PACKING SPHERES AND FRACTAL STRICHARTZ ESTIMATES IN R d FOR d 3 Daniel M. Oberlin Department of Mathematics, Florida State University January 005 Fix a dimension d and for x R d and r > 0, let Sx, r) stand for the sphere in R d with center x and radius r. Identifying the collection of all such spheres with S =R d 0, ) R d+, it makes sense to talk about the dimension of a Borel) set of spheres. Since the dimension of any Sx, r) is d, it is natural to conjecture that if E is the union of the spheres in a collection whose dimension exceeds one, then E, the d-dimensional Lebesgue measure of E, is positive. When d = this is a difficult question, answered in the affirmative in Wolff s paper [7]. When d > we will establish the same result much more easily by giving an elementary proof of the following estimate for the spherical average operator T fx, r) = Σ d ) fx rσ)dσ. Theorem. Suppose d >. Fix α, d+) and suppose that µ is a nonnegative Borel measure on a compact subset K of S. Let E α µ) = K K dµs )dµs ) S S α denote the α-dimensional energy of µ, where S S = y y + r r when S j = Sy j, r j ). Then, for Borel sets E R d, we have the restricted weak type estimate ) T χ E L α, µ) C E α µ) α E, where the constant depends only on d, α, and the inf and sup of r on K. Since any set whose Hausdorff dimension exceeds one carries a measure µ as in Theorem for some α >, the following corollary is immediate. Corollary. Suppose E R d and T S are Borel sets. Suppose that T has dimension exceeding one and that for every x, r) T, Sx, r) E has positive d )-dimensional measure. Then E > 0. We mention that the paper [] contains an analogue of Corollary in the case where the projection of T onto R d has dimension greater than one. It should also be true that for 0 < α < the union of an α-dimensional set of spheres has dimension at least d + α. We will prove this if the points x, r) corresponding to the spheres either lie on a curve or comprise an appropriate Cantor Typeset by AMS-TEX
DANIEL M. OBERLIN set. It seems likely that these extra hypotheses are redundant, but we have not yet been able to eliminate them. Estimates like ) are closely connected with the wave equation. Thus it is not surprising that ) implies the following estimate of Strichartz type, again a higher-dimensional analogue of certain results in [7]. Theorem. Suppose α and µ are as in Theorem. Suppose ux, t) is a solution of the wave equation u tt u = 0 on R d [0, ) with ux, 0) = gx), u t x, 0) = hx). For q < α and ɛ > 0 there is the estimate u L q µ) g W,d )/+ɛ + h W,d 3)/+ɛ). This note is organized as follows: we begin with the proof of Theorem. The argument here was first used in [3] and [4], and then, in a form essentially identical to that employed here, in [5]. After the proof of Theorem we sketch proofs for the statements following Corollary. We conclude with the proof of Theorem. Proof of Theorem : For small 0 < δ < r, let T δ be the operator which maps f to its average T δ fx, r) over the annulus Ax, r, δ) of radii r δ and r + δ centered at x. The estimate ) is obviously a consequence of similar estimates, uniform in δ, for the operators T δ. A simple-minded strategy, introduced in [3], for obtaining such estimates for operators like T δ starts from the inequality ) E N E n E m E n, n= where the E n s are subsets of E. In the present case, the E n s will be intersections of E with annuli Ax n, r n, δ). The measures of the E m E n will be controlled by the integral E α µ) and the following crude observation whose proof we include for the sake of completeness). Lemma. Suppose 0 < r 0 < R 0. There is a constant C = CR 0 ) such that if 0 < δ < r 0 < r, r < R 0 and y, y R d, then 3) Ay, r, δ) Ay, r, δ) Cδ δ + y y + r r. Proof of Lemma : Fix y and y and, with no loss of generality, suppose that y = 0, 0,..., 0) and y = t, 0, 0,..., 0). For s, s > 0, the spheres Sy, s ) = {x,..., x d ) : x + + x d = s } and Sy, s ) = {x,..., x d ) : x t) + x + + x d = s }
PACKING SPHERES AND FRACTAL STRICHARTZ ESTIMATES IN R d FOR d 3 3 intersect in or in a d )-sphere {x 0, x,..., x d )} where Then and s s = tx 0 t, x + + x d = ρ, and ρ = s x 0 ). x 0 = x 0 s, s ) = t + s s t ρ = ρs, s ) = 4s t t + s s ) 4t. Consider the map s, s, σ) x 0, ρσ) of 0, ) Σ d ) into R d and the corresponding formula 4) fx 0, ρσ)js, s ) ds ds dσ = f. 0 0 Σ d ) R d Suppose Sy, s ) Sy, s ). If f approximates the function χ {s=s 0,s=s0 } and c d is the volume of the unit d )-sphere, it is easy to see that c d Js 0, s 0 ) = c d ρs 0, s 0 ) d. Some algebra then shows that s + s + t s + s t s s + t s s t 5) Js, s ) = t We will use 4) to estimate 6) Ay, r, δ) Ay, r, δ) c d { s r <δ} { s r <δ} ) d. Js, s ) ds ds. If Ay, r, δ) Ay, r, δ) then it follows that r r y y + δ. Now recall that y y = t to observe that if t δ, then 3) follows from r j < R 0. On the other hand, if t > δ then s r, s r < δ and r r t + δ give s s ± t δ + r r + t t + 4δ 6 t. With 5) this shows that Js, s ) CR 0 ) and so 6) gives 3). Let r 0 be the inf of r on the compact) support of µ. The analogue of ) for T δ is the inequality 7) λ µ{t δ χ E S) > λ} α CEα µ) α E for λ > 0 and 0 < δ < r 0. Here T δ fs) is the average of f over the annulus Ay, r, δ) if S = Sy, r). If T δ χ E S) > λ for S = Sy, r), then E Ay, r, δ) cλδ where c depends only on r 0. Thus if then ) and Lemma give 8) E cnλδ S,..., S N {T δ χ E S) > λ} =T, n<mn Cδ S n S m.
4 DANIEL M. OBERLIN Control of the sum in 8) is provided by Lemma. Let µ be as in Theorem. Then given n N and a Borel T S with µt ) > 0, one can choose S n T, n N, such that S m S n Eα µ) ) /α N. µt ) /α Proof of Lemma : Suppose S,..., S N are chosen independently and at random from the probability space T, ). Then, for m < n N, µt ) T T ) E S m S n by the hypothesis on µ. Thus and the lemma follows. µ µt ) = µt ) T T dµs m )dµs n ) S m S n ) /α dµs m ) dµs n ) ) /α Eα µ) ) /α dµs m )dµs n ) K K S m S n α µt ), /α E With Lemma, 8) becomes ) Eα µ) ) /α N S m S n µt ) /α 9) E cnλδ Cδ E α µ) ) /α N /µt ) /α. Let N 0 = cλµt ) /α /CE α µ) /α δ). Noting that N = N 0 makes the RHS of 9) equal to 0, we consider two cases: Case I: Assume N 0 > 0. In this case choose N N such that λcµt ) /α Then it follows from 9) that C E α µ) ) /α δ N λcµt )/α 3C E α µ) ). /α δ λcµt ) /α E c 3C E α µ) ) λδ Cδ E /α α µ) ) /α λc µt )4/α δ 4CE α µ) /α ) δ µt ) /α = κλ µt ) /α / E α µ) ) /α for κ = c /C). This gives 7). Case II: Assume N 0 0. In this case unless T is empty) we estimate E cλδ λ c µt ) /α 0CE α µ) /α
PACKING SPHERES AND FRACTAL STRICHARTZ ESTIMATES IN R d FOR d 3 5 which again yields 7), concluding the proof of Theorem. The following result will allow us to deal with the case 0 < α <. Proposition. Suppose 0 < γ β α < and suppose that µ is a finite nonnegative Borel measure on a compact subset K of S satisfying the following condition: given N N and Borel T S, one can choose S,..., S N T such that 0) +β γ)/β CN S m S n µt ). +γ)/β Suppose Then there is the estimate p = + β γ + β γ, q = + γ + β γ, η = γ + β γ. for Borel E R d and δ 0, ). T δ χ E L q, µ) E /p δ η It follows from the proof of Lemma.5 in [] that the estimate T δ χ E L q, µ E /p δ η implies a lower bound of n pη for the Hausdorff dimension of a Borel set containing positive-measure sections of each sphere S in the support of µ. Plugging in the values for p and η which are given in Proposition yields the lower bound n γ)/ + β γ). We will see below that if T S has dimension α 0, ) and either lies on a curve or is a Cantor set, then T supports measures µ allowing choices of β and γ arbitrarily close to α and so leading to the desired lower bound of n + α for the dimension of S T S. First, though, we indicate the proof of the proposition. Proof of Proposition : The proof follows the proof of Theorem in [5] and is only a slight modification of the proof of Theorem. Using 0) instead of Lemma, the analogue of 9) is ) E cnλδ Cδ N +β γ)/β µt ) +γ)/β. The two cases are now defined by comparing N 0. cλ ) β/+β γ)µt +γ = ) +β γ Cδ and 0. In case N 0 > 0, choosing N in ) such that N 0 / N N 0 /3 gives E λ +β γ +β γ δ γ +β γ µt ) +γ +β γ κ
6 DANIEL M. OBERLIN where β +β γ C +β γ κ = c +β γ 3 +β γ)/β ) > 0. This gives the desired estimate λµt ) /q E /p δ η if N 0 > 0. On the other hand, the inequality N 0 0 gives λµt ) +γ)/β δ and so ) λ A µt ) A+γ)/β δ A if A > 0. Since E cλδ unless T is empty), there is also the inequality 3) λ A E A δ A as long as 0 < A <. Multiplying ) and 3) gives λµt ) A+γ)/β E A δ A. Then the choice A = β/ + β γ) yields λµt ) /q E /p δ η again, completing the proof of Proposition. When µ is supported on a curve, the following lemma verifies the hypotheses of Proposition. Thus it follows from standard facts about Hausdorff dimension that if the Borel set K S lies on a curve as in Lemma 3, if E R d is Borel, and if S K S E, then 4) α. = dimk) 0, ) implies dime) n + α. Lemma 3. Suppose α 0, ). Suppose µ is a nonnegative measure on a compact interval J R which satisfies the condition µi) I α for subintervals I J. Let µ be the image of µ under a one-to-one and bi-lipschitz mapping of J into S. Suppose 0 < γ < β < α <. Then given N N and Borel T S, one can choose S,..., S N T such that +β γ)/β Cα, β, γ)n. S m S n µt ) +γ)/β We omit the proof of Lemma 3 since it is technical, of little intrinsic interest, and completely parallel to the proof of Lemma in [5]. The next lemma is an analogue of Lemma 3 in case K is a Cantor set. Lemma 4. Suppose k, l are positive integers and set α = log k/ log l. Suppose that K S supports a Borel measure µ such that E α ɛ µ) < for 0 < ɛ < α and having the property that for each M N, K is the union of k M l M -separated compact sets K M m of diameter l M such that µk M m ) = k M. Then, given β 0, α), N N, and Borel T S, one can choose S,..., S N T such that +β)/β Cβ)N S m S n µt ). +β)/β
PACKING SPHERES AND FRACTAL STRICHARTZ ESTIMATES IN R d FOR d 3 7 Again, it follows from Proposition that 4) holds for such Cantor sets K. Proof of Lemma 4: Fix β, ɛ 0, α) such that 5) + α + ɛ α = + β β. For a given N N and Borel T S, choose M N such that 6) N k M µt ). Since µk M m ) = k m, T must intersect N of the K M m s. Therefore we can choose N points S m T such that distinct S m s lie in distinct K M m s. Then, using the hypotheses on the K M m s, k M K M m S m S n K M n Since l = k /α, it follows from 6) and 5) that as desired. K M m K M n dµt )dµt ) T T dµt )dµt ) T T α ɛ l M ) α+ɛ E α ɛµ)l M ) α+ɛ. N ) +α+ɛ)/α N ) +β)/β S m S n km ) +α+ɛ)/α = µt ) µt ) We will give the proof of Theorem in case d is even. The proof for odd d is similar but slightly less complicated. Before beginning, we recall the well-known formula for the solution of the wave equation for even d: ux, t) = γ d [ t for some constant γ d. t ) t ) d )/ t d gx + ty) )+ t dy B0,) y ) d )/ t d hx + ty) ) ] t dy B0,) y Proof of Theorem 4: Recalling that T fx, t) = Σ d ) fx + tσ)dσ, we define Sfx, t) = B0,) dy fx + ty). y Let t 0 and T 0 be the inf and sup of t on the support of µ. For β > 0, let I β be a potential operator on R d with smooth and nonzero multiplier p β ξ) equal to ξ β for ξ /T 0 ). The next lemma is essentially the difference between the proofs of Theorem for even and odd d.
8 DANIEL M. OBERLIN Lemma 5. If µ is as in Theorem then for q < α and 0 < β < / we have 7) Sf Lq µ) I β f L, R d ). Proof of Lemma 5: Suppose γξ) is a smooth function on R d which vanishes on B0, ) and which equals ξ d/ for large ξ. Let mξ) be the multiplier e πi ξ γξ) and suppose that K is the kernel on R d whose Fourier transform is m. Define the operator S 0 by S 0 fx, t) = fx + ty) Ky) dy. R d By the asymptotic expansion of the Fourier transform of y ) / +, the operator S is the sum of two operators like S 0 and nicer terms. We will explain why 8) S 0 f Lq µ) I β f L, R d ). Define m by p β ξ) mξ) = mξ) so that mξ) = e πi ξ γξ) where γξ) is smooth and equal to ξ β d/ for large ξ. Let the kernel K have Fourier transform m and put Sfx, t) = fx + ty) Ky) dy. R d Since S 0 f, t)ξ) = mtξ) fξ) = p β tξ) mtξ) fξ) = t β mtξ)p β ξ) fξ) = t β SI β f, t)ξ), 8) will follow from t 0 t T 0 on the support of µ and 9) Sf L q µ) f L, R d ). Let ɛ = / β. According to the considerations on p. 46 of [6], the kernel K satisfies Ky) y +ɛ for y and Ky) y d+)/ for large y. Thus K = K j where K 0 L R d ) and for j =,,... j=0 K j jɛ j χ A0,, j ). We recall that A0,, j ) is the annulus centered at 0 with radii j and + j.) If the operators S j correspond to the kernels K j, then S 0 maps L R d ) into L µ) since t is bounded away from 0 on the support of µ). Also, for j =,,... we have S j f L q µ) jɛ T j f L q µ) jɛ f L, R d )
PACKING SPHERES AND FRACTAL STRICHARTZ ESTIMATES IN R d FOR d 3 9 by the analogue of Theorem for the operators T δ. Since S = Sj, 9) follows. Returning to the proof of Theorem 4, we can write ux, t) = α =d/ c α t)sd α g)x, t) + α =d/ d α t)sd α h)x, t) where the coefficients c α and d α are bounded on the support of µ. Fix ɛ > 0 and let β = / ɛ/. If α = d/ then Lemma 5 gives SD α g) Lq µ) I β D α g L, R d ) I β D α g W,ɛ/ g W,d/+ɛ/ β = g W,d )/+ɛ. Analogous considerations for h complete the proof of Theorem. References. J. Bourgain, Besicovitch type maximal operators and applications to Fourier analysis, Geometric and Functional Analysis 99), 47 87.. T. Mitsis, On a problem related to sphere and circle packing, J. London Math. Soc. 60 999), 50 56. 3. D. Oberlin, L p L q mapping properties of the Radon transform, Springer Lecture Notes in Mathematics 995 983), 95 0. 4., An estimate for a restricted X-ray transform, Canadian Mathematical Bulletin 43 000), 47 476. 5., Restricted Radon transforms and unions of hyperplanes, preprint. 6. E.M. Stein, Harmonic Analysis: Real-Variable Methods, Orthogonality, and Oscillatory Integrals, Princeton University Press, 993. 7. T. Wolff, Local smoothing estimates on L p for large p, Geom. Funct. Anal. 0 000), 37 88.