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 Introduction If f x , g x ≥ 0, satisfying < ∞f ∞ f x dx < ∞ and < x g y dx dy < π x y ∞ f x dx ∞ g x dx < ∞, then we have see ∞ 1/2 g x dx , 1.1 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, In recent years, by using the way of weight functions, a number of extensions of 1.1 were given by Yang 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 Recently, some inequalities with the homogenous kernels of degree and nonhomogenous kernels have been studied see 5–9 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: ∞ ∞ f x g y dx dy < π x y −∞ e −∞ ∞ e−x f x dx −∞ 1/2 e−x g x , 1.2 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 Some Lemmas Lemma 2.1 If |λ| < 1, < α1 < α2 < π, define the weight functions ω x and −∞, ∞ as follow: ∞ ω x : −∞ i∈{1,2} |x|1 2xy cos αi x2 y2 y x, y ∈ λ dy, y λ y 1−λ 2.1 ∞ y : Then we have ω x y k λ : k : −∞ i∈{1,2} sin λα1 π sin λπ sin α1 lim k λ Proof For x ∈ −∞, , setting u second integrals, we have ω x x2 −∞ ∞ x2 ∞ u2 |x|−λ y2 dx x, y / , where k λ λ→0 2xy cos αi x2 sin λ π − α2 sin α2 α1 sin α1 2.2 π − α2 sin α2 −y/x, respectively, in the following first and y/x, u 2xy cos α1 y2 2xy cos α2 u−λ 2u cos α1 < |λ| < ; y2 −y λ dy −x λ dy yλ · ∞ 1 λ −x · du u−λ u2 − 2u cos α2 2.3 du Journal of Inequalities and Applications Setting a complex function as f z 1/ z2 2z cos α1 , where z1 −eiα1 and z2 −e−iα1 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 < |λ| < that ∞ u2 u−λ du 2u cos α1 ∞ u 1−λ −1 du u2 2u cos α1 2πi Re s z−λ f z , z1 − e2π 1−λ i −λ z−λ z2 − z1 z1 2πi 2π 1−λ i z − z 1−e −π · −1 −λ sin π − λ · −1 1−λ Re s z−λ f z , z2 cos −λ α1 i sin −λ α1 −2i sin α1 cos λα1 i sin λα1 2i sin α1 π sin λα1 sin πλ sin α1 2.4 For λ 0, we can find by the integral formula that ∞ u2 2u cos α1 du α1 sin α1 2.5 Obviously, we find that for < |λ| < 1, ∞ u−λ u2 − 2u cos α2 ∞ du u2 u−λ 2u cos π − α2 π · sin λ π − α2 ; sin λπ · sin α2 ∞ Hence we find ω x u2 − 2u cos α2 k λ du x ∈ −∞, π − α2 sin α2 for λ du 0, 2.6 Journal of Inequalities and Applications For x ∈ 0, ∞ , setting u second integrals, we have ω x x2 −∞ ∞ x2 ∞ −y/x, u 2xy cos α2 y2 2xy cos α1 u−λ u2 − 2u cos α2 y/x, respectively, in the following first and y2 · λ dy x1 λ dy yλ ∞ 1 2xy cos αi x2 i∈{1,2} λ −y du By the same way, we still can find that lemma is proved Note 1 It is obvious that ω that x1 · 2.7 u2 u−λ 2u cos α1 y ω x 0; If α1 y2 x2 du y, x / 0; |λ| < The k λ α2 2xy cos α k λ α ∈ 0, π , then it follows y2 , 2.8 and by Lemma 2.1, we can obtain y ω x π cos λ α − π/2 cos λπ/2 sin α y, x / 2.9 Lemma 2.2 If p > 1, 1/p 1/q 1, |λ| < 1, < α1 < α2 < π, and f x is a nonnegative measurable function in −∞, ∞ , then we have ∞ J: −∞ y p 1−λ −1 ∞ −∞ i∈{1,2} x2 2xy cos αi p y2 f x dx dy 2.10 ≤ kp λ ∞ −∞ |x|−pλ−1 f p x dx Journal of Inequalities and Applications Proof By Lemma 2.1 and Holder’s inequality 13 , we have ă 2xy cos i x2 −∞ i∈{1,2} ∞ ≤ ∞ −∞ i∈{1,2} × |x| −λ /q y2 x2 y2 2xy cos αi x2 −∞ i∈{1,2} y y y2 x2 −∞ i∈{1,2} y f x p λ/p |x| −λ /q dx f p x dx λ y ∞ p λ−1 λ/p |x| 1−p λ 2xy cos αi ∞ kp−1 λ y y2 2xy cos αi x2 −∞ i∈{1,2} p f x p−1 q−1 λ |x| −λ 2xy cos αi 2.11 dx |x| 1−p λ y2 y λ f p x dx Then by Fubini theorem, it follows that ∞ J ≤ kp−1 λ −∞ i∈{1,2} −∞ ∞ kp−1 λ kp λ ∞ −∞ ∞ −∞ x2 |x| 1−p λ 2xy cos αi y2 y λ f p x dx dy ω x |x|−pλ−1 f p x dx 2.12 |x|−pλ−1 f p x dx The lemma is proved Main Results and Applications Theorem 3.1 If p > 1, 1/p 1/q 1, |λ| < 1, < α1 < α2 < π, f, g ≥ 0, satisfying < ∞ ∞ |x|−pλ−1 f p x dx < ∞ and < −∞ |y|qλ−1 g q y dy < ∞, then we have −∞ ∞ I: −∞ i∈{1,2} ∞ by < p < 1, we have the equivalent reverses of 3.1 and 3.2 with the best constant factors Proof By the reverse Holder’s inequality 13 , we have the reverse of 2.10 and 3.3 It is ă easy to obtain the reverse of 3.2 In view of the reverses of 3.2 and 3.3 , we obtain the reverse of 3.1 On the other hand, suppose that the reverse of 3.1 is valid Setting the same g y as Theorem 3.1, by the reverse of 2.10 , we have J > 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 2u cos α1 u2 u2 1 − 2u cos α2 u−λ 2ε/p du 3.14 ∞ 2u cos α1 u2 1 u2 − 2u cos α2 u −λ−2ε/q du > K For ε → , by the Levi’s theorem 14 , we find 1 2u cos α1 u2 1 u2 − 2u cos α2 u−λ 2ε/p du 3.15 −→ 1 2u cos α1 u2 For < ε < ε0 , q < 0, such that |λ u2 − 2u cos α2 1 u2 2u cos α1 u−λ du 2ε0 /q| < 1, since u−λ−2ε/q ≤ u−λ−2ε0 /q , ∞ 1 u2 − 2u cos α2 u ∈ 1, ∞ , u−λ−2ε0 /q du ≤ k λ 2ε0 q 3.16 < ∞, then by Lebesgue control convergence theorem 14 , for ε → , we have ∞ −→ u2 2u cos α1 − 2u cos α2 u−λ−2ε/q du 3.17 ∞ 1 u2 u2 2u cos α1 1 u2 − 2u cos α2 −λ u du By 3.14 , 3.15 , and 3.17 , for ε → , 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 > by < p < 1, we have the equivalent reverses of 3.12 with the best constant factors Journal of Inequalities and Applications 11 References G H Hardy, J E Littlewood, and G Polya, Inequalities, The University Press, Cambridge, UK, 2nd ´ edition, 1952 D S Mitrinovi´ , J E Peˇ ari´ , and A M Fink, Inequalities Involving Functions and Their Integrals c c c and Derivatives, vol 53 of Mathematics and Its Applications (East European Series), Kluwer Academic Publishers, Dordrecht, The Netherlands, 1991 B Yang, The Norm of Operator and Hilbert-Type Inequalities, Science Press, Beijing, China, 2009 B C Yang, “A survey of the study of Hilbert-type inequalities with parameters,” Advances in Mathematics, vol 38, no 3, pp 257–268, 2009 B C Yang, “On the norm of an integral operator and applications,” Journal of Mathematical Analysis and Applications, vol 321, no 1, pp 182–192, 2006 J Xu, “Hardy-Hilbert’s inequalities with two parameters,” Advances in Mathematics, vol 36, no 2, pp 189–202, 2007 B C Yang, “On the norm of a Hilbert’s type linear operator and applications,” Journal of Mathematical Analysis and Applications, vol 325, no 1, pp 529–541, 2007 D M Xin, “A Hilbert-type integral inequality with a homogeneous kernel of zero degree,” Mathematical Theory and Applications, vol 30, no 2, pp 70–74, 2010 B C Yang, “A Hilbert-type integral inequality with a homogeneous kernel of degree zero,” Journal of Shandong University Natural Science, vol 45, no 2, pp 103–106, 2010 10 B C Yang, “A new Hilbert-type inequality,” Bulletin of the Belgian Mathematical Society, vol 13, no 3, pp 479–487, 2006 11 Z Zeng and Z Xie, “On a new Hilbert-type integral inequality with the integral in whole plane,” Journal of Inequalities and Applications, vol 2010, Article ID 256796, pages, 2010 12 Y Ping, H Wang, and L Song Jr., Complex Function, Science Press, Beijing, China, 2004 13 J Kuang, Applied Inequalities, Shangdong Science and Technology Press, Jinan, China, 2004 14 J Kuang, Introudction to Real Analysis, Hunan Educiton Press, Changsha, China, 1996  ...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:... 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... 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