Table Of ContentHAUSDORFF DIMENSION OF A CLASS OF THREE-INTERVAL
EXCHANGE MAPS
D. Karagulyan*
7
Abstract. In [5] Bourgain proves that Sarnak’s disjointness conjecture holds for
1
0 a certain class of Three-interval exchange maps. In the present paper we estimate
2 the constants in Bourgain’s proof and subsequently estimate the measure of the
parameter set in his result. We show, that it has positive, but not full Hausdorff
n
a dimension. This,inparticular,impliesthattheLebesguemeasureofthissetiszero.
J
8 1. Introduction
intro
]
S Let µ denote the Mo¨bius function, i.e.
D (cid:40)
(−1)k if n = p p ···p for distinct primes p ,
. µ(n) = 1 2 k k
h
0 otherwise.
t
a
m In [25], [24] Sarnak introduced the following conjecture. Recall that a topological
[ dynamical system (Y,T) is a compact metric space Y with a homeomorphism T :
1 Y → Y , and the topological entropy h(Y,T) of such a system is defined as
v
1
4 h(Y,T) = lim lim logN((cid:15),n),
8 (cid:15)→0n→∞ n
9
where N((cid:15),n) is the largest number of (cid:15)-separated points in Y using the metric
1
0 d : Y ×Y → R+ defined by
n
.
1 d (x,y) = max d(Tix,Tiy).
0 n
0≤i≤n
7
1 A sequence f : Z → C is said to be deterministic if it is of the form
:
v f(n) = F(Tnx),
i
X
for all n and some topological dynamical system (Y,T) with zero topological entropy
ar h(Y,T) = 0, a base point x ∈ Y, and a continuous function F : Y → C.
con1 Conjecture 1. Let f : N → C be a deterministic sequence. Then
n
1 (cid:88)
ff00 (1.1) S (T(x),f) = µ(k)f(k) = o(1).
n
n
k=1
The conjecture, called Moebius orthogonality, is known to be true for several dy-
namical systems. For a Kronecker flow (that is a translation on a compact abelian
group) it is proved in [27] and [8] and these results predate Sarnak’s conjecture and
the methods are number-theoretical. When (X,T) is a translation on a compact
nilmanifold it is proved in [17]. In [4] it is established also for horocycle flows. For
*DepartmentofMathematics,RoyalInstituteofTechnology,S-10044Stockholm,Sweden. Email:
[email protected]
1
2 HAUSDORFF DIMENSION OF A CLASS OF THREE-INTERVAL EXCHANGE MAPS
orientation preserving circle-homeomorphisms and continuous interval maps of zero
entropy the conjectures is proved in [20]. The conjecture has also been proved to hold
inseveralothercases([1], [2], [3]). Anothernaturalclassofdynamicalsystemsarethe
interval exchange maps. In [5] Bourgain, using the dynamical description of trajecto-
ries in ([10]-[12]) and the Hardy-Littlewood circle method, shows that under a certain
diophantine condition Conjecture 1 holds for a certain class of three-interval exchange
maps. In this paper we slightly improve the diophantine condition obtained by Bour-
gain and estimate the Hausdorff dimension of the new parameter set (Theorem 6).
We want to note, that using the criterion of Bourgain in [5] and the generalization
of the self-dual induction defined in [15], for each primitive permutation, Ferenczi
and Mauduit([14]) construct a large family of k-interval exchanges satisfying Sarnaks
conjecture.
RecentlyEskinandChaika([6])provedtheMoebiusorthogonalityforthreeinterval
exchange maps satisfying a certain mild diophantine condition. Even though their
resultholdsforalmostallthreeintervalexchangemaps, butthediophantinecondition
considered in their paper is essentially complementary to the one considered here. In
[6] the continued fractions are required to have certain bound from above, while in
Bourgains method they need to be uniformly large. We note that Eskin and Chaika,
in fact, give two proofs of the fact that the Moebius orthogonality holds almost surely
for three-interval exchange maps, however the second proof does not provide explicit
Diophintine condition. Their proof is based on the Katai-Bourgain-Sarnak-Ziegler [4]
criterion, while Bourgain uses a direct approach.
The present paper is a part of the authors Ph.D. thesis.
2. Three interval exchange maps
sec:2
The three-interval exchange transformation T with probability vector (α,β,1 −
(α+β)), α,β > 0,0 < α < 1, 0 < β < 1−α, and permutation (3,2,1) is defined by
x+1−α if x ∈ [0,α)
T (x) = x+1−2α−β if x ∈ [α,α+β)
α,β
x−α−β if x ∈ [α+β,1).
T depends only on the two parameters 0 < α < 1 and 0 < β < 1−α. We note that
T is continuous except at the points α and α+β.
In order to present the main result of the paper we need to recall some facts and
definitions from [10] and [5]. Set
1−α
A(α,β) =
1+β
and
1
B(α,β) = .
1+β
More precisely, T is obtained from the 2-interval exchange map R on [0,1) given by
([22],[21])
(cid:40)
x+A(α,β), if x ∈ [0,α)
iinndd (2.1) R(x) =
x+A(α,β)−1 if x ∈ [1−A(α,β),1),
HAUSDORFF DIMENSION OF A CLASS OF THREE-INTERVAL EXCHANGE MAPS 3
by inducing (according to the first return map) on the subinterval [0,B(α,β)] and
then renormalizing by scaling by 1 + β. We say T satisfies the infinite distinct or-
bit condition (or i.d.o.c. for short) of Keane [22] if the two negative trajectories
{T−n(α)} and {T−n(α + β)} of the discontinuities are infinite disjoint sets.
n≥0 n≥0
Under this hypothesis, T is both minimal and uniquely ergodic; the unique invariant
probability measure is the Lebesgue measure µ on [0,1) (and hence (X,T,µ) is an
ergodic system).
Let I denote the open interval (0,1), D ⊂ R2 the simplex bounded by the lines
0
y = 0, x = 0, and x+y = 1, and D the triangular region bounded by the lines x = 1,
2
x+y = 1, and 2x+y = 1. Note that
D = {(α,β) : α,β > 0,0 < α < 1,0 < β < 1−α}.
0
We define two mappings on I ×I
(cid:18) (cid:19)
2x−1 y
F(x,y) = , and G(x,y) = (1−x−y,y).
x x
One checks that if (α,β) ∈ D is not in D and is not on any of the rational lines
0
pα + qβ = p − q, pα + qβ = p − q + 1, pα + qβ = p − q − 1 then there exists a
unique finite sequence of integers l ,l ,...,l such that (α,β) is in H−1D, where H is
0 1 k
a composition of the form Gt ◦Fl0 ◦G◦Fl1 ◦G...◦G◦Flk ◦Gs,s,t ∈ {0,1}. Let
H = {Gt ◦Fl0 ◦G◦Fl1 ◦G...◦G◦Flk ◦Gs : s,t ∈ {0,1} and l ,l ,...,l ∈ N}.
0 1 k
Clearly H is a countable set.
The function H(α,β) is computed recursively as follows: we start with α(0) = α,
β(0) = β. Then, given (α(k),β(k)), we have three mutually exclusive possibilities: if
(α(k),β(k)) is in D, the algorithm stops; if α(k) < 1, we apply G; if 2α(k) +β(k) < 1,
2
we apply F.
Associated to each point (α,β) ∈ D there is a sequence (n ,m ,ε ) , where n
0 k k k+1 k≥1 k
and m are positive integers, and ε is ±1. This sequence is called the three-interval
k k+1
expansion of (α,β); it is constructed as follows:
• For (α,β) in D let
1−α−β 1−2α
x = and y = ,
0 0
1−α 1−α
and define for k ≥ 0
(cid:16)(cid:110) (cid:111)(cid:17)
yk , xk if x +y > 1
(x +y )−1 (x +y )−1 k k
(xk+1,yk+1) = (cid:16)(cid:110) k k k k (cid:111)(cid:17)
1−yk , 1−xk if x +y < 1.
1−(x +y ) 1−(x +y ) k k
k k k k
(cid:16)(cid:104) (cid:105) (cid:104) (cid:105)(cid:17)
yk , xk if x +y > 1
(x +y )−1 (x +y )−1 k k
(nk+1,mk+1) = (cid:16)(cid:104) k k (cid:105) (cid:104) k k (cid:105)(cid:17)
1−yk , 1−xk if x +y < 1,
1−(x +y ) 1−(x +y ) k k
k k k k
where {a} and [a] denote the fractional and integer part of a respectively. For
k ≥ 0 set
(cid:15) = sgn(x +y −1).
k+1 k k
4 HAUSDORFF DIMENSION OF A CLASS OF THREE-INTERVAL EXCHANGE MAPS
• For (α,β) ∈/ D we let H be the function above for which (α,β) ∈ H−1D and
put
¯
(α¯,β) = H(α,β),
¯
and define (n ,m ,(cid:15) ) as in the previous case, starting from (α¯,β) ∈ D.
k k k+1
In [10] the authors also prove the following propositions and theorem
propos1 Proposition 1 ([10, Proposition 2.1, (2)]). An infinite sequence (n ,m ,(cid:15) ) is the
k k k+1
expansion of at least one pair (α,β) defining a T satisfying the i.d.o.c. condition,
if and only if n and m are positive integers, (cid:15) = ±1, (n ,(cid:15) ) (cid:54)= (1,+1) for
k k k+1 k k+1
infinitely many k and (m ,(cid:15) ) (cid:54)= (1,+1) for infinitely many k.
k k+1
¯
propos Proposition 2 ([10, Proposition 2.1, (4)]). For (α,β) ∈ D , let (α¯,β) = H(α,β) as
0
above, then
1−α¯ 1
¯
A(α¯,β) = = .
¯
1+β 1
2+
(cid:15)
2
m +n −
1 1
(cid:15)
3
m +n −
2 2 m +n − ...
3 3
Ft Theorem 1 ([10, Theorem 2.2]). LetT satisfythei.d.o.c. condition, andlet(n ,m ,(cid:15) ) ,
k k k+1 k≥1
be the three-interval expansion of (α,β). Then there exists an infinite sequence of
nested intervals J , k ≥ 1, which have exactly three return words, A , B and C ,
k k k k
given recursively for k ≥ 1 by the following formulas
ffoorrmm11 (2.2) A = Ank−1C Bmk−1A ,
k k−1 k−1 k−1 k−1
ffoorrmm22 (2.3) B = Ank−1C Bmk
k k−1 k−1 k−1
ffoorrmm33 (2.4) C = Ank−1C Bmk−1
k k−1 k−1 k−1
if (cid:15) = +1, and
k+1
ffoorrmm44 (2.5) A = Ank−1C Bmk ,
k k−1 k−1 k−1
ffoorrmm55 (2.6) B = Ank−1C Bmk−1A
k k−1 k−1 k−1 k−1
ffoorrmm66 (2.7) C = Ank−1C Bmk−1A
k k−1 k−1 k−1 k−1
if (cid:15) = −1. The initial words A ,B ,C satisfy ||A |−|B || = 1.
k+1 0 0 0 0 0
Let a = |A |, b = |B |, c = |C |. Note that |a − b | = |a − b | = 1 and
k k k k k k k k k−1 k−1
c ≤ 2a .
k k
Remark 2. In [6] Eskin and Chaika show, that for three interval exchange maps
satisfying the conditions (A0)–(A9) (page 3), Sarnak’s conjecture holds. We want
to note, that their approach is also based on the fact that three interval exchange
maps can be induced from a two interval exchange map. The numbers {a }∞ , in
k k=1
conditions (A0)–(A9), are the continued fractions of the rotation number of the two
intervalexchangemapwhichinducesthethreeintervalexchangemapwithparameters
(α,β). In our case this corresponds to the number A(α,β) (see (2.1)), hence the
HAUSDORFF DIMENSION OF A CLASS OF THREE-INTERVAL EXCHANGE MAPS 5
continued fractions of A(α,β) are the numbers {a }∞ and from (2) it is easy to
k k=1
see, that they are related to the numbers {(n + m )}∞ . However in Bourgain’s
k k k=1
approach this numbers are required to be sufficiently large (see Theorem 4), while
the conditions (A0)–(A9) essentially give upper bounds.
Using the dynamical descriptions of trajectories in [10], Bourgain [5], proves Sar-
nak’s disjointness conjectures for a certain class of three interval exchange maps. Now
we recall the statement of his theorem.
Consider a symbolic system on the alphabet V with finitely many symbols and
with order-n words W ∈ W of the form
n
(cid:91)
ff1111 (2.8) W = Wk1Wk2···Wkr for some W ,W ,··· ,W ∈ W(cid:48) = W ,
1 2 r 1 2 r n−1 m
m<n
where it is assumed that r remains uniformly bounded, r < C. It is also assumed the
following property for the system {W }. For W ∈ W , which is expressed in words
n n
W(cid:48) ∈ W , 0 < s ≤ n, by iteration of (2.8) we have,
n−s
|W|
ff1122 (2.9) > β(s),
max|W(cid:48)|
where
ff1133 (2.10) β(s) > Cs,
0
for some large s and for some sufficiently large constant C .
0
thm Theorem 3 ([5, Theorem 2, page 126]). Let {W ;n ≥ 1} be a symbolic system with
n
(cid:83)
properties (2.8)-(2.10) and σ be the shift on the system. Then, if W ∈ W and
n
|W| = N, one has
(cid:90) N
(cid:88)
|P (θ)|| µ(k)e(kθ)|dθ = O (N(logN)A),
W A
T
k=1
for any A > 0, where
N
(cid:88)
P (θ) = f(k)e(kθ),
W
k=1
and f(k) = f(σk(x)).
To seehowthisimpliesSarnak’s conjecture we recallthefollowinginequality, which
immediately follows from Parseval’s identity
(cid:12) (cid:12)
(cid:12)(cid:88)N (cid:12) (cid:90) (cid:88)N (cid:88)N
iiddeenntt (2.11) (cid:12) µ(k)f(k)(cid:12) ≤ | f(k)e(kθ)|| µ(k)e(−kθ)|dθ.
(cid:12) (cid:12)
(cid:12) (cid:12) T
k=1 k=1 k=1
and according to Theorem 3
(cid:12) (cid:12)
(cid:12)(cid:88)N (cid:12) (cid:90) (cid:88)N
iiddeenntt (2.12) (cid:12) µ(k)f(k)(cid:12) ≤ |P (θ)|| µ(k)e(−kθ)|dθ ≤ O (N(logN)A),
(cid:12) (cid:12) W A
(cid:12) (cid:12) T
k=1 k=1
which implies Sarnak’s conjecture. For three interval exchange maps
W = {A ,B ,C }.
k k k k
6 HAUSDORFF DIMENSION OF A CLASS OF THREE-INTERVAL EXCHANGE MAPS
One can see, that if m and n are uniformly large, then the conditions (2.9)–(2.10)
k k
are satisfied and as a corollary from (3) one gets the following result:
th1 Theorem 4 ([5, Theorem 3]). Assume T is a three-interval exchange transforma-
α,β
tion satisfying the Keane condition and such that the associated three-interval expan-
sion sequence
(n ,m ,(cid:15) )
k k k+1 k≥1
of integers fullfills the conditions
bboouu11 (2.13) min(n ,m ) ≥ C , for k ≥ k .
k k 0 α,β
Then T satisfies Sarnak’s disjointness conjecture.
α,β
We note, that Bourgain’s theorem is in fact more general than the way it is stated
in Theorem 4. As it was mentioned above, in Theorem 4 it is assumed, that there is
uniform expansion at each step, but as one can see from the conditions (2.9)-(2.10)
it is sufficient to have this expansion after s steps, for some fixed s. Next we prove a
proposition, which will allow us to rewrite the diophantine condition (2.24) below in
more general form.
main-prop Proposition 3. In Theorem 4 the condition (2.13) can be replaced by
tt11 (2.14) m +n ≥ 2C , for all k ≥ k .
k k 0 0
Proof. As we said for three interval exchange maps we have W = {A ,B ,C }.
k k k k
From (2.9)–(2.10) it follows, that it suffices to show, that for any W ∈ W and
k k
W ∈ W , one has
k−1 k−1
|W | m +n
k k k
LL1122 (2.15) ≥ .
|W | 2
k−1
One can check from (2.2)–(2.4), that
ss11 (2.16) |W | ≥ (n −1)a +c +(m −1)b .
k k k−1 k−1 k k−1
Therefore
ee22 (2.17)
|W | (n −1)a +c +(m −1)b (n −1)a c (m −1)b
k k k−1 k−1 k k−1 k k−1 k−1 k k−1
≥ ≥ + + .
|W | |W | |W | |W | |W |
k−1 k−1 k−1 k−1 k−1
We have, that |a −b | = 1 and 2a ≥ c and 2b ≥ c . Hence
k k k k k k
a 1 b 1
k k
ss22 (2.18) ≥ , ≥ .
c 2 c 2
k k
Similarly
b 1 b 1
k−1 k−1
ss33 (2.19) ≥ , ≥ .
a 2 c 2
k−1 k−1
HAUSDORFF DIMENSION OF A CLASS OF THREE-INTERVAL EXCHANGE MAPS 7
Assume W = C , then from (2.17) and (2.19)
k−1 k−1
|W | (n −1)a +c +(m −1)b
k k k−1 k−1 k k−1
≥
|C | c
k−1 k−1
(n −1)a (m −1)b
k k−1 k k−1
≥ +1+
c c
k−1 k−1
(n −1) (m −1) n +m
k k k k
≥ +1+ = .
2 2 2
So from now on we can assume, that W ∈ {A ,B }. First, let W = A .
k−1 k−1 k−1 k−1 k−1
Then
|W | (n −1)a +c +(m −1)b
k k k−1 k−1 k k−1
≥
|A | a
k−1 k−1
(m −1)b
k k−1
≥(n −1)+ .
k
a
k−1
We want to show, that
(m −1)b n +m
k k−1 k k
(n −1)+ ≥ .
k
a 2
k−1
Denote α = b /a . Then
k k−1 k−1
n +m
k k
(n −1)+(m −1)α ≥ ,
k k k
2
or
n 1
k
LL1111 (2.20) +m (α − ) ≥ 1+α .
k k k
2 2
Since we have |a −b | = 1, it follows that
k k
(2.21) lim α = 1.
k
k→∞
Now we assume, that m ,n (cid:54)= 1. Then n ,m ≥ 2. If n ≥ 3, then from (2.20) we
k k k k k
get
n 1 3 1
k
+m (α − ) ≥ +α − ≥ 1+α .
k k k k
2 2 2 2
If m ≥ 3, then for large k’s
k
n 1 2 1
k
+m (α − ) ≥ +3(α − ) ≥ 1+α .
k k k k
2 2 2 2
In the same way we can show (2.15) assuming W = B . Now consider the case
k−1 k−1
m = n = 2. From (2.17) we need to show
k k
a +c +b
k−1 k−1 k−1
pprroo::11 (2.22) ≥ 2.
|W |
k−1
Since |W | = a or |W | = b , c is large for large k and |a −b | = 1,
k−1 k−1 k−1 k−1 k−1 k−1 k−1
then
pprroo::22 (2.23) a +c ≥ |W |, and b +c ≥ |W |,
k−1 k−1 k−1 k−1 k−1 k−1
8 HAUSDORFF DIMENSION OF A CLASS OF THREE-INTERVAL EXCHANGE MAPS
which implies (2.22). Now we assume, that one of the m ’s and n ’s is 1. First let
k k
m = n = 1. In this case, according to Proposition 1, (cid:15) can not be positive for
k k k+1
infinitely many values of k, so we assume, that we have (m ,n ,(cid:15) ) = (1,1,−1).
k k k+1
Hence W is defined by the formulas (2.5)-(2.7) and in view of (2.23) we will have
k
|W | min{|A |,|B |} min{c +b ,c +a }
k k k k−1 k−1 k−1 k−1
≥ = ≥ 1,
|W | |W | |W |
k−1 k−1 k−1
as W ∈ {A ,B }.
k−1 k−1 k−1
Again from Proposition 1, it remains to show (2.15) for (1,m ,−1) or (n ,1,−1),
k k
where n ,m ≥ 2. In the case of (1,m ,−1), from (2.5)-(2.7), we have
k k k
|W | min{c +m b ,c +(m −1)b +a }
k k−1 k k−1 k−1 k k k−1
≥ .
|W | |W |
k−1 k−1
Since c +a ≥ b , then one has
k−1 k−1 k−1
|W | min{c +m b ,m b } m b m +1
k k−1 k k−1 k k−1 k k−1 k
≥ = > ,
|W | |W | |W | 2
k−1 k−1 k−1
for large values of k, as bk−1 tends to 1, when k → ∞, and m > (m + 1)/2, for
|W | k k
k−1
m > 1. Similarly, for (n ,1,−1)
k k
|W | min{(n −1)a +c +b ,(n −1)a +c +a }
k k k−1 k−1 k−1 k k−1 k−1 k−1
≥ ,
|W | |W |
k−1 k−1
and
|W | min{n a ,n a +c } n a n +1
k k k−1 k k−1 k−1 k k−1 k
≥ = > .
|W | |W | |W | 2
k−1 k−1 k−1
for large values of k. (cid:3)
From this proposition we arrive at the following Theorem:
th2 Theorem 5. Assume T is a three-interval exchange transformation satisfying the
α,β
Keane condition and such that the associated three-interval expansion sequence
(n ,m ,(cid:15) )
k k k+1 k≥1
of integers fullfills the conditions
bboouu (2.24) (n +m )(n +m )···(n +m ) ≥ (2C )s,
k k k−1 k−1 k−s+1 k−s+1 0
for all k ≥ k and for some s ≥ 0. Then T satisfies Sarnak’s disjointness
α,β α,β
conjecture.
Now we turn to the estimation of the constant C . In the proof of Theorem 3
0
Bourgain, first estimates the L1 norm of the polynomials P , namely the Lemmas
W
3 and 4 in [5]. For the result it is also essential to slow down the growth of the
L1 norm of the polynomial P , whenever |W | → ∞ (see lemma 4 in [5]). This
W k
condition is achieved by assuming that the lengths |W | of the words in the symbolic
k
representations (2.8) grow sufficiently fast, i.e. conditions (2.9) and (2.10). One of
the key places where this is used is Lemma 4. To estimate how big the constant
C has to be we will follow Bourgains steps and give more quantitative proof of this
0
lemma.
HAUSDORFF DIMENSION OF A CLASS OF THREE-INTERVAL EXCHANGE MAPS 9
First we note the following. Let W → W ··· → W be a sequence of words with
1 2 n
W ∈ W , where each W participates in the symbolic representation of W . Then
k k k k+1
according to the assumptions (2.9) and (2.10) one has
|W |
qq11 (2.25) n ≥ β(s) ≥ Cs.
|W | 0
n−s
In the same way
|W | |W |
qq22 (2.26) n−s ≥ Cs, n−2s ≥ Cs,···
|W | 0 |W | 0
n−2s n−3s
We note that there may be only finitely many indexes where the above inequalities do
not hold, but that will not affect our estimates. Multiplying together the inequalities
in (2.25) and (2.26) we will have
|W | ≥ C(cid:48)Cn,
n 0
which leads to
|W |1 ≥ (C(cid:48))1/nC .
n n 0
Hence
1
liminf|W | ≥ C .
n n 0
n→∞
Now back to Lemma 4. We observe that the proof of the lemma is based on the
inequality (2.13), i.e.
(cid:90) 2π k
(cid:88)
mmaaiinniinneeqq (2.27) |P (θ)|| e(jlθ)| ≤ Clog(k +2)(cid:107)P (cid:107) .
W W 1
0
j=0
The proof of (2.27), in its turn, is based on Lemma 3. We note, that in the proof
of Lemma 3, Bourgain doesn’t use any particular property of the polynomial P
W
and since the inequality (2.27) holds for any polynomial P (θ), we can assume, that
W
P (θ) ≡ const. So we will end up with the following inequality
W
(cid:90) 2π k
(cid:88)
| e(jlθ)| ≤ Clog(k +2).
0
j=0
We know
N
(cid:88) sin((N +1)θ/2)
eikθ = eiNθ/2 .
sin(θ/2)
k=0
Hence
(cid:90) 2π (cid:88)k (cid:90) 2π(cid:12)(cid:12)sin((k +1)lθ/2)(cid:12)(cid:12)
| e(jlθ)| = (cid:12) (cid:12)dθ =
(cid:12) sin(lθ/2) (cid:12)
0 0
j=0
(cid:90) 2π(cid:12)(cid:12)sin((k +1)θ/2)(cid:12)(cid:12) (cid:90) 2π
(cid:12) (cid:12)dθ = |D (θ)|dθ,
k
(cid:12) sin(θ/2) (cid:12)
0 0
where D is the Dirichlet kernel for which one has (e.g., see [28])
k
(cid:90) π sint 8
(cid:107)D (cid:107) ≥ 4 dt+ logk.
k L1 t π
0
10 HAUSDORFF DIMENSION OF A CLASS OF THREE-INTERVAL EXCHANGE MAPS
So we conclude, that
(cid:90) π sint 8
4 dt+ logk ≤ Clog(k +2).
t π
0
Dividing both sides by logk and tending k to infinity we get, that
8
(2.28) ≤ C.
π
Next we estimate the ε in (2.15) of [5]. For this we refer to the inequality (2.26) in
[5]. In (2.23) and (2.24) Bourgain defines the minor and major arcs, which depend
on parameters K and Q. Lemma 6 yields the inequality (2.25). For arcs which are
sufficiently close to rational numbers with large denominators (i.e. when Q is large)
or sufficiently far from the rationals with small denominator (i.e. when K is large)
applying Lemma 6 one gets the inequality (2.25). Furthermore applying Lemma 4
and the inequality (2.25) we get
(cid:90) N
(2.29) (cid:88) |P (θ)||(cid:88)µ(n)e(mθ)|dθ (cid:28) (Q−41 +N−τ/4)N1+(cid:15).
W (cid:15) 0
max(Q,K)>Q0 VQ,K 1
Or dividing both sides by N
(cid:90) N
(2.30) 1 (cid:88) |P (θ)||(cid:88)µ(n)e(mθ)|dθ (cid:28) (Q−41 +N−τ/4)N(cid:15).
N W (cid:15) 0
max(Q,K)>Q0 VQ,K 1
Now from (2.12) Bourgain assumes, that Q,K < Nε. But this is possible only, if
τ
(2.31) > ε,
4
and from Lemma 6 we see, that 0 < τ < 1/3. So we conclude, that
ε < 1/12.
Now we return to the proof of Lemma 6 in [5]. To obtain the formula (2.15) Bourgain
uses the inequality (2.27) to iterate the formulas (2.2)-(2.7), i.e. for W ∈ W one gets
n
iitteerr (2.32) (cid:107)P (cid:107) ≤ Clog(2+k )(cid:107)P (cid:107) +···+Clog(2+k )(cid:107)P (cid:107) .
W L1 1 W1 L1 r Wr L1
Iterating further the polynomials P ,P ,··· ,P at step n we will get at most 4n
W1 W2 Wr
many members of the form
Cnlog(2+k )···log(2+k )
1 n
(we say 4n since in the formulas (2.2)-(2.7) at most 4 subwords appear). Now using,
the geometric arithmetic-mean inequality one gets
