𝜆 ( 𝑠 ) + 𝑜 ( 1 ) 𝜀 ⟶ 0 + . ( 2 . 2 7 )

Proof. (1) By 𝛼 > 0 and 𝜀 > 0 , one has 0 < 𝐼 1 = 𝜀 ∞ 𝑚 = 0 ( 𝑚 + 𝛼 ) − 1 − 𝜀 1 / 𝑝 ∞ 𝑛 = 0 ( 𝑛 + 𝛼 ) − 1 − 𝜀 1 / 𝑞 1 = 𝜀 𝛼 1 + 𝜀 + ∞ 𝑛 = 1 1 ( 𝑛 + 𝛼 ) 1 + 𝜀 1 < 𝜀 𝛼 1 + 𝜀 + ∞ 0 1 ( 𝑥 + 𝛼 ) 1 + 𝜀 1 𝑑 𝑥 = 𝜀 𝛼 1 + 𝜀 − 1 𝜀 ( 𝑥 + 𝛼 ) − 𝜀 | | | ∞ 0 , ( 2 . 2 8 ) which implies that inequality (2.26) holds.
(2) By 𝑦 ≥ 1 − 𝛼 , letting 0 < 𝜀 < 𝑝 𝜆 / 2 𝑟 , one has ( 𝑦 + 𝛼 ) − 1 − 𝜀 ≤ ( 𝑦 + 𝛼 ) − 1 . And setting 𝑢 = ( ( 𝑥 + 𝛼 ) / ( 𝑦 + 𝛼 ) ) 𝜆 , with l i m 𝑢 → 0 + ( l n 𝑢 / ( 𝑢 − 1 ) ) 𝑢 1 / 2 𝑟 = 0 ( 𝑟 > 1 ) , | ( l n 𝑢 / ( 𝑢 − 1 ) ) 𝑢 1 / 2 𝑟 | ≤ 𝐿 1 ( 𝑢 ∈ ( 0 , 1 ) , 𝐿 1 > 0 ) , one has 𝐼 2 = 𝜀 𝜆 2 ∞ 1 − 𝛼 ( 𝑦 + 𝛼 ) − 1 − 𝜀 ∞ ( 1 / ( 𝑦 + 𝛼 ) ) 𝜆 l n 𝑢 𝑢 𝑢 − 1 1 / 𝑟 − 𝜀 / 𝑝 𝜆 − 1 = 𝜀 𝑑 𝑢 𝑑 𝑦 𝜆 2 ∞ 1 − 𝛼 ( 𝑦 + 𝛼 ) − 1 − 𝜀 ∞ 0 l n 𝑢 𝑢 𝑢 − 1 1 / 𝑟 − 𝜀 / 𝑝 𝜆 − 1 𝑑 𝑢 − ( 1 / ( 𝑦 + 𝛼 ) ) 𝜆 0 l n 𝑢 𝑢 𝑢 − 1 1 / 𝑟 − 𝜀 / 𝑝 𝜆 − 1 = 𝐵 𝑑 𝑢 𝑑 𝑦 2 ( 1 / 𝑟 − 𝜀 / 𝑝 𝜆 , 1 / 𝑠 + 𝜀 / 𝑝 𝜆 ) 𝜆 2 − 𝜀 𝜆 2 ∞ 1 − 𝛼 ( 𝑦 + 𝛼 ) − 1 − 𝜀 ( 1 / ( 𝑦 + 𝛼 ) ) 𝜆 0 l n 𝑢 𝑢 𝑢 − 1 1 / 𝑟 − 𝜀 / 𝑝 𝜆 − 1 ≥ 𝐵 𝑑 𝑢 𝑑 𝑦 2 ( 1 / 𝑟 − 𝜀 / 𝑝 𝜆 , 1 / 𝑠 + 𝜀 / 𝑝 𝜆 ) 𝜆 2 − 𝜀 𝐿 1 𝜆 2 ∞ 1 − 𝛼 ( 𝑦 + 𝛼 ) − 1 ( 1 / ( 𝑦 + 𝛼 ) ) 𝜆 0 𝑢 1 /