Hindawi Publishing Corporation Journal of Inequalities and Applications Volume 2007, Article ID 27962, 17 pages doi:10.1155/2007/27962 Research Article On a Multiple Hilbert-Type Integral Inequality with the Symmetric Kernel Wuyi Zhong and Bicheng Yang Received 26 April 2007; Accepted 29 August 2007 Recommended by Sever S Dragomir We build a multiple Hilbert-type integral inequality with the symmetric kernel K(x, y) and involving an integral operator T For this objective, we introduce a norm x n (x ∈ α Rn ), two pairs of conjugate exponents (p, q) and (r,s), and two parameters As applica+ tions, the equivalent form, the reverse forms, and some particular inequalities are given We also prove that the constant factors in the new inequalities are all the best possible Copyright © 2007 W Zhong and B Yang 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, notations, and lemmas If p > 1, 1/ p + 1/q = 1, f (x),g(x) ≥ 0, f ∈ L p (0, ∞), g ∈ Lq (0, ∞), < ( ∞ ∞, and < ( g q (y)d y)1/ p < ∞, then ∞ f (x)g(y) π dx d y < x+y sin(π/ p) ∞ 1/ p f p (x)dx ∞ ∞ f p (x)dx)1/ p < 1/q g q (y)d y , (1.1) where the constant factor π/sin(π/ p) is the best possible Equation (1.1) is the famous Hardy-Hilbert’s inequality proved by Hardy-Riesz [1] in 1925 By introducing the norms f p , g q , and an integral operator T : L p (0, ∞) → L p (0, ∞), Yang [2] rewrite (1.1) as (T f ,g) < π f sin(π/ p) p g q, (1.2) Journal of Inequalities and Applications where (T f ,g) is the formal inner product of T f and g For f ∈ L p (0, ∞) (or g ∈ Lq (0, ∞ ∞)), the integral operator T is defined by (T f )(y) := ( f (x)/(x + y))dx (or (Tg)(x) := ∞ ∞ ∞ p 1/ p q 1/q (g(y)/(x + y))d y) and f p := { | f (x)| dx } , g q := { |g(y)| d y } , then (T f ,g) := ∞ ∞ 0 ∞ f (x) dx g(y)d y = x+y f (x)g(y) dx d y x+y (1.3) Inequality (1.2) posts the relationship of Hilbert inequality and the integral operator T Recently, inequality (1.1) has been extended by [3–6] by using the way of weight function and introducing some parameters A reverse Hilbert-Pachpatte’s inequality was first proved by Zhao in [7] Yang and Zhong [8–10] gave some reverse inequalities concerning some extensions of Hardy-Hilbert’s inequality (1.1) Because of the requirement of higher-dimensional harmonic analysis and higherdimensional operator theory, multiple Hardy-Hilbert integral inequalities have been studied by some mathematicians (see [11–15]) Our major objective of this paper is to build a multiple Hilbert-type integral inequality with the symmetric kernel K(x, y) and involving an integral operator T In order to fulfil the aim, we introduce the norm x n (x ∈ Rn ), two pairs of conjugate exponents (p, q), α + (r,s), and two parameters α, λ As applications, the equivalent form, the reverse forms, and some particular inequalities are given We also prove that the constant factors in the new inequalities are all the best possible For these purposes, we introduce the following notations If p > 1, 1/ p + 1/q = 1, r > 1, 1/r + 1/s = 1, α > 0, λ > 0, and n ∈ Z+ , we set Rn := x = x1 , ,xn : x1 , ,xn > , + x α α α := x1 + · · · + xn 1/α (1.4) If f (x) and ω(x) > are measurable in Rn , define the norm of f with the weight + function ω(x) as f p,ω p := 1/ p ω(x) f (x) dx n R+ (1.5) p If < f p,ω < ∞, it is marked by f ∈ Lω (Rn ) (for < p < or q < 0, we still use (1.5) + with this formal mark in the following) Suppose that K(x, y) is a measurable and symmetric function satisfying K(x, y) = K(y,x) > (for all (x, y) ∈ Rn × Rn ) For f ,g ≥ 0, define an integral operator T as + + (T f )(y) := Rn + K(x, y) f (x)dx y ∈ Rn , + (1.6) K(x, y)g(y)d y x ∈ Rn + (1.7) or (Tg)(x) := Rn + W Zhong and B Yang Then we have the formal inner product as (T f ,g) = (Tg, f ) = Rn + K(x, y) f (x)g(y)dx d y (1.8) We also define the following weight functions: Cα,λ,n (s,x) := K(x, y) Rn + C α,λ,n (q,s,ε,x) := K(x, y) n R+ x λ/r α d y, n y α−λ/s x y (1.9) λ/r+ε/q α n−λ/s+ε/q d y, α (1.10) and the notation as C := x α >1 x −n −ε α 0< y α ≤1 x K(x, y) y λ/r+ε/q α n−λ/s+ε/q dx d y, α (1.11) where ε > in (1.10) and (1.11) are small enough Lemma 1.1 (cf [16]) Assume that p > 0, 1/ p + 1/q = 1, F,G ≥ 0, and F ∈ L p (E), G ∈ Lq (E) We have the following Hălders inequalities: o (1) if p > 1, then 1/ p E F(t)G(t)dt ≤ E 1/q F p (t)dt E Gq (t)dt ; (1.12) , (1.13) (2) if < p < 1, then 1/ p E F(t)G(t)dt ≥ E 1/q F p (t)dt E Gq (t)dt where equality holds if and only if there exists nonnegative real numbers A and B (A2 + B2 = 0) such that AF p (t) = BGq (t) a.e in E Lemma 1.2 (cf [17]) If pi > (i = 1,2, ,n), α > 0, and Ψ(u) is a measurable function, then ··· p −1 α α {(x1 , ,xn )∈Rn ;(x1 +···+xn )≤1} + = Γ p1 /α · · · Γ pn /α αn Γ p1 + · · · + pn /α α α Ψ x1 + · · · + xn x1 1 p n −1 · · · xn dx1 · · · dxn (1.14) Ψ(u)u((p1 +···+pn )/α)−1 du, where Γ(·) is the Gamma function By (1.14), it is easy to obtain following result 4 Journal of Inequalities and Applications Lemma 1.3 If pi > (i = 1,2, ,n), α > 0, and Ψ(u) is a measurable function, then p −1 Rn + α α Ψ x1 + · · · + xn x1 p n −1 · · · xn ∞ Γ p1 /α · · · Γ pn /α αn Γ p1 + · · · + pn /α = dx1 · · · dxn (1.15) Ψ(u)u((p1 +···+pn )/α)−1 du Proof In view of (1.14), setting t = ρα u, we have p −1 Rn + α α Ψ x1 + · · · + xn x1 = lim ρ p1 +···+pn ··· ρ→∞ × Ψ ρα p n −1 · · · xn {(x1 , ,xn )∈Rn ;((x1 /ρ)α +···+(xn /ρ)α )≤1} + α x1 ρ dx1 · · · dxn xn ρ + ··· + α x1 ρ p −1 ··· p n −1 xn ρ d x1 xn ···d ρ ρ Γ p1 /α · · · Γ pn /α Ψ ρα u u((p1 +···+pn )/α)−1 du ρ→∞ p1 + · · · + pn /α ∞ Γ p1 /α · · · Γ pn /α = n Ψ(t)t ((p1 +···+pn )/α)−1 dt, α Γ p1 + · · · + pn /α = lim ρ p1 +···+pn αn Γ (1.16) and (1.15) holds The lemma is proved By (1.14) and (1.15), we still have the following lemma Lemma 1.4 If pi > (i = 1,2, ,n), α > 0, and Ψ(u) is a measurable function, then ··· p −1 α α {(x1 , ,xn )∈Rn ;(x1 +···+xn )>1} + α α Ψ x1 + · · · + xn x1 Γ p1 /α · · · Γ pn /α = n α Γ p1 + · · · + pn /α ∞ p n −1 · · · xn dx1 · · · dxn (1.17) Ψ(u)u (p1 +···+pn )/α−1 du Lemma 1.5 For ε > small enough and n ∈ Z+ , we have x α >1 x −n −ε α dx = Γn 1/α ε · αn−1 Γ n/α (1.18) Proof By using Lemma 1.4, we have x α >1 = = x −n −ε α ··· dx α α {(x1 , ,xn )∈Rn ;(x1 +···+xn )>1} + Γn 1/α αn Γ n/α ∞ α α x1 + · · · + xn u−(n+ε)/α un/α−1 du = Hence (1.18) is valid The lemma is Γn 1/α αn Γ n/α −(n+ε)/α 1−1 x1 ∞ 1 · · · xn−1 dx1 · · · dxn u−ε/α−1 du (1.19) W Zhong and B Yang Main results Theorem 2.1 Suppose that p > 1, 1/ p + 1/q = 1, r > 1, 1/r + 1/s = 1, α,λ > 0, n ∈ Z+ , p(n−λ/r)−n , (y) = f ,g ≥ 0, K(x, y) is a measurable and symmetric function, ω(x) = x α q(n−λ/s)−n pλ/s−n y α , h(y) = y α , and the integral operator T is defined by (1.6) (or (1.7)) If Cα,λ,n (s,x) = Cα,λ,n (s) = Cα,λ,n (r), (2.1) ε −→ 0+ C α,λ,n (q,s,ε,x) = Cα,λ,n (s) + o(1) (2.2) are all constants independent of x, and C = O(1) ε − 0+ , → (2.3) where Cα,λ,n (s,x), C α,λ,n (q,s,ε,x) and C are defined by (1.9), (1.10), and (1.11), respectively We have the following: p q (1) if f ∈ Lω (Rn ), g ∈ L (Rn ), then + + (T f ,g) = Rn + K(x, y) f (x)g(y)dx d y < Cα,λ,n (s) f p p,ω g q, ; (2.4) p (2) if f ∈ Lω (Rn ), then T f ∈ Lh (Rn ) and + + Tf p,h = Rn + y 1/ p p pλ/s−n α K(x, y) f (x)dx n dy R+ < Cα,λ,n (s) f p,ω , (2.5) where the same constant factor Cα,λ,n (s) in (2.4) and (2.5) is the best possible Inequalities (2.4) and (2.5) are equivalent Proof (1) Since p > 1, we use Hă lders inequality (1.12) in the following: o (T f ,g) = Rn + ≤ K 1/ p (x, y) f (x) K(x, y) n Rn + × R+ Rn + Rn + x y (1/q)(n−λ/r) α (1/ p)(n−λ/s) α x λ/r α dy n y α−λ/s K(x, y) y x x λ/s α n−λ/r dx α K 1/q (x, y)g(y) (1/ p)(n−λ/s) α (1/q)(n−λ/r) x α 1/ p y dx d y p(n−λ/r)−n p f (x)dx α q(n−λ/s)−n q g (y)d y α y 1/q (2.6) By (1.9), (2.1), and notations (1.5), (1.8), it follows (T f ,g) ≤ Cα,λ,n (s) f p,ω g q, (2.7) Journal of Inequalities and Applications If (2.6) takes the form of equality, then by Lemma 1.1, there exist real numbers A and B (A2 + B2 = 0), such that A x (p−1)(n−λ/r) α n y α−λ/s f p (x) = B (q−1)(n−λ/s) α g q (y), n x α−λ/r y a.e in Rn × Rn + + (2.8) It follows that there exists a constant E, such that A x p(n−λ/r) p f (x) = B α y q(n−λ/s) q g (y) = E, α a.e in Rn × Rn + + (2.9) Without lose of generality, suppose that A = We have p(n−λ/r)−n p f (x) = α x E A x n α , a.e in Rn , + (2.10) p which contradicts the fact that f ∈ Lω (Rn ) Hence, (2.6) takes the form of strict inequal+ ity; so does (2.7) Then we obtain (2.4) Suppose there exists a number < C ≤ Cα,λ,n (s), such that (2.4) is still valid if we replace Cα,λ,n (s) by C In particular, for ε > small enough, setting ⎧ ⎨ x fε (x) = ⎩ 0, ⎧ ⎨ y gε (y) = ⎩ 0, −(n−λ/r)−ε/ p α −(n−λ/s)−ε/q α , x ∈ x α > ∩ Rn , + x ∈ < x α ≤ ∩ Rn ; + , y ∈ y α > ∩ Rn , + y ∈ < y α ≤ ∩ Rn , + (2.11) it follows T fε ,gε < C Rn + =C x x α >1 p(n−λ/r)−n p fε (x)dx α x −n −ε α dx = C 1/ p Rn + y Γn (1/α) ε · αn−1 Γ n/α q(n−λ/s)−n q gε (y)d y α 1/q (2.12) (by (1.18)) But by (2.2), (1.18), and (2.3), we have T fε ,gε = Rn + = x α >1 − = K(x, y) fε (x)gε (y)dx d y x −n −ε α 0< y α ≤1 Γn (1/α) Rn + K(x, y) x K(x, y) x ε · αn−1 Γ(n/α) λ/r+ε/q α y λ/r+ε/q α y −(n−λ/s)−ε/q α −(n−λ/s)−ε/q α Cα,λ,n (s) + o(1) + o(1) dy (2.13) d y dx (ε − 0+ ) → In view of (2.12) and (2.13), we have [Cα,λ,n (s) + o(1)](1 + o(1)) < C, and then Cα,λ,n (s) ≤ C (ε → 0+ ) Hence the constant factor C = Cα,λ,n (s) is the best possible W Zhong and B Yang pλ/s−n (2) Setting g(y) = y α ( Rn K(x, y) f (x)dx) p−1 (y ∈ Rn ), then we have g(y) ≥ + + Using the notation (1.5), by Hă lders inequality (1.12) (as (2.6)) and (2.1), we have o Tf p p,h q q, = g = = Rn + p pλ/s−n α y q(n−λ/s)−n q g (y)d y α y Rn + d y = (T f ,g) ≤ Cα,λ,n (s) f K(x, y) f (x)dx n R+ p,ω g q, , (2.14) which is equivalent to p p,h Tf = g p q q, p ≤ Cα,λ,n (s) f p p,ω q (2.15) p In view of f ∈ Lω (Rn ), it follows that g ∈ L (Rn ) and T f ∈ Lh (Rn ) Using the result of + + + (2.4), we can find that inequality (2.14) takes the strict form; so does (2.15) Hence we obtain (2.5) On the other hand, if inequality (2.5) holds, then by using the Hă lders inequality o (1.12) again, we nd (T f ,g) = = ≤ Rn + K(x, y) f (x)g(y)dx d y y Rn + Rn + λ/s−n/ p α y pλ/s−n α Rn + K(x, y) f (x)dx y n/ p−λ/s g(y) α 1/ p p K(x, y) f (x)dx n R+ dy dy Rn + y q(n−λ/s)−n q g (y)d y α 1/q (2.16) By (2.5), we have (2.4) It follows that (2.5) is equivalent to (2.4) If the constant factor Cα,λ,n (s) in (2.5) is not the best possible, then by (2.16), we can get a contradiction that the constant factor Cα,λ,n (s) in (2.4) is not the best possible The theorem is proved Theorem 2.2 Let < p < (q < 0), 1/ p + 1/q = 1, r > 1, 1/r + 1/s = 1, α,λ > 0, and n ∈ Z+ Assume that f ,g ≥ 0, K(x, y), ω(x), (y), h(y) are all defined as in Theorem 2.1, qλ/r −n , the integral operator T is defined by (1.6) (or (1.7)), and the weight setting φ(x) = x α functions Cα,λ,n (s,x) and C α,λ,n (q,s,ε,x) satisfy (2.1) and (2.2) Then we have the following: p q (1) if f ∈ Lω (Rn ) and g ∈ L (Rn ), then + + (T f ,g) = Rn + K(x, y) f (x)g(y)dx d y > Cα,λ,n (s) f p,ω g q, ; (2.17) p (2) if f ∈ Lω (Rn ), then + Tf p,h = Rn + y pλ/s−n α 1/ p p Rn + K(x, y) f (x)dx dy > Cα,λ,n (s) f p,ω ; (2.18) Journal of Inequalities and Applications q q (3) if g ∈ L (Rn ), then Tg ∈ Lφ (Rn ), and + + Tg q q,φ = x Rn + q qλ/r −n α K(x, y)g(y)d y n R+ q dx < Cα,λ,n (s) g q q, , (2.19) q where the constant factors Cα,λ,n (s) and Cα,λ,n (s) are the best possible Inequalities (2.18) and (2.19) are all equivalent to inequality (2.17) Proof (1) Since < p < (q < 0), we can use the reverse Hă lders inequality (1.13) Using o the combination as (2.6) and notation (1.8), we have (T f ,g) = Rn + ≥ K 1/ p (x, y) f (x) K(x, y) n Rn + × R+ Rn + (1/q)(n−λ/r) α (1/ p)(n−λ/s) α x y x λ/r α dy n y α−λ/s K(x, y) n R+ y x x p(n−λ/r)−n p f (x)dx α x λ/s α n−λ/r dx α y K (1/q) (x, y) (1/ p)(n−λ/s) α (1/q)(n−λ/r) α 1/ p q(n−λ/s)−n q g (y)d y α y dx d y 1/q (2.20) By (1.9), (2.1), and notation (1.5), we have (T f ,g) ≥ Cα,λ,n (s) f p,ω g q, (2.21) If (2.20) takes the form of equality, then by using the conclusions of (2.8)–(2.10), we still p q can get a result which contradicts the condition of f ∈ Lω (Rn ) (or g ∈ L (Rn )) It means + + that (2.20) takes the form of strict inequality; so does (2.21) The form (2.17) is valid If there exists a positive number C ≥ Cα,λ,n (s), such that (2.17) is still valid if we replace Cα,λ,n (s) by C, then in particular, for ε > small enough, setting fε (x) and gε (y) as (2.11), we have T fε ,gε > C fε p,ω gε =C q, x α >1 x −n −ε α dx (2.22) But by (1.10) and (2.2), we have T fε ,gε = ≤ Rn + K(x, y) fε (x)gε (y)dx d y x α >1 x −n −ε α = Cα,λ,n (s) + o(1) Rn + K(x, y) x x α >1 x λ/r+ε/q α −n −ε α dx y −(n−λ/s)−ε/q α d y dx (2.23) W Zhong and B Yang In view of (2.22) and (2.23), we find C < Cα,λ,n (s) + o(1), and then C ≤ Cα,λ,n (s) (ε → 0+ ) Hence the constant C = Cα,λ,n (s) is the best possible pλ/s−n (2) Setting g(y) = y α ( Rn K(x, y) f (x)dx) p−1 (y ∈ Rn ), it follows g(y) ≥ By + + Notation (1.5) and in view of (2.21), we have Tf p p,h = Tf p p,h q q, = g = y Rn + q q, = g p Rn + q(n−λ/s)−n q g (y)d y α y p pλ/s−n α d y = (T f ,g) ≥ Cα,λ,n (s) f K(x, y) f (x)dx n R+ p ≥ Cα,λ,n (s) g , q, (2.24) f q p,ω p p,ω (2.25) p If T f p,h = g q, = ∞, by f ∈ Lω (Rn ), (2.25) takes the form of strict inequality (2.18) + p q holds If T f ∈ Lh (Rn ) (g ∈ L (Rn )), this tells us that the condition of (2.17) is satisfied, + + then by using (2.17), it follows that both (2.24) and (2.25) keep the strict forms and (2.18) holds On the other hand, if (2.18) is valid, using the reverse Hă lders inequality (1.13) again, o we have (T f ,g) = ≥ λ/s−n/ p α y Rn + Rn + y Rn + K(x, y) f (x)dx y K(x, y) f (x)dx n dy R+ dy 1/ p p pλ/s−n α n/ p−λ/s g(y) α Rn + y q(n−λ/s)−n q g (y)d y α 1/q (2.26) By (2.18), we have (2.17) This means that (2.18) is equivalent to (2.17) qλ/r −n ( Rn K(x, y)g(y)d y)q−1 (x ∈ Rn ), then it follows (3) Firstly, setting f (x) = x α + + f (x) ≥ Using the notation (1.5) and in view of (1.9), (2.1), and (2.20), we have Tg q q,φ = f = Rn + p p,ω x = Rn + x p(n−λ/r)−n p f (x)dx α q qλ/r −n α K(x, y)g(y)d y n R+ d y = (Tg, f ) ≥ Cα,λ,n (s) f p,ω g q, (2.27) It follows Tg q,φ p/q p,ω = f = Rn + x p(n−λ/r)−n p f (x)dx α 1/q ≥ Cα,λ,n (s) g q, (2.28) , and by q < 0, we have 0< Tg q q,φ = f p p,ω = Rn + x qλ/r −n α q K(x, y)g(y)d y n R+ q d y ≤ Cα,λ,n (s) g q q, < ∞ (2.29) 10 Journal of Inequalities and Applications q p This follows that Tg ∈ Lφ (Rn ), f ∈ Lω (Rn ) And by (2.17), we find that (2.27)–(2.29) are + + strict inequalities Thus inequality (2.19) holds Secondly, if (2.19) is valid, using the reverse Hă lders inequality (1.13) again, in view o of (T f ,g) = = ≥ Rn + K(x, y) f (x)g(y)dx d y n/q−λ/r α x Rn + Rn + x f (x) λ/r −n/q α x p(n−λ/r)−n p f (x)dx α Rn + K(x, y)g(y)d y dx 1/ p Rn + x qλ/r −n α q Rn + K(x, y)g(y)d y dx 1/q , (2.30) by (2.19) and q < 0, it follows that (2.17) holds, and (2.19) is equivalent to (2.17) q If the constant factor Cα,λ,n (s) (or Cα,λ,n (s)) in (2.18) (or in (2.19)) is not the best possible, then by (2.26) (or (2.30)), we can get a contradiction that the constant factor Cα,λ,n (s) in (2.17) is not the best possible The theorem is proved Applications to some particular cases Corollary 3.1 Let p > 0, 1/ p + 1/q = 1, r > 1, 1/r + 1/s = 1, α > 0, < λ < 1, n ∈ Z+ , p(n−λ/r)−n q(n−λ/s)−n , (y) = y α , and f ,g ≥ Then ω(x) = x α p q (1) if p > 1, f ∈ Lω (Rn ), and g ∈ L (Rn ), then + + f (x)g(y) Rn + x α− y λ dx d y < Cα,λ,n (s) f p,ω g q, ; (3.1) α p (2) if p > 1, f ∈ Lω (Rn ), then + Rn + y pλ/s−n α p f (x) Rn + x α− y p λ dx p d y < Cα,λ,n (s) f p p,ω ; (3.2) α q (3) if < p < 1, f ∈ Lω (Rn ), and g ∈ L (Rn ), then + + f (x)g(y) Rn + x α− y λ dx d y > Cα,λ,n (s) f p,ω g q, ; (3.3) α p (4) if < p < and f ∈ Lω (Rn ), then + Rn + y pλ/s−n α p f (x) Rn + x α− y λ dx p p p,ω ; (3.4) q q q, (3.5) d y > Cα,λ,n (s) f α q (5) if < p < and g ∈ L (Rn ), then + Rn + x qλ/r −n α q g(y) Rn + x α− y λ dy α dx < Cα,λ,n (s) g , W Zhong and B Yang 11 where the constant factors Cα,λ,n (s) = (Γn (1/α)/αn−1 Γ(n/α))[B(λ/s,1 − λ) + B(λ/r,1 − λ)] p q (B(·, ·) is the Beta function) and Cα,λ,n (s), Cα,λ,n (s) are the best possible Inequality (3.2) is equivalent to (3.1); inequalities (3.4) and (3.5) are all equivalent to (3.3) Proof Setting K(x, y) = 1/(| x α − y α |λ ), it is a measurable function, satisfying K(x, y) = K(y,x) > (for all (x, y) ∈ Rn × Rn ) In view of Theorems 2.1 and 2.2, just need to + + prove that conditions (2.1)–(2.3) are all satisfied (a) When p > 0, by (1.9) and (1.15), setting t = u1/α , we have Cα,λ,n (s,x) = x λ/r α Rn + −(n−λ/s) y λ α− y x α ∞ u−(n−λ/s)/α x = x λ/r α Γn (1/α) αn Γ(n/α) = x λ/r α Γn (1/α) αn−1 Γ(n/α) dy α x α −u u 1/α λ x du (3.6) ∞ t λ/s−1 α n/α−1 λ dt + α −t x t λ/s−1 α t− x λ dt α Setting v = t/ x α , we have x λ/r α Γn (1/α) αn−1 Γ(n/α) x t λ/s−1 α α −t x λ dt = Γn (1/α) αn−1 Γ(n/α) vλ/s−1 dv (1 − v)λ (3.7) uλ/r −1 du (1 − u)λ (3.8) Setting u = x α /t, it follows that dt = − x α u−2 du and x λ/r α Γn (1/α) n−1 Γ(n/α) α ∞ x t λ/s−1 α t− x λ dt = α Γn (1/α) n−1 Γ(n/α) α In view of (3.7), (3.8), and < λ < 1, it follows Cα,λ,n (q,s,x) = Cα,λ,n (s) = Cα,λ,n (r) = (Γn (1/α)/αn−1 Γ(n/α))[B(λ/s,1 − λ) + B(λ/r,1 − λ)], and condition (2.1) is satisfied (b) When p > 1, by (1.10) and (1.15), setting t = u1/α , for < ε < qλ/s, we have C α,λ,n (q,s,ε,x) = x λ/r+ε/q α = x λ/r+ε/q α = x λ/r+ε/q α Rn + x α− Γn (1/α) αn Γ(n/α) ∞ y y −(n−λ/s)−ε/q α 1/α α −u x Γn (1/α) x dy α u−(n−λ/s+ε/q)/α αn−1 Γ(n/α) λ α λu t λ/s−ε/q−1 x α −t n/α−1 du ∞ λ dt + x t λ/s−ε/q−1 α t− x λ dt α (3.9) 12 Journal of Inequalities and Applications Setting v = t/ x or u = x α /t, respectively, as (3.7) or (3.8), we find α C α,λ,n (q,s,ε,x) = Γn (1/α) αn−1 Γ(n/α) vλ/s−ε/q−1 dv + (1 − v)λ uλ/r+ε/q−1 du (1 − u)λ (3.10) Γn (1/α) λ ε λ ε B − ,1 − λ + B + ,1 − λ = n −1 α Γ(n/α) s q r q It follows that condition (2.2) is satisfied Note When < p < (q < 0), setting < ε < −qλ/r, the constant C α,λ,n (q,s,ε,x) satisfies (2.2) as well (c) If p > 1, by (1.11), (1.14), and (1.17), respectively, setting t = u1/α and v = t/ x α , for < ε < qλ/s and < λ < 1, we have 01 x α 0< y Γn (1/α) αn−1 Γ(n/α) Γn (1/α) n−1 Γ(n/α) α Γn (1/α) = n −1 α Γ(n/α) λ/r+ε/ p α λ n−λ/s+ε/q dx d y y α α x −n −ε x α >1 x α >1 x α >1 α ≤1 x −n −ε α x −n x −n α α α− x y 1/ x dx 1/ x dx α 1/ x dx vλ/s−ε/q−1 dv (1 − v)λ α α vλ/s−ε/q−1 dv 1−v (3.11) ∞ v k+λ/s−ε/q−1 dv k =0 ∞ = Γn (1/α) αn−1 Γ(n/α) k=0 k + λ/s − ε/q = Γn (1/α) n−2 Γ(n/α) α k + λ/s − ε/q k =0 x α >1 x −(n+k+λ/s−ε/q) α dx ∞ It follows that C satisfies (2.3) In view of (3.7)–(3.11), by Theorems 2.1 and 2.2, Corollary 3.1 is proved Corollary 3.2 Suppose that p > 0, 1/ p + 1/q = 1, r > 1, 1/r + 1/s = 1, α,λ > 0, n ∈ Z+ , p(n−λ/r)−n q(n−λ/s)−n , (y) = y α , and f ,g ≥ Then ω(x) = x α p q (1) if p > 1, f ∈ Lω (Rn ), and g ∈ L (Rn ), then + + ln Rn + y α/ x y λ− x α α λ α f (x)g(y)dx d y < Cα,λ,n (s) f p,ω g q, ; (3.12) W Zhong and B Yang 13 p (2) if p > 1, f ∈ Lω (Rn ), then + Rn + y y α / x α f (x) dx y λ− x λ α α ln pλ/s−n α Rn + p p p p p,ω ; (3.13) ; (3.14) p p p,ω ; (3.15) q q q, (3.16) d y < Cα,λ,n (s) f q (3) if < p < 1, f ∈ Lω (Rn ), and g ∈ L (Rn ), then + + ln y Rn + y α/ x λ− x α α λ α f (x)g(y)dx d y > Cα,λ,n (s) f p,ω g q, p (4) if < p < and f ∈ Lω (Rn ), then + Rn + y pλ/s−n α y α / x α f (x) dx y λ− x λ α α ln Rn + p d y > Cα,λ,n (s) f q (5) if < p < and g ∈ L (Rn ), then + Rn + x qλ/r −n α ln Rn + y α / x α g(y) dy y λ− x λ α α q dx < Cα,λ,n (s) g , where the constant factor Cα,λ,n (s) = (Γn (1/α)/λ2 αn−1 Γ(n/α))B2 (1/s,1/r) (B(·, ·) is Beta p q function) and Cα,λ,n (s), Cα,λ,n (s) are all the best possible Inequality (3.13) is equivalent to (3.12); inequalities (3.15) and (3.16) are all equivalent to (3.14) Proof Setting K(x, y) = ln( y α / x α )/( y λ − x λ ), it is a measurable function, satε α isfying K(x, y) = K(y,x) > As in Corollary 3.1, we just need to prove that conditions (2.1)–(2.3) are all satisfied Setting t = u1/α and v = (t/ x α )λ , respectively, we can find the results in the following (a) When p > 0, by (1.9) and (1.15), we have ln y α/ x y λ− x α Cα,λ,n (s,x) = x λ/r α = x λ/r α Γn (1/α) αn Γ(n/α) = x λ/r α Γn (1/α) n−1 Γ(n/α) α = Rn + Γn (1/α) λ2 αn−1 Γ(n/α) ∞ y −(n−λ/s) α dy lnu1/α − ln x α u−(n−λ/s)/α n/α−1 u du uλ/α − x λ α ∞ α λ α ∞ lnt − ln x α t λ/s−1 dt tλ − x λ α (3.17) lnv 1/s−1 dv v v−1 It follows Cα,λ,n (q,s,x) = Cα,λ,n (s) = (Γn (1/α)/λ2 αn−1 Γ(n/α))B2 (1/s,1/r) satisfies (2.1) 14 Journal of Inequalities and Applications (b) When p > 1, for < ε < qλ/s, by (1.10) and (1.15), we have C α,λ,n (q,s,ε,x) = x = x λ/r+ε/q α = = y α/ x y λ− x α ln λ/r+ε/q α Rn + ∞ Γn (1/α) αn Γ(n/α) ∞ Γn (1/α) λ2 αn−1 Γ(n/α) Γn (1/α) λ2 αn−1 Γ(n/α) α λ α y −(n−λ/s)−ε/q α dy ln u1/α − ln x α u−(n−λ/s+ε/q)/α n/α−1 u du uλ/α − x λ α lnv 1/s−ε/qλ−1 dv v v−1 B2 ε ε − , + s qλ r qλ (3.18) It follows that (2.2) is valid Note When < p < (q < 0), setting < ε < −qλ/r, the constant C α,λ,n (q,s,ε,x) satisfies (2.2) as well (c) If p > 1, then for < ε < qλ/s, by (1.11), (1.14), and (1.17), we have 01 0< y α ≤1 = Γn (1/α) λ2 αn−1 Γ(n/α) ≤ Γn (1/α) αn−1 Γ(n/α) λ = Γn (1/α) αn−1 Γ(n/α) λ = Γn (1/α) αn−1 Γ(n/α) λ y α/ x y x α >1 x α >1 x α >1 x x x λ− α −n −ε α −n α −n α λ α x −(n−λ/r)−ε/ p α dx d n−λ/s+ε/q α x α y 1/ x dx 1/ x dx λ α λ α lnv 1/s−ε/qλ−1 dv v v−1 ∞ (− lnv) k =0 ∞ dx k + 1/s − ε/qλ k =0 ∞ λ k + 1/s − ε/qλ k =0 x α >1 x = 2Γn (1/α) λ3 αn−2 Γ(n/α) (k + 1/s − ε/qλ)2 k =0 1/ x λ α (− lnv)dvk+1/s−ε/qλ −[n+λ(k+1/s−ε/qλ)] α ∞ + vk+1/s−ε/qλ−1 dv x α >1 x ln x α dx −[n+λ(k+1/s−ε/qλ)] α dx ∞ (k + 1/s − ε/qλ)3 k =0 (3.19) It is obvious that C is a bounded quantity and satisfies (2.3) In view of (3.17)–(3.19), by Theorems 2.1 and 2.2, Corollary 3.2 is proved Similarly, by setting K(x, y) = 1/( x λ + y λ ) and K(x, y) = 1/(Max{ x α , y α })λ , α α respectively, we have Corollaries 3.3 and 3.4 in the following In order to compress the length of the paper, the proof for Corollaries 3.3 and 3.4 are here omitted W Zhong and B Yang 15 Corollary 3.3 Suppose that p > 0, 1/ p + 1/q = 1, r > 1, 1/r + 1/s = 1, α,λ > 0, n ∈ Z+ , p(n−λ/r)−n q(n−λ/s)−n , (y) = y α , and f ,g ≥ Then ω(x) = x α p q (1) if p > 1, f ∈ Lω (Rn ), and g ∈ L (Rn ), then + + f (x)g(y) dx d y < Cα,λ,n (s) f x λ+ y λ α α Rn + p,ω g q, ; (3.20) p (2) if p > 1, f ∈ Lω (Rn ), then + Rn + y p f (x) pλ/s−n α Rn + x λ+ α p y λ α dx p p p,ω ; d y < Cα,λ,n (s) f (3.21) q (3) if < p < 1, f ∈ Lω (Rn ), and g ∈ L (Rn ), then + + f (x)g(y) dx d y > Cα,λ,n (s) f x λ+ y λ α α Rn + p,ω g q, ; (3.22) p (4) if < p < and f ∈ Lω (Rn ), then + Rn + y pλ/s−n α Rn + f (x) λ+ y x α p dx λ α p p p,ω ; (3.23) q q q, (3.24) d y > Cα,λ,n (s) f q (5) if < p < and g ∈ L (Rn ), then + Rn + qλ/r −n α x q g(y) Rn + x λ+ α y λ α dy dx < Cα,λ,n (s) g , where the constant factors Cα,λ,n (s) = (Γn (1/α)/λαn−1 Γ(n/α))B(1/s,1/r) (B(·, ·) is the Beta p q function) and Cα,λ,n (s), Cα,λ,n (s) are all the best possible Inequality (3.21) is equivalent to (3.20); inequalities (3.23) and (3.24) are all equivalent to (3.22) Corollary 3.4 Suppose that p > 0, 1/ p + 1/q = 1, r > 1, 1/r + 1/s = 1, α,λ > 0, n ∈ Z+ , p(n−λ/r)−n q(n−λ/s)−n , (y) = y α , and f ,g ≥ Then ω(x) = x α p n ), and g ∈ Lq (Rn ), then (1) if p > 1, f ∈ Lω (R+ + f (x)g(y) Rn + max x α, y λ dx d y < Cα,λ,n (s) f p,ω g q, ; (3.25) α p (2) if p > 1, f ∈ Lω (Rn ), then + Rn + y pλ/s−n α p f (x) Rn + max x α, y p λ dx p d y < Cα,λ,n (s) f p,ω ; (3.26) α q (3) if < p < 1, f ∈ Lω (Rn ), and g ∈ L (Rn ), then + + f (x)g(y) Rn + max x α, y λ dx d y α > Cα,λ,n (s) f p,ω g q, ; (3.27) 16 Journal of Inequalities and Applications p (4) if < p < and f ∈ Lω (Rn ), then + Rn + y pλ/s−n α p f (x) Rn + max x α, y λ dx p p p,ω ; (3.28) q q q, (3.29) d y > Cα,λ,n (s) f α q (5) if < p < and g ∈ L (Rn ), then + Rn + x qλ/r −n α q g(y) Rn + max x α, y λ dy dx < Cα,λ,n (s) g , α p q where the constant factors Cα,λ,n (s) = srΓn (1/α)/λαn−1 Γ(n/α) and Cα,λ,n (s), Cα,λ,n (s) are all the best possible Inequality (3.26) is equivalent to (3.25); inequalities (3.28) and (3.29) are all equivalent to (3.27) Remark 3.5 For n = 1, the inequalities in Corollaries 3.1–3.4 reduce to the correspondent inequalities in the 2-dimensional space References ´ [1] G H Hardy, J E Littlewood, and G Polya, Inequalities, Cambridge University Press, Cambridge, UK, 2nd edition, 1952 [2] B Yang, “On the norm of an integral operator and applications,” Journal of Mathematical Analysis and Applications, vol 321, no 1, pp 182–192, 2006 [3] I Brneti´ and J Peˇ ari´ , “Generalization of inequalities of Hardy-Hilbert type,” Mathematical c c c Inequalities & Applications, vol 7, no 2, pp 217–225, 2004 [4] W Zhong and B Yang, “A best extension of Hilbert inequality involving seveial parameters,” Jinan University Journal (Natural Science and Medical Edition), vol 28, no 1, pp 20–23, 2007 (Chinese) [5] B Yang and L Debnath, “On the extended Hardy-Hilbert’s inequality,” Journal of Mathematical Analysis and Applications, vol 272, no 1, pp 187–199, 2002 [6] B Yang and M Z Gao, “An optimal constant in the Hardy-Hilbert inequality,” Advances in Mathematics, vol 26, no 2, pp 159–164, 1997 (Chinese) [7] C.-J Zhao and L Debnath, “Some new inverse type Hilbert integral inequalities,” Journal of Mathematical Analysis and Applications, vol 262, no 1, pp 411–418, 2001 [8] B Yang, “A reverse of the Hardy-Hilbert’s type inequality,” Journal of Southwest China Normal University (Natural Science), vol 30, no 6, pp 1012–1015, 2005 [9] W Zhong, “A reverse Hilbert’s type integral inequality,” International Journal of Pure and Applied Mathematics, vol 36, no 3, pp 353–360, 2007 [10] W Zhong and B Yang, “On the extended forms of the reverse Hardy-Hilbert’s integral inequalities,” Journal of Southwest China Normal University (Natural Science), vol 29, no 4, pp 44–48, 2007 [11] B Yang, “A multiple Hardy-Hilbert integral inequality,” Chinese Annals of Mathematics, vol 24, no 6, pp 743–750, 2003 [12] I Brneti´ and J Peˇ ari´ , “Generalization of Hilbert’s integral inequality,” Mathematical Inequalc c c ities & Applications, vol 7, no 2, pp 199–205, 2004 [13] I Brneti´ , M Krni´ , and J Peˇ ari´ , “Multiple Hilbert and Hardy-Hilbert inequalities with nonc c c c conjugate parameters,” Bulletin of the Australian Mathematical Society, vol 71, no 3, pp 447– 457, 2005 [14] B Yang and T M Rassias, “On the way of weight coefficient and research for the Hilbert-type inequalities,” Mathematical Inequalities & Applications, vol 6, no 4, pp 625–658, 2003 W Zhong and B Yang 17 [15] Y Hong, “On multiple Hardy-Hilbert integral inequalities with some parameters,” Journal of Inequalities and Applications, vol 2006, Article ID 94960, 11 pages, 2006 [16] J Kuang, Applied Inequalities, Shangdong Science and Technology Press, Jinan, China, 2004 [17] G M Fichtingoloz, A Course in Differential and Integral Calculus, Renmin Education, Beijing, China, 1957 Wuyi Zhong: Department of Mathematics, Guangdong Institute of Education, Guangzhou, Guangdong 510303, China Email address: wp@bao.ac.cn Bicheng Yang: Department of Mathematics, Guangdong Institute of Education, Guangzhou, Guangdong 510303, China Email address: bcyang@pub.guangzhou.gd.cn ... an integral operator and applications,” Journal of Mathematical Analysis and Applications, vol 321, no 1, pp 182–192, 2006 [3] I Brneti´ and J Peˇ ari´ , “Generalization of inequalities of Hardy-Hilbert... Education, Guangzhou, Guangdong 510303, China Email address: wp@bao.ac.cn Bicheng Yang: Department of Mathematics, Guangdong Institute of Education, Guangzhou, Guangdong 510303, China Email address:... Hilbert and Hardy-Hilbert inequalities with nonc c c c conjugate parameters,” Bulletin of the Australian Mathematical Society, vol 71, no 3, pp 447– 457, 2005 [14] B Yang and T M Rassias, ? ?On the way