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Hindawi Publishing Corporation Journal of Inequalities and Applications Volume 2011, Article ID 401428, 11 pages doi:10.1155/2011/401428 Research Article A Hilbert-Type Integral Inequality in the Whole Plane with the Homogeneous Kernel of Degree −2 Dongmei Xin and Bicheng Yang Department of Mathematics, Guangdong Education Institute, Guangzhou, Guangdong 510303, China Correspondence should be addressed to Dongmei Xin, xdm77108@gdei.edu.cn Received 20 December 2010; Accepted 29 January 2011 Academic Editor: S. Al-Homidan Copyright q 2011 D. Xin 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. By applying the way of real and complex analysis and estimating the weight functions, we build a new Hilbert-type integral inequality in the whole plane with the homogeneous kernel of degree −2 involving some parameters and the best constant factor. We also consider its reverse. The equivalent forms and some particular cases are obtained. 1. Introduction ∞ ∞ If f x , g x ≥ 0, satisfying 0 < f 2 x dx < ∞ and 0 < g 2 x dx < ∞, then we have see 0 0 1 ∞f ∞ ∞ 1/2 xg y f 2 x dx g 2 x dx 1.1 dx dy < π , xy 0 0 0 where the constant factor π is the best possible. Inequality 1.1 is well known as Hilbert’s integral inequality, which is important in analysis and in its applications 1, 2 . In recent years, by using the way of weight functions, a number of extensions of 1.1 were given by Yang 3 . Noticing that inequality 1.1 is a Homogenous kernel of degree −1, in 2009, a survey of the study of Hilbert-type inequalities with the homogeneous kernels of degree negative numbers and some parameters is given by 4 . Recently, some inequalities with the homogenous kernels of degree 0 and nonhomogenous kernels have been studied see 5–9 .
2 Journal of Inequalities and Applications All of the above inequalities are built in the quarter plane. Yang 10 built a new Hilbert-type integral inequality in the whole plane as follows: ∞ ∞ ∞ 1/2 fxg y e−x f 2 x dx e−x g 2 x 1.2 dx dy < π , xy −∞ 1 e −∞ −∞ where the constant factor π is the best possible. Zeng and Xie 11 also give a new inequality in the whole plane. By applying the method of 10, 11 and using the way of real and complex analysis, the main objective of this paper is to give a new Hilbert-type integral inequality in the whole plane with the homogeneous kernel of degree −2 involving some parameters and a best constant factor. The reverse form is considered. As applications, we also obtain the equivalent forms and some particular cases. 2. Some Lemmas Lemma 2.1. If |λ| < 1, 0 < α1 < α2 < π , deﬁne the weight functions ω x and y x, y ∈ −∞, ∞ as follow: ∞ |x | 1 λ 1 ωx : min dy, x2 y2 λ 2xy cos αi −∞ i∈{1,2} y 2.1 1−λ ∞ y 1 y: min dx. |x|−λ x2 y2 2xy cos αi −∞ i∈{1,2} Then we have ω x x, y / 0 , where y kλ sin λ π − α2 π sin λα1 0 < |λ | < 1 ; kλ : sin λπ sin α1 sin α2 2.2 π − α2 α1 k0 : lim k λ . sin α1 sin α2 λ→0 Proof. For x ∈ −∞, 0 , setting u −y/x, respectively, in the following ﬁrst and y/x, u second integrals, we have 1λ 0 −x 1 · ωx dy x2 y2 λ 2xy cos α1 −y −∞ ∞ −x 1 λ 1 2.3 · dy x2 y2 yλ 2xy cos α2 0 ∞ ∞ u−λ u−λ du du. u2 − 2u cos α2 u2 2u cos α1 1 1 0 0
Journal of Inequalities and Applications 3 1/ z2 2z cos α1 1 , where z1 −eiα1 and z2 −e−iα1 Setting a complex function as f z are the ﬁrst-order poles of f z , and z ∞ is the ﬁrst-order zero point of f z , in view of the theorem of obtaining real integral by residue 12 , it follows for 0 < |λ| < 1 that ∞ ∞ u−λ du u 1−λ −1 du u2 u2 2u cos α1 2u cos α1 1 1 0 0 2πi Re s z−λ f z , z1 Re s z−λ f z , z2 1 − e2π 1−λ i −λ z−λ z1 2πi 2 2π 1−λ i z − z z2 − z1 1−e 1 2 −λ −π · −1 cos −λ α1 i sin −λ α1 cos λα1 i sin λα1 −2i sin α1 1−λ 2i sin α1 sin π 1 − λ · −1 π sin λα1 . sin πλ sin α1 2.4 For λ 0, we can ﬁnd by the integral formula that ∞ 1 α1 du . 2.5 u2 sin α1 2u cos α1 1 0 Obviously, we ﬁnd that for 0 < |λ| < 1, ∞ ∞ u−λ u−λ du du u2 − 2u cos α2 2u cos π − α2 u2 1 1 0 0 π · sin λ π − α2 2.6 ; for λ 0, sin λπ · sin α2 ∞ π − α2 1 du . − 2u cos α2 u2 sin α2 1 0 x ∈ −∞, 0 . Hence we ﬁnd ω x kλ
4 Journal of Inequalities and Applications For x ∈ 0, ∞ , setting u −y/x, u y/x, respectively, in the following ﬁrst and second integrals, we have 0 x1 λ 1 · ωx dy x2 y2 λ 2xy cos α2 −y −∞ ∞ x1 λ 1 · dy x2 y2 yλ 2xy cos α1 2.7 0 ∞ ∞ u−λ u−λ du du kλ. u2 − 2u cos α2 u2 1 2u cos α1 1 0 0 y, x / 0; |λ| < 1 . The By the same way, we still can ﬁnd that y ωx kλ lemma is proved. α ∈ 0, π , then it follows Note 1. 1 It is obvious that ω 0 0 0; 2 If α1 α2 that 1 1 min , 2.8 x2 y2 x2 y2 2xy cos αi 2xy cos α i∈{1,2} and by Lemma 2.1, we can obtain π cos λ α − π/2 y ωx y, x / 0 . 2.9 cos λπ/2 sin α Lemma 2.2. If p > 1, 1/p 1/q 1, |λ| < 1, 0 < α1 < α2 < π , and f x is a nonnegative measurable function in −∞, ∞ , then we have ∞ ∞ p 1 p 1−λ −1 J: y min f x dx dy x2 y2 2xy cos αi −∞ i∈{1,2} −∞ 2.10 ∞ |x|−pλ−1 f p x dx. ≤ kp λ −∞
Journal of Inequalities and Applications 5 Proof. By Lemma 2.1 and Holder’s inequality 13 , we have ¨ ∞ p 1 min fx x2 y2 2xy cos αi −∞ i∈{1,2} p λ/p |x| −λ /q ∞ y 1 min fx dx |x| −λ /q x2 y2 λ/p 2xy cos αi −∞ i∈{1,2} y |x| 1−p λ ∞ 1 ≤ f p x dx min 2.11 x2 y2 λ 2xy cos αi −∞ i∈{1,2} y p−1 q−1 λ ∞ y 1 × min dx |x| −λ x2 y2 2xy cos αi −∞ i∈{1,2} |x| 1−p λ ∞ 1 p λ−1 1 kp−1 λ y f p x dx. min x2 y2 λ 2xy cos αi −∞ i∈{1,2} y Then by Fubini theorem, it follows that ∞ ∞ |x| 1−p λ 1 J ≤ kp−1 λ f p x dx dy min x2 y2 λ 2xy cos αi −∞ i∈{1,2} −∞ y ∞ ω x |x|−pλ−1 f p x dx kp−1 λ 2.12 −∞ ∞ |x|−pλ−1 f p x dx. kp λ −∞ The lemma is proved. 3. Main Results and Applications Theorem 3.1. If p > 1, 1/p 1/q 1, |λ| < 1, 0 < α1 < α2 < π, f, g ≥ 0, satisfying 0 < ∞ ∞ |x|−pλ−1 f p x dx < ∞ and 0 < −∞ |y|qλ−1 g q y dy < ∞, then we have −∞ ∞ 1 I: min f x g y dx dy x2 y2 2xy cos αi −∞ i∈{1,2} 3.1 ∞ ∞ 1/p 1/q qλ−1 q −pλ−1 p |x |
6 Journal of Inequalities and Applications ∞ ∞ p 1 p 1−λ −1 J y min f x dx dy x2 y2 2xy cos αi −∞ i∈{1,2} −∞ 3.2 ∞ |x|−pλ−1 f p x dx, < kp λ −∞ where the constant factor k λ and kp λ are the best possible and k λ is deﬁned by Lemma 2.1. Inequality 3.1 and 3.2 are equivalent. Proof. If 2.11 takes the form of equality for a y ∈ −∞, 0 ∪ 0, ∞ , then there exist constants A and B, such that they are not all zero, and A |x| 1−p λ /|y|λ f p x B |y| q−1 λ /|x| −λ g q y a.e. in −∞, 0 ∪ 0, ∞ . Hence, there exists a constant C, such that A ·|x|−pλ f p x B ·|y|qλ g q y −pλ−1 p C a.e. in 0, ∞ . We suppose A / 0 otherwise B A 0 . Then |x| C/A|x| a. e. in fx ∞ −∞, ∞ , which contradicts the fact that 0 < −∞ |x|−pλ−1 f p x dx < ∞. Hence 2.11 takes the form of strict inequality, so does 2.10 , and we have 3.2 . By the HÓlder’s inequality 13 , we have ∞ ∞ 1 1/q−λ λ−1/q I y min f x dx y g y dy x2 y2 2xy cos αi −∞ i∈{1,2} −∞ 3.3 ∞ 1/q qλ−1 q ≤ J 1/p y g y dy . −∞ By 3.2 , we have 3.1 . On the other hand, suppose that 3.1 is valid. Setting p−1 ∞ 1 p 1−λ −1 3.4 gy y min f x dx , x2 y2 2xy cos αi −∞ i∈{1,2} ∞ |y|qλ−1 g q y dy. By 2.10 , we have J < ∞. If J then it follows J 0, then 3.2 is obvious −∞ value; if 0 < J < ∞, then by 3.1 , we obtain ∞ qλ−1 q 0< y g y dy J I −∞ ∞ ∞ 1/p 1/q qλ−1 q |x|−pλ−1 f p x dx
Journal of Inequalities and Applications 7 For ε > 0, deﬁne functions f x , g x as follows: ⎧ ⎪xλ−2ε/p , x ∈ 1, ∞ , ⎪ ⎪ ⎪ ⎪ ⎨ x ∈ −1, 1 , fx : 0, ⎪ ⎪ ⎪ ⎪ ⎪ ⎩ −x λ−2ε/p , x ∈ −∞, −1 , 3.6 ⎧ ⎪x−λ−2ε/q , x ∈ 1, ∞ , ⎪ ⎪ ⎪ ⎪ ⎨ x ∈ −1, 1 , gx : 0, ⎪ ⎪ ⎪ ⎪ ⎪ ⎩ −x −λ−2ε/q , x ∈ −∞, −1 . ∞ ∞ |x|−pλ−1 f p x dx}1/p { |y|qλ−1 g q y dy}1/q Then L : { 1/ε and −∞ −∞ ∞ 1 I: min f x g y dx dy I1 I2 I3 I4 , 3.7 x2 y2 2xy cos αi −∞ i∈{1,2} where −λ−2ε/q −1 −1 −y λ−2ε/p −x I1 : dy dx, x2 y2 2xy cos α1 −∞ −∞ −1 ∞ y−λ−2ε/q λ−2ε/p −x I2 : dy dx, x2 y2 2xy cos α2 −∞ 1 3.8 −λ−2ε/q ∞ −1 −y xλ−2ε/p I3 : dy dx, x2 y2 2xy cos α2 −∞ 1 ∞ ∞ y−λ−2ε/q xλ−2ε/p I4 : dy dx. x2 y2 2xy cos α1 1 1 By Fubini theorem 14 , we obtain ∞ ∞ u−λ−2ε/q y x−1−2ε I1 I4 du u u2 x 2u cos α1 1 1 1/x ∞ ∞ u−λ−2ε/q du u−λ−2ε/q du 1 x−1−2ε dx u2 2u cos α1 1 u2 2u cos α1 1 1 1/x 1 ∞ ∞ u−λ−2ε/q du u−λ−2ε/q du 1 1 x−1−2ε dy u2 u2 2ε 2u cos α1 1 2u cos α1 1 0 1/u 1
8 Journal of Inequalities and Applications ∞ u−λ 2ε/p u−λ−2ε/q 1 1 du du , u2 u2 2ε 2u cos α1 1 2u cos α1 1 0 1 ∞ u−λ 2ε/p u−λ−2ε/q 1 1 I2 I3 du du . 2 − 2u cos α − 2u cos α2 u2 2ε u 1 1 2 0 1 3.9 In view of the above results, if the constant factor k λ in 3.1 is not the best possible, then exists a positive number K with K < k λ , such that ∞ u−λ 2ε/p u−λ−2ε/q 1 du du u2 u2 2u cos α1 1 2u cos α1 1 0 1 3.10 ∞ u−λ 2ε/p du u−λ−2ε/q du 1 εI < εK L K. − 2u cos α2 1 − 2u cos α2 1 u2 u2 0 1 By Fatou lemma 14 and 3.10 , we have ∞ ∞ u−λ u−λ kλ du du u2 − 2u cos α2 u2 2u cos α1 1 1 0 0 ∞ u−λ 2ε/p u−λ−2ε/q 1 lim du lim du u2 u2 0 ε→0 2u cos α1 1 1 ε→0 2u cos α1 1 ∞ u−λ 2ε/p u−λ−2ε/q 1 lim du lim du 3.11 u2 − 2u cos α2 u2 − 2u cos α2 0 ε→0 1 1 ε→0 1 ∞ u−λ 2ε/p u−λ−2ε/q 1 ≤ lim du du u2 u2 2u cos α1 1 2u cos α1 1 ε→0 0 1 ∞ u−λ 2ε/p u−λ−2ε/q 1 du ≤ K, du 2 − 2u cos α − 2u cos α2 u2 u 1 1 2 0 1 which contradicts the fact that K < k λ . Hence the constant factor k λ in 3.1 is the best possible. If the constant factor in 3.2 is not the best possible, then by 3.3 , we may get a contradiction that the constant factor in 3.1 is not the best possible. Thus the theorem is proved.
Journal of Inequalities and Applications 9 In view of Note 2 and Theorem 3.1, we still have the following theorem. Theorem 3.2. If p > 1, 1/p 1/q 1, |λ| < 1, 0 < α < π , and f, g ≥ 0, satisfying ∞ ∞ 0 < −∞ |x|−pλ−1 f p x dx < ∞ and 0 < −∞ |y|qλ−1 g q y dy < ∞, then we have ∞ 1 f x g y dx dy x2 y2 2xy cos α −∞ ∞ ∞ 1/p 1/q π cos λ α − π/2 qλ−1 q |x|−pλ−1 f p x dx < y g y dy , cos λπ/2 sin α −∞ −∞ 3.12 ∞ ∞ p 1 p 1−λ −1 y f x dx dy x2 y2 2xy cos α −∞ −∞ ∞ p π cos λ α − π/2 |x|−pλ−1 f p x dx, < cos λπ/2 sin α −∞ where the constant factors π cos λ α − π/2 / cos λπ/2 sin α and π cos λ α − π/2 / cos λπ/2 sin α p are the best possible. Inequality 3.12 is equivalent. In particular, for α π /3, we have the following equivalent inequalities: ∞ 1 f x g y dx dy 2 y2 −∞ x xy ∞ ∞ 1/p 1/q 2π cos λπ/6 qλ−1 q |x|−pλ−1 f p x dx 0. If J ∞, then the reverse of 3.2 is obvious value; if J < ∞, then by the reverse of 3.1 , we obtain the reverses of 3.5 . Hence we have the reverse of 3.2 , which is equivalent to the reverse of 3.1 .
10 Journal of Inequalities and Applications If the constant factor k λ in the reverse of 3.1 is not the best possible, then there exists a positive constant K with K > k λ , such that the reverse of 3.1 is still valid as we replace k λ by K . By the reverse of 3.10 , we have 1 1 1 u−λ 2ε/p du − 2u cos α2 u2 u2 2u cos α1 1 1 0 3.14 ∞ 1 1 −λ−2ε/q u du > K. u2 − 2u cos α2 u2 2u cos α1 1 1 1 For ε → 0 , by the Levi’s theorem 14 , we ﬁnd 1 1 1 u−λ 2ε/p du u2 − 2u cos α2 u2 2u cos α1 1 1 0 3.15 1 1 1 u−λ du. −→ u2 − 2u cos α2 u2 2u cos α1 1 1 0 For 0 < ε < ε0 , q < 0, such that |λ 2ε0 /q| < 1, since u−λ−2ε/q ≤ u−λ−2ε0 /q , u ∈ 1, ∞ , 3.16 ∞ 1 1 2ε0 u−λ−2ε0 /q du ≤ k λ < ∞, u2 − 2u cos α2 u2 q 2u cos α1 1 1 1 then by Lebesgue control convergence theorem 14 , for ε → 0 , we have ∞ 1 1 u−λ−2ε/q du − 2u cos α2 u2 u2 2u cos α1 1 1 1 3.17 ∞ 1 1 −λ −→ u du. u2 − 2u cos α2 u2 2u cos α1 1 1 1 By 3.14 , 3.15 , and 3.17 , for ε → 0 , we have k λ ≥ K , which contradicts the fact that k λ < K . Hence the constant factor k λ in the reverse of 3.1 is the best possible. If the constant factor in reverse of 3.2 is not the best possible, then by the reverse of 3.3 , we may get a contradiction that the constant factor in the reverse of 3.1 is not the best possible. Thus the theorem is proved. By the same way of Theorem 3.3, we still have the following theorem. Theorem 3.4. By the assumptions of Theorem 3.2, replacing p > 1 by 0 < p < 1, we have the equivalent reverses of 3.12 with the best constant factors.
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