On a Problem in Geometric Measure Theory Related to Sphere and Circle Packing Citation Mitsis, Themistoklis (1998) On a Problem in Geometric Measure Theory Related to Sphere and Circle Packing. Dissertation (Ph.D.), California Institute of Technology. doi:10.7907/992h-9263. https://resolver.caltech.edu/CaltechTHESIS:08012025-171149569 Abstract In this thesis we prove that a Borel set which contains spheres centered at all points of a Borel set of Hausdorff dimension greater than 1 must have positive Lebesgue measure, and, using the same method, we rederive a special case of Stein's spherical means maximal inequality. We also prove the corresponding result for circles, provided that the set of centers has Hausdorff dimension greater than 3/2. Item Type: Thesis (Dissertation (Ph.D.)) Subject Keywords: (Mathematics) Degree Grantor: California Institute of Technology Division: Physics, Mathematics and Astronomy Major Option: Mathematics Thesis Availability: Public (worldwide access) Research Advisor(s): Wolff, Thomas H. Thesis Committee: Wolff, Thomas H. (chair) Kahn, Jeremy Kechris, Alexander S. Last, Y. Defense Date: 16 March 1998 Record Number: CaltechTHESIS:08012025-171149569 Persistent URL: https://resolver.caltech.edu/CaltechTHESIS:08012025-171149569 DOI: 10.7907/992h-9263 ORCID: Author ORCID Mitsis, Themistoklis 0000-0003-4512-1853 Default Usage Policy: No commercial reproduction, distribution, display or performance rights in this work are provided. ID Code: 17583 Collection: CaltechTHESIS Deposited By: Benjamin Perez Deposited On: 04 Aug 2025 23:08 Last Modified: 04 Aug 2025 23:18 Thesis Files PDF - Final Version See Usage Policy. 8MB Repository Staff Only: item control page

ON A PROBLEM IN GEOMETRIC MEASURE THEORY RELATED TO SPHERE AND CIRCLE PACKING Thesis by Themistoklis Mitsis In Partial Fulfillment of the Requirements for the Degree of Doctor of Philosophy California Institute of Technology Pasadena, California 1998 (Submitted March 16, 1998) ii Our revels are now ended ... These our actors, As I foretold you, were all spirits, and Are melted into air, into thin air, And, like the baseless fabric of this vision, The cloud-capped towers, the gorgeous palaces, The solemn temples, the great globe itself, Yea, all which it inherit, shall dissolve, And, like this insubstantial pageant faded, Leave not a rack behind: we are such stuff As dreams are made on; and our little life Is rounded with a sleep ... William Shakespeare iii Acknowledgment I thank my advisor, professor Thomas Wolff, for his patience, his guidance and his invaluable support throughout all these years. IV Abstract In this thesis we prove that a Borel set which contains spheres centered at all points of a Borel set of Hausdorff dimension greater than 1 must have positive Lebesgue measure, and , using the same method, we rederive a special case of Stein's spherical means maximal inequality. We also prove the corresponding result for circles, provided that the set of centers has Hausdorff dimension greater than 3/2. V Contents Introduction 1 Chapter 1. Background 5 Chapter 2. The higher dimensional case 10 Chapter 3. The two-dimensional case 22 Chapter 4. Possible improvements 37 Appendix 44 Bibliography 49 vi List of Figures Figure l. The circles of Appolonius 7 Figure 2. Construction of the parameter set E 18 Figure 3. Tangential and transversal intersection of two annuli 23 INTRODUCTION 1 Introduction By the term packing problem we usually refer to the following type of question: given a collection of geometric objects in a Euclidean space is it possible to find a set of Lebesgue measure zero containing a translate, or a congruent copy of dilates of every object in the collection? The beginning of the study of packing problems can be traced to Besicovitch who, as early as 1919, solved the "Kakeya problem" by constructing a compact set of plane measure zero containing a line segment in every direction. The monograph by Falconer [5] contains a fairly extensive account of the history of the subject, as well as its applications to other areas of mathematics. In this thesis we are concerned with the problem of investigating the relation between the measure of the union of a set of spheres or circles and the metric properties of the set of their centers. To begin with, a partial answer to the above question comes from the work of Stein [14] on the spherical means maximal operator. INTRODUCTION 2 For f E Cc(IRd), define where sd-l is the (d-1)-dimensional sphere and dais surface measure. Then we have the following: THEOREM 0.1. Suppose that d 2". 3, p > d~l . Th en th ere exists a constant C > 0 that depends only on d and p such that Using this result one can easily prove the following: THEOREM 0.2. Let F C JRd, d 2". 3, be a set of positive Lebesgue m easure. If E C ]Rd is a set which contains spheres cent ered at all points of F , then E has positive Lebesgue measure. The two-dimensional case was settled by Bourgain [2] , who proved that the conclusion of the above theorem still holds if d = 2, and, independently, by Marstrand [9], who used purely geometric methods . THEOREM 0.3. Let F C IR 2 be a set of positive Lebesgue measure. If E c IR 2 is a set that contains circles centered at all points of F , then E has positive plane measure. INTRODUCTION 3 This should be contrasted with a construction due to Talagrand [15] . THEOREM 0.4. There is a set of plane measure zero containing for each .T on a given straight line, a circle centered at .1:. It is , therefore, natural to ask whether one can weaken the condition that the set Fin the above theorems should be of postive measure. The main results in this thesis are the following: THEOREM 1. Let F C !Rd, d ~ 3, be a Borel set of Hausdorff dimension s, s > l. If E C ]Rd is a Borel set that contains spheres centered at each point of F, then E has positive Lebesgue m easure . THEOREM 2. Let F C IR 2 be a Borel set of Hausdorff dimension s, s > 3/2. If E C IR 2 is a Borel set that contains circles centered at each point of F, then E has positive Lebesgue m easure. The thesis is organized as follows: In Chapter 1 we state some geometric lemmas needed later on, in Chapt er 2 we prove Theorem 1 and construct a counterexample related to it, in Chapter 3 we prove Theorem 2, and finally, in Chapter 4, we discuss the possibility of weakening the condition s > 3/2 in Theorem 2. INTRODUCTION Throughout this thesis, stant A , a, ;S and similarly with measure by I • J. a, b means 2:, and b a, a, : : ; ~ 4 A b b . for some absolute con- We will denote Lebesgue 1. BACKGROUND 5 CHAPTER 1 Background We start with some notation: B(x, r) is the open disk (or ball) with center .T and radius r. C(.r , r) is the circle (or sphere) with center .T and radius r. C 8 (x, r) is the 5-neighborhood of the circle (or sphere) C(.r, r ), i.e., the set {y E IR.d: r - 5 < Ix - YI < r + 5}. If C (x , r) and C (y, s) are circles, then define d((.r , y), (r, s)) = Ix - YI+ Ir - sl ~((x,y), (r, s)) = llx -yl - Ir - s11 Note that~ = 0 if and only if the circles are internally tangent, that is, they are tangent and one is contained in the bounded component of the complement of the other. In what follows, we assume that the centers of all circles (or spheres) in question are contained in the disk B(O, 1/4) and that their radii are in the interval [1/2, 2]. The following lemma gives estimates on the size of the intersection of two annuli in terms of their relative position and their degree of tangency. The reader is referred to Wolff [17] for a proof. 1. BACKGROUND 6 Suppose that r =/- s, 0 < 8 < l. Then there exists an LEMMA 1.1. absolute constant A 0 such that: 1. C 0 (.r✓ , r) n C 0 (y, s) is contained in a 8-neighborhood of arc length IX • t .r✓ cen t,ered a,t th . ,e pow,, ll+8 _ I+" y V - r • sgn (r - s ) -Xx-y _y . 1 1 2. The area of intersection satisfies 0 52 0 IC (x, r) n C (y, s)I < Ao--;::==== - J (8 + ~) (8 + d) Furthermore we have the higher dimensional analogue whose proof we omit. LEMMA 1. 2. Suppose that C 0 ( x, r), C 0 (y, s) are spheres in ffi.d, d 2:: 3, such that r =/- s. Then for O < 8 < l 0 0 IC (x,r)nC (y,s)l~(j r + x-y I The following lemma is essentially Marstrand's three circle lemma [9]. It is a quantitative version of the following fact known as the circles of Apollonius: Given three circles which are not internally tangent at a single point, there are at most two other circles that are internally tangent to the given ones. 1. BACKGROUND FIGURE 1. LEMMA 1.3. satisfy ,\ 2: A1 7 The circles of Apollonius. There exists a constant A 1 such that if E, t, ,\ E (0, 1) J1 then for three fixed circles C(.Ti, ri), i = 1, 2, 3 and for 8 :::; E the set {(x,r) E ffi. 2 x IR: ~((x,r), (.Ti,ri)) < E Vi, d((x,r), (xi,ri)) > t Vi, C 8 (x, r) n C 8 (xi, ri) =/- (/J Vi, dist(C 8 (.T, r) n C 8 (xi, ri), C 8 (.T, r) n C 8 (xj, rj)) 2: ,\ Vi, j : i =/- j} l. BACKGROUND 8 is contained in the union of two ellipsoids in IR 3 each of diam eter ;S >..E2 . A proof of the preceding result can be found in Wolff [17]. We conclude this chapter with some facts from Geometric Measure Theory. We refer the reader to Falconer [5] for definitions , proofs and details . In what follows, Hs denotes s-dimensional Hausdorff measure. THEOREM 1.1. Let E be a Borel set in !Rd and let s > 0. As- sume that H s(E) > 0. Th en there exists a nontrivial finite measure /J, supported on E such that p,(B (x, r)) :S rs for :r E !Rn and r > 0. If E is an s-set, i.e., 0 < H s( E) < oo, then a point .r, E E is called regular if the upper and the lower densities at .r are equal to one; otherwise x is called irregular. An s-set E is said to be irregular if Hsalmost all of its points are irregular. Irregular 1-sets are characterized by the following: THEOREM 1. 2. A 1-set in IR 2 is irregular if and only if it has pro- jections of lin ear Lebesgue measure zero in two distin ct directions. In fact, one can say much more. THEOREM 1.3. Let E be an irregular 1-set in IR 2 . Th en proj0(E) has lin ear Lebesgue measure zero for almost all 0 E [O, 1r) , where proj0 1. BACKGROUND 9 denotes orthogonal projection from JR. 2 onto the line through the origin making angle 0 with some fixed axis. 2. THE HIGHER DIMENSIONAL CASE 10 CHAPTER 2 The higher dimensional case In this chapter we assume that d 2'. 3. For f: JRd---+ IR, t5 > 0 small, we define M8: B(O, 1/4) ---+ IR by Theorem 1 will be a consequence of the following L 2 ---+ L 2 maximal inequality: PROPOSITION 2 .1. Let F C B(O, 1/ 4) be a compact set in ]Rd such that th ere exist s > 1 and a finit e measure µ, supported on F with µ,(B(.1; ,r)):::; r 8 for x E ]Rd and r > 0. Then there exists a constant A that depends only on s such that ( l 1/2 (Md(.1;))2 dµ,(.-r) ) :::; AIIJll2 for small t5 > 0 and all f. In order to prove Proposition 2.1 we need the following weighted version of Schur's Lemma: 2. THE HIGHER DIMENSIONAL CASE LEMMA 2.1. 11 Let (S, µ,) be a measure space, and let C > 0. Suppose K is a measurable kernel on S x S and w is a measurable function on S such that ls ls IK(x, y)w(y)ldµ,(y) :s; Clw(x)I Vx ES IK(x, y)w(.r,)ldµ,(x) :s; Clw(y)I Vy ES Then for J E L 2 ( S) the function T J defined by TJ(x) = ls K(.r,,y)J(y)dµ,(y) is well defined almost everywhere, is in L 2 (S), and satisfies IITJll2 :s; PROOF. Let I= ls Then by Holder 's inequality Therefore IK(x,y)f(y)ldµ,(y) 2. THE HIGHER DIMENSIONAL CASE 12 (1 IK(.cr, y)J lwty)J lf(y)J 2d11,(y)) dµ,(x) = C fs 1J (y)J Jwty) J (1 IK(x, y)w(x)Jdµ,(.cr)) dµ,(y) ~ C fs if(y )J2d11,(y). =Ch w(x) 2 □ PROOF OF PROPOSITION 2.1. Decompose IR_d into disjoint cubes {Qj} of the form where k1 , ... , kd E Z. Then Note that if x, y E Qj then Therefore, if ,r,j is the center of the cube Q j and if we let a.j = µ,(Q j) then l (Mof(x))2dµ,(.cr) ;S L aj(M 2of(xj)) 2 J Since JC 8 (x, r)I ~ 5, we can choose rj E [1/2, 2] such that (1) 2. THE HIGHER DIMENSIONAL CASE 13 It follows that By duality, there exists {bj} with I:,jb] = l such that 2 = :, ( ;;= b;a )f' L •(x, ,,,) J(y)dy) = J, (J (f(y) ;;= b;a;;'xc"{x,,,, J(Y)) dy) 2 2 '.o :, II JIil j (;;= b;a]I' Xc"{x, ,,,) (y) ) dy = J2 IIJII ~ J(:z= 12 2 bibja; aY xc26(xi,ri)nC26(xjn)(Y) ) dy 1.,J 1.,J (3) where t he last inequality follows from Lemma 1.2. Now let 1/2 1/2 Q,. K (i, j)= 1 • Q,. 1 !xi - .x j l + 8 2. THE HIGHER DIMENSIONAL CASE 14 Then, by Holder's inequality, (4) For each i,j let ii,j E Qi be such that 1 . 1 = min ---lii,j - .Tjl + D xEQ; 1.1: - .Tjl + D Then for all j we have L K(i,j)w(i) = w(j) LI-Ti_ 1, :JI+ D 1, :s:; Cw(j) where the last inequality follows from the fact that a measure satisfying the condition in the statement of Theorem 1 has uniformly bounded potential (see [5]). 2. THE HIGHER DIMENSIONAL CASE 15 By symmetry L K(i,j)w(j) ;S Cw(i), Vi j Hence, by Lemma 2.1, we have (5) It follows from 1, 2, 3, 4, and 5 that where A depends only on the measure of F and on s. □ Note that if we let then, using Fatou's Lemma, we obtain the 8-free version of Proposition 2.1 as follows: J (Af (x))Pdµ,(x) = f supsupinf ( 1 8/ )I f . lf (y) ldy ) p dµ(x) t £ 8<£ C x, t J cb(x,t) ::::;: supinff sup (1c8/ E :S: 8<£ t j (M8f( .1; ))Pdµ,(.r) :S 11!11~ X, t )I f . }Cb(x,t ) lf (y) ldy)p dµ(x) 2. THE HIGHER DIMENSIONAL CASE PROOF OF THEOREM 1. We may assume that F C B(0, ¼)- 16 Sup- pose that IEl=O and choose t so that 1 < t < s. Then there exist a compact set E 1 C E, a compact set F 1 C F with Ht(F1 ) > 0, and a positive number r, such that , for each x E F 1 , there is a sphere centered at .T with radius r(.-r) E (r, 2r) which intersects E 1 in set of surface measure at least rd-l (.T). Without loss of generality we may assume that r = l. It follows that for all x E Fi where Ef is the 8-neighborhood of the set E 1 . By Theorem 1.1, there exists a nontrivial finite measureµ, supported on F 1 such that µ(B(x, r)) :::; rt, for x E IRd, r > 0. Therefore, by Proposition 2.1 , we have The right-hand side of the above inequality tends to zero as 8 ---+ 0, so □ we get a contradiction. We will show that we cannot drop the condition s > 1 in Theorem 1. PROPOSITION 2.2. There exists a set of d-dimensional Lebesgue measure zero containing for each x E [O, 1] x {0} x --- x {0}, a sphere d-l centered at x. 2. THE HIGHER DIMENSIONAL CASE PROOF. 17 The idea, which goes back to Davies [4], is to parametrize the set of radii using a suitable irregular 1-set. Divide the unit square [O, 1] x [O, 1] C ~ 2 into 16 disjoint squares of side 1/ 4. Let S(i,j) l :::; i, j :::; 4 be those squares (indexed from bottom to top, left to right). Let Apply the same procedure to each of s(l,2), s(1,4), s(4,1), s(4,3), and let E 2 be the union of the new squares. Continuing in the same manner we obtain a decreasing sequence of compact sets {En}- Let E = n~=l En. Then E is a 1-set such that where I· I is linear Lebesgue measure, and proj 0 , proj1r; 2 , proj1r; 4 denote, respectively, orthogonal projection onto the x-axis, the y-axis, and the line through the origin making angle 1r / 4 with the x-axis. It follows from Theorem 1.2 that E is irregular. Let A= LJ {(.r1, ... ,xd): (.r1-a) +.r~+-··+.1:~=a,2+b} 2 (a,b)EE = LJ {(x1, ... ,.1:d): x~ = 2ax1 + b- .Ti - • • • - .1:t 1} (a,b)EE 2. THE HIGHER DIMENSIONAL CASE FIGURE 2. 18 Construction of the parameter set E. Since proj 0 (E) = [O, 1], A contains a sphere centered at each point of {(a, O, .. . , 0) : a, E [O, 1]}. Fix s 1, . .. , sd-l· Then 2. THE HIGHER DIMENSIONAL CASE 19 = {s1} X { s2} X • • • X { Sd-1} X B where B has measure zero if and only if {2as 1 + b: (a, b) E E} has measure zero. But £ 1( {2as 1 + b : (a, b) E E}) = 0 for almost all s 1 E IR by Theorem 1.3. Therefore, by Fubini, A has d-dimensional measure □ zero. It is interesting to note that using Proposition 2.1, we obtain a geometric proof of the following special case of the spherical means maximal theorem in JRd: COROLLARY 2.1. There exists an absolute constant C such that for small 8 > 0 and all f. PROOF. Decompose ]Rd into disjoint cubes {Q 1 } of diameter 1/2. Then by Proposition 2.1, we have that for all j 2. THE HIGHER DIMENSIONAL CASE 20 For each j let Then there exists an integer N such that card(J(j)) < N \:/j L XI(j)(k) < N \:/k j Also, for each k let f k = f XQk, so that Note that if k (/_ I(j) then XQ 1 Mofk = 0. It follows that { (Mof(x)) 2 dx JN.3 = L 1Q ·(Mof(x))2d.T J J ~ k, (~M,J,(x)) dx =LL r Mofk(x)Mofz(.T)dx 2 'o j k,l }Qj 112 ::; L L (1 .(Mofk(x))2dx) j k ,lEI(j) L llfklbllf1ll2 ;SL j Q1 k,lEI(j) ::; NL L llfkll~ j kEI(j) (1. 112 2 (Mof1(x)) dx) Q1 2. THE HIGHER DIMENSIONAL CASE 21 = N L L XI(j)(k) llfk ll~ k :::; N 2 j L llfkll~ k 2 = N llfll~ □ 3. THE TWO-DIMENSIONAL CASE 22 CHAPTER 3 The two-dimensional case Before we proceed with the proof of Theorem 2, it might be instructive to discuss briefly the underlying ideas. It turns out that the two- dimensional problem can be reduced to estimating the measure of a family of thin annuli . By Lemma 1.1, the measure of the intersection of two annuli is large when the corresponding circles are internally tangent. It is, therefore, essential that we be able to control the total number of such tangencies. To this end, we employ Marstrand's three circle lemma together with a suitable counting argument. This approach was first used in Kolasa and Wolff [8], and, subsequently, by Schlag [10], [11], [12]. We should, however, mention that , in contrast with the afforementioned authors, we do not make any cardinality estimates since these are not particularly useful in the case of general Hausdorff measures. The motivation for the combinatorial part of the proof is the following observation (see [6] for more details): 3. THE TWO-DIMENSIONAL CASE FIGURE 23 3 . Tangential and transversal intersection of two annuli. PROPOSITION 3 .1. Let { Cj }f=, 1 be a family of distinct circles such that no three are tangent at a single point. Then where Ci II Cj means that Ci and Cj are internally tangent. PROOF. Let Q = {(i,j1,J2 ,j3): Ci I Cjk, k = 1,2,3}. Fix j 1 ,32,j 3 . Then, by the circles of Apollonius, there are at most two choices for j. Therefore , card(Q) ~ 2N(N - l)(N - 2) < 2N 3 3. THE TWO-DIMENSIONAL CASE 24 On the other hand, if we let n(i) = card( {j : C II Cj} ), then N Q= U({i} x {(j1 ,J2,J3): ci II cjk, k = 1,2,3}) i=l Hence N card(Q) 2 L n(i)(n(i) - l)(n(i) - 2) i=l N 2 I:(n(i) - 2) 3 i=l It follows that N card({(i,j): C II Cj}) = I:n(i) i=l N = L(n(i) - 2) + 2N i=l :S (card( Q)) 1/ 3 N 2/ 3 + 2N < N5/3 ~ □ The proof of Theorem 2 will be a quantitative version of Proposition 3.1, with Lemma 1.3 playing the role of the restriction imposed by the circles of Apollonius. 3. THE TWO-DIMENSIONAL CASE 25 PROOF OF THEOREM 2. We may assume that F C B(O, ¾)- Suppose that IEI = 0 and choose s 1 so that 3/2 < s 1 < s. Then there exist a compact set E 1 C E , a compact set F 1 c F with '}-{ 51 (F1 ) > 0 and a positive number r, such that, for each x E F1 , there is a circle centered at x with radius r (x) E (r, 2r) which intersects E 1 in a set of angle measure at least 1r. Without loss of generality we may assume that r = 1. By Theorem 1.1, there exists a nontrivial finite measureµ, supported on F 1 such that µ(B(x ,r)) ~ r 51 , for x E JR. 2 , r > 0. Let { .ri}iEI be a maximal 5-separated set of points in Fi, and let ai = µ,(B (xi, D)). Choose ri > 0 such that (6) where Ef is the ()-neighborhood of E 1 . Let "' be the infimum of those .,\ > 0 such that there exists J C I satisfying and for all j E J l{xE C (.Tj,rj)nEf: I:aiXC6(xi,ri)(.r) ~>-} I~l1C'5(.Tj,rj)I 8 iEI 3. THE TWO-DIMENSIONAL CASE 26 Choose N large enough so that 3 S1-2 > 7 + 2s 1 - 1 --N-- and N2 - N 2s1 + 1 2 N - N - 2 < 3 Let C2 > 1 be a large constant to be determined later on. Define /3 : [8, 1] x [8, 1] ---+ IR by if t 2s1 +1 3 /3(t, E) = 2__JJ__ < C2 N- 2 E Then, for small 8, /3 has the following properties: (7) (8) L /3(82\ 8i) < M 62k<l 62 1~1 where M is a constant that depends only on N and on s 1 . Define for all i, j E I and t, E E [8, 1] (9) 3. THE TWO-DIMENSIONAL CASE 8 At,E(j) = {XE C (xj, rj) : L 27 a,iXC6(xi,r i )(x) 2'. !(3(t, E)i} iESt,, (j) CLAIM 3.1. There exist t, EE [8, 1] and a set of indices J such that and PROOF. lo= Let {j EI: l{-r E C 8 (xj,rj) n Ef: L a,iXC6(xi,r;_)(x) :S i}I iEJ By the minimality of"", we have Therefore, if l' is the complement of 10 , then L jEJ' and for all j E J' a,j 21µ,(F1) (10) 3. THE TWO-DIMENSIONAL CASE 28 Hence, using 6 we obtain (11) For each j E J' let 8 Bj = {XE C (.Tj, rj) n Ef: L a,iXC6(xi,ri)(-T) > ~} (12) iEI Then for all j E J' Bj c LJ A 62 k, 62 1 (j) k,l Indeed, suppose there existed j E J' , x E Bj such that for all k, l with :::; ! ~ I: 13( 82\ 821) k ,l < K, 2, contradicting 12. It follows that for all j E J' there exist k, l such that (13) In fact, if this were not the case, we would have that for some j E J' IBjl:::; I LJk,l Ao2k,021(j) I : :; L IA02k,021U)I k,l 3. THE TWO-DIMENSIONAL CASE < 1 29 8 4IC (xj, rj)I, which contradicts 11. Finally, let Then, by 13 J' = LJ J(k, l) k,l We claim that there exist t = 82k ,E = 821 such that If not , then we would have L°'j::; I: I: °'j jEJ' k,l jEJ(k,l) 1 < µ(F1), contradicting 10. 2 □ Fix t, EE [8, 1] as above. Then there are two cases: 3. THE TWO-DIMENSIONAL CASE 30 It follows from the definition of r,, that there exists a set of indices J C I such that and I{ XE C 8(xj , rj) n Ef: L aiXC6(x;,r;)(.x):::; 2K,} I ~ i1c0(.xj, rj)I iE I for all j E J. Let Q = {(j, j1,J2,j3): j E J, J1,J2,J3 E J, J1,J2,J3 E St,f(j) · 8 8 8 8 /3(t, E) d1st(C (xj, rj) n C (xJk' rJk), C (xj, rj) n C (xj1, rj 1)) ~ CiM \/k, l k =/= l} where C 1 > 1 is a constant to be determined before C2. Further, define the following sets of indices: Q1 = {(j1 , J2,J3): ::lj such that (j,j1,J2,J3) E Q} 3. THE TWO-DIMENSIONAL CASE 31 For J1 E Q2 let = {j3 : :3j2 such that (j1 , J2, j3) E Q1} Let L R= °'J°'h ahaj3 (j,j1 ,h,ja)EQ Note that if C 2 is large enough, then where A1 is the constant in Lemma 1.3. It follows that if (j 1, j 2, j 3) E Q 1 then the set {x j : (j, j 1 , j 2 , j 3 ) E Q} is contained in the union of two E ellipsoids of diameter ;S /3 2( ) . Hence t ,E Furthermore, if j 1 E Q 2 and j 2 E Q(j 1) then there exists j such that It follows that for fixed J1 E Q2 the set { Xj 2 : j 2 E Q(j 1)} is contained in a disk with center Xj 1 and radius 4t . Hence L j2EQ(h) ah ;S ts1 3. THE TWO-DIMENSIONAL CASE 32 Therefore, (14) Now fix j E J. CLAIM 3.2. There are three subsets D 1 , D 2 , D 3 of At,E(j) such that and provided that C1 is large enough. PROOF. We use complex notation. If O ::; 01 ::; 02 ::; 21r let Then there exist O = 01 < · · · < 07 = 21r such that Let 3. THE TWO-DIMENSIONAL CASE 33 Note that for all l Therefore, It follows that if we choose C 1 large enough, then we have □ For each k let Then K, (3 2( t, E)!:u < 1 ~ Dk (J(t, E) !5:.d X M 2 s; L ailDk n C 6 (xi, ri)I iESk s; L ailC6 (xj , rj) n C 6 (xi , ri)I iESk 52 < ~a -- rz: ~ L 1. sk iE vtE where the last inequality follows from Lemma 1.1. Therefore, 2 Lai 2: iK,(3 (t, E)~ iEDk 3. THE TWO-DIMENSIONAL CASE 34 By Lemma 1.1 , if i E St,E(j) then Therefore, i1 E Sk, i 2 E S 1, k i=- l implies that 2A(J(t, E) C2 > (J(t, E) - C1M provided that C2 is sufficiently large. It follows that if Jk E Sk, k = 1, 2, 3, then (j,j 1,j2,j3) E Q. Hence R 2'. L L aj jEJ aj 1 ahaj 3 JIES1 j2ES2 j3ES3 If we compare the above equation with 14, and then use 7, we obtain Fix j E J. Then we have 3. THE TWO-DIMENSIONAL CASE = 35 r K,f3(t, E) dx At,.(j) M J ;S L ai[C"(xi,ri)nC"(xj , ri)l iESt,e(j) where we have used Lemma 1.1 and the definition of At,E (j). Note that the set {xi : i E St,E(j)} is contained in a disk of radius 2t. Therefore, L ai ;S tsi iESt,e (j) It follows that t si -1/2 K, < 5 - - - - ~ El/2j32(t, E) Using 8, we obtain r;, ;S 5. We conclude that, in either case (15) To complete the proof, notice that ;S 1 Lai l {.r E C"(.Tj,rj) nEf: L aiXco(x;,r;)(x) :S 2r;,} I jEJ iE / 3. THE TWO-DIMENSIONAL CASE l 36 8 ~ 8',,IE1 I ~ IEfl (16) where the last inequality follows from 15. If we let 8 ---+ 0 then the right-hand side of 16 tends to zero, which is a contradiction. □ 4. POSSIBLE IMPROVEMENTS 37 CHAPTER 4 Possible improvements As we discussed at the beginning of Section 3, the proof of Theorem 2 was motivated by a result of combinatorial nature, namely Proposition 3.1, which asserts that if one is given a family of N circles such that no three of them are internally tangent at a point, then there is a bound of the form CN 513 on the total number of tangencies. This, however, is far from being sharp. Clarkson, Edelsbrunner, Guibas, Sharir and Welzl [3] developed a technique which leads to a bound of the form CEN 3/2+E VE > 0, suggesting that it might be possible to weaken the conditions > 3/2 in Theorem 2. Indeed, Wolff [16] proved the following L 3 ---+ L 3 maximal inequality: THEOREM 4.1. For x 1 E JR, let Mof (x1) = sup rE[l/2,2] 1 IC8 (x, r)I 1 IJI C6(x ,r) x2EIR Using this, he proved, in the same paper, the following: 38 4. POSSIBLE IMPROVEMENTS THEOREM 4.2. If a S 1 and if E is a set in the plane which con- tains circles centered at all points of a set with Hausdorff dimension at least a, then E has Hausdorff dimension at least 1 + a. The preceding result suggests that a set E as in the statement of Theorem 2 has to be fairly large. In view of this and the analogy between Proposition 2.1 and the spherical means maximal theorem, it seems reasonable to make the following conjecture which would imply that Theorem 2 is true for all s > 1. CONJECTURE 1. For 8 > 0 small, f IR. 2 ---+ IR., define Ms B(O, 1/ 4) ---+ IR., by Let F c B(O, 1/4) be a compact set in IR. 2 such that there exists > 1 and a finite measure µ supported on F with µ( B (x, r)) S r 8 , for .T E IR. 2 and r > 0. Th en there exists a constant A that depends only on s, such that ( r j F (Mof(x))P(s)dµ,(.x) ) 1/p(s) S Allfllp(s)· (15) Note that in order for the above inequality to hold , it is necessary that p(s) ?::'. 4 - s. To see that, let I= [-1/8 , 1/8], and let EC I be a CantorsetofHausdorffdimensions-1. ThenH 8 - 1 (En B(0,8)) ~ 5s-l_ 4. POSSIBLE IMPROVEMENTS 39 Define Fs =IX (En B(O, 0112 )) and Notice that Therefore, using 15 = i53p(s) /2 On the other hand Hence 1 (52 _3_ O~ 2p(s) < (5 2p(s) ~ which is possible only if p(s) 2: 4 - s. Now we will present a heuristic argument affording evidence that the techniques in [3] might be used to prove that Theorem 2 is true for all s > 1. Assume that E is compact, IEI = 0, and suppose that F 4. POSSIBLE IMPROVEMENTS 40 satisfies the following regularity condition: There exists a non-trivial finite measure µ, supported on F such that µ( B (x, r)) ~ r5, for all .T E F, r > 0. Let {xJf=1 be a maximal o-separated subset of F. B = {(i,j): C(xi,ri), C(xj,rj) intersect at an angle;: 1r/100} Now suppose that if i # j then Choose E > 0 such that E > ½(1 - ¼). It follows from the remarks at the beginning of this chapter that IAI ,S N~+E. By an argument similar to the one used in the proof of Theorem 1, one shows that 4. POSSIBLE IMPROVEMENTS 41 Letting 8 ----t 0, we get a contradiction. We conclude by presenting an alternate way to attack the twodimensional problem (see Schlag[ll]). the space of Bessel potentials, with norm llf lla,p = llgllp• Here G°' is the Bessel kernel, i.e., the inverse Fourier transform of the function The Bessel capacity of a set E c !Rd is defined as Ba,p(E) = inf{IIJll~,P: J ~ l on E} The relation between capacity and Hausdorff dimension is given by the following result due to Havin and Maz 'ya [7]: THEOREM 4.3. Let EC !Rd be a Borel set. If p > l , etp::; d, then We refer the reader to the Appendix for a proof of a generalization of the preceding result. Furthermore, for f E C~(IR 2), define where dCJ is arc measure. 4. POSSIBLE IMPROVEMENTS 42 Sogge [13] made the following conjecture regarding the above operator: CONJECTURE 2. For all E > 0, there exists AE > 0, such that (18) for all i E C~(IR 2). 111 We will show how 18 would imply that the conclusion of Theorem 2 holds for all s > 1. As before, suppose that IEl=0 and choose s 1 and b > 0, so that 1 < 1 + b < s 1 < s. Then there exist a compact set E 1 C E and a compact set F 1 C F with 'H 81 (Fi) > 0 such that for each x E F 1 there is a circle centered at x with radius r(x) E (1, 2) which intersects E 1 in a set of angle measure at least 1r. Let .E1 = {(x, r( x)) : x E Fi}. Then there exists a sequence {in} in C~(IR 2) such that in?::: 1 on E1 and llinll4---+ 0 as n---+ oo. Choose 'ljJ E C~(IR 2) such that 'I/J = 1 on (1, 2) and define 9n(.x, t) = 9J1fn (.x, t)'lj)(t) Then 9n ,2: 1 on E1. Hence, by 18 I 4. POSSIBLE IMPROVEMENTS Letting n ----+ oo we obtain B 1 rem 5.1 contradicting the choice of [J. ; 2 _ 818 , (E 4 ) 1 43 = 0, and therefore, by Theo- APPENDIX 44 Appendix Here we prove a generalization of Theorem 4.3. Let V 1 ,P 2 (IR.d 1 x 1R_d2 ), p 1 > 1, p 2 > 1 be the space of all functions with finite II · llp1 ,p2 norm, where For a > 0, define the space The mixed-norm capacity of E C IR_d = 1R_d1 x 1R_d2 is defined as THEOREM. Let EC IR_d = 1R_d1 x 1R_d 2 be a Borel set. APPENDIX 45 PROOF. Without loss of generality we may assume that EC [O, l]d. Letµ be a finite measure supported on E, and let u be a non-negative CC: function such that u ~ 1 on E. Then µ,(E) :SJ u(.r)dµ(x) J =J = Ga* D°'u(x)dµ(.r) D°'u(y) J G0 (x - y)dµ,(.r)dy where q1 , q2 are the conjugate exponents of p 1 , P2 respectively, and D°'u is the fractional derivative operator acting on u, defined as the inverse Fourier transform of the function (1 + l~l 2 )°'12u(O. For each n ~ 0 we subdivide ]Rd into disjoint dyadic cubes of side 2-n , so that each cube of side 2-k is split into 2d cubes of side 2-(k+l). If Q is such a dyadic cube then l(Q) denotes its sidelength and Q the cube with the same center as Q and sidelength 3l(Q). Let lxl°'-d, if O < 1-rl :S 1, o, if lxl > 1. APPENDIX 46 It follows from the properties of the Bessel kernel (see, e.g., [1]) that there exist constants a and A such that and Therefore, =B+B' B' is easy to estimate. By Minkowski's inequality for integrals, we have On the other hand APPENDIX 47 n=O where b is a positive number. Suppose that P1 S P2 and that H_s+e( E) > 0 for some E > 0. Then there exists a nontrivial finite measure µ, supported on E such that µ,(B(x, r)) S rs+e for all x E !Rd, r > 0. It follows that 00 = cLt1(q2(d-a+8)-d1~-d2) n=O l(Q)=2- 11 oo n(q2(d-a+8)-d1 ~-d2) L 2n(q2-l)(s+e) < C ,(E) ~ 2 ~ µ µ,(Q)µ,(Qt2-l L n=O ql < CX) APPENDIX 48 provided that 8 has been chosen so that p 2 8 < E. Now suppose that p 2 s; p 1 and that H_HE(E) > 0 for some E > 0. Then, as above, there exists a nontrivial finite measure supported on E such that µ,(B(x, r)) s; rt+£ for all x E IRd, r > 0. It follows that cI::: L µ,(Q)µ(Q)q1-l 00 = 2n(qi(d-a+8)-d2~-d1) n=O l(Q)=2-n oo < Cµ,(E) ~ L 2 n(q1(d-a+8)-d 2 ll-d 1 ) q2 -- - - < oo 2n(q1 -l)(t+E) n=O provided that p18 < E. It follows that µ(E) ;S llulla,p1 ,p2 • By assumption, Ba,p 1 ,p2 (E) = 0. Therefore µ,(E) = 0 which is a contradiction. □ BIBLIOGRAPHY 49 Bibliography [l] D. R. Adams, L. I. Hedberg, Fnnction Spaces and Potential Theory, SpringerVerlag, 1996. [2] J. Bourgain, Averages over convex curves and maximal operators, J. Analyse Math. 47 (1986), 69-85. [3] K. L. Clarkson, H. Edelesbrunner, L. J . Guibas, M. Sharir, E. Welz!, Combinatorial complexity bounds for arrangements of curves and spheres, Discrete Comput. Geom. 5 (1990), 99-160. [4] R.O. Davies, Another thin set of circles. Journal of the London Mathematical Society (2), 5 (1972), 191-2. [5] K. J. Falconer, The geometry of fractal sets, Cambridge University Press, 1985. [6] R. L. Graham, B. L. Rothschild, J. H. 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