Sin e
Λ ˆ λ × ˆ Λ λ = [ ∞ k=0
[
i
R ˆ k i1 (λ) × ˆ R i−1 k (λ)
we anpro eedasfollows
Z
I
Z
Ω ˆ λ
Z
Ω ˆ λ
χ {|y 1 −z 1 |<r} dˆ µ λ,s,ˆ SBR x(λ) (z)dˆ µ λ,s,ˆ SBR x(λ) (y)dλ
= X p t=1
X ∞ k=0
X
i
T i,t,k
≤ X p t=1
X ∞ k=0
X
i
c 3 r(γ −1 + ε) −k (γ − ε) −k µ ˆ λ SBR t ,s,ˆ x(λ t ) ( ˆ R k i1 (λ t ))
≤ X p t=1
X ∞ k=0
c 3 r(γ −1 + ε) −k (γ − ε) −k ≤ X p t=1
c 4 r = c 5 r.
Hen e
lim inf
r→0
1 r
Z
I
Z
Ω ˆ λ
Z
Ω ˆ λ
χ {|y 1 −z 1 |<r} dˆ µ λ,s,ˆ SBR x(λ) (ˆ z)dˆ µ λ,s,ˆ SBR x(λ) (ˆ y)dλ ≤ c 5 .
(3.6)Thisisindependentofthe hoi e of
ˆ x : I → Q
so thisprovesTheorem3.3.0.2.andahyperboli produ tstru ture;
Θ = [
ω s ∈Γ s
ω s
!
\ [
ω u ∈Γ u
ω u
! ,
where
Γ s
andΓ u
are olle tionsofstableandunstable urvesrespe tively,and areturntimeR : Θ → N
su h thatf R(·) ( ·) : Θ → Θ
.For
ω ∈ Γ u
letν ω
denotethe onditionalmeasureoftheLebesguemeasureν
withrespe ttothe urveω
. Ifthe onditionsν ω (ω ∩ Θ) > 0,
forea hω ∈ Γ u
ν {x ∈ Θ | R(x) > n} < Cθ n ,
forsomeC
andθ < 1, (f n , Q)
isergodi forea hn > 0,
andsomeotherregularity onditionsaresatisedthenthisissu ientto
on- ludeexponentialde ayof orrelationsforHölder ontinuousfun tions.
Amongotherexamplesin[38℄,Youngshowshowtondtheset
Θ
andthereturntime
R
for a lass ofpie ewiseC 2
hyperboli mapsin twodimensions.This lassofmaps isdierentfromour lassbutstillthemethod anbeused
to onstru t
Θ
andR
. Wegivethe onstru tionofΘ
andR
below.The fa t that
(λ, γ, κ) ∈ P
makes the onstru tion ofΘ
easier. This isbe ausethereisauniform estimateof thelengthofthelo alstable manifolds
forthese parameters. From Se tion3.4 we on ludethat
W α s (x)
existsfor allx ∈ Λ
. Theorem 3.5.0.3istruealsoif(λ, γ, κ) 6∈ P
but thenwehaveto workwiththeset
D δ + = {x | d(f n (x), S) > δγ −n ∀n ≥ 0},
whi hwouldhavemadethe onstru tionof
Θ
somewhatmorete hni al.Wewill nowpro eedwith the onstru tionof
Θ
andR
. Itisimportanttogatherenough expansionin the unstabledire tion. Forthis purpose wetake
an
N ∈ N
su hthatγ N > 2e 1 e
.Choose
δ > 0
sothat forany urveω
in theunstabledire tion withlengthnotgreaterthan
δ
thesetf N (ω)
onsistsofatmosttwo onne ted omponents.This an be done sin ewe an hoose
δ
to be the smallestdistan e betweenthelinesintheset
∪ N n=0 f −n (S h )
.Take
0 < δ 0 < 1 6 δ
tobesosmallthat thesetA δ 0 = {x ∈ Θ | W 2δ u 0 (x)
exists}
has positive Lebesgue measure. For any
x ∈ A δ 0
we deneΩ(x) = W δ u 0 (x)
.Put
Γ s (x) = {W s (y) | y ∈ Ω(x)},
Γ u (x) = {W u (z) | z ∈ A δ 0 , W u (z) ∩ W s 6= ∅, ∀W s ∈ Γ s (x) }.
Welet
Θ(x)
bethehyperboli setwiththeprodu tstru turedenedbyΓ s (x)
and
Γ u (x)
.For any
x ∈ A δ 0
we letQ(x)
be the smallest open re tangle ontainingΘ(x)
. Theopensets{Q(x)} x∈A δ0
form anopen overingofA δ 0
. Sin eA δ 0
isompa twe ansele tanite sub over
{Q(x i ) } r i=1
. Thesets{Θ(x i ) } r i=1
willthen over
A δ 0
. WewillwriteΘ i
forΘ(x i )
.We dene thereturn time
R
to∪Θ i
ona subsetof the setsΘ i
. Leti
bexed. We iteratethe map
f N
and onsider the onne ted omponentsof theset
(f N ) n (Ω(x i ))
.We onstru tforea h
n ∈ N
apartitionP n i
ofΩ(x i ) \{R ≤ n}
into onne tedurveswiththepropertythatif
ω ∈ P n i
then(f N ) n (ω)
isa onne ted urveoflength
< 6δ 0
.Let
P 0 i
bethetrivialpartition,P 0 i = {Ω(x i ) }
. AssumethatP n−1 i
isdened.Let
ω ∈ P n−1 i
. Sin e(f N ) n−1 (ω)
isa onne ted urveoflength< 6δ 0
theset(f N ) n (ω)
onsistsofatmosttwo onne ted omponents. Let{ω ′ j } j=1,2
betheorresponding omponents of
ω
. If(f N ) n (ω j )
haslength< 6δ 0
then weputω j
inP n i
. Ifhowever(f N ) n (ω j )
haslength≥ 6δ 0
then there isak
su hthat(f N ) n (ω j )
rossesQ(x k )
with segments of length≥ δ 0
sti king out on ea hsideof
Q(x k )
. Sin e|(f N ) n (ω j ) | > 6δ 0
wehave(f N ) n (ω j ) ∈ Γ u (x k )
and thisimpliesthat
(f N ) n (ω j ) ∩Q(x k ) ⊂ Θ k
. WedeneR = n
onω j ∩(f N ) −n (Θ k )
. Inthiswaywegettwopie esoflength
> δ 0
onea hsideof(f N ) n (ω j \ {R = n})
.Wepartition
ω \ {R = n}
into{ω j,k } m k=1
su hthatδ 0 < (f N ) n (ω j,k ) < 6δ 0
andput
{ω j,k }
inP n i
.Finally,if
ω ∈ P n−1
thenwedeneR = n
onthesetS ω = [
ω s ∈Γ s (ω)
ω s
!
\ [
ω u ∈Γ u
ω u
!
⊂ Θ i ,
where
Γ s (ω)
isthesetoflo alstablemanifoldsinΓ s (x i )
whi hhasnon-emptyinterse tion with
ω
. The uniform estimate on the length ofω s ∈ Γ s
impliesthat
(f N ) n (S ω ) ⊂ Θ k
.Lemma3.5.0.4. Let
Ω = Ω(x i )
for somex i
. There existsC > 0
andθ 1 < 1
su hthat
ν Ω {R > n} ≤ Cθ n 1 .
Proof. Let
T 1 (x)
bethesmallestn ≥ 1
su hthatifω ∈ P n−1
isthe omponentontaining
x
then|(f N ) n (ω) | ≥ 6δ 0
. If there is no su hn
then wesay thatT 1 (x)
isnotdened. NotethatT 1 ≤ R
Suppose that
T k (x)
has been dened. Then we deneT k+1 (x)
to be thesmallest
n > T k (x)
su hthat ifω ∈ P n−1
is the omponent ontainingx
then|(f N ) n (ω) | ≥ 6δ 0
. LetT k = {T k
isdened}
.Ea htimeasegmentisstret hedtoalengthover
6δ 0
apie eoflength2δ 0
returnstooneofthesets
Θ i
. Thisimpliesthatifω ∈ P n−1
andT 1 | ω = n
then|(f N ) n (ω) | < γ N 6δ 0
andsoν Ω (ω ∩ {R = T 1 })
ν Ω (ω) > 2δ 0
6δ 0 γ N = 1 3γ N .
Hen e
ν Ω (ω ∩ {R > T 1 })
ν Ω (ω) < 1 − 1 3γ N = θ 2 .
Thismeansthat
ν Ω (T 2 )
ν Ω (T 1 ) < θ 2
. Similarlywegetν Ω ( T k+1 ) ν Ω ( T k ) < θ 2 .
Thisimpliesthat
ν Ω ( T k ) < θ k 2
andν Ω {R > T k } < θ k 2
.Forany
k
{R > n} ⊂ {T k ≥ n} ∪ {T k < n < R }.
(3.7)There is a number
M
su h that ifω ∈ P n−1
thenω \ {R = n}
an beoveredbylessthan
M
elementsfromP n
.Let
K p = {k i } p i=1
wherek 1 < k 2 < · · · < k p
withk p ≥ n
andk p−1 < n
.Considerthe set
A K p = {T i = k i , i = 1, 2, . . . , p }
. It anbe overedby lessthan
2 k p M p
elementsfromP k p
. Sin ethelengthofΩ
is2δ 0
wehaveν Ω (A K p ) ≤ 2 k p M p 6δ 0 (γ N ) −k p 2δ 0
= 3M p 2 γ N
k p
.
If
p ≪ n
thenbyStirling'sformulan p
≈ n n+ 1 2 e −n
p!(n − p) n−p+ 1 2 e −n+p = e −p n p p!
1 − p n
p−n
andsoif
ε
issmallenoughX [εn]
p=0
n p
≤ C 1 X [εn]
p=0
e −p n p p!
1 − p n
p−n
< C 1 (1 − ε) −n X [εn]
p=0 n
e (1 − ε) p
p! < C 1 (1 − ε) −n e n e (1−ε) .
Choose
ε
sosmall thatθ 3 = e
1−ε e M ε 2
(1−ε)γ N < 1
. This anbedone be auseof thehoi eof
N
. Thenν Ω {T [εn] > n } ≤
X [εn]
p=0
X
K p
ν Ω (A K p ) ≤ X [εn]
p=1
n p
3M p
X ∞ k p =n
2 γ N
k p
≤ C 2 e 1−ε e M ε 2 (1 − ε)γ N
! n
= C 2 θ n 3 .
Weuse(3.7)toapproximate
ν Ω {R > n}
. We hoosek = εn
andgetν Ω {R > n} ≤ C 2 θ n 3 + θ εn 2 ≤ Cθ n 1 .
Wehaveproved that thereturn time
R
de ays exponentially. Thisis not quitewhatwewant.R
isthetimeneededforapie eofΘ i
toreturntosomeΘ j
, but we would needR
to be the return time fromΘ i
toΘ i
. Arguingasin [38℄, we an hoose
Θ ∗ = Θ i
for somei
su h that the return timeR ∗
ofΘ ∗
satisesν ω {R > n} < Cθ n
for someθ < 1
. This is su ientto on ludeTheorem3.5.0.3. Notethatthe
SBR
-measure onstru tedwiththemethodin [38℄isthesamemeasureastheSBR
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