The power law for the Buffon needle probability of the four-corner Cantor set

0.2 MB·English

by  Fedor Nazarov

THE POWER LAW FOR THE BUFFON NEEDLE PROBABILITY
OF THE FOUR-CORNER CANTOR SET
8
0 FEDOR NAZAROV,YUVALPERES, AND ALEXANDERVOLBERG
0
2
Abstract. Let Cn be the n-th generation in the construction of the middle-
n
half Cantor set. The Cartesian square K of C consists of 4n squares of side-
a n n
J length4−n. Thechancethatalongneedlethrownatrandomintheunitsquare

8 will meet K is essentially the average length of the projections of K , also
n n
1
known as the Favard length of Kn. A classical theorem of Besicovitch implies

] that the Favard length of K tends to zero. It is still an open problem to
A n
determine its exact rate of decay. Until recently, theonly explicit upperbound
C
was exp(−clog n), due to Peres and Solomyak. (log n is the number of times
. ∗ ∗
h
oneneedstotakelogtoobtainanumberlessthan1startingfromn). Weobtain
t
a
a power law bound by combining analytic and combinatorial ideas.
m
[

1
v 1. Introduction
2
4 The four-corner Cantor set is constructed by replacing the unit square by
9 K
2 four sub-squares of side length 1/4 at its corners, and iterating this operation in a
.
1 self-similar manner in each sub-square. More formally, consider the set that is
n
0 C
8 the union of 2n segments:
0
: n n
v
i n = aj4−j, aj4−j +4−n ,
X C
r aj∈{0,[3},j=1,..,nhXj=1 Xj=1 i
a and let the middle half Cantor set be
∞
:= .
n
C C
n\=1
It can also be written as C = { ∞n=1an4−n : an ∈ {0,3}}. The four corner Cantor
set is the Cartesian square P .
K C ×C
Since the one-dimensional Hausdorff measure of satisfies 0 < 1( ) <
K H K ∞
and the projections of in two distinct directions have zero length, a theorem of
K
1991 Mathematics Subject Classification. Primary: 28A80; Secondary: 28A75, 60D05, 28A78.
Research of the authors was supported in part by NSF grants DMS-0501067 (Nazarov and
Volberg) and DMS-0605166 (Peres).
1
2 FEDORNAZAROV,YUVALPERES,ANDALEXANDERVOLBERG
Figure 1. , the third stage of the construction of .
3
K K
Besicovitch (see[3, Theorem 6.13]) yields that the projection of to almost every
K
linethroughtheorigin haszerolength. Thisisequivalent tosaying thattheFavard
length of equals zero. Recall (see [1, p.357]) that the Favard length of a planar
K
set E is defined by
1 π
Fav(E) = Proj E dθ, (1.1)
θ
π Z | R |
0
where Proj denotes the orthogonal projection from R2 to the horizontal axis,
θ
R
is the counterclockwise rotation by angle θ, and A denotes the Lebesgue measure
| |
of a measurable set A R. The Favard length of a set E in the unit square has a
⊂
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, and
then locate the needle in a uniformly chosen position in that direction, at distance
at most √2 from the center of the unit square).
The set = 2 is a union of 4n squares with side length 4 n (see Figure 1
Kn Cn −
for a picture of ). By the dominated convergence theorem, Fav( ) = 0 implies
3
K K
lim Fav( ) = 0. We are interested in good estimates for Fav( ) as n .
n n n
→∞ K K → ∞
A lower bound Fav( ) c for some c > 0 follows from Mattila [8, 1.4]. Peres
Kn ≥ n
and Solomyak [10] proved that
BUFFON NEEDLE PROBABILITY OF THE FOUR-CORNER CANTOR SET 3
Fav( ) Cexp[ alog n] for all n N,
n
K ≤ − ∗ ∈
where
log n = mink 0: loglog...logn 1 .
∗  ≥ ≤ 
k
This resultcan beviewed as an attempt|to m{azke a q}uantitative statement out of
a qualitative Besicovitch projection theorem [1], [11], using this canonical example
of the Besicovitch irregular set.
It is very interesting to see what are quantitative analogs of Besicovitch theorem
in general. The reader can find more of that in [11].
We now state our main result, which improves this upper bound to a power law.
Theorem 1. For every δ > 0, there exists C > 0 such that
Fav( ) Cnδ 1/6 for all n N.
n −
K ≤ ∈
Remarks.
The 1/6 in the exponent is certainly not optimal, and, indeed, can be
•
improved slightly with the methods of this paper. However, a bound
decaying faster than O n 1/4 would require new ideas.
−
(cid:16) (cid:17)
In [10], Theorem 2.2, a random analog of the Cantor set is analyzed,
• K
and it is shown that, with high probability, the Favard length of then-th
stage in the construction has upper and lower bounds that are constant
multiples of n 1. However, it is not clear to us whether Fav( ) also
− n
K
decays at this rate.
It follows from the results of Kenyon [5] and Lagarias and Wang [6]
•
that Proj = 0 for all θ such that tanθ is irrational. As noted in
θ
| R K|
[10], this information does not seem to help obtain an upper bound for
Fav( ).
n
K
The set was one of the first examples of sets of positive length and
• K
zero analytic capacity, see [2] for a survey. The asymptotic behavior of
the analytic capacity of was determined in 2003 by Mateu, Tolsa and
n
K
Verdera [7], it is equivalent to 1 .
√n
It will be convenient to translate so that its convex hull is the unit square
n
K
centered at the origin. Due to the symmetries of the square, one can average over
4 FEDORNAZAROV,YUVALPERES,ANDALEXANDERVOLBERG
θ (0, π) in the definition (1.1) of Fav( ). After translation, the projection
∈ 4 Kn
of (1, 1) to the horizontal axis is the union of 4n intervals of length
Rθ Kn − 2 2
(cid:16) (cid:17)
4−n(cosθ+sinθ) centered at the points kn=−014−kξk, where
P
3√2 π 3√2 π
ξ cos( θ), sin( θ) .
k
∈ ± 8 4 − ± 8 4 −
n o
Let now t = tan(π θ) [0,1]. Since √2 cos(π θ) 1 on (0, π), the
4 − ∈ 2 ≤ 4 − ≤ 4
length of the projection Proj ( ) is comparable to the length of the union of
Rθ Kn
4n intervals of length 4−nρ centered at the points kn=−014−kξk with ξk ∈ {±1,±t},
where ρ= ρ(θ) = 8 (1+tan(π θ)). The exact vPalue of ρ(θ) is of no importance,
3√2 4 −
the only thing that matters is that it is separated from both 0 and + . We shall
∞
also need the function f that is the product of 1 and the sum of the characteristic
n ρ
functions of these intervals. In other words,
4n
f = ν(n) χ
n ∗ ρ [ ρ4−n,ρ4−n],
−2 2
where
1
ν(n) = n 1ν , and ν = [δ +δ +δ +δ ].
∗k=−0 k k 4 −4−k −4−kt 4−kt 4−k
Geometrically, f is (up to minor rescaling) the number of squares whose pro-
n
jections contain a given point. Finally, since dt = 1 is between 1 and 2
for all θ [0, π), we can replace averaging over|dθθ|withcotsh2(aπ4t−oθv)er t.
∈ 4
2. Fourier-analytic part.
In what follows, we will use and ., & to denote, respectively, equality or the
≍
corresponding inequality up to some positive multiplicative constant. Let K,S be
large positive numbers. Our first aim is to show that there exists a power p > 0
(we’ll see that any p > 4 fits) such that the measure of the set
E = t [0,1] : max f2 K
(cid:26) ∈ 1 n (KS)pZR n ≤ (cid:27)
≤ ≤
is at most 1. Supposenot. Let N be the least even integer exceeding 1(KS)p. For
S 2
every t E, we must have
∈
4N/2
K f (x)2dx f (y)2dy & ν(N)(y)2dy,
≥ ZR| N | ≍ ZR| N | Z1 | |
b b
BUFFON NEEDLE PROBABILITY OF THE FOUR-CORNER CANTOR SET 5
because ψ = 4Nχ satisfies ψ(y) & 1 for all y < 4N/2 if N is suffi-
ρ [ ρ4−N,ρ4−N] | |
−2 2
ciently large. Thus b
1 N/2 4n
ν(N)(y)2dy dt K
|E| ZEhnX=1Z4n−1| | i ≤
b
and for each m N/2, there exists n N/2 satisfying
≤ ≤
1 4n 4Km
ν(N)(y)2dy dt .
E ZEhZ4n−m| | i ≤ N
| |
Thus b
4n 8Km
E = t E : ν(N)(y)2dy
∗ n ∈ Z4n−m| | ≤ N o
satisfies E E /2. Our assumption bon E implies that E 1 . Now for
| ∗| ≥ | | | ∗| ≥ 2S
y [4n m,4n], we have
−
∈
n cos4 ky+cos4 kty 2
ν(N)(y)2 − − ,
| | ≍ kY=0(cid:12)(cid:12) 2 (cid:12)(cid:12)
b (cid:12) (cid:12)
since the remaining terms (that correspond to k [n + 1,N]) in the product
∈
converge geometrically to 1. Making the change of variable y 4ny, we get
7→
4n 1 n cos4ky+cos4kty 2
ν(N)(y)2dy 4n dy.
Z4n−m| | ≍ Z4−m(cid:12)(cid:12)kY=0 2 (cid:12)(cid:12)
b (cid:12) (cid:12)
Now split the last product into
m cos4ky+cos4kty n cos4ky+cos4kty
P (y) = and P (y) = .
1 2
2 2
kY=0 k=Ym+1
Consider the integral
1
P (y)2dy
2
Z | |
4−m
first. Writing the cosines as sums of exponentials, we have
4n−m
P2(y) = 4m−n eiλjy,
Xj=1
where λ 4n−m are the sums of all subsets of 4k, 4kt : k [m+1,n] . For
{ j}j=1 {± ± ∈ }
t E E, the definition of E yields that
∈ ∗ ⊂
6 FEDORNAZAROV,YUVALPERES,ANDALEXANDERVOLBERG
1
−1 −2/L−1/L 0 1/L 2/L 1
Figure 2. Triangle kernel function.
2
χ K 4n
ZR(cid:16)Xj [λj−ρ24m,λj+ρ24m](cid:17) ≤ ·
(this is equivalent to f2 K). The last inequality can be viewed as a
R n−m ≤
separation condition onRthe spectrum, so one can hope that a variation of Salem’s
trick should allow us to conclude that
1
P (y)2dy & 4m n,
2 −
Z | |
4−m
providedthatL = 4m ischosenappropriately. Weshallchoosemsuchthat4m = L
is a large constant multiple of K. Since P ( y) = P (y), we can integrate over
2 2
| − | | |
[ 1,1] [ L 1,L 1]. Consider the function g given by
− −
− \ −
g(y) = (1 y ) 2(1 L 1)(1 L y ) +(1 2L 1)(1 L y ) .
−| | + − − − − 2| | + − − − | | +
Note that g is even, 0 g 1, suppg [ 1,1] [ L 1,L 1] and 1 g 1 if L
≤ ≤ ⊂ − \ − − − 1 ≥ 2
is not too small. Now, let h denote “the triangle function” that is 1Ra−t 0, vanishes
on R ( 1,1) and is linear on [ 1,0] and on [0,1]. Then
\ − −
g(y) = h(y) 2(1 L 1)h(Ly)+(1 2L 1)h(Ly),.
− − − 2 − −
As h(λ) = 21 cosλ [0, C ], we get
−λ2 ∈ 1+λ2
b
C 1
g(λ) .
≥ −L · 1+(λ/L)2
b
BUFFON NEEDLE PROBABILITY OF THE FOUR-CORNER CANTOR SET 7
So we got the estimate
L
g(λ) C
≥ − λ2+L2
with some numerical constant Cb.
Denote M = 4n m. Let us call k 1,...,M good if
−
∈ { }
L 1
C .
L2+(λ λ )2 ≤ 8
Xj j − k
Then
2 2
eiλky dy g(y) eiλky dy
Z[−1,1]\[−L−1,L−1](cid:12)(cid:12)Xk (cid:12)(cid:12) ≥ ZR (cid:12)(cid:12)Xk (cid:12)(cid:12)
(cid:12) (cid:12) (cid:12) (cid:12)
1 2 L
+ g(y) eiλky dy 2 C
≥ k:kXisgood 2 ZR (cid:12)(cid:12) k:kXisbad (cid:12)(cid:12) − k:kXisgood Xj L2+(λj −λk)2
{ } (cid:12){ } (cid:12) { }
1
# k : k is good .
≥ 4 { }
Now we need only to show that the number of good indices is comparable to M.
To this end, note that we have the condition
2
χ (λ) dλ MLK.
ZR(cid:16)Xj [λj−ρ2L,λj+ρ2L] (cid:17) ≤
Convolving with the Poisson kernel (λ) = 1 L and taking into account
PL πL2+λ2
that
χ cL ( λ )
[λj−ρ2L,λj+ρ2L] ∗PL ≥ PL ·− j
with c > 0 (here we use that ρ stays bounded away from 0 and + ), we get
∞
2
L2 (λ λ ) dλ C MLK,
ZRhXj PL − j i ≤ ′
but
(λ λ ) (λ λ )dλ c (λ λ ).
ZRPL − j PL − k ≥ ′PL j − k
Thus
c (λ λ ) C MKL 1
′ PL j − k ≤ ′ −
Xj,k
and
8CπC M
# k :k is bad ′(KL 1)M ,
−
{ }≤ c ≤ 2
′
provided that L 16CπC′K. Therefore, indeed, 1 P (y)2dy & 4m n.
≥ c′ 4−m| 2 | −
R
8 FEDORNAZAROV,YUVALPERES,ANDALEXANDERVOLBERG
Thedangeristhatthislargeintegralcanbecompletelykilledwhentheintegrand
is multiplied by P 2. Note that
1
| |
cos4ky+cos4kty
= cos2 14k(y+ty)cos2 14k(y ty)
− −
2 −
so
m
P (y) = cos2 14k(y+ty)cos2 14k(y ty).
1 − −
−
kY=0
Using the formula
2m
u
2 4msin( ) cos2ℓ 1u= sin4mu.
−
· 2
ℓY=0
we conclude that
P (y) & 4 2m sin4m(y+ty) sin4m(y ty) .
1 −
| | | |·| − |
This can be small only if sin4m(y+ty) or sin4m(y ty) is small. For δ (0,1),
− ∈
denoteby theunionof intervals of length 4 mδ centered at thepoints πℓ,ℓ Z.
Iδ − 4m ∈
Define ω(t;δ) by
ω(t;δ) = y (4 m,1) : y+ty or y ty .
−
{ ∈ ∈ Iδ − ∈Iδ}
We would like to estimate P (y)2dy from above. This may be a hard
ω(t;δ)| 2 |
task for an individual t E , bRut we can bound the average fairly easily. We have
∈ ∗
1 1
P (y)2dy dt 2S P (y)2dy dt
2 2
|E∗| ZE∗(cid:16)Zω(t;δ,)| | (cid:17) ≤ Z0 (cid:16)Zω(t;δ,)| | (cid:17)
n n
du
. 2S cos22 14ku cos22 14kv dv
− −
Z[4−m,1]∩Iδ(cid:16)k=Ym+1 (cid:17)u+v Z[0,1](cid:16)k=Ym+1 (cid:17)
n n
du
+2S cos22 14ku cos22 14kv dv
− −
Z[4−m,1](cid:16)k=Ym+1 (cid:17) u+v Z[0,1]∩Iδ(cid:16)k=Ym+1 (cid:17)
where u = y+ty and v = y ty.
−
Using the formula cos2α = 1(1+cos2α) and the inequality 1 Ldudv, we
2 u+v ≤
can estimate the last expression by
n n
E := CSL 4m n (1+cos4ku)du (1+cos4kv)dv .
−
· hZIk=Ym+1 i·hZ[0,1]k=Ym+1 i
BUFFON NEEDLE PROBABILITY OF THE FOUR-CORNER CANTOR SET 9
Above we observed that
n n
(cos22 14ku)= 2m n (1+cos4ku) =:2m nR(u).
− − −
k=Ym+1 k=Ym+1
We want to see now that
E cSL 4m n√δ.
−
≤ ·
To this end we notice that the Riesz product R(u) is π -periodic.
4m
Note also that, for any interval of length 4 m π (j Z ), we have
J − 4j ∈ +
R(u)du = R (u)R (u)du,
1 2
Z Z
J J
whereR (u) = m+j (1+cos4ku)andR (u) = n (1+cos4ku). Observe
1 k=m+1 2 k=m+j+1
that R (u) 2jQfor all u and R (u) is π -perioQdic, so
1 ≤ 2 4m+j
1 π π
R (u)du = R (u)du = .
Z 2 4m+j Z 2 4m+j
0
J
Thus R(u)du π4 m. Choose j in such a way that δ 4 j.
≤ 2j − ≍ −
It fRoJllows that, for each constituting interval of , we have R(u)du .
J Iδ
4 m√δ. RJ
−
R(u)du . 4m 4 m√δ . √δ.
−
Z ·
[4−m,1]
∩Iδ
In conjunction with the estimate R(v)du . 1, we finally get
[0,1] η
R ∩I
E cSL 4m n√δ.
−
≤ ·
The resulting estimate is much less than 4m n if δ is much less than S 2L 2.
− − −
Thus, for at least one t E, we must have (recall that L = 4m)
∈
P (y)2dy c4m n
2 −
Z[L−1,1] Ω(t)| | ≥
\
and, thereby, (if we remember that K was a small constant times 4m)
P (y)2 P (y)2dy 4 4m(S 2L 2)4 4m n cS 8K 114 n.
1 2 − − − − − − −
Z[L−1,1]| | | | ≥ · ≥
10 FEDORNAZAROV,YUVALPERES,ANDALEXANDERVOLBERG
Thus, if p > 12 then our choice of N at the beginning of the proof gives N >
2KlogK
(KS)12+ε/2 > (KS)12+ε/2, hence is much less than S 8K 11, and we
− −
N
get a contradiction.
However, we promised to show that N > (KS)4+ε already leads to a contradic-
tion. To do this, we make our considerations more elaborate, but we follow the
same lines. In fact, let us consider
Ω(t;δ,η) = y (4 m,1) : y+ty and y ty .
{ ∈ − ∈Iδ − ∈ Iη}
We changed the word “or” in the definition of ω(t;δ) by the word “and” in the
definition of Ω(t;δ,η). This will allow us to make a subtler estimate. Notice that
ℓ
y : sin4m(y+ty) sin4m(y ty) 2 l Ω(t;2 k,2 ℓ+k+1).
− − −
{ | |·| − |≤ }⊂
k[=0
We would like to estimate P (y)2dy from above. As before we have
Ω(t;δ,η)| 2 |
R
1 1
P (y)2dy dt 2S P (y)2dy dt
2 2
|E∗| ZE∗(cid:16)Zω(t;δ,)| | (cid:17) ≤ Z0 (cid:16)ZΩ(t;δ,η)| | (cid:17)
n n
du
. 2S cos22 14ku cos22 14kv dv
− −
Z[4−m,1]∩Iδ(cid:16)k=Ym+1 (cid:17) u Z[0,1]∩Iη(cid:16)k=Ym+1 (cid:17)
where u = y+ty and v = y ty as before.
−
WealreadyintroducedR(u)= n (1+cos4ku)andestablishedthefollowing
k=m+1
estimate Q
R(v)dv . √η.
Z
[0,1] η
∩I
Now we can estimate
du 4m
R(u) . 4 m√δ √δm.
−
Z u πj · ≤
[4−m,1]∩Iδ 1 jX14m
≤ ≤π
Therefore, we obtain
1
P (y)2dy dt . Sm δη4m n.
2 −
|E∗| ZE∗(cid:16)ZΩ(t;δ,η)| | (cid:17) p
Let us denote Ω (t) := ℓ Ω(t;2 k,2 ℓ+k+1). We know now that
ℓ k=0 − −
S
1
P (y)2dy dt . Smℓ 2 ℓ/24m n.
2 − −
|E∗|ZE∗(cid:16)ZΩℓ(t)| | (cid:17) ·