On Falconer s Distance Set Conjecture

Transcription

1 Rev. Mat. Iberoamericana (006), no., On Falconer s Distance Set Conjecture M. Burak Erdo gan Abstract In this paper, using a recent parabolic restriction estimate of Tao, we obtain improved partial results in the direction of Falconer s distance set conjecture in dimensions d Introduction Let E be a compact subset of R d. The distance set, (E), of E is defined as (E) ={ x y : x, y E}. Erdös famous distinct distances conjecture [7] states that for any ε>0and for any finite set E R d, d, # (E) C d,ε (#E) d ε. This conjecture is still open in all dimensions d. For various partial results and references see [17], [1] and [13]. Falconer s conjecture [8] is a variant of Erdös conjecture: Conjecture. Let d. Let E be a compact subset of R d. Then, dim(e) > d = (E) > 0. Here is the Lebesgue measure and dim( ) is the Hausdorff dimension. Like Erdös conjecture, Falconer s conjecture is open in every dimension. In [8], Falconer gave an example showing that d in the conjecture is optimal and proved that dim(e) > d+1 implies (E) > 0. Bourgain [3] improved this result in every dimension, and in particular proved that in R, 000 Mathematics Subject Classification: 4B10. Keywords: Distance sets, Fourier restriction estimates, Frostman measures.

2 650 M. B. Erdo gan dim(e) > 13 9 suffices. Later, Wolff [4] proved that in R, dim(e) > 4 3 suffices. In [6], the author obtained a simplified proof of Wolff s result and noted that it is possible to obtain the following improved partial result in higher dimensions using the method in [6] and a bilinear Fourier restriction estimate by Tao []. In this paper, we prove 1 Theorem 1. Let d. LetE be a compact subset of R d with Then (E) > 0. dim(e) > d(d +) (d +1). There are other positive results in the direction of Falconer s conjecture. For example, Mattila [14] proved that in R, dim(e) > 1 implies dim( (E)) 1. Recently, Bourgain [4] improved this result and proved that there exists c>0 such that in R,dim(E) > 1 implies dim( (E)) > 1 + c. Bourgain s result relies on a paper by Katz and Tao [1] which relates the Falconer s conjecture to various other problems in harmonic analysis. There are lots of variations of Falconer s problem. Notably, Mattila and Sjölin [16] proved that (E) has interior points if dim(e) > d+1. Peres and Schlag [18] considered pinned distance sets, (x, E) ={ x y : y E}, and proved that if dim(e) > d+1 then (x, E) > 0 for almost every x E. One can also consider distance sets with respect to general metrics. Let K be a convex symmetric body in R d, d. Define K (E) = {d K (x, y) :x, y E}, where d K is the distance induced by K. Iosevichand Laba [10] investigated the relation between the curvature of the boundary of K and the size of the distance sets. Hofmann and Iosevich [9] (also see [] for a similar result in higher dimensions) proved that in R if dim(e) > 1 then K (E) > 0 for almost every ellipse K centered at the origin. We note that our main result, Theorem 1, remains valid for K in the case when the boundary of K is smooth and has non-vanishing Gaussian curvature (see Remark 1 below). List of notations. χ A : characteristic function of the set A. B(x, r) :={y : x y <r}. d(a, B): the distance between the sets A and B. 1 Recently, Theorem 1 has been improved by the author. The current best known exponent in Falconer s conjecture is d/+1/3 in every dimension d.

3 On Falconer s Distance Set Conjecture 651 A R (C) :={x R d : x R C}. C: a constant which may vary from line to line. A B: A CB. A B: A B and B A. A B: A 1 B, for some large constant C. C A : length of the vector A or the measure of the set A.. Mattila s approach to distance set problem In [14], Mattila developed a method to attack the distance set problem. For a very good exposition of this method, see [6]. Mattila s approach was used in [14, 3, 4, 9, 6, ]. Let µ be a probability measure supported in E. Let ν µ be the push forward of µ µ under the distance map (x, y) x y, i.e., ν µ (A) =µ µ({(x, y) : x y A}), for Borel sets A R. It is easy to check that ν µ is a probability measure supported in (E). Note that if the Fourier transform of ν µ, ν µ (ξ) := e ix ξ dν µ (x), is an L function, then ν µ should be absolutely continuous with an L density and hence (E) Supp(ν µ ) > 0. Using this idea and the Fourier asymptotics of the surface measure of the unit sphere in R d, Mattila proved [14]: Theorem A. Let α (0,d). LetE be a compact subset of R d with dim(e)>α. Assume that there is a probability measure µ supported in E such that (.1) µ(r ) L (S d 1 ) C µ R α d, R >1. Then (E) > 0. Note that Theorem A proves the distance set conjecture for Salem sets [19, 11]. AsetE R d is called a Salem set if for each β<dim(e), there exists a probability measure µ supported in E such that µ(ξ) ξ β, ξ R d. To apply Theorem A to arbitrary compact sets, one needs Frostman s lemma (see, e.g., [15]).

4 65 M. B. Erdo gan Definition 1. A compactly supported probability measure µ is called α-dimensional if it satisfies (.) µ(b(x, r)) C µ r α, r >0, x R d. Frostman s Lemma. If E is a compact subset of R d with dim(e) >α, then there is an α-dimensional measure µ supported in E. Frostman s lemma and Mattila s theorem imply: Lemma.1. Fix α (0,d). Assume that the inequality (.1) holds for all α-dimensional measures. Then for any compact E R d dim(e) >α = (E) > 0. In view of Lemma.1, Theorem 1 is a corollary of the following: Theorem. Let d and α (0,d). Letµ be an α-dimensional measure. Then for each q> d+ d, µ(r ) L (S d 1 ) C q,µ R α q, R >1. Like Theorem 1, Theorem was first proved in [4] for d =. Under the hypothesis of Theorem, it is also known that [14, 0] (also see [1, 6]) α 1 max( (.3) µ(r ) L (S d 1 ) R, min( α, d 1 4 )), R >1. Theorem and (.3) give optimal bounds for each α (0, ) for d = (see, e.g., [0, 4, 6]). Therefore, one can not improve the result in Theorem 1 for d = using Mattila s approach. In higher dimensions, (.3) is optimal for α d 1 (see [0]); however, there is no reason to believe that Theorem and (.3) give optimal bounds for α> d 1. It is essential that in Theorem, we are averaging µ(r ) onasurface with non-vanishing Gaussian curvature. In general, the Fourier transform µ(ξ) ofanα-dimensional measure µ does not have to converge to zero as ξ.infact,foranyd 1andforanyα (0,d), there are Cantor-type measures in R d of dimension greater than α whose Fourier transform does not converge to 0 at infinity [19]. Remark 1. Mattila s approach can be modified for distance sets with respect to general metrics. Let K be a convex symmetric body. Assume that the boundary of K is smooth and has non-vanishing Gaussian curvature. Let K be the dual of K. One can modify Mattila s approach and prove that the statement of Lemma.1 remains valid if (E) is replaced with

5 On Falconer s Distance Set Conjecture 653 K (E) ands d 1 in (.1) is replaced with K (see [9, ]). We note that Theorem remains valid, too, if we replace S d 1 with K. The proof of this fact follows the same line below with minor changes in the statements and proofs of Corollary and Lemma 5.. Therefore, Theorem 1 holds for K if K has a smooth boundary with non-vanishing Gaussian curvature. 3. Tao s bilinear parabolic extension estimate In the proof of Theorem, we use a bilinear restriction estimate for elliptic surfaces by Tao []. First let us recall the definition of elliptic surfaces from [3]: Definition. We say φ : B(0, 1) R d 1 R is an (M,ε 0 )-elliptic phase if φ satisfies i) φ C <M, ii) φ(0) = φ(0) = 0, and iii) for all x B(0, 1), all eigenvalues of the Hessian φ xi x j (x) lie in [1 ε 0, 1+ε 0 ]. We say S is an (M,ε 0 )-elliptic surface if S = {(x, y) B(0, 1) R R d : y = φ(x)} for some (M,ε 0 )-elliptic phase φ. Note that in this definition the term elliptic is used in a slightly nonstandard way. In classical PDE, a non-vanishing symbol is considered to be elliptic. In the definition above, the non-vanishing of the curvature is required, too, (see II below). A model example for an elliptic phase is φ(x) = x. We recall the following properties of elliptic phases (see, e.g., [3]): I) Let φ be an (M,ε 0 )-elliptic phase and B(x 0,η) B(0, 1). Let φ(x) := 1 η (φ(xη + x 0) φ(x 0 ) ηx φ(x 0 )), x B(0, 1). Then φ is a (C d M,ε 0 )-elliptic phase. II) Let S be a smooth compact submanifold of R d with strictly positive principal curvatures. Note that for any ε 0 > 0andforanys S there is a neighborhood U s of s and an affine bijection a s of R d such that a s (U s )isan(m,ε 0 )-elliptic surface, where M depends only on d, φ C and the principal curvatures at s. Moreover, by using a partition of unity, we can write S as a union of affine images of finitely many (M,ε 0 )-elliptic surfaces. These observations are especially important for the extension of Theorem to K (see Remark 1 above).

6 654 M. B. Erdo gan The following theorem is proved in [] for d 3. The d =caseis basically the Carleson-Sjölin Theorem [5]. In [6], it was used in the proof of Theorem for d =. Theorem B. Let d. For any M>0, there exists ε 0 > 0 such that the following statement holds. Let S 1, S be compact subsets of an (M,ε 0 )-elliptic surface in R d with d(s 1,S ) > 1.Letσ j be the Lebesgue measure on S j, j =1,. Then for all, we have q> d+ d (3.1) f 1 dσ 1 f dσ L q (R d ) C M,q,d f 1 L (S 1,dσ 1 ) f L (S,dσ ), for all f j L (S j, dσ j ), j =1,. In [], this theorem is proved explicitly only for the paraboloid. The version we stated here can be proved similarly, see the last section of [] where the necessary modifications are described. We need the following scaled and mollified version of this theorem (see e.g. [3]). In view of II) above, choose N d large enough so that any subset of S d 1 of diameter 1 N d is an affine image of an elliptic surface which satisfies the hypothesis of Theorem B. Let A R (ε) denotethe set {x R d : x R ε}. Corollary 1. Fix a spherical cap U in A 1 (ε), (ε 1/N d ),ofdiameter 1/N d.ifi 1 and I are subsets of U of diameter η with d(i 1,I ) η, then for q> d+, we have d for all f j L (I j ), j =1,. f d+1 d 1 1 f L q (R d ) C q,d εη q f1 f, Proof. First note that the inequality (3.1) is invariant under translations of one or both of the surfaces S 1, S. Therefore, under the hypothesis of Theorem B, we have (3.) f 1 f L q (R d ) ε f 1 f, for all f j L (Sj ε ), j =1,, where Sj ε is the ε-neighborhood of S j. This follows easily from the definition of Lebesgue measure. Let e be the unit vector in the direction of the center of mass of I 1 I. Let {e 1 = e, e,...,e d } be an orthogonal basis for R d.lett: R d R d be the linear map which satisfies T (e 1 )= 1 η e 1, T(e j )= 1 η e j, j =, 3,...,d,

7 On Falconer s Distance Set Conjecture 655 In view of I) and II) above, C j = TI j is contained in ε -neighborhood η of an affine image of a surface S j, j =1,, where the surfaces S 1, S satisfy the hypothesis of Theorem B (with M independent of η, I 1,I ). Let g j (x) =f j (T 1 x), j =1,. Since g j is supported in C j,using(3.)we obtain (3.3) ĝ 1 ĝ q ε η g 1 g. The following elementary identities and (3.3) yield the claim of the corollary: f j (ξ) = 1 det(t )ĝj(t 1 (ξ)) = η d+1 ĝ j (T 1 (ξ)), j =1,, f 1 f q = η (d+1)( 1 q ) ĝ 1 ĝ q, f j = η d+1 gj, j =1,. The following Corollary is obtained from Corollary 1 using a dilation: Corollary. If I 1 and I are subsets of A R (ε), (ε R/N d ),ofdiameter η R/N d with d(i 1,I ) η, then for q> d+, we have d (3.4) f 1 f L q (R d ) C q,d εr 1 q η d 1 d+1 q f1 f, for all f j L (I j ), j =1,. 4. Uncertainty principle Let ϕ be a Schwartz function satisfying ϕ(ξ) =1, for ξ < andϕ(ξ) =0, for ξ > 4. Let D be a ball of radius s in R d. Fix an affine bijection a D of R d which maps D to B(0, 1). Let ϕ D = ϕ a D. Since ϕ is a Schwartz function, for each M N, wehave (4.1) ϕ D(x) = s d ϕ (sx) C M,d s d Mj χ B(0, j s 1 )(x), x R d. The following well-known corollary of the uncertainty principle (see, e.g., [6, Chapter 5]) is another important ingredient of the proof of Theorem. We give a proof for the sake of completeness.

8 656 M. B. Erdo gan Lemma 4.1. Let µ be an α-dimensional measure in R d.letd be a ball of radius s in R d. Then the function µ D := ϕ D µ satisfies i) µ D s d α, ii) µ D 1 1, iii) µ D (B) := µ B D(y)dy r α, for any ball B of radius r 100s 1. Proof. i) Fix M>100d. Using (4.1) and (.), we obtain 0 µ D (x) s d Mj χ B(0, j s 1 )(x y)dµ(y) s d Mj ( j s 1 ) α s d α. ii) follows from Young s inequality and the observation ϕ D 1 1. iii) Using (4.1), we get µ D (B) s d Mj χ B (y)χ B(0, j s 1 )(y u)dµ(u)dy Note that y Band y u B(0, j s 1 )implyu B+ B(0, j s 1 ). Using this, Fubini s theorem and then (.), we obtain µ D (B) s d Mj χ B+B(0, j s 1 )(u)χ B(0, j s 1 )(y u)dydµ(u) s d Mj (r + j s 1 ) α ( j s 1 ) d Mj (r + j s 1 ) α r α. 5. Proof of Theorem The proof is similar to the proof given in [6]. As in [4, 6], we work with the dual formulation: Lemma 5.1. Theorem follows from the following statement: For all q> d+, for all α-dimensional measures µ, for all R>1 and for all f supported d in A R (1), we have (5.1) f (u)dµ(u) C q,µ R d 1 α q f, where f is the inverse Fourier transform of f.

9 On Falconer s Distance Set Conjecture 657 Proof. ([4]) Fix q 0 > d+. Note that by duality, Fubini s theorem and the d statement of the lemma, we have µ L (A R (1)) = sup f(u) µ(u)du = sup f(u)dµ(u) f L (AR (1)) =1 A R (1) C q,µ R d 1 α q, R >1. This easily implies that for any 0 <ε 1, f L (AR (1)) =1 (5.) µ L (A R (R ε )) C q,µr d 1 α q +Cε, R >1. Take a Schwartz function φ equal to 1 in the support of µ. µ = µ φ. Letdσ R be the surface measure on RS d 1.Wehave Note that µ(r ) L (S d 1 ) = C dr (d 1) µ L (RS d 1 ) = C dr (d 1) µ φ L (RS d 1 ) (5.3) C d R (d 1) φ 1 µ φ L 1 (RS d 1 ) R (d 1) µ (u)( φ dσ R )(u)du R (d 1) µ (u)(1 + R u ) M du. The second line follows from Cauchy-Schwarz inequality (as in (5.7) below); the third line from Fubini s theorem and the last line from the Schwartz decay of φ. Here M is a large constant and the implicit constants in the inequalities depend on d, µ, φ, and M. Chooseq ((d+)/,q 0 ). Using (5.) for small ε = ε(d, α, q, q 0 )and(5.3)forlargem = M(ε, d, q, q 0,α), we obtain [ ] µ(r ) L (S d 1 ) R (d 1) µ L (A R (R ε )) A + (1 + R u ) M du R (R ε ) c R α q +Cε + R Mε/ R α q 0. This yields Theorem and hence finishes the proof of the lemma. Let f be as in Lemma 5.1 with L norm 1. Below, we prove that (5.4) f L (dµ) R d 1 α q. (5.1) can be obtained from (5.4) using Cauchy-Schwarz inequality. As in [6], we use the bilinear approach. It suffices to prove (5.4) for functions f supported in a subset of A R (1) of diameter R. Consider a dyadic decomposition of A R (1) into spherical caps, I, with dimensions n n for R 1 n R.

10 658 M. B. Erdo gan We say I has sidelength n and write l(i) = n. The unique cap of sidelength n+1 which contains I is called the parent of I. Let I and J be caps with the same sidelength. We say I and J are related, I J, if they are not adjacent but their parents are. Let f I := fχ I.Asin[6],wehave f L (dµ) (5.5) fi fj L 1 (dµ) + fi L (dµ) I I E R 1 n R =: S 1 + S. l(i)= n,i J Here I E is a set of dyadic caps with sidelengths R 1 overlapping property: (5.6) χ I 1. I IE satisfying the finite First, we obtain a bound for S. Since each I I E is contained in a ball D of radius CR 1,wehavef I = f I ϕ D (ϕ D is defined in the beginning of Section 4). Using this and Cauchy-Schwarz inequality, we have (5.7) f I ( f I ϕ D ) 1 ϕ D 1 1 ( f I ϕ D ) 1. Using this, Fubini s theorem and Lemma 4.1, we obtain (5.8) f I L (dµ) f I (x) (µ ϕ D )(x)dx f I R d α = f I R d α. Using (5.8) and (5.6), we obtain S = I I E f I L (dµ) d α R f I R d α f = R d α I I E This term is harmless since d α <d 1 α,forα (0,d)andq>d+. q d In the remaining part of the paper we prove that for q> d+, d S 1 R d 1 α q. Fix n and I J with I = J = n. First, we prove that (5.9) f I f J L 1 (dµ) C α,q,d R d 1 α q fi f J. Note that I + J is contained in a ball of radius C n. Hence, f I f J is supported in a ball D of radius C n. Using this as in (5.8), we obtain (5.10) fi fj L 1 (dµ) fi (x)fj (x) µ D (x)dx, where µ D = µ ϕ D..

11 On Falconer s Distance Set Conjecture 659 Let e be the unit vector which is in the direction of the center of mass of I J. Consider a tiling of R d with rectangles P of dimensions n R n R, the long axis being in the direction e. ForeachP,leta P be an affine bijection from R d to R d which maps P to the unit cube. Let φ be a Schwartz function satisfying (5.11) φ(x) χ B(0,1) (x), x R, and supp( φ) B(0, 1). Let φ P := φ a P and f I,P := f I φ P. Using (5.11) and the fact that the rectangles P tile R d,weobtain (5.10) fi,p(x)f J,P (x) µ D (x)φ P (x)dx P (5.1) P f I,Pf J,P q µ D φ P q, where q> d+ and q = q. d q 1 To estimate fi,p f J,P q, we use the Corollary of Tao s theorem. Let I P be the support of f I,P.NotethatI P is contained in I +supp( φ P ) I +P dual, where P dual is the dual of P centered at the origin. We have Lemma 5.. I + P dual is contained in a spherical cap of dimensions n n in A R (10) which contains I. Proof. Note that P dual is a rectangle of dimensions R n R n, the short axis being in the direction e. Foreachp P dual and x I, the angle between p e p, e and the hyperplane H x with normal x is 10 n. Therefore P R dual is contained in 1 -neighborhood of 10 H x B(0, R n ). Note that if x R, and r R 1, then x + (H x B(0,r)) is contained in a spherical cap containing x of dimensions 1 r... r in A x (1). This finishes the proof since R n R 1 n. UsingLemma5.forI and J, weseethati P and J P have diameter n ; they are contained in A R (10) and d(i P,J P ) n. Therefore, Corollary implies that (5.13) f I,Pf J,P q R 1 q n(d 1 d+1 q ) f I,P f J,P. We bound µ D φ P q by interpolating between L 1 and L. Using the Schwarz decay of φ P,wehave µ D φ P 1 Mj µ D (x)χ j P (x)dx.

12 660 M. B. Erdo gan Note that j P can be covered by R n using Lemma 4.1, we get (5.14) µ D φ P 1 balls of radius j n. Therefore, R Mj nα n R 1 α nα n R 1 α. Using Lemma 4.1 once again, we obtain (5.15) µ D φ P µ D nd nα. Using (5.15) and (5.14), we obtain (5.16) µ D φ P q µ D φ P 1/q µ D φ P 1/q 1 n d α q ( nα n R 1 α ) 1/q. Using (5.1), (5.13), (5.16) and then Cauchy-Schwarz inequality, we get f I f J L 1 (dµ) R 1 α q n(α(1 q )+d ) P f I,P f J,P R 1 α q n(α(1 )+d )[ q f I,P P ] 1 [ f J,P Using the Schwartz decay of φ, the fact that the rectangles P tile R d and Plancherel formula, we get (5.17) fi fj L 1 (dµ) R 1 α q n(α(1 q )+d ) f I f J. The exponent of n in (5.17) is non-negative and n R. Therefore (5.18) fi fj L 1 (dµ) R 1 α q R α(1 q )+d f I f J R d 1 α q fi f J. Finally, using (5.18) and L -orthogonality, as in [3] and [5], we bound S 1. Note that for each dyadic cap I, there are finitely many (depending on d) dyadic caps J related to I. Therefore, for each I, f J f I, J I for a cap I of sidelength C n which contains I. Also note that for each n, the caps {I : l(i) = n } are finitely overlapping. Thus, f I f I f. l(i)= n l(i)= n Therefore, [ ] 1/ [ ( ) ] 1/ f I f J f I f J f. l(i)= n,i J l(i)= n l(i)= n J I Using this, (5.18) and the fact that there are log(r) valuesofn in the sum for S 1 in (5.5), we obtain (for each q> d+) d S 1 R d 1 α q. P ] 1

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