ON A PROBLEM RELATED TO SPHERE AND CIRCLE PACKING

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1 ON A PROBLEM RELATED TO SPHERE AND CIRCLE PACKING THEMIS MITSIS ABSTRACT We prove that a set which contains spheres centered at all points of a set of Hausdorff dimension greater than must have positive Lebesgue measure We show by a counterexample that this is sharp We also prove the corresponding result for circles provided that the set of centers has Hausdorff dimension greater than 3/2 INTRODUCTION The following geometric result is a consequence of the work of Stein on the spherical means maximal operator Theorem Let F R d, d 3, be a set of positive Lebesgue measure If E R d is a set which contains spheres centered at all points of F, then E has positive Lebesgue measure Bourgain [], and, independently, Marstrand [7], proved the two - dimensional analogue, given as follows Theorem 2 Let F R 2 be a set of positive Lebesgue measure If E R 2 is a set that contains circles centered at all points of F, then E has positive plane measure This should be contrasted with the following construction due to Talagrand [2] Theorem 3 There is a set of plane measure zero containing for each x on a given straight line, a circle centered at x It is, therefore, natural to ask whether one can weaken the condition that the set F in the above theorems should be of postive measure The main results in this paper are the following Theorem 4 Let F R d, d 3, be a Borel set of Hausdorff dimension s, s > If E R d is a Borel set that contains spheres centered at each point of F, then E has positive Lebesgue measure Theorem 5 Let F R 2 be a Borel set of Hausdorff dimension s, s > 3/2 If E R 2 is a Borel set that contains circles centered at each point of F, then E has positive Lebesgue measure The paper is organized as follows In Section 2 we state some geometric lemmas needed later on, in Section 3 we prove Theorem 4 and construct a counterexample related to it, in Section 4 we prove Theorem 5, and finally, in Section 5, we discuss the possibility of weakening the condition s > 3/2 in Theorem 5 Throughout this paper, a b means a Ab for some absolute constant A, and similarly with a b and a b We will denote Lebesgue measure by

2 2 THEMIS MITSIS 2 BACKGROUND We start with some notation B(x, r) is the open disk (or ball) with center x and radius r C(x, r) is the circle (or sphere) with center x and radius r C δ (x, r) is the δ-neighborhood of the circle (or sphere) C(x, r), ie, the set {y R d : r δ < x y < r + δ} If C(x, r) and C(y, s) are circles, then we define d((x, y), (r, s)) = x y + r s, ((x, y), (r, s)) = x y r s 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(0, /4) and that their radii are in the interval [/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 [4] for a proof Lemma 2 Suppose that C(x, r), C(y, s) are circles with r s Then for 0 < δ < there exists an absolute constant A 0 such that () C δ (x, r) C δ (y, s) is contained in a δ-neighborhood of arc length less than + δ A 0 x y + δ, centered at the point x r sgn(r s) x y x y (2) The area of intersection satisfies C δ (x, r) C δ (y, s) A 0 δ 2 (δ + )(δ + d) The next result is essentially Marstrand s three circle lemma [7] 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 Lemma 22 There exists a constant A such that if, t, λ (0, ) satisfy λ A t then for three fixed circles C(x i, r i ), i =, 2, 3 and for δ the set {(x, r) R 2 R : ((x, r), (x i, r i )) < i, d((x, r), (x i, r i )) > t i, C δ (x, r) C δ (x i, r i ) i, dist(c δ (x, r) C δ (x i, r i ), C δ (x, r) C δ (x j, r j )) λ i, j : i j} is contained in the union of two ellipsoids in R 3 each of diameter λ 2

3 ON A PROBLEM RELATED TO SPHERE AND CIRCLE PACKING 3 A proof of the preceding result can be found in Wolff [4] We conclude this section with some facts from geometric measure theory We refer the reader to Falconer [4] for definitions, proofs and details In what follows, H s denotes s-dimensional Hausdorff measure Theorem 2 Let E be a Borel set in R d and let s > 0 Assume that H s (E) > 0 Then there exists a nontrivial finite measure µ supported on E such that µ(b(x, r)) r s for x R n and r > 0 If E is an s-set, ie, 0 < H s (E) <, then a point x E is called regular if the upper and the lower densities at x are equal to one; otherwise x is called irregular An s-set E is said to be irregular if H s -almost all of its points are irregular Irregular -sets are characterized by the following: Theorem 22 A -set in R 2 is irregular if and only if it has projections of linear Lebesgue measure zero in two distinct directions In fact, one can say much more Theorem 23 Let E be an irregular -set in R 2 Then proj θ (E) has linear Lebesgue measure zero for almost all θ [0, π), where proj θ denotes orthogonal projection from R 2 onto the line through the origin making angle θ with some fixed axis 3 THE HIGHER DIMENSIONAL CASE In this section we assume that d 3 For f : R d R, δ > 0 small, we define M δ : B(0, /4) R by M δ f (x) = C δ f (y) dy (x, r) sup /2 r 2 C δ (x,r) Theorem 4 will be a consequence of the following L 2 L 2 maximal inequality Proposition 3 Let F B(0, /4) be a compact set in R d such that there exist s > and a finite measure µ supported on F with µ(b(x, r)) r s for x R d and r > 0 Then there exists a constant A that depends only on the measure of F and on s such that ( for small δ > 0 and all f F (M δ f (x)) 2 dµ(x)) /2 A f 2 The proof generally follows the lines of the proof of Theorem in [6] Even though the necessary modifications are not immediately obvious, we choose not to include the proof A similar argument will appear in a forthcoming paper on the Kakeya maximal function Proof of Theorem 4 We may assume that F B(0, 4 ) Suppose that E =0 and choose t so that < t < s Then there exist a compact set E E, a compact set F F with H t (F ) > 0, and a positive number r, such that, for each x F, there is a sphere centered at x with radius r(x) (r, 2r) which intersects E in set of surface measure at least r d (x) Without loss of generality we may assume that r = It follows that for all x F M δ χ Eδ(x),

4 4 THEMIS MITSIS where E δ is the δ-neighborhood of the set E By Theorem 2, there exists a nontrivial finite measure µ supported on F such that µ(b(x, r)) r t, for x R d, r > 0 Therefore, by Proposition 3, we have µ(f ) (M δ χ E δ(x)) 2 dµ(x) E F δ The right-hand side of the above inequality tends to zero as δ 0, so we get a contradiction We will show that we cannot drop the condition s > in Theorem 4 Proposition 32 There exists a set of d-dimensional Lebesgue measure zero containing for each x [0, ] {0} {0}, a sphere centered at x } {{ } d Proof The idea, which goes back to Davies [3], is to parametrize the set of radii using a suitable irregular -set Divide the unit square [0, ] [0, ] R 2 into 6 disjoint squares of side /4 Let S (i, j) i, j 4 be those squares (indexed from bottom to top, left to right), and put E = S (,2) S (,4) S (4,) S (4,3) Apply the same procedure to each of S (,2), S (,4), S (4,), 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 {E n } Let E = n= E n Then E is a -set such that proj 0 (E) = [0, ], proj π/2 (E) = 0, proj π/4 (E) = 0 where is linear Lebesgue measure, and proj 0, proj π/2, proj π/4 denote, respectively, orthogonal projection onto the x-axis, the y-axis, and the line through the origin making angle π/4 with the x-axis It follows from Theorem 22 that E is irregular Let A = {(x,, x d ) : (x a) 2 + x x2 d = a2 + b} = (a,b) E (a,b) E {(x,, x d ) : x 2 d = 2ax + b x 2 x2 d } Since proj 0 (E) = [0, ], A contains a sphere centered at each point of {(a, 0,, 0) : a [0, ]} Now fix s,, s d Then where A {(x,, x d ) : x = s, x 2 = s 2,, x d = s d } = {s } {s 2 } {s d } B B = {x d : xd 2 = 2as + b s 2 s2 2 s2 d, (a, b) E} B has measure zero if and only if {2as + b : (a, b) E} has measure zero But L ({2as + b : (a, b) E}) = 0 for almost all s R by Theorem 23 Therefore, by Fubini, A has d-dimensional measure zero

5 ON A PROBLEM RELATED TO SPHERE AND CIRCLE PACKING 5 4 THE TWO-DIMENSIONAL CASE Before we proceed with the proof of Theorem 5, 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 2, 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 [6], and, subsequently, in Schlag [8], [9] 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 [5] for more details) Proposition 4 Let {C j } N j= be a family of distinct circles such that no three are tangent at a single point Then card({(i, j) : C i C j }) N 5/3, where C i C j means that C i and C j are internally tangent Proof Let Q = {(i, j, j 2, j 3 ) : C i C jk, k =, 2, 3} and fix j, j 2, j 3 Then, by the circles of Apollonius, there are at most two choices for i Therefore, card(q) 2N(N )(N 2) < 2N 3 On the other hand, if we let n(i) = card({ j : C i C j }), then N ( Q = {i} {( j, j 2, j 3 ) : C i C jk, k =, 2, 3} ) i= Hence N N card(q) n(i)(n(i) )(n(i) 2) (n(i) 2) 3 i= i= It follows that N N card({(i, j) : C i C j }) = n(i) = (n(i) 2) + 2N i= i= /3 N (n(i) 2) 3 N 2/3 + 2N i= (card(q)) /3 N 2/3 + 2N (2N 3 ) /3 N 2/3 + 2N N 5/3 The proof of Theorem 5 will be a quantitative version of Proposition 4, with Lemma 22 playing the role of the restriction imposed by the circles of Apollonius

6 6 THEMIS MITSIS Proof of Theorem 5 We may assume that F B(0, 4 ) Suppose that E = 0 and choose s so that 3/2 < s < s Then there exist a compact set E E, a compact set F F with H s (F ) > 0 and a positive number r, such that, for each x F, there is a circle centered at x with radius r(x) (r, 2r) which intersects E in a set of angle measure at least π Without loss of generality we may assume that r = By Theorem 2, there exists a nontrivial finite measure µ supported on F such that µ(b(x, r)) r s, for x R 2, r > 0 Let {x i } be a maximal δ-separated set of points in F, and let a i = µ(b(x i, δ)) Choose r i > 0 such that C δ (x i, r i ) E δ 2 Cδ (x i, r i ), (6) where E δ is the δ-neighborhood of E Let κ be the infimum of those λ > 0 such that there exists J I satisfying a j 2 µ(f ), j J and for all j J { x C δ (x j, r j ) E δ : a i χ C δ (x i,r i )(x) λ } 4 Cδ (x j, r j ) Choose N large enough so that s 3 2 > 7 + 2s N and N 2 N N 2 N 2 < 2s + 3 Let C 2 > be a large constant to be determined later on and define β : [δ, ] [δ, ] R by t 2 (s /2) C 3 N 2 N 2 β(t, ) = /4 2, if t /N, if t t /(N+) Then, for small δ, β has the following properties: β(t, ) C 2 t 2s + N < C 2 N 2s + N C 2 N β(t, ) < C 2 t β(t, ) = t 2 (s /2) δ2 k δ2 l where M is a constant that depends only on N and on s Now for all i, j I and t, [δ, ], we define β(t, ) = /N, (7) t/(n+) N N 2 /4 C 3 2 2, (8) β(δ2 k, δ2 l ) < M, (9) i j = max{δ, x i x j r i r j }, S t, ( j) = {i I : C δ (x i, r i ) C δ (x j, r j ), t x i x j 2t, i j 2},

7 ON A PROBLEM RELATED TO SPHERE AND CIRCLE PACKING 7 A t, ( j) = { x C δ (x j, r j ) : i S t, ( j) a i χ C δ (x i,r i )(x) M β(t, ) κ 2} Claim 4 There exist t, [δ, ] and a set of indices J such that A t, ( j) 4M β(t, ) Cδ (x j, r j ), j J, and j J a j 2M β(t, )µ(f ) Proof Let J 0 = { j I : { x C δ (x j, r j ) E δ : a i χ C δ (x i,r i )(x) 2} κ 4 Cδ (x j, r j ) } By the minimality of κ, we have j J 0 < 2 µ(f ) Therefore, if J is the complement of J 0, then j J a j 2 µ(f ) (0) and for all j J { C δ (x j, r j ) E δ : a i χ C δ (x i,r i )(x) 2} κ < 4 Cδ (x j, r j ) Hence, using (6) we obtain { C δ (x j, r j ) E δ : a i χ C δ (x i,r i )(x) > 2} κ 4 Cδ (x j, r j ) () For each j J let B j = { x C δ (x j, r j ) E δ : a i χ C δ (x i,r i )(x) > 2} κ (2) Then for all j J B j A δ2 k,δ2l( j) Indeed, suppose there existed j J, x B j such that for all k, l with δ2 k, δ2 l we had x A δ2 k,δ2l( j) Then, by (9) a i χ C δ (x i,r i )(x) = a i χ C δ (x i,r i )(x) κ β(δ2 k, δ2 l ) < κ M 2 2, i S δ2 k,δ2 l ( j) contradicting (2) It follows that for all j J there exist k, l such that A δ2 k,δ2l( j) 4M β(δ2k, δ2 l ) C δ (x j, r j ) (3) In fact, if this were not the case, we would have that for some j J B j A δ2 k,δ2 l( j) A δ2 k,δ2l( j)

8 8 THEMIS MITSIS which contradicts () Finally, let 4M Cδ (x j, r j ) β(δ2 k, δ2 l ) < 4 Cδ (x j, r j ), J(k, l) = { j J : A δ2 k,δ2l( j) 4M β(δ2k, δ2 l ) C δ (x j, r j ) } Then, by (3) J = J(k, l) We claim that there exist t = δ2 k, = δ2 l such that a j 2M β(t, )µ(f ) If not, then we would have a j j J contradicting (0) j J() j J() a j < 2M µ(f ) β(δ2 k, δ2 l ) < 2 µ(f ), So, fix t, [δ, ] as above Then there are two cases Case : β(t, ) C 2 t It follows from the definition of κ that there exists a set of indices J I such that a j 2 µ(f ), and j J { x C δ (x j, r j ) E δ : a i χ C δ (x i,r i )(x) 2κ } 4 Cδ (x j, r j ), for all j J Now let Q = {( j, j, j 2, j 3 ) : j J, j, j 2, j 3 J, j, j 2, j 3 S t, ( j) dist(c δ (x j, r j ) C δ (x jk, r jk ), C δ (x j, r j ) C δ (x jl, r jl )) k, l k l} where C > is a constant to be determined before C 2 Further, define the following sets of indices: For j Q 2 let Now consider the quantity Q = {( j, j 2, j 3 ) : j such that ( j, j, j 2, j 3 ) Q}, Q 2 = { j : j 2, j 3 such that ( j, j 2, j 3 ) Q } Q( j ) = { j 2 : j 3 such that ( j, j 2, j 3 ) Q } = { j 3 : j 2 such that ( j, j 2, j 3 ) Q } R = ( j, j, j 2, j 3 ) Q a j a j a j2 a j3 β(t, ) C M

9 ON A PROBLEM RELATED TO SPHERE AND CIRCLE PACKING 9 Note that if C 2 is large enough, then β(t, ) C M C 2 C M t A t, where A is the constant in Lemma 22 It follows that if ( j, j 2, j 3 ) Q then the set {x j : ( j, j, j 2, j 3 ) Q} is contained in the union of two ellipsoids of diameter β 2 (t, ) Hence ( ) s R β 2 a j a j2 a j3 (t, ) ( j, j 2, j 3 ) Q Furthermore, if j Q 2 and j 2 Q( j ) then there exists j such that j, j 2 S t, ( j) Therefore, x j x j2 x j x j + x j x j2 4t It follows that for fixed j Q 2 the set {x j2 : j 2 Q( j )} is contained in a disk with center x j and radius 4t Hence a j2 t s j 2 Q( j ) Therefore, ( ) s ( ) 2 s R β 2 a j a j2 a j3 (t, ) β 2 a j (t, ) a j2 ( j, j 2, j 3 ) Q j Q 2 j 2 Q( j ) ( ) s ( µ(f ) t s ) 2 β 2 (4) (t, ) Now fix j J Claim 42 There are three subsets D, D 2, D 3 of A t, ( j) such that and dist(d k, D l ) provided that C is large enough 2β(t, ), k, l k l, C M D k δβ(t, ), k, Proof We use complex notation If 0 θ θ 2 2π let G θ,θ 2 = A t, ( j) {x j + re iθ C δ (x j, r j ) : θ θ θ 2 } Then there exist 0 = θ < < θ 7 = 2π such that Let Note that for all l Therefore, G θk,θ k+ = A t,( j), k =,, 6 6 D k = G θ2k,θ 2k, k =, 2, 3 β(t, ) 24M Cδ (x j, r j ) G θl,θ l+ diam(g θl,θ l+ )δ diam(g θl,θ l+ ) β(t, )

10 0 THEMIS MITSIS It follows that if we choose C large enough, then we have dist(d k, D l ) 2β(t, ) C M, and D k β(t, )δ Then For each k let S k = {i S t, ( j) : D k C δ (x i, r i ) } κβ 2 β(t, ) κ (t, )δ D k M 2 dx a i D k C δ (x i, r i ) i S k a i C δ (x j, r j ) C δ (x i, r i ) a i δ2, i S k i S k t where the last inequality follows from Lemma 2 Therefore, i D k a i δ κβ2 (t, ) t By Lemma 2, if i S t, ( j) then diam(c δ (x j, r j ) C δ (x i, r i )) A Therefore, i S k, i 2 S l, k l implies that t Aβ(t, ) C 2 dist(c δ (x i, r i ) C δ (x j, r j ), C δ (x i2, r i2 ) C δ (x j, r j )) 2β(t, ) C M j J 2Aβ(t, ) β(t, ) C 2 C M, provided that C 2 is sufficiently large It follows that if j k S k, k =, 2, 3, then ( j, j, j 2, j 3 ) Q Hence ( R a j a j a j2 a j3 β(t, ) δ κβ2 (t, ) ) 3 t j S j 2 S 2 j 3 S 3 If we compare the above equation with (4), and then use (7), we obtain Fix j J Then we have κ 3 δ 3 s 3/2 t 2s 3/2 β 2s +7 (t, ) κδβ 2 β(t, ) (t, ) κ M A t,( j) = i S t, ( j) δ 3 Case 2: β(t, ) C 2 t A t, ( j) κ β(t, ) M dx a i C δ (x i, r i ) C δ (x j, r j ) δ2 t where we have used Lemma 2 and the definition of A t, ( j) i S t, ( j) a i,

11 ON A PROBLEM RELATED TO SPHERE AND CIRCLE PACKING Note that the set {x i : i S t, ( j)} is contained in a disk of radius 2t Therefore, a i t s It follows that i S t, ( j) t s /2 κ δ /2 β 2 (t, ) Using (8), we obtain κ δ We conclude that, in either case κ δ (5) To complete the proof, notice that 2 µ(f ) a j = a j δ δ j J j J { a j x C δ (x j, r j ) E δ δ : a i χ C δ (x i,r i )(x) 2κ } j J {x Eδ: ( a j χ δ C δ (x j,r j )(x) ) dx j J a jχcδ(x j,r j)(x) 2κ} j J δ κ Eδ Eδ, (6) where the last inequality follows from (5) If we let δ 0 then the right-hand side of (6) tends to zero, which is a contradiction 5 POSSIBLE IMPROVEMENTS As we discussed at the beginning of Section 4, the proof of Theorem 5 was motivated by a result of combinatorial nature, namely Proposition 4, 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 5/3 on the total number of tangencies This, however, is far from being sharp Clarkson, Edelsbrunner, Guibas, Sharir and Welzl [2] developed a technique which leads to a bound of the form C N 3/2+ > 0, suggesting that it might be possible to weaken the condition s > 3/2 in Theorem 5 Indeed, Wolff [3] proved the following L 3 L 3 maximal inequality Theorem 5 For x R, let where x = (x, x 2 ) Then M δ f (x ) = sup r [/2,2] C δ f, (x, r) C δ (x,r) x 2 R > 0 A : M δ f L 3 (R) A δ f 3 Using this, he proved, in the same paper, the following Theorem 52 If α and if E is a set in the plane which contains circles centered at all points of a set with Hausdorff dimension at least α, then E has Hausdorff dimension at least + α

12 2 THEMIS MITSIS The preceding result suggests that a set E as in the statement of Theorem 5 has to be fairly large In view of this and the analogy between Proposition 3 and the spherical means maximal theorem, it seems reasonable to make the following conjecture which would imply that Theorem 5 is true for all s > Conjecture 5 For δ > 0 small, f : R 2 R, define M δ : B(0, /4) R, by M δ f (x) = C δ f (y) dy (x, r) sup /2 r 2 C δ (x,r) Let F B(0, /4) be a compact set in R 2 such that there exist s > and a finite measure µ supported on F with µ(b(x, r)) r s, for x R 2 and r > 0 Then there exists a constant A that depends only on the measure of F and on s, such that ( /p(s) (M δ f (x)) dµ(x)) p(s) A f p(s) (5) F Note that in order for the above inequality to hold, it is necessary that p(s) 4 s To see that, let I = [ /8, /8], and let E I be a Cantor set of Hausdorff dimension s Then H s (E B(0, δ)) δ s Define and Notice that Therefore, using (5) On the other hand Hence F δ = I (E B(0, δ /2 )), R δ = [ δ, + δ] [ 2δ /2, 2δ /2 ] x F δ M δ χ Rδ (x) δ /2 ( δ /2 (H s (F δ )) /p(s) (M δ χ Rδ ) p(s) (x)dh s (x) F δ χ Rδ p(s) = δ 3p(s)/2 H s (F δ ) H s (E B(0, δ /2 )) δ (s )/2 δ 2 δ (s ) 2p(s) δ 3 2p(s), ) /p(s) which is possible only if p(s) 4 s We conclude by mentioning an observation made by Schlag: if the local smoothing conjecture due to Sogge [0] is correct, then Theorem 5 is true for all s > ACKNOWLEDGEMENT I thank Tom Wolff for his interest in this work, for his encouragement, and for repeated proof-reading

13 ON A PROBLEM RELATED TO SPHERE AND CIRCLE PACKING 3 REFERENCES [] J Bourgain, Averages over convex curves and maximal operators, J Anal Math 47 (986), [2] K L Clarkson, H Edelesbrunner, L J Guibas, M Sharir, E Welzl, Combinatorial complexity bounds for arrangements of curves and spheres, Discrete Comput Geom 5 (990), [3] RO Davies, Another thin set of circles J London Math Soc (2), 5 (972), 9-92 [4] K J Falconer, The geometry of fractal sets, Cambridge University Press, 985 [5] R L Graham, B L Rothschild, J H Spencer, Ramsey Theory, 2nd edition, Wiley-Interscience, 990 [6] L Kolasa, T Wolff, On some variants of the Kakeya problem, Pacific J Math, to appear [7] J M Marstrand, Packing circles in the plane, Proc London Math Soc 55 (987), [8] W Schlag, A generalization of Bourgain s circular maximal theorem, J Amer Math Soc 0 (997), [9] W Schlag, A geometric proof of the circular maximal theorem, Duke Math J, to appear [0] C Sogge, Propagation of singularities and maximal functions in the plane, Invent Math 04 (99), [] E M Stein, Maximal functions: spherical means Proc Nat Acad Sci USA 73 (976), [2] M Talagrand, Sur la mesure de la projection d un compact et certains familles de cercles, Bull Sci Math (2) 04 (980), [3] T Wolff, A Kakeya type problem for circles, Amer J Math, to appear [4] T Wolff, Recent work connected with the Kakeya problem, survey article, anniversary proceedings, Princeton 996, to appear DEPARTMENT OF MATHEMATICS, CALIFORNIA INSTITUTE OF TECHNOLOGY, PASADENA, CA 925, USA address: themis@ccocaltechedu

756 Thomas Wolff 3. Is the following estimate true? 8 >09C : kf k L n (P n;1 ) C ; kfk Ln (P n;1 ) (1) We discuss below the partial results that have

756 Thomas Wolff 3. Is the following estimate true? 8 >09C : kf k L n (P n;1 ) C ; kfk Ln (P n;1 ) (1) We discuss below the partial results that have Doc. Math. J. DMV 755 Maximal Averages and Packing of One Dimensional Sets Thomas Wolff Abstract. We discuss recent work of several authors on the Kakeya needle problem and other related problems involving