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Hindawi Publishing Corporation Journal of Inequalities and Applications Volume 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 Correspondence should be addressed to Wuyi Zhong, wp@bao.ac.cn Received February 2009; Accepted March 2009 Recommended by Nikolaos Papageorgiou p p A Hilbert-type linear operator T : φ → 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 Copyright q 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 Introduction In 1925, Hardy extended Hilbert inequality as follows p If p > 1, 1/p 1/q 1, an , bn ≥ 0, < ∞ an < ∞, and < n ∞ ∞ n 1m π am bn < m n sin π/p ∞ ∞ n m am m n p < ∞ 1/p n π sin π/p < ∞, then 1/q ∞ p an q ∞ n bn q bn , 1.1 n p ∞ p an , 1.2 n where p, q is a pair of conjugate exponents The constant factors π/sin π/p and π/sin π/p p are the best possible The expression 1.1 is the famous Hardy-Hilbert’s inequality 2 Journal of Inequalities and Applications Under the same conditions, there are the classic inequalities : ∞ ∞ n 1m π ln m/n am bn < m−n sin π/p ∞ ∞ n m ln m/n am m−n p < 1/p ∞ p q bn n π sin π/p 1/q ∞ an 1.3 , n 2p ∞ p an , 1.4 n where the constant factors π/sin π/p and π/sin π/p 2p are also the best possible The expression 1.3 is well known as a Hilbert-type inequality 1/p By setting a real space of sequences: p : {a; a {an }∞ , a p { ∞ |an |p } < ∞} n n ∞ and defining a linear operator T : p → p , T a n Cn n ln m/n am / m − n n ∈ N0 , the expressions 1.3 and 1.4 can be rewritten as T a, b < T Ta p a < T p b q, 1.5 a p, 1.6 respectively, where T π/sin π/p , b ∈ 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 These inequalities and their integral forms have been recently extended or strengthened in 4–8 Zhao and Debnath 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, an , bn ≥ 0, such that p q < ∞ an < ∞, < ∞ bn < ∞, then one has n n ∞ ∞ π α / n α am bn < m−n sin π/p ln m n 0m ∞ ∞ n m ln m α /n m−n α am p < ∞ 1/p p an n π sin π/p ∞ 1/q q bn , 1.7 n 2p ∞ p an 1.8 n It has been proved that 1.7 and 1.8 are two equivalent inequalities and their constant factors π/sin π/p and π/sin π/p 2p are the best possible When α 1, the expressions 1.7 and 1.8 can be reduced to 1.3 and 1.4 , respectively p p This paper reports the studies on a Hilbert-type linear operator T : φ → 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 Journal of Inequalities and Applications If s > 1, r, s is a pair of conjugate exponents, then the Beta function is defined as follows cf 2, Theorem 342 , ∞ 0, f i ∞ ln u 1/s−1 u du u−1 π sin π/s B 1 , s r B 1 , r s 1.9 Euler-Maclaurin’s summation formula Set f ∈ C3 0, ∞ , if −1 i f i 0, 1, 2, , then cf 19, Lemma ∞ ∞ f n < f n > f 1.11 f x dx n > 1.10 ∞ ∞ x 1 f − f 0, 12 f x dx n i Lemmas Lemma 2.1 Set r, s as a pair of conjugate exponents, s > 1, α > 0, < λ ≤ 1, and define g u : fs x : ⎧ ⎪ ln u ⎨ , u−1 ⎪ ⎩1, hm,λ x, u ∈ 0, ∪ 1, ∞ , u s 2.1 1, : g x α m α λ x α m α λ 1/s−1/λ x ∈ −α, ∞ , m ∈ N0 , 2.2 Then, one has the following: the function fs x satisfies the conditions of 1.10 and 1.11 This means i −1 i fs x > i x > −α , fs ∞ i 0, 1, 2, , 2.3 kλ s : λm α ∞ −α fs x dx B 1/s, 1/r λ π λ sin π/s Proof For α > 0, x > −α, m ∈ N0 , < λ ≤ and s > 1, set z x λ 1/s−1/λ t x x α / m α show that z x g u and fs x x α / m zx t x λ/s−1 g x 2.4 α / m λ α λ , α and u x α / m α These g u t x when u > With the settings, Journal of Inequalities and Applications −1 i g i u > 0, g t x > 0, z x z x z x ∞ i g u u > 0, i λ λ g λ u t x t x t x λ−1 3g u λ/s − x α m α m α α m α x α m α 2λ−3 2.5 < 0, λ/s−3 x α m α > 0, λ/s − λ/s − λ/s − m λ2 λ − > 0, < 0, α λ−2 λ/s−2 λ/s − λ/s − m α x α m α λ−3 x α m α α m λ−1 λ λ−1 λ−2 m λ λ−1 g u x α α m α m g u < 0, x α α m α m cf 16, Lemma 2.2 , one has z x > 0, λ−1 x α α m α m g u 0, 1, 2, 3 λ/s−4 x α m α < These are followed by fs ∞ fs x z x t x > 0, fs x z x t x z x t x < 0, fs x z x tx 2z x t x fs x z x t x 3z x t x 0, 2.6 fs ∞ 0, fs ∞ z x t x > 0, 3z x t x ∞ −α fs x dx λ λ2 ∞ −α ∞ ln x x α / m α/ m α ln u 1/s−1 u du u−1 By 1.9 , then 2.4 holds Lemma 2.1 is proved fs ∞ z x t x < 0, Then inequality 2.3 holds For x > −α, m ∈ N0 and λ > 0, s > 1, set u λ m α 0, x α λ λ −1 α / m x α m α α λ 2.7 , then one has λ 1/s−1/λ d x α m α 2.8 Journal of Inequalities and Applications Lemma 2.2 Set r, s as a pair of conjugate exponents, s > 1, α ≥ 1/2, < λ ≤ and define ∞ ωλ m, s : ln n n n α / m α λ − m α α λ m · α n λ/r m ∈ N0 , 1−λ/s α 2.9 then, one has < ωλ m, s < kλ s , < ωλ n, r < kλ r 2.10 n ∈ N0 , kλ s 2.11 where kλ s is defined by 2.4 Proof By 2.9 and 2.2 , it is evident that λ m α < ωλ m, s λ m α ∞ ln n n n α / m α λ λ −1 α /m n, s ∞ hm,λ n α λ m α n α m α λ 1/s−1/λ 2.12 ∞ fs n n In view of 2.3 , 1.10 , and 2.4 , one has ωλ m, s < ∞ λ m α ∞ λ m α kλ s − where R s, m : f −α s 1 fs − fs 12 fs x dx −α fs x dx − fs −α fs x dx − f 12 s 2.13 R s, m , λ m α x dx − 1/2 fs λ α fs z0 t0 g fs z t0 1/12 fs m ∈ N0 With 2.6 , it follows that λ/s−1 α z0 t λ−s g sα m m λ α m α α λ/s−1 α m α α , λ g α 2.14 λ α m α m 2.15 λ/s λ−1 α α Journal of Inequalities and Applications Set u x α / m α λ , with the partial integration, by the strictly monotonic increase of g u g u > and s r/ r − , it gives −α f x dx m α m x −α sα g λ sα g λ α λ λ −1 x α m α λ 1/s−1/λ d x α m α g u u1/s−1 du λ α/ m α g u du1/s s m α g λ sα g λ α / m α s m α λ > α / m λ α/ m α α λ ln x λ α m m α α α λ/s−1 α m λ α α λ/s−1 α m α m m λ α m α λ α λ/s α α λ/s−1 α m α − α/ m α s m α − λ u1/s g u du r m α g λ r−1 λ α m α/ m α α r2 m α g − λ r − 2r − − λ u1−1/r du λ α m r2α g λ r − 2r − λ α m λ α m λ 2−1/r α α λ/s λ−1 α α m α 2.16 In view of 2.13 – 2.16 , one has sα − λ R s, m > λ−s g 12sα λ α m α λ r2α − g − λ r − 2r − 12α λ/s−1 α m λ α m α α m 2.17 λ/s λ−1 α α If α ≥ 1/2, s > r > , < λ ≤ 1, g u > 0, −g u > 0, one has sα − λ λ−s 12sα 12s2 α2 − 6sαλ λ λ − s 12sαλ ≥ r2α λ − λ r − 2r − 12α 6sα s − λ − λ s − λ 12sαλ 6sα 2sα − λ − λ s − λ 12sαλ 6sα − λ s − λ > 0, 12sαλ 12r α2 − λ2 r − 2r − 12λα r − 2r − 2r 6α2 − λ2 λ2 3r − > 12λα r − 2r − 2.18 Journal of Inequalities and Applications This means that R s, m > 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, < λ ≤ 1, and ωλ m, s , kλ s are defined by 2.9 , 2.4 , respectively, then, < ηλ m < θλ r < θλ r : where ηλ m : ωλ m, s > kλ s − ηλ m , ηλ m 1/kλ s λ m m ln u −1/r u du , 0u − m −→ ∞ , α f −α s α 1 kλ s λ2 λ/2s O 2.19 2.20 2.21 x dx − 1/2 fs , fs x is defined by 2.2 Proof By 2.12 , 1.11 , and 2.4 , λ m α ωλ m, s ∞ fs n > n ∞ λ m α kλ s 1− −α ∞ λ m α 0 fs x dx − kλ s λ m fs x dx −α fs x dx fs fs 2.22 fs x dx − fs −α α kλ s − ηλ m This implies that 2.19 holds From the monotonic decrease of the function fs x see 2.3 , fs > and α ≥ 1/2, one has ηλ m > 1/kλ s λ m α αfs − 1/2 fs ≥ On the other hand, if fs > and by the computation as in 2.16 , kλ s λ m ηλ m < kλ s λ m fs x dx − fs −α α α Equation 2.20 is valid −α fs x dx kλ s λ2 α/ m α λ 2.23 ln u 1/s−1 u du ≤ θλ r < u−1 Journal of Inequalities and Applications Since limu → ln u/ u − u1/2s s > , there exists a constant L > 0, such that | ln u/ u − u1/2s | ≤ L u ∈ 0, α/ m α λ Then, This means that ηλ m λ α/ m α L kλ s λ2 < ηλ m < 2sL kλ s λ2 u1/2s−1 du O 1/ m α λ/2s λ/2s α m α 2.24 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, < ε < pλ/2r, am : λ/r−ε/p−1 ∞ I1 : ε m α n α , bn : λ/s−ε/q−1 1/p m α ∞ p 1−λ/r −1 p am 1/q n m , and kλ s is defined by 2.4 Defining α q 1−λ/s −1 q bn , n 2.25 ∞ I2 : ε x λ/r−ε/p−1 α y α x 1−α λ/s−ε/q−1 λ α − y ln x α α / y α λ dx dy, then ε α1 ε < I1 < I ≥ kλ s , αε o1 2.26 ε −→ 2.27 Proof By α > and ε > 0, one has 1/p ∞ < I1 ε m α n m ε ε ∞ 1 α −1−ε n α1 ε α1 ε 1/q ∞ −1−ε n − n x ε α α −ε which implies that inequality 2.26 holds , | ln u/ u − u1/2r | ≤ u x α / y α λ , with limu → ln u/ u − u1/2r L1 u ∈ 0, , L1 > , one has I2 ε λ2 ε λ2 ∞ y ∞ −1−ε α 1−α 1/ y α ⎡ ∞ y −1−ε α ⎣ 1−α ∞ λ ln u 1/r−ε/pλ−1 u du dy u−1 ln u 1/r−ε/pλ−1 u du − u−1 B2 1/r − ε/pλ, 1/s λ2 ε/pλ B2 1/r − ε/pλ, 1/s ≥ λ2 ε/pλ B 1/r − ε/pλ, 1/s λ ε/pλ B 1/r − ε/pλ, 1/s λ ε/pλ − ⎡ y α −1−ε ⎣ ⎡ ∞ y α −1 1/ y α ⎣ 1−α ⎤ ln u 1/r−ε/pλ−1 ⎦ u du dy u−1 ⎤ λ u1/2r−ε/pλ−1 du⎦dy ∞ εL1 λ2 1/2r − ε/pλ y 1/2r − ε/pλ α −λ 1/2r−ε/pλ −1 dy 1−α εL1 λ3 λ 1/ y α 1−α εL1 − λ ln u 1/r−ε/pλ−1 ⎦ u du dy u−1 ∞ ε − λ ⎤ λ 1/ y α 2.29 Set ε → , then the inequality 2.27 holds Lemma 2.4 is proved Main Results Firstly, the following notations are given Set p > 0, p / 1, r > 1, p, q and r, s are two pairs of conjugate exponents Let φ x : x α ϕx : x α ψ x : ϕx p 1−λ/r −1 q 1−λ/s −1 1−p , 3.1 , x α pλ/s−1 , x ∈ 0, ∞ Set p > 1, p, q is a pair of conjugate exponents Let p φ ⎧ ⎨ : a; a ⎩ {an }∞ , a n ∞ p,φ : n 1/p φ n |an |p ⎫ ⎬ 1, p, q is a pair of conjugate exponents Define a Hilbert-type linear p operator T , for all a ∈ φ , one has ∞ ln m T a n : Cn : m m For a ∈ ∞ p , φ b∈ ∞ T a, b : n q ϕ, ln m m m α α / n λ − n α α λ n ∈ N0 am 3.4 define the formal inner product of T a and b as α / n α λ ∞ α am − n λ α ∞ ln m bn n 0m m α/ n α λ − n α am bn α λ , 3.5 Then one will have some results in the following theorem Theorem 3.1 Suppose that p, q and r, s are two pairs of conjugate exponents and p > 1, r > 1, p 1/2 ≤ α ≤ 1, < λ ≤ 1, an ≥ Then for ∀a ∈ φ , one has Ta p C {Cn }∞ ∈ n p ψ 3.6 p It means that T : φ → ψ T is a bounded linear operator and T p,ψ : sup a∈ where Cn , T are defined by 3.4 , T a defined by 2.4 p,ψ p φ Ta a a/θ C p,ψ p,ψ kλ s , 3.7 p,φ is defined as by 3.3 , and kλ s is a constant Journal of Inequalities and Applications 11 and the result 2.11 , for n ∈ N0 , it is Proof If p > 1, by using Holder’s inequality cf 20 ¨ obvious that Cn ≥ and ∞ p Cn ln m m m ∞ ≤ m ∞ ∞ p−1 ≤ kλ m m λ α α ∞ λ m m n α λ α α α m λ α/ n α λ α α − n α p p−1 3.8 p−1 1−λ/r n m ωλ n, r ϕ n p−1 1−λ/r α n p am 1−λ/s λ 1−λ/r /q 1−λ/r α α α p 1−λ/s /p am 1−λ/s q−1 1−λ/s m α m am p−1 1−λ/r α n α − n ln m s m α − n α /n 1−λ/s /p λ α α /n ln m m − n α n 1−λ/r /q n α λ α λ α α /n α m α − n ln m m And if ψ n λ α ln m m × α / n α p ϕp−1 n am 1−λ/s p−1 ϕ1−p n , by 2.9 and 2.10 , it follows that Ta ∞ p p,ψ p ψ n Cn n p−1 ≤ kλ ∞ ∞ s ln m m m 0n p−1 kλ ∞ α /n α λ − n p α α m λ 1−λ/s n p ωλ m, s φ m am ≤ kλ s a s p−1 1−λ/r α m α p p,φ p am 3.9 < ∞ This means that C {Cn }∞ ∈ ψ and T p,ψ ≤ kλ s n If there exists a constant K < kλ s , such that T p,ψ ≤ K, then for < ε < pλ/2r, by the definition 3.5 , and by using Holders inequality and the result 2.26 , one has ă p ε T a, b ≤ ε T p,ψ a ∞ p,φ b q,ϕ ≤ KI1 < K ε α1 ε , αε 3.10 where a {am }∞ ∈ φ , b {bn }n ∈ ϕ and am , bn are defined as in Lemma 2.4 m On the other hand, from the strictly monotonic decrease of the function g u ln u/ u − and the exponents λ/r − ε/p − < 0, λ/s − ε/q − < and − α ≥ 0, and by α > 0, λ > 0, in view of 2.27 , one has p ε T a, b ≥ ε ∞ x α λ/r−ε/p−1 y x 1−α I ≥ kλ s q o1 α α ε −→ λ/s−ε/q−1 λ − y ln x α λ α/ y α dx dy 3.11 12 Journal of Inequalities and Applications In view of 3.10 and 3.11 , one has kλ s o < K ε/α1 ε 1/αε Setting ε → , one has kλ s ≤ K This means that K kλ s , that is, T p,ψ kλ s Theorem 3.1 is proved Theorem 3.2 Suppose that p, q and r, s are two pairs of conjugate exponents, r > 1, p > 1, 1/2 ≤ α ≤ 1, < λ ≤ 1, an , bn ≥ n ∈ N0 Then one has the following p q If a ∈ φ , b ∈ ϕ , and a p,φ > 0, b q,ϕ > 0, then T a, b < kλ s a If a ∈ p φ and a p,φ b p,φ q,ϕ 3.12 > 0, then Ta < kλ s a p,ψ p,φ , 3.13 where the mark T a p,ψ is defind as in Theorem 3.1 The inequality 3.13 is equivalent to 3.12 , and 1/λ B 1/s, 1/r kλ r is the best possible the constant factor kλ s Proof In view of 3.9 and < a p,φ < ∞, one has Ta p p,ψ p p p,φ < kλ s a 3.14 And by p > 1, 3.13 holds By using Holder’s inequality and 3.13 , one has ă T a, b ln m n 0m ≤ Ta p,ψ m b α /n α λ − n α < kλ s a q,ϕ m α α 1−λ/r /q n α 1−λ/s /p λ p,φ am n α m α 1−λ/s /p b 1−λ/r /q n 3.15 q,ϕ b The inequality 3.12 is obtained From 3.12 and a p,φ > 0, there exists k0 ∈ N, such that K m 0φ p m am > and p−1 λ K ψ n − n α λ > when K > k0 By a bn K m ln m α / n α am / m α combination as in 3.15 and by 1/2 ≤ α ≤ 1, < λ ≤ 1, and with 2.10 and 2.11 , then, K ϕn 0< q bn K K ψ n K n K n K n 0m ln m m m m α /n α ln m λ α am bn K − n α λ α / n α λ − n α am p λ α 1/p K < kλ s φ n p an n 1/q K ϕn q bn K < ∞ n 3.16 By p > and q > 1, it follows that K 0< n q p ϕ n bn K < k λ s ∞ n p φ n an < ∞ 3.17 Journal of Inequalities and Applications 13 Letting K → ∞ in 3.17 , this means < ∞ ϕ n bn ∞ < ∞, that is, b {bn ∞ }∞ ∈ ϕ n n and b q,ϕ > Therefore the inequality 3.16 keeps the form of the strict inequality when q p K → ∞ So does 3.17 In view of ∞ ϕ n bn ∞ T a p,ψ , the inequality 3.13 holds, n and 3.12 is equivalent to 3.13 By T p,ψ kλ s , it is obvious that the constant factor kλ r is the best possible This completes the proof of Theorem 3.2 kλ s q q Theorem 3.3 Set p, q and r, s as two pairs of conjugate exponents, < p < q < , r > 1, 1/2 ≤ α ≤ 1, < λ ≤ 1, an , bn ≥ Let ∞ ∞ ln m H a, b α /n m n 0m α λ α am bn − n α λ 3.18 Then the reverse inequalities can be established as follows If < a p,φ < ∞ and < b q,ϕ < ∞, then 1/p ∞ p − ηλ m φ m am H a, b > kλ s b q,ϕ 3.19 m If < a ∞ p,φ < ∞, then ∞ ψ n n ln m m If < b ∞ m q,ϕ α / n m α λ p α am − n α ∞ p > kλ s λ p − ηλ m φ m am 3.20 m < ∞, then q−1 φ−1 m − ηλ m ∞ n ln m m α / n α λ − n α bn α λ q q < kλ s b q q,ϕ , 3.21 where the marks a p,φ and b q,ϕ < p < as two formal norms are still defined like in 3.3 and the factor ηλ m in 3.19 – 3.21 is defined in Lemma 2.3 The inequalities 3.20 and 3.21 are p q equivalent to 3.19 The constant factors kλ s , kλ s and kλ s in 3.19 , 3.20 , and 3.21 are all the best possible Proof By < p < q < , with the reverse Holder’s inequality, one has the following ă By the combination as in 3.15 for 3.18 , then one has H a, b ≥ 1/p ∞ ωλ m, s φ m m p am 1/q ∞ ωλ n, r ϕ n q bn 3.22 n If 1/2 ≤ α ≤ 1, < λ ≤ 1, the expressions 2.19 and 2.11 are established for ωλ m, s and ωλ n, r , respectively And by q < 0, 3.19 holds 14 Journal of Inequalities and Applications Setting a constant K ≥ kλ r , 3.19 is still valid if we replace kλ r by K, then for < ε < −qλ/r, by 3.19 and 2.21 , one will have ∞ H a, b > K 1/p − ηλ m φ m p am b q,ϕ m ∞ K m m ∞ α n n α ε ∞ n ⎧ ⎨ K 1/p − ηλ m n α ∞ m 0O 1− ε⎩ 1/q 3.23 ε 1/ m ∞ m 1/ α m α ∞ m α λ/r−ε/p−1 , bn n where a {am }∞ , b {bn }n , am m that < a p,φ < ∞, < b q,ϕ < ∞ On the other hand, by 3.18 , 2.2 , 2.12 , and 2.10 , ∞ ∞ H a, b ln n α / m α n m 0n ∞ m α −1−ε m ∞ m α −1−ε m < m α λ m α λ m α B2 1/s − ε/qλ, 1/r λ2 ∞ α λ λ/r−ε/p−1 − m ln n α α / m n ∞ hm n, n λ/s−ε/q−1 and it is apparent λ/s−ε/q−1 α λ n α /m α/ m α λ λ/s−ε/q−1 α −1 ε − s qλ ∞ n α α , ⎭ ε λ n ε/qλ n ⎫1/p ⎬ ε λ/2s n α ε 3.24 In view of 3.23 and 3.24 , one has ⎧ ⎨ K ⎩ 1− ∞ m 0O 1/ m ∞ m 1/ m α ε λ/2s α ε ⎫1/p ⎬ ⎭ < Setting ε → , one has K ≤ kλ s , which means K is the best possible B2 1/s − ε/qλ, 1/r λ2 ε/qλ 3.25 kλ s The constant factor kλ s in 3.19 Journal of Inequalities and Applications By < a 15 ∞ n 0ψ < ∞, one has p,φ ∞ m n ln m α / n α am / m p α λ− p−1 λ ∞ > Setting bn : ψ n − n α λ , n αλ m ln m α / n α am / m α q making the calculations as in 3.8 and 3.9 , one has < b q,ϕ < ∞ By using 3.19 in the following: b ∞ q q,ϕ ∞ q ϕ n bn ∞ ψ n n n ln m m m α /n α λ α am − n p λ α 3.26 1/p ∞ p − ηλ m φ m am H a, b > kλ s b q,ϕ , m one has ∞ ∞ ψ n n ln m α / n m m α λ α am − n α p ∞ p > kλ s λ p − ηλ m φ m am 3.27 m Therefore, 3.20 holds On the other hand, if 3.20 is valid, by < p < q < and by using the reverse Holder’s inequality, it has ¨ ∞ H a, b n n ≥ ∞ λ/s−1/p α ln m m m ∞ ∞ ψ n n ln m m m α / n α α / n α λ − n λ − n α am α α am α λ p 1/p n α bn 1/q ∞ n λ 1/p−λ/s α q 1−λ/s −1 q bn 3.28 n 1/p ∞ − ηλ m φ m > kλ s p am b q,ϕ m Then 3.19 holds It means that 3.20 is equivalent to 3.19 By < b q,ϕ < ∞, it is obvious that there exist n0 ∈ N, such that K m φ−1 m − ηλ m q−1 ⎡ ⎢ ⎣ ⎤q K n ln m m α bn ⎥ ⎦ > 0, α λ− n α λ α / n K q ϕ n bn > when K > n0 n 3.29 16 Journal of Inequalities and Applications λ K φ−1 m / 1− ηλ m n ln m α / n α bn / m α − n α p m am K < ∞ By 3.19 , Setting am K , one has < K m 0φ K λ q−1 > p − ηλ m φ m am K m q−1 φ−1 m − ηλ m K m K ln m α / n m n α λ α bn − n q λ α 3.30 K K ln m α /n m m 0n K > kλ s α λ α bn am K − n α λ 1/p − ηλ m φ m p am 1/q K ϕn K m q bn n Further one has K 1/q − ηλ m φ m p am > kλ s K 1/q K m ϕn q bn > 3.31 n By q < 0, one has K p 0< m q − ηλ m φ m am K < kλ s K n q q ϕ n bn < k λ s ∞ q ϕ n bn < ∞ 3.32 n Setting K → ∞ in 3.32 , via 2.20 , one has ∞ 0< p φ m am ∞ < m 1 − θλ r ∞ p − ηλ m φ m am ∞ < ∞ 3.33 m It means that < a p,φ < ∞ a : {am ∞ }∞ The conditions for 3.19 are satisfied m Equation 3.30 keeps a strict form when K → ∞ So does 3.31 By q < 0, the inequality 3.21 holds Journal of Inequalities and Applications 17 Also, from 3.21 , by q < and by using the reverse Holders inequality, ă H a, b m ≥ − ηλ m φ−1 m ⎧ ⎨ φ−1 m ⎩ − ηλ m 1/p am ⎧ ∞ 1− ηλ m φ m m ∞ > kλ s p am 1/p⎨ ∞ ⎩m ⎫ ln m α / n α bn ⎬ m α λ− n α λ ⎭ 1/p ∞ n φ−1 m 1− ηλ m q−1 ∞ n ln m m α / n α λ− n α bn α λ ⎫ q ⎬1/q ⎭ 1/p p − ηλ m φ m am b q,ϕ m 3.34 Therefore, 3.19 holds Equation 3.21 is equivalent to 3.19 p q If the constant factor kλ s or kλ s in 3.20 or in 3.21 is not the best possible, by 3.28 or by 3.34 , then it leads to a contradiction in which the constant factor kλ s in 3.19 is not the best possible Theorem 3.3 is proved Remark 3.4 Set r q, s p, λ and 1.8 , respectively So 3.12 1, the inequalities 3.12 and 3.13 can be reduced to 1.7 or 3.13 is an extension of 1.7 or 1.8 Acknowledgments The work is supported by the National Natural Science Foundation of China no.10871073 The author would like to thank the anonymous referee for his 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