Buffon's needle landing near Besicovitch irregular self-similar sets
aa r X i v : . [ m a t h . A P ] J a n BUFFON’S NEEDLE LANDING NEAR BESICOVITCHIRREGULAR SELF-SIMILAR SETS
MATTHEW BOND AND ALEXANDER VOLBERG
Abstract.
In this paper we get an upper estimate of the Favard length ofan arbitrary neighborhood of an arbitrary self-similar Cantor set. Consider L closed discs of radius 1 /L inside the unit disc. By using linear maps of the disconto the smaller discs we can generate a self-similar Cantor set G . One suchprocess is to let G n be the union of all possible images of the unit disc under n -fold compositions of the similarity maps. Then G = T n G n . One may thenask the rate at which the Favard length – the average over all directions of thelength of the orthogonal projection onto a line in that direction – of these sets G n decays to zero as a function of n . Previous quantitative results for the Favardlength problem were obtained by Peres–Solomyak [19] and Tao [21]; in the latterpaper a general way of making a quantitative statement from the Besicovitchtheorem is considered. But being rather general, this method does not give agood estimate for self-similar structures such as G n . In the present work weprove the estimate Fav( G n ) ≤ e − c √ log n . While this estimate is vastly improvedcompared to [19] and [21], it is worse than the power estimate Fav( G n ) ≤ Cn p proved for specific sets G n with additional product structures in Nazarov-Peres-Volberg [17] and Laba-Zhai [11]. The power estimate still appears to be relatedto a certain regularity property of zeros of a corresponding linear combination ofexponents (we call this property analytic tiling ). We consider also the Sierpinskigasket, where this regularity of zeros exists, resulting in an improvement to apower estimate. Introduction
Let E ⊂ C , and let proj θ denote orthogonal projection onto the line having angle θ with the real axis. The average projected length or Favard length of E , Mathematics Subject Classification.
Primary: 28A80. Fractals, Secondary: 28A75,Length, area, volume, other geometric measure theory 60D05, Geometric probability, stochas-tic geometry, random sets 28A78 Hausdorff and packing measures.Research of the authors was supported in part by NSF grants DMS-0501067, 0758552 .
Fav( E ), is given by Fav( E ) = 1 π Z π | proj θ ( E ) | d θ. For bounded sets, Favard length is also called
Buffon needle probability , sinceup to a normalization constant, it is the likelihood that a long needle dropped withindependent, uniformly distributed orientation and distance from the origin willintersect the set somewhere.Consider L closed discs of radius 1 /L inside the unit disc. By using linear mapsof the unit disc onto the smaller discs we can generate a self-similar Cantor set G .A partial construction G n of G consists of the union of all possible images of theunit disc under n -fold compositions of these similarity maps. Then G = T n G n .One may then ask the rate at which the Favard length of these sets G n decays tozero as a function of n ( ). Observe that G n is in some sense comparable to an L − n neighborhood of G ( ), so Fav( G n ) is comparable to the likelihood that “Buffon’sneedle” will land in a L − n -neighborhood of G .The first quantitative results for the Favard length problem were obtained in[19],[21]; in the latter paper a general way of making a quantitative statement fromthe Besicovitch theorem is considered. But being rather general, this method doesnot give a good estimate for self-similar structures such as G n .Indeed, vastly improved estimates have been proven in these cases: in [17], it wasshown that for 1 / p < /
6, such that
F av ( K n ) ≤ c p n p ,and in [5], [6] the same type power estimate was proved for the Sierpinski gasket S n for some other p >
0. These results cannot possibly be improved to p = 1: F av ( K n ) ≥ c log nn . (This is [1]( ), and the argument and result also apply to S n .)Compare this with [19], in which it was shown that certain random sets of which K n is a special case almost surely decay in Favard length like n in the liminf.Crucial to [1] was a tiling property: namely, under orthogonal projection on theline with slope 1 /
2, the squares composing K n tile a line segment. Oddly enough,such a property can be used to prove upper bounds as well: under the assumptionthat some orthogonal projection in some direction contains an interval, Laba and Such decay must occur by the Besicovitch projection theorem and by continuity of measures,since one takes the Lebesgue measure of decreasing sets in the parameter space of { directions } ×{ projected x values } . an L − n -neighborhood of G is contained in several small translates of G n , while G n is containedin a neighborhood of size ≈ L − n of G the method is stable under “bending the needle” slightly - see [7]. UFFON’S NEEDLE AND BESICOVITCH IRREGULAR SELF-SIMILAR SETS 3
Zhai [11] showed that the result of [17] holds for Cantor-like product sets of finite H measure (but with a smaller exponent). Their argument uses tiling resultsobtained in Kenyon [10] and Lagarias-Wang [12] to fill in a gap where [17] fails togeneralize (more on this shortly).With the exception of [19] and [21], the above papers all extract their resultsfrom information about L norms of the projection multiplicity function, whichcounts how many squares (or discs) project to cover each point. The function f n,θ : R → N is defined by f n,θ = X discs T of G n χ proj θ ( T ) . Note that
F av ( G n ) = π − R π | supp( f n,θ ) | dθ . In [17] and [5], the L norm of theanalog of this function for squares was studied to obtain Buffon needle probabilityestimates for K n – in [5], p = 1 ,
2, were related to χ supp( f n,θ ) via the Cauchyinequality, while in [17], p = 2 was studied via Fourier transforms and related tothe measure of the level sets of f n,θ = f .Consider some heuristics. Forgetting for the moment about angles, let f :[0 , → N be any sum of measurable characteristic functions such that || f || L = 1.If the mass is concentrated on a small set, the L p norm should be large for p > L p norm should indicate that the support of a function is small,and vice versa. Let K >
0, let A = supp { f } , and let A K = { x : f ≥ K } .1 = R f ≤ || f || p || χ A || q , so m ( A ) ≥ || f || − qp , a decent estimate. The other basicestimate is not so sharp: m ( A ) ≤ − ( K − m ( A K ). However, a combinato-rial self-similarity argument of [17] shows that for the Favard length problem, itbootstraps well under further iterations of the similarity maps - this argument isrevisited in Section 4. Hence, up to some loss of sharpness, it has been shown thatto study Favard length of these self-similar sets, it is sufficient to study the L norms of f n,θ .( )One must average | supp f n,θ | over the parameter θ to get Favard length of G n .For K n and S n , there are some directions for which the orthogonal projections donot even decay to length zero with n (i.e., the L norms of f n,θ are bounded forthese angles), and this countable dense set of directions is to a large extent classifiedin [10]. In [17], a method for controlling the measure of a set of angles E on whichthe projections fail to decay rapidly was found: one takes the Fourier transform of So far, only L p for p = 1 , , or ∞ have played any useful role, to our knowledge. MATTHEW BOND AND ALEXANDER VOLBERG f n,θ in the length variable, and takes a sample integral of | ˆ f n,θ ( x ) | over a chosensmall interval I where R E × I | ˆ f n,θ ( x ) | dθdx is small. One then shows that there is a θ ∈ E such that R I | ˆ f n,θ ( x ) | dx is large relative to | E | , and so | E | must be small.In all cases, ˆ f n,θ is a decay factor times a self-similar product Q k ϕ θ ( L − k y ) oftrigonometric polynomials ϕ θ . The danger is that the low-frequency zeroes mightkill off the better-behaved high-frequency terms. In [17], the four frequencies of ϕ θ were symmetric around 0, allowing the terms to simplify to two cosines, andtrigonometric identities allowed the whole product to be estimated by a single sineterm. In [11], an analogous role was played by tilings of the line on the non-Fourierside by proj θ ( G n ) in the special direction θ , and the product structure of G n allowed for a change and separation of variables.Separating variables is more difficult when there is no product structure. Thesimplest case without the product structure is the Sierpinski gasket S consideredin Section 2. We give there a sketch of the power estimate (proven in detail in[5]), which is based on the fact that zeroes of ϕ (3 k · ) are separated away from eachother for different values of k . This special structure of zeros (we call it “analytictiling” after [11]) is not always available for all angles. We have not yet found anadequate substitute for it in the general case, and this is why the for the generalcase we still only have F av ( G n ) ≤ e − ǫ √ log n .Rather strangely, a claim in the spirit of the Carleson Embedding Theorem,in the form of Lemma 22, plays an important part in our reasoning. Becausethe Fourier transform turns stacks of discs (i.e., sums of overlapping characteristicfunctions) into clusters of frequencies, this lemma provides important upper boundswhen θ belongs to E .The main result of this article is the following estimate. Theorem 1.
For all n ∈ N , Fav( G n ) ≤ C e − ǫ √ log n . For the Sierpinski gasket, the result is exactly that of [17]:
Theorem 2.
For all p < / , there exists C p > such that for all n ∈ N , Fav( S n ) ≤ C p n − p . Acknowledgements . We are greatly indebted to Fedja Nazarov for many valu-able discussions, we also express our deep gratitude to Izabella Laba for many
UFFON’S NEEDLE AND BESICOVITCH IRREGULAR SELF-SIMILAR SETS 5 useful conversations and to John Garnett who introduced the subject to one of theauthors.2.
Definitions and result for Sierpinski gasket. Sketch of the prooffor Sierpinski gasket
We give in the first section the sketch of the estimate for a special self-similar set–the Sierpinski gasket. The structure of zeros of a certain trigonometric polynomialrelated to this set plays the crucial part in this “good” (meaning power) estimateof the type ≤ n − c . We put this sketch here for the reader to be able to compareit with the general case, where we get only the estimate ≤ e − c √ log n due to thelack of understanding of these zeros. It is elaborated in more detail in [5]. Itmay be instructive to compare the general and the special cases. If the reader isinterested only in the general case then he/she can skip the present section and godirectly to the next one. Conversely, the reduced detail and difficulty may benefitthe first-time reader looking for the general overview of the method. B ( z , r ) := { z ∈ C : | z − z | < r } . For α ∈ {− , , } n let z α := n X k =1 ( 13 ) k e iπ [ + α k ] , S n := [ α ∈{− , , } n B ( z α , − n ) . This set is our approximation of a partial Sierpinski gasket; it is strictly larger. Wemay still speak of the approximating discs as “Sierpinski triangles.”The result for the Sierpinski gasket is the following:
Theorem 3.
For some c > , Fav( S n ) ≤ Cn c . We will simplify the proof by picking specific values for constants; at the endof this paper, a short remark shows how to recover the full range c < / S n is 3 − n approximation to a Besicovitch irregular set (see [8] for defi-nition) called the Sierpinski gasket. Recently one detects a considerable interest inestimating the Favard length of such ǫ -neighborhoods of Besicovitch irregular sets,see [19], [21], [17], [11]. In [19] a random model of such Cantor set is consideredand estimate ≍ n is proved. But for non-random self-similar sets the estimatesof [19] are more in terms of ··· log n (number of logarithms depending on n ) and MATTHEW BOND AND ALEXANDER VOLBERG more suitable for general class of “quantitatively Besicovitch irregular sets” treatedin [21].As in the introduction, let f n,θ := X Discs D of S n χ proj θ ( D ) . Self-similarity allows us to write f n,θ in a form well-suited to Fourier analysis: f n,θ = ν n ∗ n χ [ − − n , − n ] , where ν n := ∗ nk =1 e ν k and e ν k := 13 [ δ − k cos ( π/ − θ ) + δ − k cos ( − π/ − θ ) + δ − k cos (7 π/ − θ ) ] . For
K >
0, let A K := A K,n,θ := { x : f n,θ ≥ K } . Let L θ,n := proj θ ( G n ) = A ,n,θ .For our result, some maximal versions of these are needed: f ∗ N,θ := max n ≤ N f n,θ , A ∗ K := A ∗ K,n,θ := { x : f ∗ n,θ ≥ K } . Also, let E := E N := { θ : | A ∗ K | ≤ K − } for K = N ǫ , where ǫ > ϕ θ ( x ) := 13 [ e − i cos( π/ − θ ) x + e − i cos( − π/ − θ ) x + e − i cos(7 π/ − θ ) x ]plays the central role: c ν n ( x ) = Q nk =1 ϕ θ (3 − k x ).2.1. General philosophy.
Fix θ . If the mass of f n,θ is concentrated on a small set,then || f n,θ || p should be large for p > R f ≤ || f n,θ || p || χ L θ,n || q ,so m ( L θ,n ) ≥ || f || − qp , a decent estimate. The other basic estimate is not so sharp: m ( L θ,N ) ≤ − ( K − m ( A K,N,θ ) (2.1)However, a combinatorial self-similarity argument of [17] and revisited in [1] showsthat for the Favard length problem, it bootstraps well under further iterations ofthe similarity maps:
Theorem 4. If θ / ∈ E N , then |L θ,NK | ≤ CK . This is proved in full detail as Theorem 17. Note that the maximal version f ∗ N is used here. A stack of K triangles at stage n generally accounts for morestacking per step the smaller n is. Thus the maximal function f ∗ N captures thisinformation by recording a large level set at height K whenever f n attains height K for a small value of n . For fixed x ∈ A ∗ K,N,θ , the above theorem considers thesmallest n such that x ∈ A K,n,θ , and uses self-similarity and the Hardy-Littlewood
UFFON’S NEEDLE AND BESICOVITCH IRREGULAR SELF-SIMILAR SETS 7 theorem to prove its claim by successively refining an estimate in the spirit of (2.1).Of course, now Theorem 3 follows from the following:
Theorem 5.
Let ǫ < / log (169) , sufficiently, ǫ ≤ / . . Then for N >> , | E | < N − ǫ . It turns out that L theory on the Fourier side is of great use here. The followingis later proved as Theorem 26: Theorem 6.
For all θ ∈ E N and for all n ≤ N , || f n,θ || L ≤ CK . One can then take small sample integrals on the Fourier side and look for lowerbounds as well. Let K = N ǫ , and let m = 2 ǫ log N . Theorem 6 easily impliesthe existence of ˜ E ⊂ E such that | ˜ E | > | E | / n , N/ < n < N/ θ ∈ ˜ E , Z n n − m n Y k =0 | ϕ θ (3 − k x ) | dx ≤ CKmN ≤ ǫ N ǫ − log N. Number n does not depend on θ ; n can be chosen to satisfy the estimate in theaverage over θ ∈ E , and then one chooses ˜ E . Let I := [3 n − m , n ] . Now the main result amounts to this (with absolute constant α large enough): Theorem 7. ∃ θ ∈ ˜ E : Z I n Y k =0 | ϕ θ (3 − k x ) | dx ≥ c m − · αm = cN − ǫ (2 α − . The result: 2 ǫ log N ≥ N − ǫ (4 α − , i.e., N ≤ N ∗ . Now we sketch the proofof Theorem 7. We split up the product into two parts: high and low-frequency: P ,θ ( z ) = Q n − m − k =0 ϕ θ (3 − k z ), P ,θ ( z ) = Q nk = n − m ϕ θ (3 − k z ). Theorem 8.
For all θ ∈ E , R I | P ,θ | dx ≥ C m . Low frequency terms do not have as much regularity, so we must control thedamage caused by the set of small values , SSV ( θ ) := { x ∈ I : | P ( x ) | ≤ − ℓ } , ℓ = α m . In the next result we claim the existence of E ⊂ ˜ E , |E| > | ˜ E | / Theorem 9. Z ˜ E Z SSV ( θ ) | P ,θ ( x ) | dx dθ ≤ m − ℓ/ ⇒ ∀ θ ∈ E Z SSV ( θ ) | P ,θ ( x ) | dx ≤ c K m − ℓ/ . MATTHEW BOND AND ALEXANDER VOLBERG
Then Theorems 8 and 9 give Theorem 7; since ℓ = αm and K = 3 m , we seethat any α > α to be largersoon.2.2. Locating zeros of P . We can consider Φ( x, y ) = 1 + e ix + e iy . The keyobservations (see also the discussion section at the end of the paper) are | Φ( x, y ) | ≥ a ( | x − | + | y − | ) , sin 3 x sin x = 4 cos x − . (2.2)Actually, we will set α = a − in the end. Changing variable we can replace 3 ϕ θ ( x )by φ t ( x ) = Φ( x, tx ).Consider P ,t ( x ) := Q nk = n − m φ t (3 − k x ), P ,t ( x ) := Q n − mk =0 13 φ t (3 − k x ).We need SSV ( t ) := { x ∈ I : | P ,t ( x ) | ≤ − ℓ } . One can easily imagine it ifone considers Ω := { ( x, y ) ∈ [0 , π ] : |P ( x, y ) | := | Q mk =0 Φ(3 k x, k y ) | ≤ m − ℓ } .Moreover, (using that if x ∈ SSV ( t ) then 3 − n x ≥ − m , and using xdxdt = dxdy )we change variable in the next integral: Z ˜ E Z SSV ( t ) | P ,t ( x ) | dxdt = 3 − n +2 m · n Z ˜ E Z − n SSV ( t ) | n Y k = m Φ(3 k x, k tx ) | dxdt ≤ − n +3 m Z Ω | n Y k = m Φ(3 k x, k y ) | dxdy . Now notice that by our key observationsΩ ⊂ { ( x, y ) ∈ [0 , π ] : | sin 3 m +1 x | + | sin 3 m +1 y | ≤ a − m m − ℓ ≤ − ℓ } . (2.3)The latter set Q is the union of 4 · m +2 squares Q of size 3 − m − ℓ/ × − m − ℓ/ . Fixsuch a Q and estimate Z Q | n Y k = m Φ(3 k x, k y ) | dxdy ≤ ℓ Z Q | n Y k = m + ℓ/ Φ(3 k x, k y ) | dxdy ≤ ℓ · (3 − m − ℓ/ ) Z [0 , π ] | n − m − ℓ/ Y k =0 Φ(3 k x, k y ) | dxdy ≤ ℓ · (3 − m − ℓ/ ) · n − m − ℓ/ = 3 − m · n − m − ℓ/ . Therefore, taking into account the number of squares Q in Q and the previousestimates we get Z E Z SSV ( t ) | P ,t ( x ) | dxdt ≤ m − ℓ/ . Theorem 9 is proved.
UFFON’S NEEDLE AND BESICOVITCH IRREGULAR SELF-SIMILAR SETS 9
Remarks . The rest of the paper is devoted to general self-similar sets, where wecan get only a e − c √ log n result due to the lack of structure (possible lack of “analytictiling”) of the zeros of trigonometric polynomials, which are “telescopic products”of one trigonometric polynomial. See the last section of this work for the discussion.It is true that α depends on the constant a in (2.2), since it appears in (2.3).One can use a = , attained at ( x, y ) = (0 , π ). Then from (2.3), we get α = m/ℓ ≥ log (162) ≈ .
631 as our last condition on α . We need this to compute thebest exponent p .Note that in our argument, we cut a couple corners. To get the best exponentcurrently available, let γ >
1. Let m = γǫ log N . Then the argument works aslong as ǫ < [2 γα + 1 − γ ] − , i.e., ǫ <
12 log (169) . Using the sharper exponent β > p = ǫ − + β < (169)] − +2 in the estimateFav( S n ) ≤ C p n − p . In particular, p < . is small enough.This can be improved if more care is taken, but not beyond p = 1 / The Fourier-analytic part
The setup.
The goal of this section is to prove Theorem 11, which shows thatfor most directions, a considerable amount of stacking occurs when the discs areprojected down. Throughout the paper, the constants c and C will vary from line toline, but will be absolute constants not depending on anything. The symbols c and C will typically denote constants that are sufficiently small or large, respectively.Everywhere we use the definition B ( z , ε ) := { z ∈ C : | z − z | < ε } .Let G := L [ j =1 B ( r j e iθ j , L ) . Then one constructs G and G n using the similarity maps of the unit disc onto thediscs forming G . For convenience, we will now rescale G n by a factor absolutelycomparable to 1 and bound the discs of G n by slightly larger discs and study thisset instead.Recall that f n,θ := L X Discs D of G n χ proj θ ( D ) , Observe that f n,θ = ν n ∗ L n χ [ − L − n ,L − n ] , where ν n := ∗ nk =1 e ν k and e ν k = 1 L [ L X l =1 δ L − k r l cos( θ − θ l ) ] . We will now slightly modify f for convenience. Note thatˆ f n,θ ( x ) = L n ˆ χ [ − L − n ,L − n ] ( x ) · n Y k =1 φ θ ( L − k x ) , where φ θ ( x ) = L [ P Ll =1 e − ir l cos( θ l − θ ) x ]. By factoring and changing the variable, wemay instead write in place of φ θ the function ϕ t ( x ) = 1 L [1 + e ix + e itx + L X l =4 e a l x + b l tx ] , t ∈ [0 , . (3.1)We assumed here that r = 0, r = r = 1, θ = 0, θ = π/
2. We can do this byaffine change of variable.For numbers
K, N >
0, define the following: f ∗ N ( s ) := f ∗ N,t sup n ≤ N f n,t ( s ) (3.2) A ∗ K := A ∗ K,N,t := { s : f ∗ N ( s ) ≥ K } (3.3) E := { t : | A ∗ K | ≤ K } . (3.4) E is essentially the set of pathological t such that || f n,t || L ( s ) is small for all n ≤ N , as in [17]. In fact, we have this result, proved in Section 7: Theorem 10.
Let t ∈ E . Then max ≤ n ≤ N k f n,t k L ( s ) ≤ c K . The aim of Section 3 is to prove the following:
Theorem 11.
Let ǫ be a fixed small enough constant. Then for N >> , | E |
Initial reductions.
Because of Theorem 10, we have ∀ t ∈ E , K ≥ || f N,t || L ( s ) ≈ || d f N,t || L ( x ) ≥ C Z L N/ | c ν N ( x ) | dx (3.5)Let m ≈ ( ǫ log N ) / , K ≈ log N . Split [1 , L N/ ] into N/ L k , L k +1 ]and take a sample integral of | c ν N | on a small block I := [ L n − m , L n ], with n ∈ [ N/ , N/
2] chosen so that1 | E | Z E Z L n L n − m | c ν N ( x ) | dx dt ≤ CKm/N .
This choice is possible by (3.5). Define˜ E := { t ∈ E : Z L n L n − m | c ν N ( x ) | dx ≤ CKm/N } . It then follows that | ˜ E | ≥ K .Note that c ν N ( x ) = Q Nk =1 ϕ ( L − k x ) ≈ Q nk =1 ϕ ( L − k x ) for x ∈ [ L n − m , L n ].So for t ∈ E , Z L n L n − m n Y k =1 | ϕ t ( L − k x ) | dx ≤ CKmN ≤ ǫ N ǫ − log N. Recall that m ≈ ( ǫ log N ) / . Later, we will show that ∃ t ∈ E and absoluteconstant α such that Z L n L n − m n Y k =1 | ϕ t ( L − k x ) | dx ≥ cL m − · αm ≥ cN − αǫ . (3.6)The result: 2 ǫ log N ≥ N − α ǫ − ǫ , i.e., N ≤ N ∗ if ǫ is small enough. In otherwords: Proposition 12.
Inequality (3.6) is sufficient to prove Theorem (11) . Further,inequality 3.6 can be deduced from Propositions 13 and 14, as will be seen shortly.
So let us prove inequality (3.6).First, let us write Q nk =1 ϕ t ( L − k x ) = P t ( x ) = P ,t ( x ) P ,t ( y ), where P is the lowfrequency part, and P is has medium and high frequencies: P ,t ( x ) := n − m Y k =1 ϕ t ( L − k x ) = [ ν n − m ( x ) P ,t ( x ) = n Y k = n − m ϕ t ( L − k x ) = c ν m ( L m − n x ) We want the following:
Proposition 13.
Let t ∈ E be fixed. Then R L n L n − m | P ,t ( x ) | dx ≥ C L m . Recall that we defined the set ˜ E, | ˜ E | > | E | /
2, and we assume that | E | > /K . (3.7)Recall that we denoted I = [ L n − m , L n ] . We also want a proportion of the contribution to the integral separated away fromthe complex zeroes of P ,t : Proposition 14.
Let
SSV ( t ) := { x ∈ I : | P ,t ( x ) | ≤ L − αm } . Suppose also that E is unable to hide, that is (3.7) is valid. Then there exists a subset E ⊂ ˜ E, |E| ≥ / K, such that for every θ ∈ E one has Z SSV ( t ) | P ,t ( x ) | dxdt ≤ c L m , where c is less than the C from Proposition 13. In particular, | ˜ E | Z ˜ E Z SSV ( t ) | P ,t ( x ) | dxdt ≤ c L m , Remarks.
1) The set
SSV ( t ) is so named because it is the set of small values of P on I . Combining this with Proposition 13 , Z L n L n − m | P ,t ( x ) | | P ,t ( x ) | dx ≥ Z I \ SSV ( t ) | P ,t ( x ) | · L − αm dx ≥ c L m − αm , which gives (3.6)–exactly what we promised to obtain from Propositions 14, 13.2) Thus Propositions 13 and 14 suffice to prove Theorem 11, and Proposition 12has been demonstrated.3) All this holds if (3.7) holds. But if we have the opposite: | E | ≤ /K = L − m = e − C ( L ) ǫ (log N ) / , (3.8)the main result is automatically proved because we have only a small set of singulardirections.First, let us fix t ∈ E and prove Proposition 13. UFFON’S NEEDLE AND BESICOVITCH IRREGULAR SELF-SIMILAR SETS 13
Proof.
We are using first Salem’s trick on Z L n | P ( x ) | dx :Let h ( x ) := (1 − | x | ) χ [ − , ( x ), and note that ˆ h ( α ) = C − cosαα >
0. Then if wewrite P = L m − n − P L n − m j =0 e iα j x , we get Z L n | P ( x ) | dx ≥ Z L n − L n h ( L − n x ) | P ( x ) | dx ≥ C ( L m − n ) [ L n · L n − m + L n − m X j = k ; j,k =1 L n ˆ h ( L n ( α j − α k ))] ≥ CL m . To show that this is not concentrated on [0 , L n − m ], we will use Theorem 10 andLemma 22. We get Z L n − m | P ( x ) | dx = Z L n − m | [ ν n − m ( x ) | dx = L m − n ) Z L n − m | n − m X j =0 e iα j x | dx ≤ CK ≤ CL m . (cid:3) So now we have Proposition 13. The greater challenge will be Proposition 14.3.3.
The proof of Proposition 14.
Recall that
SSV ( t ) := { x ∈ I = [ L n − m , L n ] : | P ,t ( x ) | ≤ L − αm } .To get Proposition 14, we will split P ,t into two parts, P ♯ ,t ( x ) and P ♭ ,t ( x )corresponding to medium and high frequencies.A straightforward application of Lemma 22 to high frequency part P ♯ ,t ( x ) willget us part of the way there, see Proposition 16 (for fixed t , the size of SSV ( t )does not overwhelm the average smallness of P ♯ ,t ( x )), and the claim 15 applied tomedium frequency term P ♭ ,t ( x ) will further sharpen the final estimate to what weneed.Naturally, P ♭ ,t ( x ) and P ♯ ,t ( x ) are defined as the medium and high frequencyparts of P ,t ( x ). Below, ℓ := αm : P ♭ ,t ( x ) := n − m − Y k = n − m − ℓ ϕ t ( L − k x ) = d ν ℓ − ( L m + ℓ − n x ) ,P ♯ ,t ( x ) := n − m − ℓ − Y k =1 ϕ t ( L − k x ) = ˆ ν n − m − ℓ − ( x ) . Here is the first claim of this subsection
Proposition 15.
For all sufficiently small positive numbers τ ≤ τ and for allsufficiently large m and ℓ = α m there exists an exceptional set H of directions t such that | H | ≤ L − ℓ/ , (3.9) ∀ t / ∈ H ∀ x ∈ [ L n − m , L n ] , | P ♭ ,t ( x ) | ≤ e − τ ℓ . (3.10) Proof.
Notice that φ θ ( r ) = Φ( r cos θ, r sin θ ) , where for x = ( x , x ), Φ( x ) := Φ( x , x ) = 1 L L X l =1 e πi h a l ,x i . As some pair of vectors a l − a , l ∈ [1 , L ] must span a two-dimensional space, wecan assume without the loss of generality (make an affine change of variable) that a = (0 , , a = (1 , , a = (0 , . Then Φ( x , x ) = 1 L (1 + e πix + e πix + L X l =4 e πi h a l ,x i ) . (3.11)We make the change of variable y = ( y , y ) = L − ( n − m ) x . Let R t denote the ray y = ty . Then we need to prove that there exists a small set H of t ′ s such that if y ∈ R t ∩ { y : | y | ∈ [1 , L m ] } , t / ∈ H then | Φ( y ) · · · · · Φ( L ℓ y ) | ≤ e − τℓ . (3.12)We consider only the case t ∈ [0 , y ’s will be such that 0 < y ≤ y ,and as | y | ≥ y ≥ √ .It is very difficult if at all possible for function Φ to satisfy | Φ( y ) | = 1. In fact,looking at (3.11) we can see that | Φ( y ) | ≤ − b dist( y, Z ) ≤ e − b dist( y, Z ) . (3.13)Therefore, we are left to understand that there are few t ’s such that ∃ y ∈ R t , : y ∈ [ 1 √ , L m ] : b · ℓ X k =0 dist( L k y, Z ) ≤ τ ℓ . (3.14)Fix y ∈ R t as above. If (3.14) holds then for 90 per cent of k ′ s one has UFFON’S NEEDLE AND BESICOVITCH IRREGULAR SELF-SIMILAR SETS 15 dist( L k y, Z ) ≤ τ ℓ . (3.15)Denote Z y := { k ∈ [0 , ℓ ] : dist( L k y, Z ) ≤ τ ℓ } . We know that | Z y | ≥ . ℓ . Let us call scenario the collection s := { m ; k , ..., k . ℓ } , where m = 0 , .., m ; 0 ≤ k < ... < k . ℓ .Every t such that there exists y such that (3.14) holds generates several scenariosaccording to y ∈ [ L m − , L m )and according to what is the set [0 , ℓ ] \ Z y —this is the set k , ..., k . ℓ of the scenario.We will calculate the number of scenarios later. Now let us fix a scenario s = { m ; k , ..., k . ℓ } , and let us estimate the measure of the set T ( s ), T ( s ) := { t ∈ (0 ,
1) : ∃ y, y = ty , y ∈ [ L m − , L m ) such that [0 , ℓ ] \ Z y = { k , . . . , k . ℓ } . To dothat for this fixed scenario we fix a net . To explain what is a net we fix a := (cid:20) log η log L (cid:21) + 1 , where η = C τ and C is an absolute constant to be chosen soon.A net is a collection N ( s ) := { n , . . . , n j } , n < n < . . . , where every n i is notamong k j included in the scenario, j ≥ ℓ a + 1, and n i +1 − n i ≥ a . Given a scenario it is always possible to built a net. In fact we just delete from[0 , ℓ ] the numbers k , ..., k . ℓ belonging to the scenario, we are left with at least0 . ℓ numbers. We choose an arithmetic progression with step a (enumerating themanew first). This arithmetic progression will be long enough, its length j ≥ ℓ a because after eliminating k , ..., k . ℓ we still have at least 0 . ℓ numbers left. Wemark the numbers of this progression. Then we put back k , ..., k . ℓ . The markednumbers will form our net.If t ∈ T ( s ) then there exists y = ( y , ty ) as above, in particular,dist( L n i y, Z ) ≤ τ ℓ , ∀ n i ∈ N ( s ) . Let us write that then there exist integers p ≤ q : | L n y − q | < τ , | L n y − p | < τ , so (cid:12)(cid:12)(cid:12)(cid:12) t − p q (cid:12)(cid:12)(cid:12)(cid:12) = (cid:12)(cid:12)(cid:12)(cid:12) L n y L n y − p q (cid:12)(cid:12)(cid:12)(cid:12) = (cid:12)(cid:12)(cid:12)(cid:12) L n y − p + p L n y − q + q − p q (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) ( L n y − p + p ) q − ( L n y − q + q ) p ( L n y − q + q ) q (cid:12)(cid:12)(cid:12)(cid:12) ≤ | L n y − p || q | + | L n y − q || p | q − τ ) q ≤ τ q . As promised we choose C : C = 40, η := 40 τ and we get ∃ p ≤ q : (cid:12)(cid:12)(cid:12)(cid:12) t − p q (cid:12)(cid:12)(cid:12)(cid:12) ≤ η q . (3.16)Next we choose integers p ≤ q : | L n y − q | < τ , | L n y − p | < τ and obtain ∃ p ≤ q : (cid:12)(cid:12)(cid:12)(cid:12) t − p q (cid:12)(cid:12)(cid:12)(cid:12) ≤ η q . (3.17)Notice also that because of | L n y − q | < η , | L n y − q | < η , y ≥ / √ τ , and the fact that n − n ≥ a , we get q q ≥ L a ≥ η . (3.18)We continue in the same vein, i = 2 , . . . , j − ≥ ℓ a : ∃ p i ≤ q i : (cid:12)(cid:12)(cid:12)(cid:12) t − p i q i (cid:12)(cid:12)(cid:12)(cid:12) ≤ η q i . (3.19)Notice also that because of | L n y − q | < η , | L n y − q | < η , y ≥ / √ τ , and the fact that n − n ≥ a , we get q i +1 q i ≥ L a ≥ η . (3.20)Inequality (3.16) gives that | T ( s ) | ≤ η , inequalities (3.16) and (3.17) in conjunc-tion with (3.18) give | T ( s ) | ≤ (cid:18) (cid:19) η , similarly all inequalities (3.19), (3.20)together give | T ( s ) | ≤ (1 . η ) ℓ a ≥ e . ℓ L − ℓ (1 − ǫ ( η )) . Here we used of course that a := (cid:20) log η log L (cid:21) + 1. Finally, if η is sufficiently small wehave | T ( s ) | ≤ L − ℓ . (3.21)Let S denote the set of all scenarios. Now we want to calculate the number ofscenarios. This is easy: S ≤ m · (cid:18) ℓ . ℓ (cid:19) ≤ ℓ · (cid:18) (cid:19) . ℓ · . ℓ . UFFON’S NEEDLE AND BESICOVITCH IRREGULAR SELF-SIMILAR SETS 17
We just proved that the measure of the set of all t ∈ (0 ,
1) such that one has(3.14) ∃ y ∈ R t , : y ∈ [ 1 √ , L m ] : ℓ X k =0 dist( L k y, Z ) ≤ τ ℓ can be estimated as ≤ ℓ · (cid:18) (cid:19) . ℓ · . ℓ · L − ℓ ≤ L − ℓ/ . Proposition 15 is proved. We indeed have very few exceptional directions in thesense that on them | P ♭ ,t ( x ) | is not uniformly smaller than e − τℓ . (cid:3) Here is the second claim of the subsection:
Proposition 16. t ∈ E ⇒ Z SSV ( t ) | P ♯ ,t ( x ) | dx ≤ C ′′ K L m . We will see in Section 5 that for each t , SSV ( t ) is contained in C · L m neighbor-hoods of size L n − m − ℓ around the complex zeroes λ j of P .Fix t . Let I j = [ λ j − L n − m − ℓ , λ j + L n − m − ℓ ] , (3.22)where SSV ( t ) ⊆ [ j I j (3.23)Choose j for which R I j | P ♯ ,t ( x ) | dx is maximized. Then Z SSV ( t ) | P ♯ ,t ( x ) | dx ≤ CL m Z I j | P ♯ ,t ( x ) | dx ≤ CL m ( L ℓ + m − n ) Z I j | n − m − ℓ X k =0 e iα j x | . As | I j | ≤ · L n − m − ℓ , so Lemma 22 and the definition of E give us Proposition 16.The estimate for t ∈ ˜ E \ H follows. If | E | ≥ /K, K = L m/ , | ˜ E | > / K ,and we also just proved that | H | ≤ L − ℓ/ , ℓ = α m with large α , we have a set E ⊂ ˜ E \ H , E > / K , such that for every t ∈ E Z SSV ( t ) | P ( r ) | dr ≤ L − ℓ Z SSV ( t ) | P ♯ ,t ( x )( r ) | dr ≤ C ′′ K L m · L − αm . So we proved Z SSV ( t ) | P ( r ) | dr ≤ c L m (3.24)with c as small as we wish. In particular, Proposition 14 is completely proved. Combinatorial part
In this section, we show how Theorem 1 follows from Theorem 11.First, let us define L θ,N := proj θ G N . (4.1) Theorem 17.
Let β > . (We used β = 3 in the previous section). If t / ∈ E (seedefinition (3.4) ), then |L θ,NK β | ≤ CK . Proof.
Let us use θ instead of t and x for the space variable on the non-Fourierside, since we do not use Fourier analysis in this proof. Fix θ and let F := A ∗ K = { x : f ∗ N ( x ) ≥ K } . We denote by N x the line orthogonal to direction θ and passingthrough x . We can call it needle at x . For every x ∈ F there are at least K discsof size L − r , r = r ( x ) , r ≤ N , intersecting N x . Mark them. Run over all x ∈ F .Consider all marked discs. Consider all L − N -discs that are sub-discs of markedones. Call them “green”. Let U be a family of green discs.We want to show card U ≥ c · K | F | L N , (4.2) | proj ( ∪ q ∈ U q ) | ≤ CK card U L − N , (4.3)Let φ := P q ∈ U χ q . Then Z φ dx = card U L − N . Let M denote uncentered maximal function. To prove (4.3) it is enough to showthat q ∈ U ⇒ proj q ⊂ { x : M φ ( x ) > KC } , and then to use Hardy–Littlewood maximal theorem. But to prove this claim iseasy. In fact, let x ∈ proj q, q ∈ U , then there exists Q –the maximal (by inclusion)marked disc containing q . Consider I := [ x − ℓ ( Q ) , x + 10 ℓ ( Q )]. This segmentcontains the projections of at least K disjoint discs Q := Q, Q , ..., Q K , ... , of thesame sidelength, which intersect N x , where x is a point because of which Q = Q was marked. (The reader should see that x lies really well inside I .) So I containsthe projections of at least ℓ ( Q ) ℓ ( q ) · K green triangles. Whence, Z I φ dx ≥ ℓ ( q ) · ℓ ( Q ) ℓ ( q ) · K ≥ | I | K .
UFFON’S NEEDLE AND BESICOVITCH IRREGULAR SELF-SIMILAR SETS 19 So M φ ( x ) > K .
We proved (4.3).Also we proved that F ⊂ { x : M φ ( x ) ≥ K } . Therefore, by Hardy–Littlewoodmaximal theorem | F | ≤ |{ x : M φ ( x ) ≥ K }| ≤ C R φK = C card U L − N K − . This is (4.2).Let us estimate |L θ,N K α | using (4.2) and (4.3). The first step: |L θ,N | ≤ | proj ( ∪ q ∈ U q ) | + L − N ( L N − card U ) ≤ CK card U L − N + ( L N − card U ) L − N . We do not touch the first term, but we improve the second term by using self-similar structure and going to step 2 N (inside triangles which are not green thereare “green” discs of size L − N ). They are just self-similar copies of the originalgreen discs. Then we have the second step: |L θ,N | ≤ CK card U L − N + the rest ≤ CK card U L − N + ( L N − card U ) CK card U L − N + ( L N − card U ) L − N . Now we leave first two terms alone and having ( L N − card U ) triangles of size L − N we find again “green” discs inside each of those, now green triangles of size L − N . They are just self-similar copies of original green discs.Then we have the third step: |L θ, N | ≤ CK card U L − N + ( L N − card U ) CK card U L − N + the rest ≤ CK card U L − N + ( L N − card U ) CK card U L − N + ( L N − card U ) CK card U L − N +( L N − card U ) L − N . After the l -th step: |L θ,l N | ≤ CK card U L − N (1 + ( L N − card U ) L − N + ... +( L N − card U ) l − L − ( l − N ) + ( L N − card U ) l L − lN . So |L θ,l N | ≤ CK card U L − N (1 − (1 − card UL N ) l )(1 − (1 − card UL N )) + e − card ULN l =: I + II .
Notice that by (4.2) II ≤ e − K | F | l ≤ e − K if the step l is chosen to be l = 1 / | F | ≤ K β .However, we always have I ≤ CK . So Theorem 17 is completely proved. (cid:3) From Theorems 11 and 17, it is not hard to get Theorem 1.5.
Putting
SSV ( t ) into a fixed number of intervals of correct size Now we have to consider P ,t ( r ) = φ t ( r ) φ t ( L − r ) · · · · · φ t ( L − m r ). We are inter-ested in the set SSV ( t ) := { r ∈ [1 , L m ] : | P ,t ( r ) | ≤ L − Am } . We will be using so-called Turan’s lemma:
Lemma 18.
Let f ( x ) = P Ll =1 c l e λ l x , let E ⊂ I , I being any interval. Then sup I | f ( x ) | ≤ e max |ℜ λ n | | I | (cid:18) A | I || E | (cid:19) L sup E | f ( x ) | . Here A is an absolute constant. In this form it is proved by F. Nazarov [16].Now let us consider any square Q = [ x ′ − , x ′ + 1] × [ − , Q theconcentric square of half the size. Lemma 19.
With uniform constant C depending only on L one has sup Q | φ t ( z ) | ≤ C sup Q | φ t ( z ) | . Proof.
Let z = x + iy is a point of maximum in the closure of Q . We first want tocompare | f ( z ) | and | f ( x ) | . Consider f x ( y ) := φ t ( x + iy ). Notice that uniformlyin Q and x | f ′ x ( y ) | ≤ C ( L ) . This means that | f x ( y ) | ≥ | f x (0) | on an interval of uniform length c ( L ).Notice also that the exponents λ l ( t ) , l = 1 , . . . , L, encountered in φ t are alluniformly bounded. Then applying Lemma 18 we get | φ t ( z ) | = | f x ( y ) | ≤ C ′ ( L ) | f x (0) | . UFFON’S NEEDLE AND BESICOVITCH IRREGULAR SELF-SIMILAR SETS 21
Now consider F ( x ) = φ t ( x ). We want to compare F ( x ) = f x (0) = φ t ( x ) withmax [ x ′ − ,x ′ + ] | F ( x ) | . By Lemma 18 we get again | f x (0) | = | F ( x ) | ≤ sup [ x ′ − ,x ′ +1] | F ( x ) | ≤ C ′′ ( L ) sup [ x ′ − / ,x ′ +1 / | F ( x ) | ≤ C ′′ ( L ) sup Q | φ t ( z ) | Combining the last two display inequalities we get Lemma 19 completely proved. (cid:3)
Lemma 20.
With uniform constant C depending only on L (and not on m ) onehas sup Q | φ t ( L − k z ) | ≤ C sup Q | φ t ( L − k z ) | , k = 0 , . . . , m . The proof is exactly the same. We just use L − k λ l ( t ) , l = 1 , . . . , L, encounteredin φ t ( L − k · ) are all uniformly bounded.By complex analysis lemmas from Section 6 we know that Lemma 20 impliesthat every Q has at most M (depending only on L ) zeros of φ t ( z ). And if wedenote them by µ , . . . , µ M then { x ∈ Q ∩ R : | φ t ( x ) | ≤ L − Mℓ } ⊂ ∪ Mi =1 B ( µ i , L − ℓ ) . (5.1)Consider µ , . . . , µ S being all zeros of P ,t in [1 / , L m + 1] × [1 / , / S ≤ M ( L ) L m . From (5.1) it is immediate that { x ∈ [1 , L m ] : | P ,t ( L − ( n − m ) x ) | = | φ t ( x ) ·· · ·· φ t ( L − m ( x )) | ≤ L − Mℓm } ⊂ ∪
M L m i =1 B ( µ i , L − ℓ ) . (5.2)Changing the variable y = L n − m x we get the structure of the set of small valuesused above during the proof of Proposition 16: SSV ( t ) ⊂ ∪ C L m i =1 I i , (5.3)where each interval I i has the length 2 L n − m − ℓ . Some important standard lemmas. A bit of complex analysis
There are a few important lemmas which we have appealed to repeatedly. Thefirst claim, Lemma 21, uses the Carleson imbedding theorem. A stronger version,Lemma 22, uses general H theory. Its importance lies in its ability to establisha key relationship between the level sets of f n,t and the L norm of d f n,t . This isbecause the Fourier transform changes the centers of intervals into the frequenciesof an exponential polynomial.The second claim we split into Lemmas 24 and 25. Given a bounded holomorphicfunction on the disc, its supremum, and an interior non-zero value, these lemmasbound the number of zeroes and contain the set of small values within certainneighborhoods of these zeroes.6.1. In the spirit of the Carleson imbedding theorem.Lemma 21.
Let j = 1 , , ...k , c j ∈ C , | c j | = 1 , and α j ∈ R . Let A := { α j } kj =1 .Then Z | k X j =1 c j e iα j y | dy ≤ C k · sup I a unit interval { A \ I } . Proof.
Let A := { µ = α + i : α ∈ A } . Let ν := P µ ∈ A δ µ . This is a measure in C + . Obviously its Carleson constant k ν k C := sup J ⊂ R , J is an interval ν ( J × [0 , | J | ]) | J | can be estimated as follows k ν k C ≤ I a unit interval { A \ I } . (6.1)Recall that ∀ f ∈ H ( C + ) Z C + | f ( z ) | dν ( z ) ≤ C k ν k C k f k H , (6.2)where C is an absolute constant. Now we compute Z | k X j =1 c j e iα j y | dy ≤ e Z | k X j =1 c j e i ( α j + i ) y | dy ≤ e Z ∞ | k X j =1 c j e i ( α j + i ) y | dy = e Z R | X µ ∈ A c µ x − µ | , where c µ := c j for µ = α j + i . The last equality is by Plancherel’s theorem. UFFON’S NEEDLE AND BESICOVITCH IRREGULAR SELF-SIMILAR SETS 23
We continue Z R | X µ ∈ A c µ x − µ | = sup f ∈ H ( C + ) , k f k ≤ (cid:12)(cid:12)(cid:12)(cid:12) h f, X µ ∈ A c µ x − µ i (cid:12)(cid:12)(cid:12)(cid:12) =4 π sup f ∈ H ( C + ) , k f k ≤ | X µ ∈ A c µ f ( µ ) | ≤ C { A } sup f ∈ H ( C + ) , k f k ≤ X µ ∈ A | f ( µ ) | ≤ C { A } sup f ∈ H ( C + ) , k f k ≤ Z C + | f ( z ) | dν ( z ) ≤ C C { A } sup I a unit interval { A \ I } . This is by (6.7) and (6.1). The lemma is proved. (cid:3)
Now we are going to prove a stronger assertion by a simpler approach. Thisstronger assertion is what is used in the main part of the article.
Lemma 22.
Let j = 1 , , ...k , c j ∈ C , | c j | = 1 , and α j ∈ R . Let A := { α j } kj =1 .Then Suppose Z R ( X α ∈ A χ [ α − ,α +1] ( x )) dx ≤ S , (6.3)
Then there exists an absolute constant C Z | X α ∈ A c α e iαy | dy ≤ C S . (6.4)Of course, one can change variables and get:
Corollary 23.
Let j = 1 , , ...k , c j ∈ C , | c j | = 1 , and α j ∈ R . Let A := { α j } kj =1 ,and let δ > . Suppose Z R ( X α ∈ A χ [ α − δ,α + δ ] ( x )) dx ≤ S , (6.5)
Then there exists an absolute constant C Z a + δ − a | X α ∈ A c α e iαy | dy ≤ C S /δ . (6.6) Remark.
Lemma 22 is obviously stronger than Lemma 21. In fact, let S be themaximal number of points A in any unit interval. Then f ( x ) := X α ∈ A χ [ α − ,α +1] ( x ) ≤ S . Now R R f ( x ) dx ≤ kS , where k as above is the cardinality of A . We can put now S := 4 kS , apply Lemma 22 and get the conclusion of Lemma 21. The proof ofLemma 22 does not require the Carleson imbedding theorem. Here it is. Proof.
Using Plancherel’s theorem we write Z | X α ∈ A c α e iα y dy | ≤ e Z | X α ∈ A c α e i ( α + i ) y dy | ≤ e Z ∞ | X α ∈ A c α e i ( α + i ) y dy | = e Z R (cid:12)(cid:12)(cid:12)(cid:12) X α ∈ A c α α + i − x (cid:12)(cid:12)(cid:12)(cid:12) dx . Recall that H ( C + ) is orthogonal to H ( C + ) (6.7)Now we continue Z R (cid:12)(cid:12)(cid:12)(cid:12) X α ∈ A c α α + i − x (cid:12)(cid:12)(cid:12)(cid:12) dx ≤ Z R (cid:12)(cid:12)(cid:12)(cid:12) X α ∈ A c α α + i − x − X α ∈ A c α α − i − x (cid:12)(cid:12)(cid:12)(cid:12) dx = π Z R (cid:12)(cid:12)(cid:12)(cid:12) X α ∈ A c α P ( α − x ) (cid:12)(cid:12)(cid:12)(cid:12) dx , where P is the Poisson kernel in the half-plane C + at hight h = 1: P h ( x ) := 1 π hh + x . We continue by noticing that P ∗ χ [ λ − ,λ +1] ( x ) ≥ c P ( λ − x ) with absolute positive c . This is an elementary calculation, or, if one wishes, Harnack’s inequality. Nowwe can continue Z | X α ∈ A c α e iα y dy | ≤ πe c Z R (cid:12)(cid:12)(cid:12)(cid:12) ( P ∗ X α ∈ A c α χ [ α − ,α +1] )( x ) (cid:12)(cid:12)(cid:12)(cid:12) dx . Now we use the fact that f → P ∗ f is a contraction in L ( R ). So Z | X α ∈ A c α e iα y dy | ≤ πe c Z R | X α ∈ A c α χ [ α − ,α +1] ( x ) | dx ≤ C S .
The lemma is proved. (cid:3)
UFFON’S NEEDLE AND BESICOVITCH IRREGULAR SELF-SIMILAR SETS 25
A Blaschke estimate.Lemma 24.
Let D be the closed unit disc in C . Suppose φ is holomorphic inan open neighborhood of D , | φ (0) | ≥ , and the zeroes of φ in D are given by λ , λ , ..., λ M . Let C = || φ || L ∞ ( D ) . Then M ≤ log ( C ) . Proof.
Let B ( z ) = M Y k =1 z − λ k − ¯ λ k z . Then | B | ≤ D , with = on the boundary. If we let g := φB , then g is holomorphicand nonzero on D, and | g ( e iθ ) | ≤ C ∀ θ ∈ [0 , π ]. Thus | g (0) | ≤ C by the maximummodulus principle. So we have C ≥ | g (0) | = | φ (0) || B (0) | ≥ M Y k =1 | λ k | ≥ M . (cid:3) Lemma 25.
In the same setting as Theorem 24, the following is also true for all δ ∈ (0 , / : { z ∈ D : | φ | < δ } ⊆ S ≤ k ≤ M B ( λ k , ε ) , where ε := 916 (3 δ ) /M ≤
916 (3 δ ) /log ( C ) . Proof.
Let δ ∈ (0 , / z ∈ D such that | z − λ k | > ε ∀ k . Note that g isharmonic and nonzero on D with | g (0) | ≥ M . Thus Harnack’s inequality ensuresthat | g | ≥ M on D , so there | φ ( z ) | ≥ | g ( z ) B ( z ) | ≥
13 2
M M Y k =1 | z − λ k − ¯ λ k z | ≥ ( 16 ε M
13 = δ. We can conclude the proof by the contrapositive. (cid:3) Combinatorial theorem
For this section, regard the set E from Section 3 as parameterized by θ , and usethe variable x instead of s on the non-Fourier side, since we will not work on theFourier side at all during this section. Theorem 26.
Let θ ∈ E . Then max n :0 ≤ n ≤ N k f n,θ k L ( R ) ≤ C K .
To prove this we first need the following claim, which is the main combinatorialassertion of this article. It repeats the one in [17] but we give a slightly differentproof.We fix a direction θ , we think that the line ℓ θ on which we project is R . If x ∈ R then by N x we denote the line orthogonal to R and passing through point x , wecall N x a needle. By F L we denote { x ∈ R : f ∗ N ( x ) := max ≤ n ≤ N f n,θ ( x ) > L } (alsoknown as A ∗ L ). Theorem 27.
There exists an absolute constant C such that for any large K and M | F KM | ≤ C K | F K | · | F M | . (7.1) Proof.
This will be a proof by greedy algorithm. First choose y ∈ F K and considerneedle N y and discs of certain size L − j y , j y ≤ N intersecting N y . Consider anyfamily of this sort having more than 4 K elements. Fix such a family. We will“fathorize” it, i.e. we consider the father of each element in the family. Two thingsmay happen: 1) there are more than 4 K distinct fathers; 2) number of fathers isat most 4 K . In the latter case the number of fathers is at least 2 K . In fact, weslash the number of elements by fathorizing, but not more than by factor of 1 / x ∈ F K and all families of cardinality biggerthan 4 K of equal size discs intersecting needle N x we come to some awfully com-plicated set of discs. But we will consider now maximal-by-inclusion discs of thisfamily, the family of these maximal discs is called F .Choose disc Q ∈ F such that its sidelength ℓ ( Q ) is maximal possible in F .It is very important to notice that F contains at least 2 K − Q pierced by a needle N y . This is because of maximality of the lengthsize,the stack pierced by N y could not be eaten up even partially by bigger in sizediscs from some other stack. So let us call by Q , ..., Q K − , ..., Q S , S ≥ K − Q and all intersect a certain needle N y .Denote I = proj Q . Consider all q ∈ F such thatproj q ∩ I = ∅ . UFFON’S NEEDLE AND BESICOVITCH IRREGULAR SELF-SIMILAR SETS 27
Call them F ( Q ). Of course ℓ ( q ) ≤ ℓ ( Q ). For every such q consider a Cantorsquare Q , q ⊂ Q , such that ℓ ( Q ) = ℓ ( Q ). Such Q ’s form family ˜ F ( Q ). Lemma 28.
For every y ∈ R the needle N y intersects at most K discs of thefamily ˜ F ( Q ) .Proof. Suppose contrary. Then N y intersects more than 4 K of discs from ˜ F ( Q ).So y ∈ F K , and our pierced family is one of those which we considered at thebeginning. It can be fathorized. Then the square of size ≥ ℓ ( Q ) will be presentin F . Contradiction with maximality of length. (cid:3) Lemma 29. card ˜ F ( Q ) ≤ K .
Proof. card ˜ F ( Q ) · ℓ ( Q ) = X Q ∈ ˜ F ( Q ) ℓ ( Q ) ≤ Z I card { Q ∈ ˜ F ( Q ) : Q ∩ N y = ∅} dy ≤ K · ℓ ( Q ) . This is by Lemma 28. (cid:3)
Lemma 30.
There exists an interval J ⊂ I y such that | J | ≥ c · | I | with a certainabsolute positive c . And J ⊂ F K .Proof. We already noticed that Q , Q , ..., Q K − intersect needle N y . Then atleast half of them have their center of symmetry to the right of N y , or at least halfof them have their center of symmetry to the left of N y . Assume that the firstcase occurs. Then the segment [ y , c · ℓ ( Q )] obviously is contained in F K . (cid:3) Lemma 31. | F KM ∩ I | ≤ C K ℓ ( Q ) | F M | = C K | I || F M | . Proof.
Of course F KM ⊂ F K . For y ∈ F KM ∩ I the whole family of small discswhose quantity is > KM intersecting N y will be inside one of those Q ∈ ˜ F ( Q ),whose number is at most 88 K by Lemma 29. Let us enumerate Q , ..., Q s , s ≤ K elements of ˜ F ( Q ). So there exists i = 1 , ..., s such that y ∈ dilated copy of F M in proj Q i . Hence F KM ∩ I ⊂ ∪ Ki =1 dilated copy of F M in proj Q i . So | F KM ∩ I | ≤ K X i =1 ℓ ( Q i ) | F M | ≤ K ℓ ( Q ) | F M | . (cid:3) Lemma 32. | F KM ∩ I | ≤ c − K | F m | · | J | . Now we want to repeat all steps for F K := F K \ I . So we fathorize discspierced by needles N x , x ∈ F K . As before we get families F , maximal sidelengthtriangle Q , families F ( Q ), ˜ F ( Q ). Notice that F < F in the sense that forevery q ∈ F there exists q ∈ F such that q is contained in Q . It is also clear that ℓ ( Q ) ≤ ℓ ( Q ) . Obviously Q , Q , ... are not in F , their projections even do not intersect R \ I .There are at least 2 K − Q : Q , ..., Q K − , ... in F such thatthey are of the same size ℓ ( Q ) and they (and Q ) intersect the same needle N y , y ∈ R \ I . This is again the maximality of the sidelength among F discs. Let I := proj Q . Notice that I ∩ I = ∅ . In fact, y ∈ I , y / ∈ I , Q size is much smaller than 20 | I | . We consider all q ∈ F such that proj q ∩ (20 I \ I ) = ∅ . Call this family F ( Q ). For every q ∈ F ( Q ) consider Cantor disc Q containing q and of the size ℓ = ℓ ( Q ). Maximal-by-inclusion among such Q ’s form ˜ F ( Q ). Lemma 33.
For any y ∈ R \ I , N y intersects at most K discs of ˜ F ( Q ) .Proof. Suppose contrary. Then there exists y ′ ∈ F K ∩ ( R \ I ), and a subfamilyof ˜ F ( Q ) of cardinality bigger than 4 K intersects N y ′ . It can be fathorized. Thendiscs of size ≥ ℓ ( Q ) would belong to F . This contradicts the maximality of ℓ ( Q ). (cid:3) Lemma 34.
For any z ∈ R , N z intersects at most K discs of ˜ F ( Q ) . UFFON’S NEEDLE AND BESICOVITCH IRREGULAR SELF-SIMILAR SETS 29
Proof.
Suppose contrary. Then there exists z ∈ F K , and a subfamily of ˜ F ( Q ) ofcardinality bigger than 4 K intersects N z . Now there is an end-point of 20 I \ I (call it a ), which is closest to z . Let it be on the right of z . Then another end-point is also on the right but farther away. As every triangle from the family hasa) z in its projection, and b) a certain point to the right of a in its projection(their projections intersect 20 I \ I –by definition), then all of them have a inits projection. Let us be lavish and say that 50 percent of them have a in theirprojection (the fact is that it is not lavishness, it is necessity: next step will be toconsider in the future 20 I \ (20 I ∪ I ), and their can be 2 closest points to z :one on the left, say, b , and one on the right, say, a , and we can guarantee that 50percent of our discs have either b or a in their projections simultaneously). We usethe previous Lemma 33, and get that this 5) percent is ≤ K . So we are done. (cid:3) Lemma 35. card ˜ F ( Q ) ≤ K .
Proof. card ˜ F ( Q ) · ℓ ( Q ) = X Q ∈ ˜ F ( Q ) ℓ ( Q ) ≤ Z I card { Q ∈ ˜ F ( Q ) : Q ∩ N y = ∅} dy ≤ K · ℓ ( Q ) . This is by Lemma 28. (cid:3)
Lemma 36.
There exists an interval J ⊂ I , | J | ≤ c · | I | , such that J ⊂ F K .Proof. The same proof as for Lemma 30. (cid:3)
Lemma 37. | F KM ∩ I | ≤ C K ℓ ( Q ≤ C | , K | I | . Proof.
The same proof as for Lemma 31. (cid:3)
Combining Lemmas 36, 37 we get
Lemma 38. | F KM ∩ I | ≤ C c − K | J | . We continue by introducing F KM = F KM \ (20 I ∪ I ) . We repeat the whole procedure. There will be I , J ⊂ I ∩ F K , | J | ≥ c · | I | : I ∩ ( I ∪ I ) = ∅ , | F KM ∩ I | ≤ Cc − K | J || F M | , et cetera.Finally, | F KM | ≤ | F KM ∩ I | + | ( F KM \ I ) ∩ I | + ... + | ( F KM \ I ∪ I ∪ .... I j − ) ∩ I j | + ... ≤ C ′ K | F M | ∞ X j =0 | J j | ≤ C ′ K | F M | | F K | . We are done with Theorem 27. (cid:3)
Now we can prove Theorem 26.
Proof.
Let E j := { x : f n,θ ( x ) > (4 K ) j +1 } , j = 0 , , .... . We know by Theorem 27that | E j | ≤ ( CK ) j | E | j +1 . Hence, Z f n,θ ( x ) dx ≤ K Z f n,θ ( x ) dx + ∞ X j +0 Z E j \ E j +1 f n,θ ( x ) dx ≤ CK + (4 K ) j +2 ( CK ) j | E | j +1 . If |{ x : f ∗ N ( x ) > K }| ≤ /K τ then for all n ≤ N we can immediately read theprevious inequality as Z f n,θ ( x ) dx ≤ C ( τ ) K . (cid:3) Discussion
Difficulties for more general self-similar sets.
The reason we were ableto prove the stronger estimate for the Sierpinski gasket is exactly given by (2.2).It is a quantified version of the fact that the three-term sum ϕ ( z ) = 1 + e iz + e itz is zero if and only if the summands are e jπi , j = 0 , ,
2, and that for such z , ϕ (3 k z ) = 3 for all integers k ≥
1. An alternate argument using this fact in thisform is employed in [6]. Both versions of this fact we call by the general term“analytic tiling”.
UFFON’S NEEDLE AND BESICOVITCH IRREGULAR SELF-SIMILAR SETS 31
But there cannot be such a thing in the general case. Suppose we had 5 self-similarities, and that for for some direction θ , we had φ θ ( x ) = 1 + ( − i ) + i + e πi/ + e πi/ = 0. Then clearly, taking fifth powers of the summands results inanother zero with exactly the same summands, in complete and utter contrast tothe three-point case. Similar examples using partitions into relatively prime rootsof unity exist for numbers other than 5.Though perhaps there is some hope that for arbitrary sets, some other formof analytic tiling occurs for typical directions in the arbitrary case (with smallmeasure of exceptional directions). Ergodic theory may be of central importance.For example, if one considers K n as in [17], one gets ϕ θ ( z ) = 1+ e iπz + e iλz + e i ( λ + π ) z ,which has the zero z = 1. Then ϕ (4 k ) = 2(1 + cos (4 k λ )) for k > λ dependscontinuously on θ , and for fixed λ such an ergodic sampling results in a sequence a k := ϕ (4 k ), and either:1: a k is eventually periodic and non-zero,2: a k takes values other than 4 only finitely often,or 3 (the case for almost every λ ): 4 k λ mod 2 π evenly samples [0 , π ] over thelong term, with long-term average N P Nk =1 a k → N → ∞ . References
1. M. Bateman, A.Volberg,
An estimate from below for the Buffon needle probability of the four-corner Cantor set , arXiv:math. 0807.2953v1, 2008, pp. 1-11.2. M. Bateman, N.Katz,
Kakeya sets in Cantor directions , arXiv:math. 0609187v1, 2006, pp.1–10.3. M. Bateman,
Kakeya sets and the directional maximal operators in the plane ,arXiv:math.CA 0703559v1, 2007, pp. 1–20.4. A. S. Besicovitch,
Tangential properties of sets and arcs of infinite linear measure , Bull. Amer.Math. Soc. (1960), 353–359.5. M. Bond, A. Volberg : The power law for Buffon’s needle landing near the Sierpinski gasket ,arXiv: 0911.0233v2, 2009, pp. 1–35.6.
M. Bond, A. Volberg : Buffon needle lands in ǫ -neighborhood of a -Dimensional SierpinskiGasket with probability at most | log ǫ | − c . Comptes Rendus Mathematique, Volume 348, Issues11-12, June 2010, 653–656.7. M. Bond, A. Volberg : Estimates from below of the Buffon noodle probability for under-cooked noodles , arXiv:math/0811.1302v1, 2008, pp. 1–10.8. K. J. Falconer, The geometry of fractal sets. Cambridge Tracts in Mathematics, 85. C.U.P.,Cambridge–New York, (1986).
9. U. Keich, On L p bounds for Kakeya maximal functions and the Minkowski dimension in R , Bull. London. Math. Soc. (1999), pp. 213–221.10. R. Kenyon, Projecting the one-dimensional Sierpinski gasket, Israel J. Math. (1997), 221–238.11. I. Laba, K. Zhai, Favard length of product Cantor sets , arXiv:0902:0964v1, Feb. 5 2009.12. J. C. Lagarias and Y. Wang,
Tiling the line with translates of one tile , Invent. Math. (1996), 341–365.13. P. Mattila,
Orthogonal projections, Riesz capacities and Minkowski content , Indiana Univ.Math. J. (1990), 185–198.14. P. Mattila, Hausdorff dimension, projections, and the Fourier transform , Publ. Mat., (2004), pp. 3–48.15. P. Mattila, Geometry of Sets and Measures in Euclidean Spaces , Cambridge University Press,1995.16. F. Nazarov,
Local estimates of exponential polynomials and their applications to inequalitiesof uncertainty principle type , St Petersburg Math. J., v. 5 (1994), No. 4, pp. 3–66.17. F. Nazarov, Y. Peres, A. Volberg
The power law for the Buffon needle probability of the four-corner Cantor set , arXiv:0801.2942, 2008, pp. 1–15.18. Y. Peres, K. Simon and B. Solomyak,
Self-similar sets of zero Hausdorff measure and positivepacking measure , Israel J. Math. (2000),353–379.19. Y. Peres and B. Solomyak,
How likely is Buffon’s needle to fall near a planar Cantor set?Pacific J. Math. 204 , 2 (2002), 473–496.20. I. J. Schoenberg,
On the Besicovitch–Perron solution of the Kakeya problem , Studies in math-ematical analysis and related topics ,21. T. Tao,
A quantitative version of the Besicovitch projection theorem via multiscale analysis, pp. 1–28, arXiv:0706.2446v1 [math.CA] 18 Jun 2007.
Matthew Bond, Dept. of Math., Michigan State University. [email protected]