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Hindawi Publishing Corporation Journal of Inequalities and Applications Volume 2010, Article ID 393025, 18 pages doi:10.1155/2010/393025 Research Article A New Hilbert-Type Linear Operator with a Composite Kernel and Its Applications Wuyi Zhong Department of Mathematics, Guangdong Institute of Education, Guangzhou, Guangdong 510303, China Correspondence should be addressed to Wuyi Zhong, zwy@gdei.edu.cn Received 20 April 2010; Accepted 31 October 2010 Academic Editor: Ondˇ ej Doˇ ly r s ´ Copyright q 2010 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 A new Hilbert-type linear operator with a composite kernel function is built As the applications, two new more accurate operator inequalities and their equivalent forms are deduced The constant factors in these inequalities are proved to be the best possible Introduction In 1908, Weyl published the well-known Hilbert’s inequality as follows: if an , bn ≥ are real sequences, < ∞ a2 < ∞ and < ∞ bn < ∞, then n n n ∞ ∞ n 1m am bn 1, α ≥ 1/2, and an , p q bn ≥ 0, such that < ∞ n α p 1−λ/r −1 an < ∞ and < ∞ n α q 1−λ/s −1 bn < ∞, then it n n has ∞ ∞ ln m m n 0m α/ n α λ α am bn − n λ α 1/p ∞ < n 1/q ∞ p 1−λ/r −1 p an α n n α q 1−λ/s −1 q bn n 1.4 Letting φ x : ∞ n x p 1−λ/r −1 α ,ϕ x : φ n |an | } < ∞}, : {b; b { expression 1.4 can be rewritten as ∞ ∞ p α/ n α am bn m n 0m q 1−λ/s −1 and ln m Ta, b : α {bn }∞ , n q ϕ p 1/p x α λ − n α b λ q,ϕ , : { p φ {an }∞ , a n : {a; a ∞ n < kλ s a q 1/q ϕ n |bn | } p,φ b q,ϕ , p,φ : < ∞}, the 1.5 p where T : φ → ψ is a linear operator, kλ s T a p,φ is the norm of the sequence a with ∞ a weight function φ Ta, b is a formal inner product of the sequences Ta n : m ln m λ λ and b α / n α am / m α − n α By setting two monotonic increasing functions u x and v x , a new Hilbert-type inequality, which is with a composite kernel function K u x , v y , and its equivalent are built in this paper As the applications, two new more accurate Hilbert-type inequalities incorporating the linear operator and the norm are deduced Firstly, the improved Euler-Maclaurin’s summation formula is introduced Set f ∈ C4 m, ∞ m ∈ N0 If −1 i f i x > 0, f i ∞ i 0, 1, 2, 3, , then it has ∞ ∞ f n < f x dx m n m ∞ 1 f m − f m , 12 1.6 ∞ f n > f x dx m n m f m 2 Lemmas Lemma 2.1 Set r, s as a pair of conjugate exponents, s > 1, β ≥ α ≥ e7/12 , < λ ≤ 1, and define f y : g y : lnλ αm lnλ/r αm , lnλ βy ln1−λ/s βy y y ∈ 1, ∞ , m ∈ N, 2.1 lnλ/s βn , ln1−λ/r αy y y ∈ 1, ∞ , n ∈ N, 2.2 lnλ αy lnλ βn Journal of Inequalities and Applications ln β/ ln αm uλ/s−1 du − f uλ f 1, 12 2.3 uλ/r−1 du − g uλ g , 12 2.4 Rλ m, s : ln α/ ln βn Rλ n, r : ln β/ ln αm ηλ m, s : uλ/s−1 du − f uλ 2.5 Then, it has the following The functions f y , g y satisfy the conditions of 1.6 It means that −1 i F F i i y >0 ∞ f, g, y ∈ 1, ∞ , F F f, g, i 2.6 0, 1, 2, 3, , Rλ m, s > 0, 2.7 Rλ n, r > 0, < ηλ m, s O ln αm ρ ρ > 0, m −→ ∞ 2.8 Proof For β ≥ α ≥ e7/12 , y ≥ 1, m ∈ N, < λ ≤ 1, and s > 1, set f1 y : lnλ αm lnλ βy , f2 y : , ln1−λ/s βy f3 y : y 2.9 It has lnλ/r αmf1 y f2 y f3 y f y 2.10 when y ≥ It is easy to find that −1 i fj i i ∞ y > 0, f y > 0, −1 i f i fj y ∈ 1, ∞ , j 1, 2, 3, i 0, 1, 2, 3, , 2.11 i ∞ y ∈ 1, ∞ , i 0, 1, 2, 3, Journal of Inequalities and Applications y ≥ 1, i 0, 1, 2, 3, These Similarly, it can be shown that −1 i g i y > 0, g i ∞ tell us that 2.6 holds and the functions f y , g y satisfy the conditions of 1.6 Set t uλ With the partial integration, it has ln β/ ln αm uλ/s−1 du uλ λ ln β/ ln αm λ t1/s−1 dt t λ/s s ln β/ ln αm λ ln β/ ln αm s λ λ ln β/ ln αm s λ dt1/s t ln β/ ln αm λ t1/s lnλ β s λ lnλ αm lnλ β ≥ λ ln αm ln β λ/r lnλ β s λ lnλ αm lnλ β ln αm ln β λ/r t dt ln β/ ln αm s2 λ s λ dt1/s t ln2λ β s2 λ s lnλ αm lnλ β ln αm ln β λ/r 2.12 By 2.1 , it has f ln αm ln β λ/r − λ/r ln αm ln β f λln2λ−2 β lnλ αm lnλ β − lnλ−1 β , lnλ αm lnλ β 2.13 lnλ−1 β − λ/s lnλ−2 β − λ lnλ αm lnλ β ln αm lnλ β 2.14 In view of 2.12 ∼ 2.14 , it has Rλ m, s ≥ ln αm ln β λ/r lnλ−1 β − λ/s − − λ αm λβ 12 ln β 12 ln ln 2.15 ln2λ β lnλ αm s ln β λ λ s2 − λ s 12ln2 β lnλ β As ln β ≥ 7/12, s > r > , and < λ ≤ 1, it has − − λ/s − 12 ln β 12 1− λ s s 7s λ ln β ≥ 1− λ 12λ s 7s − 12λ 12 ln β s2 s λ s > − − λ s 12ln2 β 12ln2 β > − 1− λ 1− 12 ln β s λ s − 12 12 ln β s 1− 3ln2 β > 0, 2.16 > It means that Rλ m, s > Similarly, it can be shown that Rλ n, r > The expression 2.7 holds Journal of Inequalities and Applications By 2.5 , 2.12 , 2.13 , and < λ < s, β ≥ e7/12 , it has ηλ m, s ≥ lnλ β s λ αm lnλ β λ ln ln αm ln β ηλ m, s < λ λ/r ln β/ ln αm λ/r ln αm ln β lnλ−1 β lnλ αm lnλ β lnλ−1 β λ αm ln lnλ β − ln αm ln β λ/r s ln β ln αm − > λ ln β λ s ln β λ ln αm u1/s−1 du λ/r lnλ−1 β ln β − λ αm λβ ln ln > 0, λ/s 2.17 The expression 2.8 holds, and Lemma 2.1 is proved Lemma 2.2 Set r, s as a pair of conjugate exponents, s > 1, β ≥ α ≥ 1/2, and < λ ≤ 1, and define f1 y : g1 y : ln y β / m α λ y β λ m α y β / m α λ −1 m α ln y α / n β λ y λ n β y α / n β λ−1 n Rλ m, s : Rλ n, r : β/m α λ2 λ ηλ m, s : y ∈ 0, ∞ , m ∈ N0 , , y ∈ 0, ∞ , n ∈ N0 , λ/r−1 f , 12 ln u 1/r−1 u du − g1 u−1 g 0, 12 λ λ2 β/m α , ln u 1/s−1 u du − f1 u−1 α/n β λ2 α β λ/s−1 λ 2.18 ln u 1/s−1 du − f1 u u−1 Then, it has The functions f1 y , g1 y satisfy the conditions of 1.6 It means that −1 i F F i i ∞ y >0 F F f1 , g1 , y ∈ 0, ∞ , f1 , g1 , i 2.19 0, 1, 2, 3, , Rλ m, s > 0, Rλ n, r > 0, 2.20 Journal of Inequalities and Applications < ηλ m, s O ρ m ρ > 0, m −→ ∞ α 2.21 Proof Letting h u : ln u/ u − , u y β / m α λ , it can be proved that f1 y 1/s−1/λ 1/λ m α h u u satisfy 2.19 as in Similarly, it can be shown that g1 y satisfy 2.19 also λ y β / m α λ−1 1/ m α λ/ y Setting u0 : β/ m α λ , by u β y β / m α λ ,u λ/β u0 , and h u > 0, it has λ2 β/ m α λ ln u 1/s−1 u du u−1 λ2 s λ2 u0 h u u1/s−1 du u0 h u du1/s s h u0 u1/s − λ2 s ≥ h u0 u1/s − h u0 λ u0 u 1/s u0 u1/s h u du 2.22 du s s h u0 u1/s h u0 u1/s − 0 s λ2 , f1 ln β/ m α λ β λ λ m α β/ m α − m α f1 1/s h u0 u0 u β2 λ/s−1 h u0 u1/s , λβ 2.23 1 − h u0 s λ 2.24 With 2.22 ∼ 2.24 , it has Rλ m, s ≥ h u0 u1/s s − 2λβ λ 12β2 1 − s λ − h u0 u1/s s 1 s − 12β2 2.25 By h u0 > 0, h u0 < 0, and β ≥ 1/2, s > 1, < λ ≤ 1, it has s − λ2 2λβ s 1 − s 12β2 12β2 1 − s λ 12β2 s − s 12β2 s 6βs 2βs − λ − λ s − λ > 0, 12β2 sλ2 4β2 s − s 8β2 s − > 12βλ s So Rλ m, s > holds Similarly, it can be shown that Rλ n, r > 2.26 Journal of Inequalities and Applications In view of 2.22 , 2.23 , by h u > 0, h u < 0, it has s − λ2 2λβ ηλ m, s > h u0 u1/s h u0 u1/s 2βs − λ > 0, 2λ2 β 2.27 and by limu → ln u/ u − u1/2s 0, so there exists a constant L > 0, such that | ln u/ u − u1/2s | < L u ∈ 0, β/ m α λ Then it has ηλ m, s < λ β/ m α λ ln u 1/s−1 L u du < u−1 λ λ β/ m α u1/2s−1 du β 2sL m α λ λ/2s 2.28 It means that 2.21 holds The proof for Lemma 2.2 is finished Main Results Set λ ∈ R, p > 1,r > 1, p, q , and r, s as two pairs of conjugate exponents K x, y ≥ x, y ∈ 0, ∞ × 0, ∞ is a measurable kernel function Both u x and v x are strictly monotonic increasing differentiable functions in n0 , ∞ such that U n0 > 0,U ∞ ∞ U u, v Give some notations as follows: φ x : ϕ x : ψ x : v x ϕ x 1−p p 1−λ/r −1 q 1−λ/s −1 v x pλ/s−1 u x 1−p , v x 1−q , 3.1 x ∈ n0 , ∞ , v x set p φ and call ux p φ ⎧ ⎨ : ⎩ a; a {an }∞ n0 , n ∞ a p,φ : 1/p φ n |an | n n0 p ⎫ ⎬ U u, v , and < Theorem 3.1 Suppose that an ≥ 0, U x /U x ε ∞ ≤ ∞ n0 u n / u n ε < ∞ ε > If there exists a positive number n n0 v n / v n n kλ , such that < ωλ m, s < kλ , kλ − O then for all a ∈ p φ and a p,φ um < ϑλ n, r < kλ ≤ ωλ m, s ρ m, n ≥ n0 , ρ > 0, m −→ ∞ , 3.7 3.8 > 0, it has the following: Ta It means that T : p φ → p ψ, C {Cn }∞ n0 ∈ n p ψ, 3.9 Journal of Inequalities and Applications T is a bounded linear operator and T p,ψ : a∈ where Cn , T are defined by 3.4 , Ta Ta sup p φ a/θ C p,ψ a p,ψ p,ψ kλ , 3.10 p,φ is defined as 3.3 Proof By using Holder’s inequality and 3.6 , 3.7 , it has Cn and ă ∞ p Cn K u m ,v n ∞ ∞ v n u m um m n0 ∞ q−1 1−λ/s 1−λ/r p−1 1−λ/r um K u m ,v n 1−λ/s v n m n0 ∞ 1−λ/s v n m n0 v n 1/p q−1 p−1 p a p−1 m p−1 1−λ/r um K u m ,v n 1−λ/r /q p p−1 v n u m u m 1/q am u m v n am u m 1−λ/s /p v n p p−1 v n K u m ,v n p−1 1/q 1−λ/s v n m n0 < kλ u m p−1 1−λ/r um K u m ,v n × 1/p 1−λ/s /p v n m n0 ≤ v n 1−λ/r /q um v n u m p−1 ϑλ n, r ϕ n p ϕp−1 n am 3.11 And by ψ n Ta p p,ψ ϕ1−p n , it follows that ∞ n n0 p ∞ p−1 ψ n Cn < kλ p−1 kλ ∞ K u m ,v n m n0 n n0 ∞ ωλ m, s φ m p am m n0 < p kλ a p p,φ um v n p−1 1−λ/r 1−λ/s v n u m p−1 p am 3.12 < ∞ This means that C {Cn }∞ ∈ ψ , Ta p,ψ ≤ kλ a p,φ , and T p,ψ ≤ kλ T is a bounded linear n operator If there exists a constant K < kλ , such that T p,ψ ≤ K, then for ε > 0, setting am : p q u m λ/r−ε/p−1 u m , bn : v n λ/s−ε/q−1 v n , it has a {am }∞ n0 ∈ φ , b {bn }∞ n0 ∈ ϕ , m n and p T a, b ≤ T p,ψ a p,φ b q,ϕ ≤K ∞ m n0 1/p u m u m ε ∞ n n0 1/q v n v n ε ≤K ∞ m n0 u m um ε 3.13 10 Journal of Inequalities and Applications But on the other side, by 3.8 , it has ∞ ∞ T a, b um m n0 n n0 ∞ um 1−λ/s ε/q ∞ u m m n0 λ/r−ε/p−1 v n K u m ,v n um K u m ,v n ε u mv n v n n n0 1−λ/s ε/q By the strictly monotonic increase of v x and v n0 > 0, v ∞ such that v n > for all n > n1 So it has ∞ 0< K u m ,v n n n0 um ε/q um v n ε/q 1−λ/s ε/q ∞ K u m ,v n ∞ K u m ,v n n n1 um ε/q ωλ m, s − v n ∞, there exists n1 > n0 λ/r ε/q n n1 ≤ um 3.14 λ/r ε/q v n λ/r v n 1−λ/s ε/q um λ/r v n v n K u m ,v n n n0 um λ/r um λ/r v n 1−λ/s ε/q v n um λ/r v n 1−λ/s ε/q um λ/r v n 1−λ/s ε/q v n v n v n 1−λ/s v n n n0 n n0 n n0 n1 −1 1−λ/s K u m ,v n K u m ,v n K u m ,v n v n n1 −1 n1 −1 n1 −1 um 3.15 The series is uniformly convergent for ε ≥ 0, so it has ∞ lim ε→0 K u m ,v n n n0 λ/r ε/q um v n 1−λ/s ε/q v n ωλ m, s 3.16 and for m > n0 , there exists ε0 > 0, when < ε < ε0 , it has ∞ K u m ,v n n n0 um v n λ/r ε/q 1−λ/s ε/q v n > ωλ m, s − um 3.17 Journal of Inequalities and Applications 11 By 3.14 and 3.8 , when < ε < ε0 , it has T a, b ≥ ∞ u m m n0 ≥ kλ um ∞ u m m n0 ∞ ⎧ ⎨ u m kλ m n0 um um 1−O ε um um ωλ m, s − ε ε⎩ ∞ 1− um × u m um m n0 ε kλ u m 3.18 −1 u m m n0 ∞ − ρ ε um O kλ u m ρ In view of 3.13 and 3.18 , letting ε → , it has kλ ≤ K This means that K T p,ψ kλ Theorem 3.1 is proved ⎫ ⎬ ⎭ kλ ; that is, Theorem 3.2 Suppose that p, q and r, s are two pairs of conjugate exponents, r > 1, p > 1, λ ∈ R Let f y : Kλ u m , v y uλ/r m v y , v1−λ/s y g y : Kλ u y , v n vλ/s n u y u1−λ/r y 3.19 Here, u y , v y satisfy the conditions as in Theorem 3.1 Set v n0 /u m Kλ 1, u uλ/s−1 du − f n0 f n0 , 12 3.20 Kλ μ, μλ/r−1 dμ − g n0 g n0 , 12 3.21 R m, s : u n0 /v n R n, r : v n0 /u m η m, s : Kλ 1, u uλ/s−1 du − f n0 3.22 If (a) Kλ x, y ≥ is a homogeneous measurable kernel function of “λ” degree in R2 , such that ∞ < kλ s : Kλ 1, u uλ/s−1 du < ∞, 3.23 (b) functions f y , g y satisfy the conditions of 1.6 ; that is, −1 i F i y > y > n0 , F i ∞ F f, g, i 0, 1, 2, 3, , 3.24 12 Journal of Inequalities and Applications (c) there exists ρ > 0, such that R m, s > 0, R n, r > 0, < η m, s O uρ m m −→ ∞ , 3.25 then it has if a ∈ p , φ b∈ q ϕ, and a ∞ p,φ > 0, b q,ϕ > 0, then ∞ Kλ u m , v n am bn < kλ s a Ta, b b p,φ q,ϕ , 3.26 n n0 m n0 if a ∈ p φ and a ∞ Ta v n p,ψ p,φ > 0, then pλ/s−1 p ∞ v n n n0 K u m , v n am 1/p < kλ s a 3.27 p,φ , m n0 where inequality 3.27 is equivalent to 3.26 and the constant factor kλ s ∞ Kλ u, uλ/r−1 du is the best possible kλ r : Proof By 3.24 , 1.6 , it has ∞ n0 f y dy − f n0 < ωλ m, s ∞ Kλ u m , v n n n0 ∞ ∞ f n < n0 n n0 Kλ u m , v n m n0 ∞ ∞ g m < m n0 v n λ/r 1−λ/s v n 3.28 f y dy − f n0 ∞ < ϑλ n, r um n0 v n um f n0 , 12 λ/s 1−λ/r u m g y dy − g n0 3.29 g n0 12 Journal of Inequalities and Applications Letting ν v y /u m and μ 3.23 , it has ∞ u y /v n in the integral of 3.28 and 3.29 , respectively, by ∞ Kλ u m , v y f y dy n0 n0 ∞ ∞ uλ/r m v y dy v1−λ/s y Kλ 1, ν νλ/s−1 dν − v n0 /u m Kλ 1, ν νλ/s−1 dν v n0 /u m Kλ 1, ν νλ/s−1 dν 3.30 kλ s v n0 /u m − 13 Kλ 1, ν νλ/s−1 dν, ∞ ∞ Kλ u y , v n g y dy n0 n0 ∞ Kλ μ, μλ/r−1 dμ − − ∞ vλ/s n u y dy u1−λ/r y u n0 /v n Kλ μ, μλ/r−1 dμ u n0 /v n Kλ μ, μλ/r−1 dμ kλ s 3.31 u n0 /v n Kλ μ, μλ/r−1 dμ, ∞ λ/r−1 where, letting t 1/u, it has kλ r du Kλ u, u view of 3.28 , 3.30 , 3.20 , 3.22 , and with 3.25 , it has ∞ Kλ 1, t tλ/s−1 dt kλ s In < ωλ m, s < kλ s − Rλ m, s < kλ s , ωλ m, s > kλ s − ηλ m, s kλ s − O1 uρ m ρ > 0, m −→ ∞ 3.32 Similarly, with 3.29 , 3.31 , 3.21 , and 3.25 , it has < ϑλ n, r < kλ s 3.33 also By Theorem 3.1, it has Ta p,ψ < kλ s a p,φ , 3.34 and 3.27 holds In view of Ta, b ≤ Ta 3.26 holds also p,ψ b q,ϕ , 3.35 14 Journal of Inequalities and Applications If 3.26 holds, from 3.26 and a p,φ > 0, there exists n1 > n0 , such that p p−1 H φ m am > and bn H : ψ n > when H > n1 For m n0 Kλ u m , v n am H b : {bn H }n n0 , it has H m n0 H 0< q ϕ n bn H n n0 H p H ψ n H H Kλ u m , v n am n n0 Kλ u m , v n am bn H m n0 1/p H φ n < kλ s 3.36 n n0 m n0 1/q H p an ϕ n n n0 q bn < ∞ H n n0 By p > and q > 1, it follows that H 0< n n0 q p ϕ n bn H < k λ s ∞ p φ n an < ∞ 3.37 n n0 ∞ Letting H → ∞ in 3.37 , it has < < ∞, and it means that b n n0 ϕ n bn ∞ q ∞ {bn ∞ }n n0 ∈ ϕ and b q,ϕ > Therefore, the inequality 3.36 keeps the form of the strict q p inequality when H → ∞ In view of ∞ n0 ϕ n bn ∞ Ta p,ψ , inequality 3.27 holds n kλ s , it is obvious that the constant factor and 3.27 is equivalent to 3.26 By T p,ψ kλ r is the best possible This completes the proof of Theorem 3.2 kλ s q Applications Example 4.1 Set p, q , r, s be two pairs of conjugate exponents and p > 1, s > 1, β ≥ α ≥ e7/12 , < λ ≤ Then it has the following If < ∞ ln αn n ∞, then ∞ ∞ m 1n p 1−λ/r −1 p an /np−1 am bn π < lnλ αm lnλ βn λ sin π/s ∞ < ∞, and < ln αn n p 1−λ/r −1 p an np−1 ∞ n 1/p ln βn ∞ q 1−λ/s −1 ln βn n q bn /nq−1 < q 1−λ/s −1 q bn nq−1 1/q 4.1 If < ∞ n ln βn n ∞ n pλ/s−1 ln αn ∞ m p 1−λ/r −1 p an /np−1 am lnλ αm lnλ βn p < < ∞, then π λ sin π/s p ∞ n ln αn p 1−λ/r −1 p an , np−1 where the constant factors kλ s kλ r : π/λ sin π/s and π/λ sin π/s best possible Inequality 4.2 is equivalent to 4.1 p 4.2 are both the Journal of Inequalities and Applications 15 1/ xλ yλ , x, y ∈ R2 , it is a homogeneous measurable kernel Proof Setting Kλ x, y function of “λ” degree Letting t uλ , it has ∞ < kλ s : Kλ 1, u uλ/s−1 du ∞ uλ/s−1 du uλ λ 4.3 ∞ 1/s−1 t t 1 B , λ s r dt kλ r < ∞ Setting u x ln αx, v x ln βx, then both u x and v x are strictly monotonic increasing differentiable functions in 1, ∞ and satisfy U ∞ U > 0, ∞ 0< n ∞ v n v n ε n 1n ln βn ε ≤ ∞ ∞ n U u, v , ∞ u n un ε n 1n ln αn As β ≥ α ≥ e7/12 , < λ ≤ 1, s > 1, and n0 1, letting uλ/r m v y v1−λ/s y lnλ/r αm , lnλ βy ln1−λ/s βy y lnλ/s βn , lnλ βn ln1−λ/r αy y 4.4 f y Kλ u m , v y g y : Kλ u y , v n vλ/s n u y u1−λ/r y lnλ αm lnλ αy 4.5 y ∈ 1, ∞ , n, m ∈ N, with 2.1 ∼ 2.8 , it has v /u m Rλ m, s 0 < ηλ m, s u /v n Rλ n, r : Kλ 1, u uλ/s−1 du − f O um ρ f > 0, 12 ρ > 0, m −→ ∞ , Kλ μ, μλ/r−1 dμ − g 4.6 g > 12 When < ∞ ln αn p 1−λ/r −1 an /np−1 < ∞ and < ∞ ln βn q 1−λ/s −1 bn /nq−1 < ∞; that n n p q is, a ∈ φ , b ∈ ϕ and a p,φ > 0, b q,ϕ > 0, by Theorem 3.2, inequality 4.1 holds, so does kλ r : π/λ sin π/s 4.2 And 4.2 is equivalent to 4.1 , and the constant factors kλ s and π/λ sin π/s p are both the best possible p q Example 4.2 Set p, q , r, s be two pairs of conjugate exponents and p > 1, s > 1, β ≥ α ≥ 1/2, < λ ≤ Then it has the following 16 Journal of Inequalities and Applications If < ∞ ∞ ∞ n ln m n 0m n m α − n If < ∞ n β ∞ n ∞ pλ/s−1 n 1/p ∞ n α p 1−λ/r −1 p an β q 1−λ/s −1 q bn β q 1−λ/s −1 q bn α < ∞, then m m 4.7 1/q n p 1−λ/r −1 p an ln m ∞ n n n n λ β π < λ sin π/s ∞ n < ∞ and < β am bn α/ n λ p 1−λ/r −1 p an α α/n α λ < ∞, then p β am − n β λ 2p ∞ π λ sin π/s < n α p 1−λ/r −1 p an , n 4.8 where inequality 4.8 is equivalent to 4.7 and the constant factors kλ s π/λ sin π/s and π/λ sin π/s 2p are both the best possible kλ r : Proof Setting Kλ x, y ln x/y / xλ − yλ x, y ∈ R2 , it is a homogeneous measurable kernel function of “λ” degree Letting t uλ , it has ∞ < kλ s : λ Kλ 1, u uλ/s−1 du λ2 ∞ ∞ − ln uλ λ/s−1 u du − uλ 1 B , λ s r ln t 1/s−1 t dt t−1 4.9 π λ sin π/s kλ r < ∞ Setting u x x α, v x x β, then both u x and v x are strictly monotonic increasing differentiable functions in 0, ∞ and satisfy U ∞ U > 0, ∞ 0< n ∞ v n v n ε n n β ε ≤ ∞ ∞ n U u, v , ∞ u n un ε n n α ε As β ≥ α ≥ 1/2, < λ ≤ 1, s > 1, and n0 f1 y uλ/r m v y v1−λ/s y Kλ u m , v y ln y β / m α λ λ m α y β/m α λ − ln m m α/ y α ln y α / n β λ y λ n β y α / n β λ −1 n α β − y β m λ α λ/r y β 1−λ/s n β λ/s , ln y y λ β λ/s−1 y β m α vλ/s n u y u1−λ/r y g1 y : Kλ u y , v n 0, letting α / n α λ − n λ/r−1 , β β λ y α 4.11 1−λ/r y ∈ 0, ∞ , n, m ∈ N, with 2.18 ∼ 2.21 , it has v /u m Rλ m, s : λ2 Kλ 1, u uλ/s−1 du − f1 β/ m α λ ln u 1/s−1 u du − f1 u−1 f > 0, 12 1 Kλ μ, μλ/r−1 dμ − g1 g > 0, 12 u /v n Rλ n, r : 0 < ηλ m, s f 12 1 u m O ρ ρ > 0, m −→ ∞ When < ∞ n α p 1−λ/r −1 an < ∞ and < ∞ n β q 1−λ/s −1 bn < ∞; that is, n n q b ∈ ϕ and a p,φ > 0, b q,ϕ > 0, by Theorem 3.2, inequality 4.7 holds, so does 4.8 p a∈ p φ, 4.12 And 4.8 is equivalent to 4.7 , and the constant factors kλ s π/λ sin π/s 2p are both the best possible q kλ r : π/λ sin π/s and Remark 4.3 It can be proved similarly that, if the conditions “β ≥ α ≥ e7/12 ” in Lemma 2.1 and “β ≥ α ≥ 1/2” in Lemma 2.2 are changed into “α ≥ β ≥ e7/12 ” and “α ≥ β ≥ 1/2”, respectively, Lemmas 2.1 and 2.2 are also valid So the conditions “β ≥ α ≥ e7/12 ” in Example 4.1 and “β ≥ α ≥ 1/2” in Example 4.2 can be replaced by “β ≥ e7/12 , α ≥ e7/12 ” and “β ≥ 1/2, α ≥ 1/2”, respectively Acknowledgment This paper is supported by the National Natural Science Foundation of China (no 10871073) The author would like to thank the anonymous referee for his or her suggestions and corrections 18 Journal of Inequalities and Applications References H Weyl, Singulare Integralgleichungen mit besonderer Beriicksichtigung des Fourierschen Integraltheorems, Inaugeral dissertation, University of Gottingen, Gottingen, Germany, 1908 ă ă G Hardy, J Littiewood, and G Polya, Inequalities, Cambridge University Press, Cambridge, UK, 1934 B C Yang, “On a more accurate Hardy-Hilbert-type inequality and its applications,” Acta Mathematica Sinica, vol 49, no 2, pp 363–368, 2006 Chinese B Yang, “On a more accurate Hilbert’s type inequality,” International Mathematical Forum, vol 2, no 37–40, pp 1831–1837, 2007 W Zhong, “A Hilbert-type linear operator with the norm and its applications,” Journal of Inequalities and Applications, vol 2009, Article ID 494257, 18 pages, 2009 B Yang, The Norm of Operator and Hilbert-Type Inequalities, Science Press, Beijing, China, 2009 J Kuang, Applied Inequalities, Shangdong Science and Technology Press, Jinan, China, 2004  ... inequality,” International Mathematical Forum, vol 2, no 37–40, pp 1831–1837, 2007 W Zhong, ? ?A Hilbert-type linear operator with the norm and its applications,” Journal of Inequalities and Applications,... UK, 1934 B C Yang, “On a more accurate Hardy -Hilbert-type inequality and its applications,” Acta Mathematica Sinica, vol 49, no 2, pp 363–368, 2006 Chinese B Yang, “On a more accurate Hilbert’s... increasing functions u x and v x , a new Hilbert-type inequality, which is with a composite kernel function K u x , v y , and its equivalent are built in this paper As the applications, two new