Rv. Mat. Ibroamricana 19 (2003), 221 234 Mapping proprtis of th lliptic maximal function M. Burak Erdoğan Abstract W prov that th lliptic maximal function maps th Sobolv spac W 4,η (R 2 )intol 4 (R 2 ) for all η>1/6. Th main ingrdints of th proof ar an analysis of th intrsction proprtis of lliptic annuli and a combinatorial mthod of Kolasa and Wolff. 1. Introduction In 1986, Bourgain [1] provd that th circular maximal function M C f (x) =sup f (x + ts) dσ(s) t>0 S 1 is boundd on L p (R 2 )ifp>2. Diffrnt proofs wr givn in [7] and [10]. In [8], Schlag gnralizd this rsult and obtaind almost sharp L p L q stimats for M C. In this papr, w attmpt to gnraliz Bourgain s thorm in a diffrnt dirction; w considr a natural gnralization of th circular maximal function by taking maximal avrags ovr llipss instad of circls. Mor xplicitly, lt E b th st of all llipss in R 2 cntrd at th origin with axial lngths in [ 1, 2]. Not that w do not rstrict ourslvs 2 to th llipss whos axs ar paralll to th co-ordinat axs. Th lliptic maximal function, M, is dfind in th following way: Lt f b a ral-valud continuous function on R 2,thn 1 (1.1) Mf(x) =sup f(x + s)dσ(s), x R 2, E E E E whr dσ is th arclngth masur on E and E is th lngth of E. 2000 Mathmatics Subjct Classification: Primary 42B25. Kywords: Multiparamtr maximal functions, circular maximal function, Sobolv spac stimats.
222 M. Burak Erdoğan W ar intrstd in th L p mapping proprtis of M. Proposition 1 M is not boundd in L p for p 4. Proof. First, w prov that M is not boundd in L p for p<4. Lt f δ b th charactristic function of th δ-nighborhood of th unit circl. A simpl calculation shows that for all x B(0, 1), Mf δ (x) δ 1/4. This is bcaus of th fact that for all x B(0, 1), thr is an llips cntrd at x which is third ordr tangnt to th unit circl. Thrfor, Mf δ p δ 1/4,whras f δ p δ 1/p. Taking th limit δ 0 yilds th claim. To prov that M is not boundd in L 4, considr th function (1.2) g δ (x) =( 1 x + δ) 1/4 χ B(0,2)\B(0,1). Not that g δ 4 log(1/δ) 1/4. On th othr hand, w hav Mg δ (x) log(1/δ) for all x B(0, 1) and hnc Mg δ 4 log(1/δ) (s[8]forth dtails). In light of Proposition 1, on may conjctur that M is boundd in L p for p>4. W ar far from proving this conjctur. Howvr, w obtain som stimats for M in this dirction. W stat our rsults for th ky xponnt p =4. Th stup is th following; w work with th family of maximal functions: 1 (1.3) M δ f (x) =sup f (u) du, E E E δ x+e δ whr E δ is th δ nighborhood of th llips E and E δ is th two-dimnsional Lbsgu masur of E δ. W invstigat th L 4 mapping proprtis of M δ. Applying M δ to th functions in (1.2), w s that th inquality (1.4) M δ f 4 A (δ) f 4, δ > 0 can not hold if A (δ) =o(log(1/δ) 3/4 ). On th othr hand, stimating th right-hand sid of (1.3) by δ 1 f 1 implis that M δ f 1 δ 1 f 1 and stimating it by f implis that M δ f f. By intrpolating ths bounds, w s that (1.4) holds for A(δ) δ 1/4. Lt E δ dnot th δ-nighborhood of th llips E. W hav th following basic proprty of th lliptic annuli. It corrsponds to th fact that two distinct llipss can b at most third ordr tangnt to ach othr.
Mapping proprtis of th lliptic maximal function 223 Lmma 2 Lt E 1 and E 2 b llipss such that th distanc btwn thir cntrs is δ 2/5.Thn E1 δ E2 δ δ5/4. 1/4 W prov this lmma in Sction 3 (Corollary 10 (i)). Now, using this lmma and Cordoba s L 2 Kakya argumnt [2], w prov th simpl fact that (1.4) holds for A(δ) δ 3/16. Lmma 3 M δ f 4 δ 3/16 f 4, δ > 0. Proof. Th lmma follows by intrpolating th trivial L bound with th following rstrictd wak typ stimat: (1.5) M δ f 2, δ 3/8 f 2,1. Fix a st A in B(0, 1) and λ [0, 1]. Lt Ω = {x : M δ (χ A ) >λ}. Tak a δ-sparatd st {x 1,...,x m } in Ω. W hav (1.6) Ω mδ 2. For ach x j, choos an llips E j such that Ej δ A >λδ. Using Cauchy- Schwarz inquality, w hav m mδλ Ej δ A = j=1 A j A 1/2 χ E δ j 2 j χ E δ j (1.7) ( 1/2. = A 1/2 Ej δ Ek ) δ j,k Now, w stimat th sum j,k Eδ j Ek δ using Lmma 2. W hav E δ j E δ k δ 5/4 x j x k 1/4 givn that x j x k δ 2/5. Using this, w obtain for fixd j Ej δ Ek δ δ 2 δ 5/4 dx + δ 2 1 x j x δ x 2/5 j x 1/4 k x j x δ 2/5 δdx (1.8) δ 3/4.
224 M. Burak Erdoğan Thus, (1.9) Ej δ Ek δ mδ 3/4. j,k Using (1.9) in (1.7), w hav mδλ A 1/2 (mδ 3/4 ) 1/2. Hnc which provs (1.5). ) 2 Ω mδ 2 3/8 A 1/2 (δ, λ W hav th following improvmnt: Thorm 4 For all ε>0, inquality (1.4) holds with A (δ) =δ 1/6 δ ε. Rmark. Thorm 4 implis that M maps W 4,1/6+ε into L 4 for all ε>0. Hr W p,η is th Sobolv spac consisting of functions f such that (1 ) η/2 f p <. Thorm 4 is a corollary of th following strongr thorm, which is th main rsult of this papr. Thorm 5 M δ f 24/7, δ 1/3 log(δ) 5/4 f 2,1. Proof of Thorm 5 utilizs an analysis of th intrsction proprtis of lliptic annuli. Lmma 2 abov and th following lmma ar th basic lmnts of th proof; w prov thm in sction 3. Th following lmma can b considrd as a Marstrand s thr circl lmma typ rsult for llipss. Lmma 6 Fix δ 2/5, d δ and u 1. Tak any two llipss E 1 and E 2 such that th distanc btwn thir cntrs (c 1,c 2 rspctivly) is approximatly d. Thn th δ-ntropy of th st S := {x R 2 : x c i, i=1, 2, an llips E cntrd at x such that E δ Ei δ δ(δ/u ) 1/4,i=1, 2, } is 1 δ 2 d 1/2 log(δ) u3/4 (δ/ ) 1/4.
Mapping proprtis of th lliptic maximal function 225 Not that in th proof of Lmma 3 (inquality (1.8)), w assumd that any two llipss can b third ordr tangnt to ach othr in a givn st of llipss. Howvr, using Lmma 6 and a combinatorial mthod of Kolasa and Wolff [4], [11], w can bound th numbr of pairs of lliptic annuli which ar third ordr tangnt to ach othr. This is th main ingrdint of th proof of Thorm 5. This tchniqu was also usd in [8], [10], [9] and [6]. Notation: S 1 : th unit circl. E,f z := {x R 2 :( x 1 z 1 ) 2 +( x 2 z 2 f ) 2 =1}. E δ : δ nighborhood of th llips E. K: A constant which may vary from lin to lin. A B: A KB. A B: A B and A B. A B: A K 1 B whr K is a larg nough constant. A : cardinality or th masur of th st A or th lngth of th vctor A in R 2. 2. Proof of Thorm 5 Lt A R 2,0<λ 1andΩ={x R 2 : M δ χ A (x) >λ}. W nd to prov that ) 24/7 Ω ( log (δ) 5/4 1/3 A 1/2 δ, λ Without loss of gnrality, w can assum that A D(0, 1). Lt {x j } m j=1 b a maximally δ sparatd st in Ω. Not that (2.1) Ω mδ 2. Choos llipss E j cntrd at x j such that E δ j A >λ E δ j λδ. Whav (2.2) mδλ m Ej δ A = j=1 A 1/2 m χ E δ j A j=1 m ( m χ E δ 2 j = A 1/2 j=1 j,k=1 ) 1/2 Ej δ Ek δ
226 M. Burak Erdoğan Lt { ( δ ) 1/4 ( S,u = (j, k) : x j x k (, 2 ),δ E δ δ ) 1/4 } j E δ u k δ. 2u Using this notation, w can stimat m j,k=1 Eδ j Ek δ as (2.3) m Ej δ Ek δ j,k=1 δ 2/5 1 δ 2/5 1 ( δ ) 1/4 m S,u δ + δ min(m, δ 6/5 ) u u u j=1 ( δ ) 1/4 S,u δ + m 17/12 δ 3/10, u whr th summations ar ovr th dyadic valus of and th dyadic valus of u ( 1,δ K) (sinc th trms with u gratr than a high powr of δ 1 maks ngligibl contribution, and Lmma 2 implis that S,u is mpty if >δ 2/5 and u 1). Now, w find a bound for th cardinality of th st S,u using Lmma 6. Considr th st of tripls: { Q = (j, k 1,k 2 ): x j x ki (, 2 ), ( δ δ u ) 1/4 E δ j Eδ k i δ ( δ 2u ) 1/4,i=1, 2 }. W calculat th cardinality of Q in two diffrnt ways. Lt S j = {k : (j, k) S,u }. Not that thr ar at last S 2 j tripls in Q whos first co-ordinat is j. Hnc, w hav (2.4) S,u = m ( m ) 1/2 S j m 1/2 Sj 2 (m Q ) 1/2. j=1 j=1 On th othr hand, w can choos k 1 in m diffrnt ways, and for fixd k 1, thr ar at most min(m, d2 ) indics k δ 2 2 such that x k2 x k1 (d, 2d). For any such (k 1,k 2 ), by Lmma 6 and δ-sparatdnss, thr ar at most 1 min ( log (δ) d 1/2 u 3/4 (δ/ ) 1/4, 2) indics j such that (j, k δ 2 1,k 2 ) Q. Summing ovr dyadic d (δ, 1), w obtain Q m ) min (m, d2 1 ( ) (2.5) δ 2 δ min log (δ) d 1/2 u 3/4 (δ/ ) 1/4, 2 2 d m ( (mu) 3/4 ) δ log (δ) min 2,m 2. 1/4 (δ )
Mapping proprtis of th lliptic maximal function 227 Using (2.5) in (2.4), w hav S,u (m Q ) 1/2 m ( (mu) 3/8 ) (2.6) δ log (δ) 1/2 min (δ ) 1/8,m1/2 m ( (mu) 3/8 ) 2/3 δ ( log(δ) 1/2 m 1/2 ) 1/3 (δ ) 1/8 m17/12 δ 13/12 log(δ) 1/2 (u ) 1/4. Using (2.6) in (2.3) togthr with th fact that thr ar at most log(δ) 2 trms in th summation, w obtain m Ej δ Ek δ ( ) 1/4 δ m 17/12 (2.7) δ u δ 13/12 log(δ) 1/2 (u ) 1/4 j,k=1 u +m 17/12 δ 3/10 m 17/12 δ 1/6 log (δ) 5/2. Using (2.7), (2.2) and (2.1), w hav (2.8) Ω mδ 2 ) 24/7 ( log (δ) 5/4 1/3 A 1/2 δ, λ which yilds th claim of th thorm. 3. Proof of Lmmas 2 and 6 Lt N (A, δ) dnotthδ nighborhood of th st A. First, w find a rlationship btwn th paramtrs z 1,z 2, and f of an llips Ez,f and th masur of th st N (Ez,f,δ) N(S 1,δ). W bgin with th following basic lmma. Lmma 7 Lt N b a positiv intgr. Thr xist constants K 1 and K 2 such that for all α>0 and for all δ>0, w hav N a i α i >δ = x 1 (0,K 1 α) and x 2 ( K 1 α, 0) i=0 such that N a i x i j >K 2 δ, j =1, 2. Proof. Th statmnt is trivial if α = 1, and th gnral cas follows from this by th chang of variabl y = xα. i=0
228 M. Burak Erdoğan Lt S 1 1 b S 1 {x R 2 : x 2 > 0, x 1 < 2/3}, andd(x, y) dnots th distanc btwn th points x, y R 2. Thorm 8 Lt d(z,0) = δ 2/5.Thn i) Th arclngth of E,f z N(S 1 1,δ) is ( δ )1/4. ii) If th arclngth of th intrsction is ( δ u )1/4 for som 1 u ( /δ) 1/3, thn w hav (3.1) (3.2) (3.3) z 1 min(u 3/2 (δ ) 1/2,u 9/4 (δ/ ) 3/4 ), f 2 min((u ) 3/4 δ 1/4,u 3/2 (δ/ ) 1/2 ), z 2 + f 1 min((u ) 3/4 δ 1/4,u 3/2 (δ/ ) 1/2 ). Proof. Considr th function f(x) :=z 2 + f(1 ((x z 1 )/) 2 ) 1/2 (1 x 2 ) 1/2. Tak a point t ( 2/3, 2/3) such that f(t) <δ. Not that th st Ez,f N (S 1,δ) consists of at most four connctd componnts. Hnc, it suffics to prov that thr xists x 1 (t ( δ )1/4,t)andx 2 ( t, t + ( δ 1/4 ) ) such that f(x j ) >δfor j =1, 2, and if x 1 or x 2 ar not in th ( δ 1/4 u ) nighborhood of t for 1 u ( /δ) 1/3, thn (3.1), (3.2) and (3.3) ar valid. W considr th first fiv trms of th Taylor xpansion of f(x) around t. Lt w := (1 (t z 1 ) 2 / 2 ) 1/2 (1 t 2 ) 1/2. W can assum that w 1. f(x) =z 2 +(fw 1 1)(1 t 2 ) 1/2 [( ) ] f + (z 2 1 t)w + t (1 t 2 ) 1/2 (x t) + 1 [(1 t 2 ) (1 3/2 f )] 2 2 w3 (x t) 2 + 1 ( [(1 t 2 ) 5/2 t w 5 f )] (3.4) 2 (t z 1) (x t) 3 4 + 1 ( [(1 t 2 ) 7/2 1+4t 2 w 7 f )] 8 (1 + 4(t z 1) 2 / 2 ) (x t) 4 4 + 1 [η 3+4η2 24 (1 η 2 ) f ] 9/2 (η x 3 2 +4(η z 1 ) 2 1) (η t) 5, 8 (1 (η z 1 ) 2 / 2 ) 9/2 η (t x t,t+ x t ). =: a 0 + a 1 (x t)+a 2 (x t) 2 + a 3 (x t) 3 + a 4 (x t) 4 + Er
Mapping proprtis of th lliptic maximal function 229 Choos u such that 4 i=0 a i ( δ u )i/4 = δ, whav a i (u ) i/4 δ 1 i/4 for i =0, 1, 2, 3, 4. W hav two cass: (i) u ( /δ) 1/3. Lmma 7 shows that if w omit th rror trm Er, thn th arclngth of th intrsction is ( δ )1/3. Itisasytosusingth hypothsis δ 2/5 that th rror trm is not significant. (ii) u ( /δ) 1/3. Using th dfinitions of a 0,a 1,a 2 and a 3,wobtain (3.5) z 2 (1 t 2 ) 1/2 + fw 1 = 1+O(δ), (3.6) f (t z 1)w 2 = t + O((u ) 1/4 δ 3/4 ), (3.7) f 2 w3 = 1+O((u ) 1/2 δ 1/2 ), (3.8) f (t z 1)w 5 4 = t + O((u ) 3/4 δ 1/4 ). Substituting (3.7) into (3.8), w obtain (3.9) (f) 2/3 (t z 1 )(1 + O((u ) 1/2 δ 1/2 )) = t + O((u ) 3/4 δ 1/4 ), which implis that (3.10) ( 1/3 ) t f 1 z1 2/3 + t z 1 t = O((u ) 3/4 δ 1/4 ). Substituting (3.7) into (3.6), w obtain f 2/3 (3.11) (t z 1)(1 + O((u ) 1/2 δ 1/2 )) = t + O((u ) 1/4 δ 3/4 ), 4/3 which implis that ) (3.12) t (1 4/3 = z f 2/3 1 + O( z 1 t (u ) 1/2 δ 1/2 +(u ) 1/4 δ 3/4 ). Subtracting (3.9) from (3.11), w obtain (3.13) (t z 1 )(f 4/3 2/3 + O((u ) 1/2 δ 1/2 )) = O((u ) 3/4 δ 1/4 ). (3.14) Substituting (3.7) into (3.5), w obtain z 2 (1 t 2 ) 1/2 + ( f 4/3 2/3 1 ) = O((u ) 1/2 δ 1/2 ).
230 M. Burak Erdoğan Now, thr ar two cass z 2 or z 1. Cas a) Assum z 2. (3.14) implis that f 2. Using this in (3.13), w obtain (3.15) t z 1 = O(u 3/4 (δ/ ) 1/4 ), which implis using (3.11) that (3.16) t = O(u 3/4 (δ/ ) 1/4 ). Using th fact f 2 and (3.15) in (3.10), w obtain t z 1 t = O((u ) 3/4 δ 1/4 ). This and th dfinition of w implis that w = 1 + O((u ) 3/4 δ 1/4 ). On th othr hand, using (3.15) and (3.16) in th dfinition of w, wobtain w =1+O(u 3/2 (δ/ ) 1/2 ). Hnc, using (3.7), w hav (3.17) f 2 min((u ) 3/4 δ 1/4,u 3/2 (δ/ ) 1/2 ). Using (3.15), (3.16) and (3.17) in (3.12), w obtain (3.18) z 1 min(u 3/2 (δ ) 1/2,u 9/4 (δ/ ) 3/4 ). Finally, using (3.16) and th stimats for w 1 in (3.5), w obtain (3.19) z 2 + f 1 min((u ) 3/4 δ 1/4,u 3/2 (δ/ ) 1/2 ). Cas b) Assum z 1. Using (3.12), w obtain (3.20) f 2, t. Using (3.7), w obtain (w 2 1)(f/ 2 ) 2/3 +(f/ 2 ) 2/3 1=O((u ) 1/2 δ 1/2 ), which implis using (3.20) that (3.21) w 2 1. Using th dfinition of w, wobtainw 2 1 (t z 1 ) 2 / 2 t 2. Hnc (3.21) implis that (3.22) t z 1 t. Using (3.9), w obtain ( 1/3 ) t f 1 z1 2/3 + t z 1 = t + O((u ) 3/4 δ 1/4 ),
Mapping proprtis of th lliptic maximal function 231 which implis using (3.22) that f 2 t z 1. Hnc f 2 and (3.14) implis that z 2. Thus th stimats that w obtaind in cas a) ar valid. Applying Lmma 7 (with K 1 δ instad of th δ in th lmma, for a sufficintly larg K 1 ), w s that f(x) Er > Kδ, for som z 1 (t K(δ/(u )) 1/4, 0) and x 2 (0,t+ K(δ/(u )) 1/4 ). Now, w prov that Er = O(δ) for x (t K(δ/(u )) 1/4,t+ K(δ/(u )) 1/4 ). Not that th stimats that w obtaind in part a) imply that 1, f 1. Lt h(η) =η 3+4η3. (1 η) 9/2 W hav ( Er h(η) f ( η h z1 )) x t 5 5 ( ( η z1 ) ( η z1 = h(η) h + h )(1 f )) x t 5 ( 5 η η z ) 1 + x t 5 ( η( 1) + z 1 + ) x t 5 ( δ ) 5/4 δ. u Finally, w prov that u can not b 1. Assum that u 1. Using th dfinition of a 4 and th stimats w obtaind abov, w obtain a 4 1 f f ( 4 1 t 4 w7 t 2 z1 ) 2 ( t z1 ) 2 1 w 7 f. 4 Hnc, u can not b 1. This yilds th uppr bound for th arclngth of th intrsction. Lt min ± (A ± B) dnotmin(a + B,A B). Corollary 9 Lt d(z,0) = δ 2/5.Thn i) Th arclngth of E,f z N(S 1,δ) is ( δ )1/4, ii) if it is ( δ u )1/4, 1 u ( /δ) 1/3, thn w hav min ± ( (f)2/3 1 ± d(z,0) ) min((u ) 3/4 δ 1/4,u 3/2 (δ/ ) 1/2 ).
232 M. Burak Erdoğan Proof. W divid N (S 1,δ) into four sgmnts; N (S 1,δ)= 4 i=1n (Si 1,δ), whr N (S1 1,δ)isasbforandN (S1 i,δ) is obtaind by rotating N (S1 1,δ) around th origin iπ/2 dgrs. Not that if th intrsction of th llips with N (S 1,δ) is larg, thn its intrsction with on of N (Si 1,δ) should b larg, too. Lt Ez,f N(S1,δ) 1 > ( δ u )1/4,forsom1u ( /δ) 1/3. Triangl inquality and (3.1) imply that (3.23) min ± ( y 1 ± d(z,0) ) z 1 min(u 3/2 (δ ) 1/2,u 9/4 (δ/ ) 3/4 ). Th fact that, f [ 1, 2] and (3.2) imply that 2 Hnc, w hav f (f) 2/3 f 2 min((u ) 3/4 δ 1/4,u 3/2 (δ/ ) 1/2 ). (3.24) f 1=(f) 2/3 1+O(min((u ) 3/4 δ 1/4,u 3/2 (δ/ ) 1/2 )). Using (3.23) and (3.24) in (3.3), w obtain min ± ( (f)2/3 1 ± d(z,0) ) min((u ) 3/4 δ 1/4,u 3/2 (δ/ ) 1/2 ). Applying Thorm 8 (aftr a rotation) also in th cass whr N (S1,δ)is 1 rplacd with N (Si 1,δ), i =2, 3, 4 yilds th claim of th corollary. E,f z Th following corollary provs Lmma 2. Lt Ez,f,θ rotatd countr-clockwis by an angl θ around its cntr. Corollary 10 Lt d(z,y) = δ 2/5.Thn dnot th llips i) ThmasurofthstN (Ez,f,θ,δ) N(Ey a,b,δ) is δ( δ )1/4, ii) if it is δ( δ u )1/4, 1 u ( /δ) 1/3, thn w hav min ± ( (f)2/3 (ab) 2/3 (1±d a,b (z,y)) ) min((u ) 3/4 δ 1/4,u 3/2 (δ/ ) 1/2 ), whr d a,b ((p 1,p 2 ), (q 1,q 2 )) = ((p 1 q 1 ) 2 /a 2 +(q 1 q 2 ) 2 /b 2 ) 1/2. Proof. By a dilation, a translation and thn a rotation, w can transform Ex a,b 0,y 0 into S 1 and Ez,f,θ into E 1,f 1 w such that 1 f 1 = f (sinc th ara of ab Ez,f,θ is qual to th ara of E 1,f 1 w tims ab) andd(w, 0) = d a,b (z,y). Th claim follows by applying Corollary 9 to E 1,f 1 w and S 1.
Mapping proprtis of th lliptic maximal function 233 Proof of Lmma 6: By making th suitabl translations, rotations and dilations w can assum that E 2 = S 1 and E 1 = Ey a,b,whr y d. W can furthr assum that u ( δ )1/3, sinc th statmnt of th thorm is void if u ( δ )1/3. Dnot u 3/2 (δ/ ) 1/2 by ξ, and considr th functions F (x) = ( x 2,d a,b (x, y) 2 ), G(r, s) = min ( 1 ± r +(ab) 2/3 (1 ± s) ). ± Thorm 10 implis that th st S iscontaindinthst (3.25) S := {x R 2 : x, d(x, y),g(f (x)) ξ}. It is asy to s that th masur of th st B ξ := {(r, s) :G(r, s) ξ} is ξ (not that ξ 1). Blow, w prov that th masur of th invrs imag of a st of masur ξ undr F is at most (ξ/d) 1/2 ( log(ξ/d) + 1), which yilds th claim of th lmma. Lt B ξ b a st of masur ξ and A η b th st whr th Jacobian of F, JF,islssthnη. Co-ara formula (s,.g., [3] Thorm 3.2.3) implis that (3.26) F 1 (B ξ ) A η + ξ η. Claim. A η η log( η ) +1. d d Proof. Without loss of gnrality, w can assum that y 1 d. It is asy to calculat that JF x 1(x 2 y 2 ) x 2(x 1 y 1 ) a 2 b 2 = x b 2 a 2 1 x 2 x 1y 2 + x 2y 1 a 2 b 2 b 2 a. 2 Hnc, { A η = x R 2 : x 1 ( 2, 2), x x 1 y 2 a 2 (3.27) 2 x 1 (a 2 b 2 )+y 1 b 2 ) This shows that if a 2 b 2 d, thn A η η. d a 2 b 2 d. (3.27) implis that 2 ( A η min 2 which provs th claim. ηa 2 b 2 ) x 1 (a 2 b 2 )+y 1 b 2, 1 ηa 2 b 2 x 1 (a 2 b 2 )+y 1 b 2 }. Now, assum that dx 1 η d log(η ) +1, d Claim of th lmma follows from (3.26) and th claim abov by choosing η =(ξd) 1/2.
234 M. Burak Erdoğan Rfrncs [1] Bourgain, J.: Avrags in th plan ovr convx curvs and maximal oprators. J. Analys Math. 47 (1986), 69 85. [2] Córdoba, A.: Th Kakya maximal function and sphrical summation multiplirs. Amr. J. Math. 99 (1977), 1 22. [3] Fdrr, H.: Gomtric masur thory, Springr-Vrlag Brlin Hidldbrg, 1996. [4] Kolasa, L., Wolff, T.: On som variants of th Kakya problm. Pacific J. Math. 190 (1999), 111 154. [5] Marstrand, J. M.: Packing circls in th plan. Proc. London Math. Soc. (3) 55 (1987), 37 58. [6] Mitsis, T.: On a problm rlatd to sphr and circl packing. J. London Math. Soc. (2) 60 (1999), 501 516. [7] Mocknhaupt, G., Sgr, A., Sogg, C.: Wav front sts and Bourgain s circular maximal thorm. Ann. of Math. (2) 134 (1992), 207 218. [8] Schlag, W.: A gnralization of Bourgain s circular maximal thorm. J. Amr. Math. Soc. 10 (1997), 103 122. [9] Schlag, W.: A gomtric inquality with applications to th Kakya problm in thr dimnsions. Gom. Funct. Anal. 8 (1998), 606 625. [10] Schlag, W.: A gomtric proof of th circular maximal thorm. Duk Math. J. 93 (1998), 505 533. [11] Wolff, T.: Rcnt work connctd with th Kakya problm. In Prospcts in Mathmatics (Princton, N.J., 1996), d. H. Rossi. Amrican Mathmatical Socity, 1998. Rcibido: 25 d sptimbr d 2001 M. Burak Erdoğan Dpartmnt of Mathmatics Univrsity of California Brkly, CA 94720, USA burak@math.brkly.du This rsult is a part of th author s Ph.D. thsis at th California Institut of Tchnology writtn undr th guidanc of Prof. Thomas Wolff.