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 Advances in Pure Mathematics, 2011, 1, 84-89 doi:10.4236/apm.2011.13019 Published Online May 2011 (http://www.scirp.org/journal/apm) Copyright © 2011 SciRes. APM A Hilbert-Type Inequality with Some Parameters and the Integral in Whole Plane Zitian Xi e1, Zheng Zeng2 1Department of Mat hem at ic s, Zhaoqing University , Zhaoqing, Chi na 2 Department of Mathematics, Shaoguan University, Shaoguan, China E-mail: gdzqxzt@163.com, zz@sgu.edu.cn Received February 13, 2011; revised March 7, 2011; accepted March 10, 2011 Abstract In this paper, by introducing some parameters and estimating the weight coefficient, we give a new Hilbert’s inequality with the integral in whole plane and with a non-homogeneous and the equivalent form is given as well. The best constant factor is calculated by the way of Complex Analysis. Keywords: Hilbert-Type Integral Inequality, Weight Function, Holder’s Inequality 1. Introduction If , f x g x are non-negative functions such that 2 0 0dfxx and 2 0 0dgxx , then 12 000 0 dd πdd fxgy x yfxxgxx xy (1.1) where the constant factor π is the best possible. Ine- quality (1.1) is well-known as Hilbert’s integral inequal- ity, which has been extended by Hardy-Riesz as [2]: If 11 1, 1ppq , , f x 0gx, such that 0 0d p fxx and 0 0d q gxx , then we have the following Hardy-Hilbert’s integral inequality: 00 11 00 dd πdd sin π pq pq fxgyxy xy f xx gxx p (1.2) where the constant factor π sin πp also is the best possible. In recent years, by introducing some parameters and estimating the way of weight function, inequalities (1.1) and (1.2) have many generalizations and variants (1.1) has been strengthened by Yang and others. (including double series inequalities) [3-15]. In 2005 Yang gave a Hilbert-type Inequality [3] as follows: If 1,p 111 pq , , f x 0gx, 2min ,1pq , 1 0 0d p xfxx , and 1 0 0d q xgxx then 00 11 11 00 dd dd p q pq fxgyxy xy K x fxxxgxx (1.3) where the constant factor 22 ,1 ,1 pq KB B pq also is the best possible. In 2008 Xie gave a new Hilbert-type Inequality [4] as follows: If 11 1,1, ,0pfxgx pq 12 0 0d pp xfxx and 12 0 0d qq xgxx , then 222 00 11 12 12 00 dd dd p q pp qq fxgy xy xayxayxay K xf xxxgxx where the constant factor π Kabbcca  Z. T. XIE ET AL. Copyright © 2011 SciRes. APM 85 also is the best possible. In this paper, by using the way of weight function and the technic of real analysis and by the way of complex analysis, a new Hilbert-type inequality with the integral in whole plane is given. In the following, we always suppose that: 0,p 111 pq , 1,p 1, 2, 0πr . 2. Some Lemmas Lemma 2.1 If 2 12 0 2 22 0 12cos :ln d, 12cos 12cos :ln d, 12cos r r uu ku u uu uu ku u uu then 1 2 11 4πsin sin 22 :, 1sinπ 1 4πsinsin 1 22 :1sinπ rr krr rr krr (2.1) and 2 2 12 12cos :ln d 12cos 11π 4πsin cos 22 π 1sin 2 ruu ku u uu kk rr r r (2.2) Proof We have 2 0 12 0 1 2 0 :ln2cos1d 1ln2 cos1 1 cos 2d 12cos1 2 :1 r r r A xxxx xxx r xx x rxx B r setting 1 2 cos 2cos1 r zz fz zz , 12 , ii zeze , then 12 2π 11 11 22 2π 12 21 2πRe ,Re , 1 cos cos 2π 1 πcos 1 sin π ri rr ri i Bsfzsfz e zz zz i zz zz e r r We find 2πcos[ 1] 2 11sinπ r AB rrr . then 2 12 0 12cos ln d 12cos 11 4πsinsin 22 1sinπ ruu ku u uu rr rr and 2 22 0 2 2 0 12cos ln d 12cos 12cosπ ln d 12cosπ 1 4πsinsin 1π 22 1sinπ r r uu ku u uu uu uu uu rr rr 2 2 2 2 0 2 0 2 12 12cos ln d 12cos 12cos ln d 12cos 12cos ln d 12cos 11π 4πsin cos 22 π 1sin 2 r r r uu ku u uu uu uu uu uu uu uu kk rr r r The lemma is proved. Lemma 2.2 Define the weight functions as follow: 122 22 122 22 12 cos :ln d, 12 cos 12 cos :lnd 12 cos r r r r xxyx y wx y xyx y y yxyx y wx x xyx y x then  Z. T. XIE ET AL. Copyright © 2011 SciRes. APM 86 11π 4πsin cos 22 π 1sin 2 wx wx k rr r r (2.3) Proof We only prove that wx k for ,0x . 122 0 22 122 22 0 122 0 22 122 22 0 12 12 cos ln d 12 cos 12 cos ln d 12 cos 12 cos ln d 12 cos 12 cos ln d 12 cos r r r r r r r r xxyx y wx y xyx y y xxyx y y xyx y y xxyx y y xyx y y xxyx y y yxyxy ww Setting uxy, then 122 0 122 2 1 2 0 12 cos ln d 12 cos 12cos ln d 12cos r r r xxyx y wy xyx y y uu uuk uu Similarly, setting uxy , 2 22 2 0 12cos ln d 12cos ruu wu uk uu And 12 wxk kk , we have (2.3). Lemma 2.3 For 0 , and 22 max,1,2rpq , define both functions, f and g as follow: 2 2 , if 1,, 0, if 1,1, , if ,1 ; rp rp xx fx x xx and 2 0, if 1,, , if 1,1, 0, if ,1; rq x gx xx x then 11 11 :2dd 1 pp pq Ixxfxxxgx x (2.4) 22 22 12 cos :2 lndd 12 cos 1, for0 xyx y I fxgy xy xyx y ko (2.5) Proof Easily, 11 1 12 12 10 :2dd 1; pp Ixxxxxxx Let yY , using , f xfxgxgx , and 22 22 22 22 12 cos ln d 12 cos 12 cos ln d 12 cos xyx y f xgy y xyx y xYx Y f xgY Y xYx Y we have that 22 22 12 cos ln d 12 cos xyx y f xgy y xyx y is an even function, then 22 22 0 22 22 0 1 2 222 22 22 11 10 1 12 cos 2ln dd 12 cos 12 cos12 cos 2lnddlndd 12 cos12 cos rq rp rp rq xyx y Ifxgy yx xyx y xy xyxy xy x yyxxyyx xyx yxyx y I 2.I Setting uxy then 22 2 1 22 122 122 2 10 10 2 1 12 2122 2 10 11 12 cos12cos 2lndd2lndd 12 cos12cos 12cos 2lndd 12cos x rp rqrq x rq rq xyx yuu I xyyxxu ux xyx yuu uu xu uxxu uu 2 2 12cos lnd d 12cos uu ux uu  Z. T. XIE ET AL. Copyright © 2011 SciRes. APM 87 22 12212 22 01 22 122 22 01 2 2 2 0 12cos 12cos lnd2lnd d 12cos 12cos 12cos 12cos lnd lnd 12cos 12cos 12cos lnd ( 12cos rq rq u rq rp rp uu uu uuu xxu uu uu uu uu uuuu uu uu uu uu uu 2 122 2 0 12cos )ln d 12cos 2 12 4πsinsin 1 22 212 1sin π 2 qpr uu uuu u uu rprp rr pp There 0 lim 0 , and we have 12 I k (for 0 ). Similarly 21 I k (for 0 ). The lemma is proved. Lemma 2.4 If 1 0d pr p xfxx , we have 22 1 22 1 12 cos :lndd 12 cos d p pr p pr pp xyx y J yfx xy xyx y kxfxx (2.6) Proof By lemma 2.2, we find 22 22 22 22 1 22 22 22 2 12 cos ln d 12 cos 12 cos ln d 12 cos 12 cos ln d 12 cos 12 cos ln 12 cos p p rq rp rp rq rp p r xyx y fx x xyx y xy xyx y f xx xyx yyx x xyx yfxx xyx yy xyx y xyx y 1 1 2 1 22 1 1 22 d 12 cos lnd , 12 cos p rq r rp pr p pp r yx x x xyx y ky fxx xyx yy 1 22 1 22 1 22 1 22 1 12 cos lnd d 12 cos 12 cos lnd d 12 cos d rp pp r rp pp r pr pp J x xyx y kfxxy xyx yy x xyx y kyfxx xyx yy kxfxx (2.7) 3. Main Results Theorem If both functions, f x and g x are nonnegative measurable functions, and satisfy 1 0d pr p x fxx , and 1 0d qr q x gxx , then 22 * 22 11 11 12 cos :ln dd 12 cos dd pp pr qr pq xyx y I fxgy xy xyx y kxfxx xgxx (3.1) and 22 1 22 1 12 cos lnd d 12 cos d p pr p pr pp xyx y J yfx xy xyx y kxfxx (3.2) Inequalities (3.1)and (3.2) are equivalent, and the con- stant factors in the two forms are all the best possible. Proof If (2.7) takes the form of equality for some ,0 0,y , then there exist constants M and N, such that they are not all zero, and 11rp rq p rr xy Mfx yx a.e. in ,, , Hence, there exists a constant C, such that rp rq p M xfxNyC a.e. in ,, . We claim that 0M . In fact, if 0M , then 1rpp x fxCMx a.e. in , which con- tradicts the fact that 1 0d pr p xfxx . In the same way, we claim that 0N. This is too a con- tradiction and hence by (2.7), we have (3.2). By Holder’s inequality with weight and (3.2), we have,  Z. T. XIE ET AL. Copyright © 2011 SciRes. APM 88 22 11 * 22 1 11 12 cos lnd d 12 cos d rq rq q qrpq xyx y I yfx xygyy xyx y Jygyy (3.3) Using (3.2), we have (3.1). Setting 1 22 1 22 12 cos ln d 12 cos p pr pxyx y gy yfxx xyx y then 1()d qr q J ygyy by (2.7) we have J . if 0J then (3.2) is proved; if 0J, by (3.1), we obtain 1* 11 11 0()d dd, qr q pq pr qr pq ygyyJI kxfxxxgxx 11 11 1 dd p p qr pr qp p x gxxJkxf xx Inequalities (3.1) and (3.2) are equivalent. If the constant factor k in (3.1) is not the best possi- ble, then there exists a positive h (with hk), such that 22 22 11 11 12 cos lnd d 12 cos dd p q pr qr pq xyx y f xg yxy xyx y hxfxx xgxx (3.4) For 0 , by (3.4), using lemma 2.3, we have 11 11 1 dd p q pq pr qr Iko hxfxx xgxx (3.5) Hence we find, 1ko h , for 0 , it follows that kh, which contradicts the fact that hk . Hence the constant h in (3.1) is the best possible. As (3.1) and (3.2) are equivalent, if the constant factor in (3.2) is not the best possible, then by using (3.2), we can get a contradiction that the constant factor in (3.1) is not the best possible. Thus we complete the prove of the theorem. Remark For π 4 , π 3 in (3.1), we have the following particular result: 22 22 11 11 12 lnd d 1 dd pq prqr pq xyx y fxgy xy xyx y kxfxx xgxx (3.6) Where the constant factor 1π51π 4πsin cos 2424 π 1sin2 rr kr r also is the best possible. 4. References [1] G. H. Hardy, J. E. Littlewood and G. Polya, “Inequali- ties,” Cambridge University Press, Cambridge, 1952. [2] G. H. Hardy, “Note on a Theorem of Hilbert Concerning Series of Positive Terems,” Proceedings of London Ma- thematical Society, Vol. 23, No. 2, 1925, pp. 45-46. [3] B. C. Yang, “A New Hilbert-Type Integral Inequality with a Parameters,” Journal of Henan University (Natu- ral Science), Vol. 35, No. 4, 2005, pp. 4-8. [4] Z. T. Xie and Z. 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T. Xie and Z. Zeng, “A Hilbert-Type Integral Inequal- ity with a Non-Homogeneous Form and a Best Constant Factor,” Advances and Applications in Mathematical Sciens, Vol. 3, No. 1, 2010, pp. 61-71. [15] Z. Zeng and Z. T. Xie, “A New Hilbert-Type Integral Inequality with a Best Constant Factor,” Journal of South China Normal University (Natural Science Edition), Vol. 3, 2010, pp. 31-33. |