circular favard length of the four-corner cantor set

J Geom Anal (2011) 21: 40–55DOI 10.1007/s12220-010-9141-4

Circular Favard Length of the Four-Corner Cantor Set

Matthew Bond · Alexander Volberg

Received: 1 August 2009 / Published online: 8 July 2010© Mathematica Josephina, Inc. 2010

Abstract Let Cn be the n-th generation in the construction of the middle-half Cantorset. The Cartesian square Kn of Cn consists of 4n squares of side-length 4−n. Wedrop a circle of radius r on the plane and try to estimate from below the conditionalprobability of this circle to intersect Kn if it already intersects a disc containing Kn.If the radius is very large ≈4n then clearly this should not differ too much from theusual Buffon needle probability. But it turns out that the best known lower bound(Bateman and Volberg in arXiv:0807.2953, 2008) persists even when the radius ismuch smaller than this—r > Cnε suffices—and the intersection probability is at leastCε logn

n. This suggests that the method of Bateman and Volberg (arXiv:0807.2953,

2008) may be of use in proving a certain estimate for the lacunary circular maximalfunction from Seeger et al. (Preprint, 2005).

Keywords Fractals · Length · Area · Volume · Other geometric measure theory

Mathematics Subject Classification (2000) Primary 28A80 · Secondary 28A75

1 Introduction

The four-corner Cantor set K is constructed by replacing the unit square by four sub-squares of side-length 1/4 at its corners, and iterating this operation in a self-similar

Communicated by Loukas Grafakos.

Research of the authors was supported in part by NSF grants DMS-0758552 (Volberg).

M. Bond (�) · A. VolbergDept. of Math., MSU, East Lansing, MI, USAe-mail: [email protected]

A. Volberge-mail: [email protected]

http://arxiv.org/abs/arXiv:0807.2953


mailto:[email protected]

mailto:[email protected]

Circular Favard Length of the Four-Corner Cantor Set 41

Fig. 1 K3, the third stage of theconstruction of K

manner in each sub-square. More formally, consider the set Cn that is the union of 2n

segments:

Cn =⋃

aj ∈{0,3},j=1,...,n

[n∑

j=1

aj 4−j ,

n∑

j=1

aj 4−j + 4−n

],

and let the middle-half Cantor set be

C :=∞⋂

n=1

Cn.

The four-corner Cantor set K is the shifted Cartesian square of C × C . Namely,K = C × C − (1/2,1/2).

We will consider its approximations Kn = Cn × Cn − (1/2,1/2).The center of symmetry of Kn, K is the origin.Since the one-dimensional Hausdorff measure of K satisfies 0 < H1(K) < ∞ and

the projections of K in two distinct directions have zero length, a theorem of Besi-covitch (see [5, Theorem 6.13]) yields that the projection of K to almost every linethrough the origin has zero length. This is equivalent to saying that the Favard lengthof K equals zero. Recall (see [4, p. 357]) that the Favard length of a planar set E isdefined by

Fav(E) = 1

π

∫ π

0|Proj RθE|dθ, (1.1)

where Proj denotes the orthogonal projection from R2 to the horizontal axis, Rθ

is the counterclockwise rotation by angle θ , and |A| denotes the Lebesgue measureof a measurable set A ⊂ R. The Favard length of a set E in the unit square hasa probabilistic interpretation: up to a constant factor, it is the probability that the“Buffon’s needle,” a long line segment dropped at random, hits E (more precisely,suppose the needle’s length is infinite, pick its direction uniformly at random, andthen locate the needle in a uniformly chosen position in that direction, at distance atmost

√2 from the center of the unit square).

The set Kn = C 2n is a union of 4n squares with side-length 4−n (see Figure 1

for a picture of K3). By the dominated convergence theorem, Fav(K) = 0 implieslimn→∞ Fav(Kn) = 0. We are interested in good estimates for Fav(Kn) as n → ∞.A lower bound Fav(Kn) ≥ c

nfor some c > 0 follows from Mattila [7, 1.4]. Peres

42 M. Bond, A. Volberg

and Solomyak [9] proved that

Fav(Kn) ≤ C exp[−a log∗ n] for all n ∈ N,

where

log∗ n = min{k ≥ 0 : log log · · · log︸︷︷︸

k

n ≤ 1}.

This result can be viewed as an attempt to make a quantitative statement out of aqualitative Besicovitch projection theorem [4, 12], using this canonical example ofthe Besicovitch irregular set.

It is very interesting to see what are quantitative analogs of the Besicovitch Theo-rem in general. The reader can find more of that in [12].

In [8] the following estimate from above was obtained

Fav(Kn) ≤ Cτ

nτ, (1.2)

where τ was strictly less than 1/6. This can be slightly improved, but it is still a longway till τ = 1. It was shown in [3], using the idea of [1, 2], that τ = 1 is impossible.

Theorem 1 There exists c > 0 such that

Fav(Kn) ≥ clogn

nfor all n ∈ N. (1.3)

Remark Result (1.2) has been extended to product Cantor sets in Laba-Zhai’s pa-per [6].

In the present article we extend (1.3) to another geometric setting: when we countnot the straight needles (lines) intersecting Kn, but rather circles (noodles) intersect-ing it. It is easy to see that circles of radius 4n or bigger are just as lines for thisproblem.

In fact, let us have a compact set K of diameter ≈1 and which is at distance ≈1from the origin. Parameterize each ray from the origin by the angle θ it makes withthe positive x-axis. Let C be a circle of radius 4n, whose center lies on a ray in theθ direction, and call the line containing this ray Lθ . Let P ∈ K ∩ C. We can projectP on Lθ along the circle C; let the circular projection be A. We can also considerthe usual orthogonal projection of P on Lθ ; let it be B . Then it is easy to see that|A − B| ≤ c4−n.

Now suppose that K = Kn, which consists of squares Qnj of diameter ≈ 4−n. So

we just saw that circular and orthogonal projections of each of these squares (callthem an

j (θ), bnj (θ) correspondingly) on any Lθ satisfy an

j (θ) ⊂ �bnj (θ) and bn

j (θ) ⊂�an

j (θ) for a certain � independent of n, θ .Hence (1.3) automatically implies the same estimate for the mean value of the

length of circular projections of Kn if the radius of the circle is ≈4n or greater.But we prove that the circular version of (1.3) is true even for much smaller radii.

We do not know the answer for r growing slower than nε for all ε > 0.


2 Geometric Setting

In addition to considering the dρ dθ measure of the set of centers of circles intersect-ing Kn, we may also naturally be interested in the ρ dρ dθ measure of this set, that isits area. Indeed, since r is much larger than the diameter of Kn, ρ ≈ r , and one maymultiply by a constant whenever one wishes to find estimates in one setting ratherthan the other.

Let Cr(z) := {z + reiθ : θ ∈ [0,2π]}. We are interested in the Lebesgue planemeasure of the set Gn,r := {z : Cr(z) ∩ Kn �= ∅}.

|Gn,r | =∫ 2π

0

∫ ∞

0χGn,r (ρeiθ )ρ dρ dθ ≥ (r − 2)

∫ 2π

0

∫ r+2

r−2χGn,r (ρeiθ ) dρ dθ.

The last integrand above (excluding the r − 2) can be thought of as a small trans-lated distortion of the integrand of Fav(Kn). We will describe the distortion and showthat for r large enough, Kn is sufficiently coarse for the argument of [3] to yield thesame lower bound.

To this end, let T : C × S1 → C, where we will write for convenience

Tθ (z) := T (z, eiθ ).

Then define, for any E ⊂ C,

FavT (E) := 1

2π

∫|Projθ (Tθ (E))|dθ. (2.1)

It remains to write∫

χGn,r (ρeiθ ) dρ = |Projθ (σθ (Kn))|for the appropriate choice of σ . The following lines accomplish this, and some of theobjects defined therein are illustrated in Figure 2.

Define

Fr(y) :={

r − √r2 − y2, |y| ≤ 2,

r − √r2 − 4, otherwise.

(2.2)

Then define σ0(x, y) := (x − Fr(y), y), and σθ := R−θ ◦ σ0 ◦ Rθ , where Rθ isclockwise rotation by the angle θ (the symbol σθ suppresses F and r from the nota-tion, but we will refer to them explicitly as needed). Then (ρ + r)eiθ ∈ Gn,r if andonly if σθ carries some point of Kn to the line perpendicular to line θ at ρeiθ , i.e.,χGn,r ((ρ + r)eiθ ) = χProjθ (σθ (Kn))(ρeiθ ).

Thus

|Gn,r | ≥ 2π(r − 2)Favσ (Kn). (2.3)

For any real function g, we may define σgθ from g in the same manner that we

defined σθ from F : σg

0 (x, y) := (x − g(y), y). Then σgθ := R−θ ◦ σ0 ◦ Rθ . Any real

function g which plays such a role will be called a noodle.We are now ready to state the main result of this paper in this setting:


Fig. 2 z and z′, and theirimages under σθ .Projθ σθ (z) = Projθ σθ (z′) =ρeiθ . Note that if z ∈ Kn orz′ ∈ Kn, then (ρ + r)eiθ ∈ Gn,r

We say that the family of cookings Tn, where Tn : C × S1 → C, undercook theplane if there is some constant C such that FavTn(Kn) ≥ C

lognn

for all n. Likewise, wesay that the noodles Frn are undercooked if the cookings σFrn undercook the plane.(We choose the word “undercooked” to indicate that the noodles are very close tobeing straight lines.) We emphasize that we do not make any definition recognizingwhether any single cooking is an undercooking or not—we are interested in the decayrate of FavTn as a function of n.

Theorem 2 If rn ≥ Cnε , then the noodles Frn are undercooked.

Corollary If r ≥ Cnε , then |Gn,r | ≥ C′ lognn

· r , where C′ depends only on C and ε.

Actually, we still have some control even when rn < Cnε .

Theorem 3 If 5 < rn < C′n, then

Favσ (Kn) ≥ C log rn

n.

We will prove Theorem 2 with ε = 1 for simplicity, but a quick remark afterwardwill show how to get Theorem 3 with no extra difficulty.

3 Needle Probability Estimate of [3] Revisited

Let us review briefly the argument of [3] which proves that Fav(Kn) ≥ C lognn

. Below,we will fix an n, and none of the constants will depend on n. We rotate the axes,defining θ = 0 to be the direction arctan(1/2), because Kn projects onto this directionnicely: the projected squares together fill out a single connected interval, and theprojected squares intersect only on their endpoints. These almost-disjoint projectedintervals induce a 4-adic structure on the interval. Since 0 is a direction with a largeprojection, the idea is that as θ gets exponentially farther from 1/n ≈ 0, the average


projections of Kn should shrink—but [3] shows that this happens no more quicklythan the same exponential rate.

For each square Q of size 4−n in Kn, χQ,θ (x) is the characteristic function of theprojection onto the direction θ . Put

fn,θ (x) =∑

Q,(Q)=4−n

χQ,θ (x).

That is, fn,θ (x) denotes the number of squares of length 4−n whose orthogonal pro-jection onto the line Lθ contains the point x of this line. Let us denote the support offn,θ (x) by En,θ , and let |En,θ | denote its length.

Let Jj := (arctan(4−j ), arctan(4−j+1)). (The count starts from the special direc-tion chosen above.) The central computation of [3] centers around a partitioning ofan estimate of Fav(Kn) into conical neighborhoods Jj × R:

∫

Jj

|En,θ |dθ ≥(∫Jj

∫fn,θ dx dθ)2

∫Jj

∫f 2

n,θ dx dθ.

Here we used the Cauchy inequality on fn,θ and χEn,θ .Trivially,

∫Jj

∫fn,θ dx dθ ≥ C4−j . The interesting part of [3] amounts to showing

that our partition has been chosen such that we may conclude that∫Jj

∫f 2

n,θ dx dθ ≤Cn4−2j for the approximately logn many values of j (3 < j < logn), so that∫Jj

|En,θ |dθ > C/n, and summing over j yields Fav(Kn) ≥ C lognn

. Now

f 2n,θ =

∑

Q,Q′χQ,θχQ′,θ =

∑

Q �=Q′χQ,θχQ′,θ +

∑

Q

χ2Q,θ .

Integrating over Jj × R, the latter diagonal sum becomes C4−j ≤ Cn4−2j

(the inequality uses j < logn). When estimating the other integral, things becomecombinatorial—most of these terms are identically 0 in Jj × R. So define Aj,k to bethe set of pairs P = (Q,Q′) of Cantor squares such that in our special coordinate sys-tem, the centers q and q ′ of Q and Q′ have vertical distance 4−k−1 ≤ |yq −yq ′ | ≤ 4−k

and satisfy the condition on horizontal spacing 4−j−1 ≤ | xq−xq′yq−yq′ | ≤ 4−j . We can think

of 4−j as being tan(θ) for θ such that the squares Q,Q′ overlap in the projectiononto θ . In [3], it was proved that

|Aj,k| ≤ C42n−k−2j . (3.1)

For any (j, k) pair, it is immediate that the integral νP := ∫ 2π

0

∫R

χQ,θχQ′,θ dx dθ

satisfies νP ≤ 4k−2n, and the integrand is supported only for angles belonging toJj−1, Jj , and Jj+1. So we fix j and sum over k to get

∫

Jj ×R

∑

Q �=Q′χQ,θχQ′,θ dθ dx


≤n−j+1∑

k=1

max{νP : P ∈ Aj ′,k for some j ′ = j − 1, j, j + 1}

× (|Aj−1,k| + |Aj,k| + |Aj+1,k|)≤ Cn4−2j .

(Note above that j + k ≤ n, since 4−j−k bounds the horizontal distance betweencenters of squares from above.)

This completes the proof of the result in [3]. We will need to remember some ofthe notations for later, and estimate (3.1).

4 The Main Result

Now we show that the σθ in the integrand of Favσ (Kn) hardly disturbs the angularsorting argument described above. We will need the following estimate on |F ′

r (y)|:

|F ′r (y)| ≤ 4

r(4.1)

because it gives us

Lip(σθ − Id) ≤ 4

r(4.2)

when we conjugate σ0 by the isometry Rθ .

Lemma 4 Let ε > 0 be small enough. Let T : C → C be such that Lip(T − Id) < ε.Then ∀z,w ∈ C,

|arg(z − w) − arg(T (z) − T (w))| < 2ε

(for appropriate choices of arg).

Proof Write z − w = ρeiθ , and let α := arg(z − w) − arg(T (z) − T (w)).

arg(T (z) − T (w)) = arg((T − Id)(z) − (T − Id)(w) + (z − w)) = arg(λρeiβ + ρeiθ )

for some λ < ε,β ∈ [0,2π]. So arg(T (z) − T (w)) = arg(λeiβ + eiθ ).Then |α| ≤ α, where tan(α) = ε

1−ε⇒ |α| < 2ε. �

For any T : C × S1 → C, define Aj,k,T by P = (Q,Q′) ∈ Aj,k,T if and only if∃θ : (Tθ (Q),Tθ (Q

′)) ∈ Aj,k .Interpret this to mean that if the centers of Q and Q′ are q and q ′, then the Tθ

images of q and q ′ satisfy the Aj,k condition: 4−k−1 ≤ |yTθ (q) − yTθ (q ′)| ≤ 4−k and

4−j−1 ≤ |xTθ (q)−xTθ (q′)||yTθ (q)−yTθ (q′)| ≤ 4−j . P could belong to Aj,k,T for more than one pair j, k,

but this redundancy does nothing more than increase the constant C of the theorem.


Fig. 3 Only where the annuliintersect will we find centers ofcircles of radius r whichintersect both Cantor squares.Approximation by a rectangle issufficient to give the desiredestimate

Lemma 5 (Sorting Lemma) Let T satisfy Lip(Tθ − Id) < 18n

. Then ∀j < logn,|Aj,k,T | ≤ C42n−k−2j .

Proof Distances are multiplied by numbers in the range 1 ± Cn

under T , so for a j, k

pair, k can change by at most one under Tθ . Lemma 4 implies that angles are changedadditively by at most 1

4nunder T , so j can change by at most one if j ≤ logn. Thus

[3] (3.1) gives us

|Aj,k,T | ≤∑

−1≤l,m≤1

|Aj+l,k+m| ≤ C42n−k−2j .�

Note that T = σF satisfies Lemma 5 for r > 32n.Instead of fn,θ and νP , consider fn,θ,T := ∑

Q χTθ (Q),θ and νP,T :=∫ |Projθ (Tθ (Q)) ∩ Projθ (Tθ (Q′))|dθ . Of course, we will use T = σ in the setting

of circles.Next we need to estimate νP,σ .

Lemma 6 (νP,T Lemma for Circles) If T = σF and r ≥ C′n, then for any j, k pairP , νP,T ≤ C4k−2n.

Proof It may be useful to consult Figure 3 during this proof. Note that r dρ dθ ≈ρ dρ dθ , so we may translate a noodle probability problem into an area problem. If anarc of radius r intersects two Cantor squares, then the arc must be centered inside theintersection of two annuli whose radii are r ± 4−n, and whose centers are the centersof the two Cantor squares. So we want to prove that the area A of this intersection ofannuli satisfies A ≤ Cr4k−2n.

Without loss of generality, the squares are centered on the x-axis at 0 and at rx0.We have rx0 ≈ 4−k and we define ε by rε = 4−n. So we need to show that A ≤Cr2ε2/x0. We can scale the problem by r . Thus if we let r = 1, then we need onlyshow that if x0 < 1/2, then A ≤ Cε2/x0. It will not hurt to let the inner radius be 1rather than 1 − ε.1 Let R = 1 + ε.

1One may divide the annulus along the circle with radius one. The inner annulus can be rescaled to haveinner radius 1, and the constants change negligibly.


The area A is taken from the region bounded by y = y1 = √1 − x2, y = y+

1 =√R2 − x2, y = y2 = √

1 − (x − x0)2, and y = y+2 = √

R2 − (x − x0)2.y+

1 = y2 at a point we will call x∗ = 12x0 + 1

2x0ε(2 + ε). So a rectangle which

contains the area A has width 2(x∗ − x02 ) = 1

x0ε(2 + ε), and height y+

1 (x02 ) − y2(

x02 ).

So we need only show that y+1 (

x02 )−y2(

x02 ) ≤ Cε. To do this, we use the Mean Value

Theorem on the function s(x) = √x.

y+1

(x0

2

)− y2

(x0

2

)=

√

R2 −(

x0

2

)2

−√

1 −(

x0

2

)2

≤ s′(

1 −(

x0

2

)2)(2ε + ε2) ≤ C

ε√1 − (

x02 )2

≤ C′ε.

Thus A ≤ Cε2/x0, as desired, so that νP,σ ≤ 4k−2n. �

Now we are in a position to finish the proof of Theorem 2. We use T instead ofσ below since the proof holds for all T ’s for which the conclusions of the last twolemmas are valid.

Let Pj,T := ⋃n−j

k=0 Aj,k,T . Then∑

P∈Pj,TνP,T ≤ Cn4−2j (sorting and νP lem-

mas). Also, let En,θ,T := suppfn,θ,T . A couple of applications of the Cauchy in-equality to fn,θ,T and χEn,θ,T

give us

∫

Jj

|En,θ,T | ≥ (∫Jj

∫R

fn,θ,T dx dθ)2

(∫Jj

∫R

f 2n,θ,T dx dθ)

. (4.3)

We have

∫

Jj

∫

R

fn,θ,T dx dθ ≈ 4−j ,

∫

Jj

∫

R

f 2n,θ,T dx dθ

=∫

Jj

∫

R

fn,θ,T dx dθ +∑

Q �=Q′

∫

Jj

∫

R

χTθ (Q)χTθ (Q′) dx dθ (4.4)

≤ C4−j +∑

P∈Pj−1∪Pj ∪Pj+1

νP,T

≤ C(4−j + n4−2j ) ≤ Cn4−2j ,

where the last inequality relies on j < logn. So (4.3) and (4.4) give us, together withthe above,

∫Jj

|En,θ,T |dθ ≥ C/n. Summing over 3 < j < logn, we get the result. �


5 Theorem 3 Follows from the Proof of Theorem 2

With hardly any effort, we may obtain the following result which holds when 5 <

rn < Cn:Favσ (Kn) ≥ C

log(rn)n

. Note that the νP,σ estimate works for all rn > 5, and thatthe Aj,k,σ estimate works for all j ≤ C + log(rn). Then the argument above showsthat

Favσ (Kn) ≥log(rn)∑

j=3

∫

Jj

|En,θ,σ |dθ ≥log(rn)∑

j=3

C

n.

So in particular, if we fix rn = r ≥ 5, then Favσ (Kn) ≥ Cn

. Furthermore, we nowknow that the bound FavσFn (Kn) ≤ C

ncannot hold for all n if lim supn rn → ∞.

6 General Buffon Noodle Probabilities and the νP Lemma for ArbitraryNoodles

We will find it useful to abuse notation by sometimes allowing Projθ (E)(x) to referto χProjθ (E)(x).

Let us define general noodle probabilities now. Because an arbitrary noodle doesnot have as many symmetries as a circular arc, a general noodle probability will needto integrate over three independent parameters: two real variables for where the noo-dle lands, and one for the orientation of the noodle. This serves two purposes: first,it better conforms to our intuition about what it means to randomly toss a possiblyasymmetric noodle. Second, an extra variable of integration allows us to more readilypartition regions of integration into ones possessing symmetry. Our parameterizationwill have three real variables, ρ and θ like before, and a third parameter τ for transla-tion orthogonal to θ . Under a change of variables, the other view can be recovered. Itis clear that such a translation of a circle is again parameterizable by the two-variablescheme, while a noodle with less symmetry is not.

Let gτ (y) := g(y − τ). (If we have a family gn of noodles, then we can writegn,τ (y) := (gn)τ (y) = gn(y − τ).) For a probability distribution P on R

2 × S1, a setE ⊂ C, and noodle g, we can define

Bug(E) =∫

Projθ (σgτ

θ (E))(x) dP (x, τ, θ).

We can choose an L > 10, say, and let P be the normalized Lebesgue measure on(−2,2) × (−L,L) × (0,2π), under which

Bug(E) = 1

16πL

∫ 2π

0

∫ L

−L

|Projθ (σgτ

θ (E))|dτ dθ = 1

16πL

∫ L

−L

Favσgτθ

(E)dτ .

Having done this, we will say that the noodles gn are undercooked if Bugn(Kn) ≥C

lognn

for some constant C.For general noodle probabilities, we will prove the following result:


Theorem 7 If ‖g′n(y)‖4∞ · ‖g′′

n‖∞ ≤ 4−n and ‖g′n(y)‖ ≤ 1

100n, then the noodles gn

are undercooked.

In particular, this theorem implies that the Frn are undercooked if rn ≥ 4n/5, whichis weak compared to Theorem 2. Another example is gn(y) = 4−n/2 sin(4n/4y).

The proof will be essentially the same, with the difference being that the corre-sponding νP Lemma will be more tortuous. Define

νP,σg =∫ L

−L

∫ 2π

0|Projθ (σ

gτ

θ (Q)) ∩ Projθ (σgτ

θ (Q′))|dθ dτ .

We want νP,σg < C4k−2n.There will be two main parts of the proof of our νP lemma. If one of the two

squares, Q, were centered at the origin and τ were fixed, the computation wouldmerely amount to finding how often a needle close to the origin intersected the othersquare, Q′. We claim that this assumption can be justified if one foliates the domainappropriately and then changes variables. In fact, one can further assume that Q′ lieson the negative y-axis and that τ = 0. Having done this, we will linearly approxi-mate gn and use the structure of the shear group to get our desired estimate. The ideais that we pick one of the two squares Q and partition the integration domain ac-cording to which point along the noodle punctures the center of Q. One can imaginedropping the noodle so it intersects Q, gluing this point of intersection in place, andthen rotating the noodle around this point, asking how often the noodle hits the othersquare, Q′. Each of these positionings of the noodle can be expressed uniquely by atriple (τ, θ, x).

Let us state the formulas. Fix a j, k pair Q,Q′. We will describe the portion of thedomain of integration in which the noodle hits the center of a square Q at the samepoint −τ0 of the noodle. That is, if Q has center z = ρeiθ0 , consider g := g − g(−τ0)

and σgτ0θ . For each θ , we need to find the unique xθ and τθ such that the line centered

at xθeiθ and with positive axis in the θ + π/2 direction intersects z at y = τθ − τ0. In

fact,

xθ = |z| cos(θ − θ0), and

τθ = τ0 − |z| sin(θ − θ0).

Then when computing

∫ 2π

0

∫ xθ+a

xθ−a

Projθ (σgτθ

θ (Q))(x)Projθ (σgτθ

θ (Q′))(x) dx dθ,

without loss of generality z = 0. That is,

∫ 2π

0

∫ xθ+a

xθ−a

Projθ (σgτθ

θ (Q))(x)Projθ (σgτθ

θ (Q′))(x) dx dθ

=∫ 2π

0

∫ a

−a

Projθ (σgθ (Q − z))(x)Projθ (σ

gθ (Q′ − z))(x) dx dθ.


For z = center of Q, and for fixed τ0, define

D = {τ = τ0 − |z| sin(θ − θ0), |x − |z| cos(θ − θ0)| ≤ C4−n, θ ∈ (0,2π)}.Then if

ID(τ0) :=∫

D

Projθ (σgτθ

θ (Q′))(x) dx dθ,

νP,σg ≤ ∫ L

−LID(τ0) dτ0.

Putting this all together, we are seeking to prove that if in addition to the hy-potheses of our theorem, g(0) = 0 and Q′ is approximately at distance 4−k from the

origin, then∫ 2π

0

∫ 4−n

−4−n Projθ (σgθ (Q′))(x) dx dθ ≤ C4k−2n. Here we use that the σ -

projection of a small square centered at the origin is essentially an interval around theorigin regardless of θ .

7 Some Useful Facts About Shear Groups

A few facts about shear groups need to be stated. Below, g and h will be arbitrarynoodles. Recall:

σg

0 (x, y) := (x − g(y), y),

σgθ := R−θ ◦ σ

g

0 ◦ Rθ .

First, there is this simple fact for arbitrary functions g and h:

σgθ ◦ σh

θ = σg+hθ . (7.1)

Next, we show how shears by linear noodles behave. We can let Eθ be a family ofsubsets of C, but for our application we will fix Eθ = Kn. For g(y) = b, we get

Projθ (σgθ (Eθ )) = Projθ (Eθ ) − b. (7.2)

For g(y) = my, α := arctanm, we get

Projθ (σgθ (Eθ )) =

(Projθ−α(Eθ )

cos(α)

)= (

√1 + m2)Projθ−α(Eθ ). (7.3)

For g(y) = my + b and given x ∈ R, we can say that x ∈ Projθ (σgθ (Eθ )) if and

only if x+b√1+m2

∈ Projθ−α(Eθ ). Thus for any measurable A ⊂ R,

∫ 2π

0

∫

A

χProjθ (σgθ (Eθ ))(x) dx dθ

=√

1 + m2

∫ 2π

0

∫

1√1+m2

(A+b)

χProjθ (Eθ+α)(x) dx dθ. (7.4)


8 Proof of the νP Lemma for General Noodles

First, the rough idea: the set of parameters for which two squares are simultaneouslypunctured by the needle may be translated considerably in parameter space by theshears, but it cannot be dilated by too much. A linear shear with small slope, forexample, is exactly a dilation and a translation. Since a shear with small curvature iswell-approximated by a linear shear, the result will follow. Let λ′ = ‖g′

n(y)‖∞ andλ′′ = ‖g′′

n(y)‖∞. Let Q and Q′ be centered at (0,0) and (0,−L), respectively, whereL ≈ 4−m. Note that

νP,σgθ

≤ C4−n|{θ : σgθ Q ∩ σ

gθ Q′}| ≤ C4−n(4k−n + λ′).

We need this quantity to be < C4k−2n. This task is already done if λ′ ≤ 4k−n, soassume the opposite. Now for such θ we have |θ | < C(4k−n + λ′) < Cλ′.

For these θ , rotation Rθ(Q) is in the band L− δ ≤ y ≤ L+ δ, for δ = 4−n +L(1−cos(Cλ′)), giving δ ≤ C max{4−n,Lλ′2}. Now transform the integral using the sheargroup. Let l(y) linearly approximate g(y) at y = L − δ, with l(y) = my + b. Notethat |b| ≤ CLλ′. Let ε(y) := g(y)− l(y) on [L− δ,L+ δ] and extend ε continuouslyto be constant elsewhere. Then, with b′ := b/

√1 + m2:

νP,σg =∫

|Projθ (σgθ (Q′))Projθ (σ

gθ (Q))|dθ

≤∫ 2π

0

∫ 4−n

−4−n

χProjθ (σgθ (Q′))(x) dx dθ

=∫ 2π

0

∫

[−4−n,4−n]χProjθ (σ l

θ (σ εθ (Q′))) dx dθ

≤ C

∫ 2π

0

∫

[b′−2·4−n,b′+2·4−n]χProjθ−α(σ ε

θ (Q′)) dx dθ.

Changing variable, we see that this is at most

C

∫ 2π

0

∫

[b′−2·4−n,b′+2·4−n]χProjθ (σ ε

θ+α(Q′)) dx dθ.

Let � := {θ : Projθ (σεθ+α(Q′)) ∩ [b′ − 2 · 4−n, b′ + 2 · 4−n] �= ∅}, and let z :=

(0,−L). If θ ∈ �, then Projθ (σεθ+α(z)) ∈ [b′ − 3 · 4−n, b′ + 3 · 4−n].

Since |ε′(y)| < Cδλ′′, it follows that |ε(y)| < Cδ2λ′′ < CL2λ′4λ′′ < C4−n. So|σε

θ+α(z) − z| < c4−n, and hence |Projθ (σεθ+α(z)) − Projθ (z)| ≤ C4−n ∀θ ∈ �. So

� ⊆ {θ : Projθ (z) ∈ [b′ − C4−n, b′ + C4−n]}= {θ : L sin θ ∈ [b′ − C4−n, b′ + C4−n]},

which implies:

|�| ≤ C|{θ : sin θ ∈ [b/L − C4k−n, b/L + C4k−n]}|. (8.1)


Since b < CLλ′ and 4k−n ≤ λ′ ≤ Cn

, sin θ ≈ θ , and we get |�| ≤ C4k−n, completingthe proof of the νP lemma.

Theorem 6 follows as well.

9 Application to Circular Lacunary Maximal Function

Let cm := {ζ : |z − ζ | = 4−m}.

Mf (z) := supm≥0

4m

∫

cm

|f (ζ )| |dζ |.

In [11] it is proved that

M : L(log logL)1+ε → L1,∞

is bounded if ε > 0. In [10] it is proved that

M : L(log logL) → L1,∞

is bounded.The question is raised in these papers: is this sharp? In [11] the idea is presented

how to try to prove this sharpness, in particular how to prove that

M : L(log logL)1−ε → L1,∞

is not bounded. To show the latter statement it would be enough to construct setsEn, |En| � 1, such that

supt

t |{z : MχEn(z) ≥ t}| � |En|(

log log1

|En|)1−ε

. (9.1)

We cannot do that yet, but the result of this paper produces a family of inequalitieswhich point the way to a type of conjecture about Kn which would give us (9.1).In this paragraph, m may be any number ≤n, thanks to the result of this paper. Themain idea is as follows: take a contraction En of Kn (by the factor 4−n), and thentake an ε ≈ 4−2n-neighborhood of this, called En. On a certain set of distance about4−m from En, there is a relatively large set of centers of circles of radius 4−m whichintersect En, so that on this set MχEn is relatively large. Then if we go to a largercircle, we get another such set, but since the maximal function averages by dividingthis intersection length by the length of an arc 4 times as long as before, we lose afactor of 4 from MχEn . At the same time, it is the dr dθ measure which is uniformlylarge, so the r dr dθ measure quadruples. If for each step up to a larger radius, wecould sacrifice the quadrupling of the measure and in return intersect four times asmany squares, we could prove (9.1) by unioning over n sets, canceling out the n

in the denominator of logn/n. Then logn would stand exactly for log log 1|En| , as

|En| ≈ 4n · 4−4n.


Let us be formal about this. Let μn,A := {z : MχEn ≥ A/(2π)}. Let En be Kn

rescaled by the factor 4−n. Let En be the 12 4−2n-neighborhood of En. Observe that

|En| ≤ 4−3n+1, since En is contained in a union of 4n squares of side-length 2 · 4−2n.Any long arc which punctures En must intersect En in a set of length at least 4−2n.

So if we rescale Gn,4n−m (see Sect. 2 for this definition) by the factor 4−n and call

this Hn,m, we get a set of points on which MχEn ≥ 4−2n

2π4−m . That is, Hn,m ⊂ μn,4m−2n ,and further, Hn,m ∩ Hn,m′ = ∅.

|Hn,m| ≥ C4−m Favσ (En) ≥ C4−m logn

n· 4−n = C

logn

n4−n−m.

Thus∣∣∣∣∣

n⋃

m=0

μn,4m−2n

∣∣∣∣∣ ≥n∑

m=0

|Hn,m| ≥n∑

m=0

Clogn

n4−n−m.

It would be nice if we could instead write the following for, say, M = αn, for someconstant α > 0:

|μn,4M−2n | ≥M∑

m=0

Clogn

n4−n−M ≥ Cα logn4−n4−M, (9.2)

because then

4M−2n

2π|{z : MχEn ≥ 4M−2n/(2π)}| ≥ C4−3n logn

≥ C|En| log log1

|En| � |En|(

log log1

|En|)1−ε

.

Let us state how one might get this. We can call by Qj , (j = 1, . . . ,4n), thesquares composing En, and let Hn,m,M := {z : (#j : C4−m(z) ∩ Qj �= ∅) ≥ 4M−m}.Then

Hn,m,M ⊂ μn,4M−2n .

Relation (9.2) would then follow if we had |Hn,m,M | ≥ Clogn

n4−n−M . Translating

back to a needle probability problem, we get the following

Conjecture There exist α,C > 0 such that for infinitely many n,

|{(x, θ) ∈ R × [0,2π] : fn,θ (x) ≥ 4m}| ≥ Clog n

n4−m (9.3)

for all m ≤ αn.

If this could be proved, then one might hope that the proof would carry over toprove the corresponding conjecture for a noodle probability problem, namely thesame estimate for fn,θ,σ4m , thus truly proving (9.1).


Due to a conversation with Michael Bateman, we now know that this family ofinequalities cannot be true. It follows by the reasoning of Sect. 3 that

∫

R×[0,2π]f 2

n,θ dx dθ ≤ Cn.

Thus m{(x, θ) : fn,θ (x) ≥ K} ≤ K−2‖fn,θ‖2L2(x,θ)

≤ Cn

K2 . So (9.3) is false for K >

n2, whereas we needed (9.3) to be true for logK ≥ αn.Michael Bateman also suggested that perhaps randomizing the Cantor set as in [9]

or in some other way may be the remedy. Though the lower bound c lognn

for the levelset of height 1 is almost surely false (at least in the liminf) in this case, perhaps withsome probability the decay of higher level sets can be made slower to compensate.

It would be interesting if the 4-corner Cantor set were indeed the desired coun-terexample due to the level sets fn,θ,σ being much larger than those of fn,θ , in whichcase it would be seen that the result of this paper takes advantage of hypotheses op-posite to those one actually needs.

Acknowledgements We would like to thank the anonymous referee for bringing to our attention that inTheorem 2, our proof immediately shows that the condition r > Cnε is sufficient, where ε > 0. Previously,Theorem 2 had only been stated for ε = 1.

Secondly, thanks to a conversation with Michael Bateman, we now know that our conjectured approachto proving the bound on the lacunary circular maximal operator cannot work. If such a bound on theoperator exists, its proof cannot be for the exact reason given. See Sect. 9.

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University Press, Cambridge (1986)6. Laba, I., Zhai, K.: Favard length of product Cantor sets. arXiv:0902.0964 (2009)7. Mattila, P.: Orthogonal projections, Riesz capacities and Minkowski content. Indiana Univ. Math. J.

39, 185–198 (1990)8. Nazarov, F., Peres, Y., Volberg, A.: The power law for the Buffon needle probability of the four-corner

Cantor set. arXiv:0801.2942 (2008)9. Peres, Y., Solomyak, B.: How likely is Buffon’s needle to fall near a planar Cantor set? Pac. J. Math.

204, 473–496 (2002)10. Seeger, A., Tao, T., Wright, J.: Pointwise convergence of lacunary spherical means. In: Mt. Holyoke

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circular favard length of the four-corner cantor set

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