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` Negative theorem of nearly monotone approximation in Lp , o p 1 Sukeina Abdulla Al-Bermani Babylon University Science College For Women Computer Science Department Abstract In [2] our direct theorem we see that , when we approximate an increasing function f in LU > 1,1@ , we wish sometimes that the approximating polynomials be increasing also. However, this constraint, restricts very much the degree of approximation, that the polynomials can achieve, namely ; only the rate of ZM2 . In[3] Kopotun, Leviatan and Prymak proved that relaxing the monotonicity requirement in intervals of measure zero near the end points allows the polynomials to achieve the rate of ZM3 . On the other hand we show in this paper , that even when we relax the requirement of monotinicity of the polynomials on sets of measure approach zero, ZM4 is not reachable . ﺍﻟﻣﺳﺗﺧﻠﺹ 0 p 1 ﻋﻧ�ﺩﻣﺎL p [ ﺍﻟﻧﻅﺭﻳ�ﺔ ﺍﻟﻣﺑﺎﺷ�ﺭﺓ ﻓ�ﻲ ﺍﻟﺗﻘﺭﻳ�ﺏ ﺍﻟﺭﺗﻳ�ﺏ ﻟﻠ�ﺩﻭﺍﻝ ﺍﻟﻘﻳﺎﺳ�ﻳﺔ ﻓ�ﻲ ﺍﻟﻔﺿ�ﺎء1] ﺑﻭﺍﺳﻁﺔ ﻣﺗﻌﺩﺩﺓ ﺣ�ﺩﻭﺩ ﺃﻳﺿ�ﺎf ﺗﺑﻳﻥ ﺍﻧﻪ ﻣﻥ ﺍﻟﻣﻣﻛﻥ ﺃﻥ ﻧﻘﺭﺏ ﺃﻟﺩﺍﻟﻪ ﺍﻟﻣﺗﺯﺍﻳﺩﺓ، ZM2 ﺑﺩﻻﻟﺔ ﻣﻘﻳﺎﺱ ﺍﻟﻧﻌﻭﻣﺔ ﺇﻥ ﻗﻳﺩ ﺍﻟﺭﺗﺎﺑﺔ ﻓﻲ ﺍﻟﻧﺗﻳﺟﺔ ﺃﻋﻼﻩ ﺳﻳﺣﺩﺩ ﺩﺭﺟﺔ ﺍﻟﺗﻘﺭﻳﺏ ﺍﻷﻓﺿﻝ ﺑﻭﺍﺳﻁﺔ ﺍﻟﻣﺗﻌﺩﺩﺍﺕ ﺍﻟﺟﺑﺭﻳﺔ ﺑﺩﻻﻟﺔ.ﻣﺗﺯﺍﻳﺩﺓ ﻟ�ﺫﻟﻙ، Z M3 ﺑﺎﻟﻣﻘﻳﺎﺱf ﻟﻠﺩﺍﻟﺔZ M2 ﻭﻟﺗﺄﻛﻳﺩ ﺫﻟﻙ ﺑﺭﻫﻧﺎ ﺍﻧﻪ ﻻ ﻳﻣﻛﻥ ﺍﺳﺗﺑﺩﺍﻝ ﺍﻟﻣﻘﻳﺎﺱ، ﻓﻘﻁZM2 ﺍﻟﻣﻘﻳﺎﺱ .ﻗﺩﻣﻧﺎ ﺍﻟﻧﻅﺭﻳﺔ ﺍﻟﻧﻘﻳﺿﺔ ﻟﻠﻧﻅﺭﻳﺔ ﺍﻟﻣﺑﺎﺷﺭﺓ ﺑﺭﻫﻥ ﺍﻧﻪ ﻋﻧﺩﻣﺎ ﻳﺟﻌﻝ ﻣﺗﻌﺩﺩﺓ ﺍﻟﺣﺩﻭﺩ ﺍﻟﺟﺑﺭﻳﺔ ﺍﻟﺗﻘﺭﻳﺑﻳﺔ ﺗﻛ�ﻭﻥK.A. Kopotun [ ﺍﻟﻌﺎﻟﻡ2] > ﺍﻟﺗﻲ ﺗﺣﺗﻭﻱ ﻋﻠﻰ ﺍﻟﻧﻘﺎﻁ ﺍﻟﺣﺭﺟﺔ1,1@ ﻏﻳﺭ ﺭﺗﻳﺑﺔ ﻣﻊ ﺩﺍﻟﺔ ﺍﻟﻬﺩﻑ ﻓﻲ ﺑﻌﺽ ﺍﻟﻔﺗﺭﺍﺕ ﺍﻟﺟﺯﺋﻳﺔ ﻣﻥ ﺍﻟﻔﺗﺭﺓ > ﻭﺍﻟﺗ �ﻲ ﺗﻛ �ﻭﻥ ﺃﻁﻭﺍﻟﻬ �ﺎ1,1@ ﺑﺎﻹﺿ �ﺎﻓﺔ ﺇﻟ �ﻰ ﺍﻟﻔﺗ �ﺭﺍﺕ ﺍﻟﺟﺯﺋﻳ �ﺔ ﺍﻟﺗ �ﻲ ﺗﺣﺗ �ﻭﻱ ﻧﻬﺎﻳ �ﺎﺕ ﺍﻟﻔﺗ �ﺭﺓf ﻟﻠﺩﺍﻟ �ﺔ . Z M3 ﻣﻘﺗﺭﺑﺔ ﺇﻟﻰ ﺍﻟﺻﻔﺭ ﻳﻣﻛﻥ ﺇﻥ ﻳﺣﺻﻝ ﻋﻠﻰ ﺍﻟﺗﻘﺭﻳﺏ ﺑﺩﻻﻟﺔ ﻣﻘﻳﺎﺱ ﺍﻟﻧﻌﻭﻣﺔ ﺣﺗ�ﻰ ﻟ�ﻭ ﻛﺎﻧ�ﺕ ﺍﻟﻣﺗﻌ�ﺩﺩﺓ ﺍﻟﺟﺑﺭﻳ�ﺔZ M4 ﺑﺎﻟﻣﻘﻳ�ﺎﺱZ M3 ﺑﺭﻫﻧﺎ ﺍﻧﻪ ﻻ ﻳﻣﻛﻥ ﺍﺳ�ﺗﺑﺩﺍﻝ ﻣﻘﻳ�ﺎﺱ ﺍﻟﻧﻌﻭﻣ�ﺔ . > ﺗﻘﺗﺭﺏ ﺃﻁﻭﺍﻟﻬﺎ ﺇﻟﻰ ﺍﻟﺻﻔﺭ1,1@ ﻏﻳﺭ ﺭﺗﻳﺑﺔ ﻣﻊ ﺩﺍﻟﺔ ﺍﻟﻬﺩﻑ ﻓﻲ ﻣﺟﻣﻭﻋﺎﺕ ﺟﺯﺋﻳﻪ ﻣﻥ ﺍﻟﻔﺗﺭﺓ 1 1. Introduction and Main result Let f LU , 0 U 1 be increasing function on I > 1,1@ in [2] we proved that there exists an increasing polynomials such that f Un dC U LU I Z M 2 § 1· ¨ f, ¸ , © n ¹U 1.1 Where C U is an absolute constant depending on U , and ZM2 f ; o denote the Ditzain Totik modulus of smoothness of order two of f and ZM3 is the secand order Ditizin_Totik modulus of smoothness of f defined by Z § 1· ¨ f, ¸ © n ¹P M 2 sup 0 h dG ' 2 hI . § 1· ¨ f , ¸ , G t 0 ,[1] © n¹ p In [4] Devore proved that if f C > 1,1@ be non decreasing on I > 1,1@ there exist non decreasing polynomials such that f Un c( I ) § 1· d CZ2 ¨ f , ¸ . © n ¹U 1.2 However , even this improvement comes to ahalt; it can not be extended to wM4 ; and thus not to Z k for any K > 3 . In order to state our theorem we need some notation . Given H > 0 and a increasing function f LU , 0 U 1 ,We denote f U n LU I . E n1 f , H U inf pn3 n '1 I Where the infimum is taken over all polynomials U n of degree not exceeding n satisfying meas x : U cn x t 0 I t 2 H Then our main result in this paper is : Theorem I: for each sequence H ^H n`fn 1 , of non negative numbers tending to Zero , and each 0 U 1 , there exists an increasing function f f LU I such that ^ Lim n of ` E sup 1 n f ;H n U § 1· Z 4 ¨© f , n ¸¹ U M 1.3 f 2. A counter example (proof of theorem I) In this article we have used C as an absolute constant which may differ on different occurrences , in this paper we will have to keep track of the constants , there fore we denote them by C1, C2 , …… ., We begin by recalling some simple properties of the Chebyshev polynomials for x the interval [-2 , 2] , for v > 1 let t v x cos v cos 1 , x > 2 ,2 @ . 2 2 Denote to the Chebyshev polynomial and let its extrema , In fact : tv x t vc x § cos¨ v cos 1 © 2 cos iS , j v 0 , .... , v , be x· ¸ 2¹ § · ¨ ¸ ¨ ¸ 1 1 x § · § · sin ¨ v cos 1 ¸¨ v ¸ ¨ ¸ 2 2 ¹¨ © § x · © 2 ¹¸ ¨ 1 ¨ ¸ ¸ ©2¹ © ¹ x· § v sin ¨ v cos 1 ¸ 2¹ © § x· 2 1 ¨ ¸ ©2¹ t vc x Zj 2 0 x· § v sin ¨ v cos 1 ¸ 0 2¹ © x· § sin ¨ v cos 1 ¸ 0 2¹ © x sin 1 0 2 x v cos 1 jS , j 0, 1, 2 , 3 ,...... 2 x jS x jS cos 1 , cos 2 v 2 v jS Zj 2 cos , j 0, 1, 2 ,....... v 1 Given 0 b , we take two points on both sides of Zj , j 1,......, v 1, 2 v cos 1 namely ,we set Z j ,l § jb S · 2 cos¨ ¸ and Z j ,r v © ¹ § j b S · 2 cos¨ ¸, v © ¹ 3 t v Z j ,l tv Z j ,r cos S b . and § jS b S · § jS bS · 2 cos ¨ ¸ 2 cos ¨ ¸ v ¹ v ¹ © v © v jS bS 4 sin sin v v Z j , r Z j ,l Then since sin u d u , u >0 , S @ , so that Z j ,r Z j ,l 4S b v 2.1 We truncate the Chebyshev polynomial by setting * v t x t * v ,b x ­cos Sb ° ® cos Sb °t x ¯v t v x ! cos Sb t v x cos Sb O.W 1 cos Sb 2 sin 2 sin ce bS 5b 2 2 2.2 For any x I ,it follows by the monotonicity of the areas as we go away from the origin , and the alternation in sign of these areas ,that Zª v º Zª v º « » ,r ¬2¼ ³t v « » ,r ¬2¼ * v u t u du d Zª v º ³t v u t v* u du Zª v º « » ,l ¬2¼ « » ,l ¬2¼ d Z ª v º Z ª v º sup t v u t v* u « 2 » ,r ¬ ¼ « 2 » ,l ¬ ¼ b t v u t v* u v b d 4S 1 cos Sb v b 4S 5b 2 v b3 c1 , v ª º Now sin ce >0, x @ « Z ª v º , Z ª v º » I > 1,1@ «¬ «¬ 2 »¼ ,l «¬ 2 »¼ ,r »¼ Then d 4S x ³ 0 Zª v º * t v u t v u du d 2.3 « »,r ¬2¼ ³ Zª v º b b3 * t v u t v u du 4S 5b 2 c1 v v « » ,l ¬2¼ 4 2.4 Where we applied (2.1) Now , given n t1 and 0 b ª 34 º «bn » 2 , where >a @ ¬ ¼ 1 , let v 2 denotes the largest integer not exceeding a .put t v ,b t v cos Sb , and t v~ t v*,b cos Sb . Finally Tv ,b x x x ³ t v,b u du and f n,b x ³t 0 ~ v ,b u du , x I 0 let x1 x 2 f n ,b x1 x1 ³t x2 v ,b 0 u du ³ t v , b u du f n ,b x 2 . 0 Obviously f n,b is a increasing function on I and it readily by 2.4 that f n , b Tv , b §1 x x ¨ t v~, b u du ³ t v , b u du ³ ³ ¨ 1 0 0 © p Now f n ,b Tv ,b p · du ¸ ¸ ¹ 1 p x ~ ³ t v ,b u t v ,b u du 0 x * ³ t v ,b u cos Sb t v ,b u cos Sb du 0 x * ³ t v ,b u t v ,b u du 0 d C1 3 3 3 C 1b b dC1 3 ª 3º «bn4 » 2 bn4 « » ¬ ¼ b v And f n ,b Tv ,b 9 §1 ¨ b4 p d ¨ ³ C1 n ¨ 1 © 9 b4 d C1 n p §1 · ¨ ³ du ¸ ¨ ¸ © 1 ¹ · ¸ du ¸ ¸ ¹ 1 p 1 p 1 p 9 b4 d C1 2 , n Then 5 C1 9 b4 n , f n ,b Tv ,b 1 p 1 1 9 b3 b3 b4 d C1 2 p C1 2 p 3 p d C1 2 v n ª 4 º «b n » 2 ¬ ¼ 2.5 Where we denote by . Lp J the quas – norm taken on the interval J , and when the norm is on J . We suppress the subscript , If we set ~ Z j ,L §§ b· ¨ ¨ j ¸S 2¹ 2 cos ¨ © ¨ v ¨ © · ¸ ¸ , and ¸ ¸ ¹ ~ Z j ,r Then we have for all j for which Z j I ~ Z j ,r Z j §§ b· · ¨ ¨ j ¸S ¸ 2¹ ¸ 2 cos ¨ © 2 cos ¨ ¸ v ¨ ¸ © ¹ § j S bS · 2 cos ¨ ¸ 2 cos © v 2v ¹ §§ b· · ¨ ¨ j ¸S ¸ 2¹ ¸ 2 cos ¨ © , ¨ ¸ v ¨ ¸ © ¹ jS v jS v § j S bS bS · § jS bS bS · ¸¸ 2 cos ¨ 2 cos ¨¨ ¸ 4v 4v ¹ © v 4v 4v ¹ © v §§ · §§ · b· b· ¨ ¨ j ¸S ¸ ¨ ¨ j ¸S ¸ bS ¸ bS ¸ 4¹ 4¹ © © ¨ ¨ 2 cos 2 cos ¨ ¨ v 4v ¸ v 4v ¸ ¨ ¸ ¨ ¸ © ¹ © ¹ §§ b· · ¨ ¨ j ¸S ¸ bS 4¹ ¸ 4 sin ¨ © sin ¨ ¸ v 4v ¨ ¸ © ¹ Then since sin bS / 4 ! 3b / 4 for b satisfying 1 2 ª Sº and sin u t u , u «0, » S 2 ¬ 2¼ b· j ¸S bS 4¹ sin 4v v bS / 4 S / 6 if 0 b ~ Z j ,r Z j § ¨ 4 sin © §§ b· ¨ ¨ j ¸S 2 4¹ t4 ¨© ¨ v S ¨ © · b· § ¸ 6¨ j ¸b b 4¹ © ¸ 3b ! ¸ 4 v v ¸ ¹ 6 2.6 And ~ Z j Z j ,l b· · §§ ¨ ¨ j ¸S ¸ jS 2¹ ¸ 2 cos 2 cos ¨ © ¨ ¸ v v ¨ ¸ © ¹ § jS b S b S · § jS b S b S · 2 cos ¨ ¸ 2 cos ¨ ¸ © v 4v 4v ¹ © v 4v 4v ¹ b· b· §§ · §§ · ¨ ¨ j ¸S ¸ ¨ ¨ j ¸S ¸ b b S S 4 4 © ¹ © ¹ ¸ 2 cos ¨ ¸ 2 cos ¨ ¨ ¨ v v 4v ¸ 4v ¸ ¨ ¸ ¨ ¸ © ¹ © ¹ § j b ·S ¨ ¸ bS 4¹ 4 sin © sin v 4v b· b · §§ 6 §¨ j ·¸ b ¸ ¨ ¨ j ¸S 2 4 ¹ 3b 4¹ ¸ b © t4 ¨© ! ¸ v S¨ v v 4 ¨ ¸ © ¹ 4.2. 6 / Let j be odd and since sin bS / 4 ! 3b / 4 For b satisfying bS / 4 S / 6 , we have Tvc , b x ~ ª~ º t v , b x d cos bS / 2 cos bS , x «Z j , l , Z j , r » . ¬ ¼ For example t v,b x § jS b S · 2 cos¨ ¸ v 2v ¹ §~ · © 1 t v x cos Sb d t v ¨ Z j , l ¸ cos Sb cos v cos cos Sb 2 © ¹ § jS b S · cos v ¨ ¸ cos Sb © v 2v ¹ bS · cos§¨ jS ¸ cos Sb 2 ¹ © bS cos cos Sb 2 And cos Sb / 2 cos Sb sin bS / 4 sin 3bS / 4 7 In fact, bS 3bS cos Sb / 2 cos Sb cos §¨ 4 © 4 bS 3bS 2 sin sin 4 4 · cos § bS 3bS · ¸ ¨ ¸ 4 ¹ ¹ © 4 ~ 3b 3b 9b 2 ª~ º 2 , x «Z j , l , Z j , r » 4 2 4 ¬ ¼ Then Tvc,b x t v ,b x d cos bS / 2 cos bS 2 sin bS / 4 sin 3bS / 4 2 3b 3b 9b 2 4 2 4 ª~ , x «Z ¬ ~ ,Z j ,l j ,r º », ¼ 2.7 Then since I > 2,2@, it follows by the Bernstein inequality § · ¨ ¸ ¨ ¸ ¨ ¸ cn pn Lp> 2 ,2 @ ¸ that ¨ pcn L P[ 2 , 2 ]d 2 ¨ ¸ §x · ¨ ¸ 1 ¨¨ ¸¸ ¨ ¸ 2 © ¹ © ¹ C2 V 3 4 3 3 Tv,b lp>2,2@ t v,b Llp >2,2@ t v Lp>2,2@ d t v Lp>2,2@ C3 v 3 , 3 § § 1 ·2 · ¨1 ¨ ¸ ¸ 2 ¨ ©2¹ ¸ © ¹ In fact Tv 4,b Lp> 2 , 2 @ · d §x 3 ¨ ³t u du ¸¸ dx ¨© 0 v ,b ¹ t v3 u L p> 2 , 2 @ Lp> 2 , 2 @ d C 2 d t v3,b V3 1 2 § ¨ 1§x· ¨ ¸ ¨ ©2¹ © C2 V 3 § 1 ¨ 1 §¨ ·¸ ¨ 2 © ¹ © Hence by (2.5) C2 V § ¨ 1 §¨ 1 ·¸ ¨ ©2¹ © 1 d C3 V 3 4 P 8 Lp> 2 , 2 @ 2 · ¸ ¸ ¹ 3 3 2 · ¸ ¸ ¹ 3 tv · ¸ ¸ ¹ 3 tv Lp > 2 , 2 @ Lp> 2 , 2 @ § 2 x ¨¨ ³ cos v cos 1 2 © 2 p · dx ¸¸ ¹ 1 p 1· § Z 4M ¨ f n ,b , ¸ n¹ © p 1· § Z 4M ¨ f n ,b Tv ,b Tv ,b , ¸ p n¹ © 1· 1· § § d Z 4M ¨ f n ,b Tv ,b , ¸ p Z 4M ¨ Tv ,b , ¸ n¹ n¹ © © C p d C p f n ,b Tv ,b p 4 Tv ,4b p n p 9 b 4 C p v3 dC p n n4 §ª 3 º · ¨ «b 4 n » 2 ¸ 9 ¨ ¸ b4 ¬ ¼ ¹ C p © C p n n4 § 34 · ¨b n¸ ¨ ¸ b © ¹ dC p C p 4 n n 3 3 9 4 9 9 9 9 b4 n b4 C p C p n n4 3 b4 b4 C p C p n n 9 b4 C ( p) n 9 b4 , C1 p n 2.8 Next we need a simple lemma Lemma (2.9) there exists a constant C4 such that For any interval J I , We have the following . For any measurable sets E I , if Then p nc x t 0 , x J \ E , 2.10 5 § 9 · 4¸ ¨ J b C b 4 4 f n,b pn (2.11) ! ¨b b E ¸. L p( J ) n n ¨ n ¸ © ¹ Proof : let J R denote the middle third of J . We consider two cases . First 2 we assume that J R contains at most one of the definition of v we get 9 Z , j s .Then by the 3 1 b4 J C5 C5 v n and b2 J n 2.12 3 5 b2 b 4 C5 n2 b4 C5 2 n 5 b2 J b4 C5 2 0 n n Then by (2.11) we have b2 J 5 4 b , 2.13 n n2 On the other hand , if J R contains at least two extremes of Z j , then it f n ,b p n lp J t0! C5 contains at least 2C 6 v J extrema , for some constant C 6 ,these extremes satisfy 4.2.6 and 4.2.6 / and about half of them (and at least one ) have odd indices , then together with 4.2.6 we conclude that § meas ¨¨ J R © ­ 9b 2 ½ · 1 2b C6 v J ® x :Tvc,b x ¾¸ t 4 ¿ ¸¹ 2 v ¯ C6 b J , 2.14 Now ,if C 6 b J d E , then f n,b p n t0!b 2 J n L p( J ) Other wise , C 6 b J ! E . bE nC6 (2.15) , Then by 4.2.14 there is a point xR J R \ E , for which Tvc,b xR 9b 2 4 Hence , 4.2.10 yields , 9b 2 2 d Tvc,b xR p cn xR Tvc,b xR d 1 4 Jp d 2 J n §1· 1 ¨ ¸ ©3¹ n §1· 1 ¨ ¸ ©3¹ 2 2 Where we used the Bernstein inequality Therefore from 4.2.6 10 p n Tv ,b p n Tv ,b Lp J Lp J 9b 2 d 4 2n §1· 1 ¨ ¸ ©3¹ J 2 Pn Tv ,b 2n p n Tv ,b 2 2 J 3 Lp J Lp J J 2 9b 2 . d p n Tv ,b Lp J 3n 4 2 3 2b J d p n Tv ,b Lp J n 4 Now b2 J 3 2 b2 J d d p n Tv ,b Lp J n n 4 d Pn f n ,b Lp J f n ,b TV ,b d Pn f n ,b Lp J b . n C1 p Then f n ,b p n Lp J t b2 J n C1 p ­ Taking C 4 max ®C5 , LP J 9 4 9 4 b , n 2.16 ½ 1 , C1 U ¾ , 2.11 now follows by combining C6 ¿ ¯ 2.13 , 2.15 and 2.16 ♣ We are now in a position to define f H f , for a given sequence H 2 § ­ 2 1 ½· 5 ¨ max ®H n , ¾ ¸ , and set d0 1 and n ¿¹ ¯ © Let bn 9 bn j 4 dj ^H n `. nj 9 bnv 4 , j t 1, 3 v 1 nv j d j 1 Where the sequence ^nv ` is defined by induction as follows .First ,we choose n1 so large that ^ b 1 8 n1 1 (as needed in 2.18 12 ` below ) and J R I . suppose that n1 , n2 , ..........., nV 1 and JV 2 JV 3 .......... J 0 ,V t 2 , have been defined Then put F V 1 V 1 ¦d j 1 And let Fc V 1 x j 1 J f nj , bnj , V 1 be an interval such that 0 , x J V 1 J V 1 J V 2 and 2.17 11 Let N 1.V be such that J And let V 1 N 1 8 n t b , n t N 1,V § ¨ ¨ ¨ © 2 ,V Finally ,we take F 2 V 1 LU d V 1 2.18 J V 1 · ¸ ¸ ¸ ¹ 10 2.19 nV ! max ^nV 1 , N1,V , N 2,V ` So big that the function f nc ,b Oscillates a few times inside the interval ( J V 1 ) and since it vanishes on some interval in each Oscillation , that is , inside J V 1 , there exists an interval JV JV 1 as required in (2.17) Now denote V nV f ¦ d j 1 f nj , b nj , )V j V Where the convergence of the series is justified by the definition of the d j 's and f n, x ³ t v ,bn u du bn 0 x d ³ t v ,bn u du 0 x ³ t v u cos Sb du o x 1 0 0 d 2 ³ du d 2 ³ du 2 Now ) V l p JR d 8d V 1 ) V p 2.2 0 f LU J R ¦d j V j 1 f nj , bnj LU J R f d 2 sup ¦ d j 1 f nj , b nj j V In fact f d 2 sup ¦ d j 1 f nj , bnj j V § 9/4 9/4 § bnV ¨ ¨ bnV d ¨ d V 1 ¨1 nV nV ¨ ¨ © © 12 9/4 bnV 1 nV 1 ·· ¸¸ .......... ¸ ¸(2 2 ) ¸¸ ¹¹ f d V 1 ¦ 2 j 2 2 j 0 8d V 1 So we define f fH f ¦ d j 1 f nj , bnj j 1 And we prove Lemma(2.21) For each V t 1 we have Z M 4 § 1 · ¨¨ f , ¸¸ d C 2 U © nV ¹ U d 2.22 V Proof : First , by (2.20) § ZM4 ¨¨ ) V 1 , © 1 nV · ¸¸ d C 3 U ) V 1 Lp J d C 3 U dV 0 ¹U At the same time , (4.2.8) yields (2.23) 9 /4 § bn 1 · ¸¸ d dV 1C 4 V Z M4 ¨¨ dV 1 f nV , b nV , nV ¹ U nV © (2.24) C 4 dV Finally Z M 4 § 1 ¨¨ FV 1 , nV © · ¸¸ d C4 U ¹U Z M 2 C5 U d F n 2 V F C5 U C5 U C5 U § 1 ¨¨ FV 1 , nV © · ¸¸ ¹U 2 V 1 LU J 0 2 V 1 LU J 0 dV 1 1 10 1 10 2 ,V V N n 1 10 1 10 V V n n nV 1 / 1 0 § 1 ¨ 2/5 ¨n © V bnV § 1 ¨ 2/5 ¨n © V bnV § ¨ 1 d C5 U ¨ ¨ 1 bnV ¨ © bnV C5 U d V § 1 ¨ 2/5 ¨n bnV © V · 9/ 4 ¸dV ¸ ¹ 9/ 4 · ¸ dV ¸ ¹ 9/ 4 · ¸ dV ¸ ¹ 9/ 4 · ¸ ¸ d ¸ V ¸ ¹ (2.25) By virtue of (2.19) and the definition of d nV , dV and nV . § ZM4 ¨¨ f , © 1 nV · § § § 1 · 1 · 1 ¸¸ d ZM4 ¨¨ ) V 1 , ¸¸ ZM4 ¨¨ dV 1 f n V , b ¸¸ ZM4 ¨¨ FV 1 , , nV nV ¹ nV ¹ U nV ¹U © © U 1©3 d C 3 U dV C 4 dV C 5 U dV · ¸¸ ¹U Now Then Lemma (2.21) follows by combining (2.23) , (2.24) and (2.25) ♣ The last Lemma that we need is Lemma (2.26) There is an absolute constant C 7 such that whenever E I is a measurable set satisfying E d H nV , 2.27 And U n is a polynomial satisfying V Uc x t 0, x I \ E , nV Then f U nV t LU I \ E b 1 / 8 nV 2.28 C7 dV , Proof : by (2.14) we have f x 2.29 F V 1 is constant on d V 1 f nV , b nV x ) V 1 x M , x J V 1 Let Qn 1 V J V 1 , we may write 2.30 U nV M dV 1 Then it follows from (2.28) Qcn x t 0 , x JV 1 \E V Thus by virtue of Lemma (2.9) 5/4 · § b n2V JV 1 C 4 ¨ § 9 / 4 · b nV ¸ ¨ ¸¸ b b E t Q nV f nV , bnV V n ¨ ¸ nV LU J V 1 nV nV ¨ ¨© ¹ nV ¸ © ¹ The definition of n6 and (2.17) § bn2V J V 1 · J V 1 ¸ bn ¨ t bn 1/ 8 ¸ V V ¨ bn ¸ © V ¹ 17 / 8 ¨ 17 / 8 On the other hand , (2.24) and the definition 5/4 Of bn imply H nV d bnV and V bnV E d bnV H nV d bnV § 1 Since bnV t ¨¨ © nV Then · ¸¸ ¹ 9 4 2/5 14 5/4 bnV 15 / 4 9 /4 d b nV b nV nV Hence (2.28) implies QnV f nV , b n V LU JV 1 words . t § 17 / 8 9/4 1 ¨ b 3 C4 bnV nV ¨© nV 9/4 bnV § 1 / 8 · ¨¨ bnV 3 C 4 ¸¸ In other nV © ¹ · ¸¸ ¹ 9/4 UnV M dV 1 f nV , b n V LU J V 1 t dV 1 bnV § 1 / 8 · ¨ bnV 3C 4 ¸ nV © ¹ § 1 / 8 · dV ¨ bnV 3C 4 ¸ , © ¹ In view of (2.30) , follows from (2.20) that f U nV LU I t f U nV § § 1 / 8 t ¨¨ ¨¨ b nV 3C 4 8 ©© · ¸¸dV ¹ LU JV 1 t U nV M dV 1 f nV , b n V LU J V 1 ) V 1 LU J V 1 · ¸, ¸ ¹ And Lemma (2.26) is proved with C 7 3C 4 8 The proof of (.1.3) now follows from Lemmas (2.21)and(2.26) since f U nV E n1V f , H nV U E n1 f , H n U LU s u s u s u t t U U U Lim lim lim 1 C 2 U dV § · nof nV o f nV o f M §¨ f , 1 ·¸ Z M4 ¨ f , ¸ Z 4 ¨ ¸ © n ¹U © nV ¹ U t Lim 1 nV o f C 2 U b nV1 / 8 C 7 f References [1] E. 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