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De ay of orrelations

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

and

Γ u

are olle tionsofstableandunstable urvesrespe tively,and areturntime

R : Θ → N

su h that

f R(·) ( ·) : Θ → Θ

.

For

ω ∈ Γ u

let

ν ω

denotethe onditionalmeasureoftheLebesguemeasure

ν

withrespe ttothe urve

ω

. Ifthe onditions

ν ω (ω ∩ Θ) > 0,

forea h

ω ∈ Γ u

ν {x ∈ Θ | R(x) > n} < Cθ n ,

forsome

C

and

θ < 1, (f n , Q)

isergodi forea h

n > 0,

andsomeotherregularity onditionsaresatisedthenthisissu ientto

on- ludeexponentialde ayof orrelationsforHölder ontinuousfun tions.

Amongotherexamplesin[38℄,Youngshowshowtondtheset

Θ

andthe

returntime

R

for a lass ofpie ewise

C 2

hyperboli mapsin twodimensions.

This lassofmaps isdierentfromour lassbutstillthemethod anbeused

to onstru t

Θ

and

R

. Wegivethe onstru tionof

Θ

and

R

below.

The fa t that

(λ, γ, κ) ∈ P

makes the onstru tion of

Θ

easier. This is

be ausethereisauniform estimateof thelengthofthelo alstable manifolds

forthese parameters. From Se tion3.4 we on ludethat

W α s (x)

existsfor all

x ∈ Λ

. Theorem 3.5.0.3istruealsoif

(λ, γ, κ) 6∈ P

but thenwehaveto work

withtheset

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

and

R

. Itisimportantto

gatherenough expansionin the unstabledire tion. Forthis purpose wetake

an

N ∈ N

su hthat

γ N > 2e 1 e

.

Choose

δ > 0

sothat forany urve

ω

in theunstabledire tion withlength

notgreaterthan

δ

theset

f N (ω)

onsistsofatmosttwo onne ted omponents.

This an be done sin ewe an hoose

δ

to be the smallestdistan e between

thelinesintheset

N n=0 f −n (S h )

.

Take

0 < δ 0 < 1 6 δ

tobesosmallthat theset

A δ 0 = {x ∈ Θ | W 2δ u 0 (x)

exists

}

has positive Lebesgue measure. For any

x ∈ A δ 0

we dene

Ω(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 turedenedby

Γ s (x)

and

Γ u (x)

.

For any

x ∈ A δ 0

we let

Q(x)

be the smallest open re tangle ontaining

Θ(x)

. Theopensets

{Q(x)} x∈A δ0

form anopen overingof

A δ 0

. Sin e

A δ 0

is

ompa twe ansele tanite sub over

{Q(x i ) } r i=1

. Thesets

{Θ(x i ) } r i=1

will

then over

A δ 0

. Wewillwrite

Θ i

for

Θ(x i )

.

We dene thereturn time

R

to

∪Θ i

ona subsetof the sets

Θ i

. Let

i

be

xed. We iteratethe map

f N

and onsider the onne ted omponentsof the

set

(f N ) n (Ω(x i ))

.

We onstru tforea h

n ∈ N

apartition

P n i

of

Ω(x i ) \{R ≤ n}

into onne ted

urveswiththepropertythatif

ω ∈ P n i

then

(f N ) n (ω)

isa onne ted urveof

length

< 6δ 0

.

Let

P 0 i

bethetrivialpartition,

P 0 i = {Ω(x i ) }

. Assumethat

P 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

j } j=1,2

bethe

orresponding omponents of

ω

. If

(f N ) n (ω j )

haslength

< 6δ 0

then weput

ω j

in

P n i

. Ifhowever

(f N ) n (ω j )

haslength

≥ 6δ 0

then there isa

k

su hthat

(f N ) n (ω j )

rosses

Q(x k )

with segments of length

≥ δ 0

sti king out on ea h

sideof

Q(x k )

. Sin e

|(f N ) n (ω j ) | > 6δ 0

wehave

(f N ) n (ω j ) ∈ Γ u (x k )

and this

impliesthat

(f N ) n (ω j ) ∩Q(x k ) ⊂ Θ k

. Wedene

R = n

on

ω j ∩(f N ) −n (Θ k )

. In

thiswaywegettwopie esoflength

> δ 0

onea hsideof

(f N ) n (ω j \ {R = n})

.

Wepartition

ω \ {R = n}

into

{ω j,k } m k=1

su hthat

δ 0 < (f N ) n (ω j,k ) < 6δ 0

and

put

{ω j,k }

in

P n i

.

Finally,if

ω ∈ P n−1

thenwedene

R = n

ontheset

S ω = [

ω s ∈Γ s (ω)

ω s

!

\ [

ω u ∈Γ u

ω u

!

⊂ Θ i ,

where

Γ s (ω)

isthesetoflo alstablemanifoldsin

Γ s (x i )

whi hhasnon-empty

interse tion with

ω

. The uniform estimate on the length of

ω s ∈ Γ s

implies

that

(f N ) n (S ω ) ⊂ Θ k

.

Lemma3.5.0.4. Let

Ω = Ω(x i )

for some

x i

. There exists

C > 0

and

θ 1 < 1

su hthat

ν Ω {R > n} ≤ Cθ n 1 .

Proof. Let

T 1 (x)

bethesmallest

n ≥ 1

su hthatif

ω ∈ P n−1

isthe omponent

ontaining

x

then

|(f N ) n (ω) | ≥ 6δ 0

. If there is no su h

n

then wesay that

T 1 (x)

isnotdened. Notethat

T 1 ≤ R

Suppose that

T k (x)

has been dened. Then we dene

T k+1 (x)

to be the

smallest

n > T k (x)

su hthat if

ω ∈ P n−1

is the omponent ontaining

x

then

|(f N ) n (ω) | ≥ 6δ 0

. Let

T k = {T k

isdened

}

.

Ea htimeasegmentisstret hedtoalengthover

6δ 0

apie eoflength

2δ 0

returnstooneofthesets

Θ i

. Thisimpliesthatif

ω ∈ P n−1

and

T 1 | ω = n

then

|(f N ) n (ω) | < γ N 6δ 0

andso

ν 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

. Similarlyweget

ν Ω ( T k+1 ) ν Ω ( T k ) < θ 2 .

Thisimpliesthat

ν Ω ( T k ) < θ k 2

and

ν 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 if

ω ∈ P n−1

then

ω \ {R = n}

an be

overedbylessthan

M

elementsfrom

P n

.

Let

K p = {k i } p i=1

where

k 1 < k 2 < · · · < k p

with

k p ≥ n

and

k p−1 < n

.

Considerthe set

A K p = {T i = k i , i = 1, 2, . . . , p }

. It anbe overedby less

than

2 k p M p

elementsfrom

P k p

. Sin ethelengthof

is

2δ 0

wehave

ν 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'sformula

n 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

issmallenough

X [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 that

θ 3 = e

1−ε e M ε 2

(1−ε)γ N < 1

. This anbedone be auseof the

hoi eof

N

. Then

ν 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 hoose

k = εn

andget

ν 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 eof

Θ i

toreturntosome

Θ j

, but we would need

R

to be the return time from

Θ i

to

Θ i

. Arguingas

in [38℄, we an hoose

Θ ∗ = Θ i

for some

i

su h that the return time

R ∗

of

Θ ∗

satises

ν ω {R > n} < Cθ n

for some

θ < 1

. This is su ientto on lude

Theorem3.5.0.3. Notethatthe

SBR

-measure onstru tedwiththemethodin [38℄isthesamemeasureasthe

SBR

-measure onstru tedinSe tion 3.2.

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