Proof of the theorem

I dokument The Family of Belykh Maps Persson, Tomas (sidor 49-57)

theset

{y ∈ W s (x) | d(x, y) < δ}

isa onne ted urveoflength

thendene

W δ s (x) = {y ∈ W s (x) | d(x, y) < δ}

tobethestablemanifoldof

x

withlength

. Otherwisewesaythat

W δ s (x)

doesnotexist.

Sin e

f

isnotinvertiblewe annotdene thelo alunstablemanifoldof

x

intheusualway. Let

D δ = {x | d(f −n (x), S) > δγ −n ∀n ≥ 0}.

ThissethaspositiveLebesguemeasureif

δ

istakensu ientlysmall,see[27℄.

Itiseventruethat

µ SBR

 ∞

[

l=1

D l 1



= 1.

If

x = (x 1 , x 2 ) ∈ D δ

thenforany

y = (y 1 , y 2 )

with

y 1 = x 1

and

|x 2 −y 2 | < δ

we

have

d(f −n (x), f −n (y)) < δγ −n

forall

n ≥ 0

. We anthusdenetheunstable

manifoldof

x = (x 1 , x 2 ) ∈ D δ

oflength

δ

tobetheset

W δ u (x) = {y = (y 1 , y 2 ) | y 1 = x 1 , |x 2 − y 2 | < δ}.

If

x 6∈ D δ

thenwesaythat

W δ u (x)

doesnotexist.

Lemma3.4.0.1. Let

ε > 0

bexedand

I ⊂ (γ −1 +ε, 0.64)

. Thereisa onstant

c

su hthatfor any

k ∈ N

andany

{a i } i=0 ∈ {−1, 0, 1} N

thefollowingestimate

isvalid.

Z

I

χ {λ||λ k +P

i=k+1 a i λ i |<r} ≤ c(γ −1 + ε) −k r.

Let

γ

,

κ

bexedand

I ⊆ (0.5, 0.64)

anyintervalsu hthat

{γ}×I×{κ} ⊂ P

,

wheretheset

P

willbe hosenlater.

Let

B r (t) = [t − r, t + r]

andlet

D(µ λ,s,x SBR , t) = lim inf

r→0

µ λ,s,x SBR (B r (t)) 2r

denotethelowerderivativeofthemeasure

µ λ,s,x SBR

the onditionalmeasureof

µ λ SBR

withrespe ttothelo alstablemanifold

W s (x)

. Thepartitionof

Q

into

lo al stable manifolds is learly measurable so the onditional measures exist

andwe anusethemasfollows. Wewanttoprovethatfora.e.

λ

thereisaset

Ω λ

su hthat

µ λ SBR (Ω λ ) > 0

and

Z

Ω λ

D(µ λ,s,x SBR | λ , y)dµ λ,s,x SBR (y) < ∞

(3.1)

holds for a.e.

x

. This implies that the measure

µ λ,s,x SBR

restri ted to the set

Ω λ

is absolutely ontinuousfora.e.

x

. Sin ethe onditionalmeasures on the unstablemanifoldsareabsolutely ontinuouswithrespe ttoLebesguemeasure,

this implies that

µ λ SBR | λ

is absolutely ontinuous with respe t to Lebesgue

measure. Sin e

µ λ SBR (Ω λ ) > 0

, ergodi itythen impliesthatthis alsoholds for

themeasure

µ λ SBR

.

Fatou'slemmaimpliesthatin ordertoprove(3.1)itsu estoprovethat

lim inf

r→o

1 r Z

Ω λ

µ λ,s,x SBR (Ω λ ∩ B r (y))dµ λ,s,x SBR (y) < ∞.

Wemayrewritethisas

lim inf

r→0

1 r

Z

Ω λ

Z

Ω λ

χ {|y 1 −z 1 |<r} dµ λ,s,x SBR (z)dµ λ,s,x SBR (y) < ∞.

(3.2)

Wewill hoose a lassof fun tions

x : I → Q

su h that

x(λ) ∈ Ω λ

andprove

thatthefollowingestimateisvalid

lim inf

r→0

1 r Z

I

Z

λ

Z

λ

χ {|y 1 −z 1 |<r} dµ λ,s,x(λ) SBR (z)dµ λ,s,x(λ) SBR (y)dλ < ∞.

(3.3)

−1 −0.8 −0.6 −0.4 −0.2 0 0.2 0.4 0.6 0.8 1

−1

−0.8

−0.6

−0.4

−0.2 0 0.2 0.4 0.6 0.8 1

γ=1.858, λ=0.54, k=0

Figure3.3: Someinverseimagesof

S v

.

This thenimplies that

µ SBR ≪ ν

for a.e.

λ ∈ I

. Instead ofproving(3.3) we

usethat

µ SBR = ˆ µ SBR ◦ π −1

andprovetheequivalent ondition

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 < ∞.

(3.4)

Toprove(3.4)thesymboli oding

Σ λ,γ,κ

willbeused. Sin ewehave

γ

and

κ

xedand vary

λ

it isthedynami s taking pla ein the horizontaldire tion

thatis ru ial. Itishen ethedynami albehaviouroftheverti al omponent

ofthesingularity

S v

thatisimportant. Below,wewill hooseaset

P

inwhi h

S v

behavesasifitdoesnotexist.

One he ks withanumeri al al ulationthat for

γ = 1.858

,

λ = 0.54

and

κ = 0

there are only nitely many points in the omplete ba kwardorbit of

theverti alpie e ofthesingularity,

S v

. Numeri sareonlyused tosee thisin

aneasyway. Itdoesnotinuen eontherigorousityoftheproof. Inthis ase,

when

κ = 0

, wehave

S v = {0}

. Figure3.3 and 3.4illustrate this byshowing

thetwopossiblepathsof

{0}

startingin

Q 1

. Thepointsintheba kwardorbit

of

S v

are marked with

×

in the gures. The dashed lines are the border of

thesets

f (Q 1 )

and

f (Q −1 )

. Thearrowsshowshowthepointsaremappedby

f −1

. Ea hpathdrawnintheguresterminatesafternitelymanysteps,after

whi hthere arenomoreinverseimages.

−1 −0.8 −0.6 −0.4 −0.2 0 0.2 0.4 0.6 0.8 1

−1

−0.8

−0.6

−0.4

−0.2 0 0.2 0.4 0.6 0.8 1

γ=1.858, λ=0.54, k=0

Figure 3.4: Yet someinverseimagesof

S v

.

The numeri al evaluations of the points are in Table 3.1 and Table 3.2.

Comparingthese valueswith

f (Q 1 ) = [1 − 2λ, 1] × [1 − γ, 1] = [ −0.08, 1] × [−0.858, 1], f (Q −1 ) = [ −1, 2λ − 1] × [−1, γ − 1] = [ −1, 0.08] × [−1, 0.858],

showsthattherearenomorepointsintheba kwardorbitof

S v

andthatthese

pointsareboundedawayfromtheset

f (S)

. Thisallowustodrawthefollowing

on lusion. Thereisanopenneighborhood

U

of

S v

su hthat

f −N (U ) = ∅

for

N

su ientlylarge.

Bythe ontinuousdependen eontheparametersofthenitelymanypoints

in the ba kward orbit of

S v

, there exists an open ball

P ⊂ {γ, λ, κ}

around

(γ, λ, κ) = (1.858, 0.54, 0)

su h that theba kwardorbit of

S v

behavesin the

same way for any

(γ, λ, κ) ∈ P

in the followingsense. Forany

(γ, λ, κ) ∈ P

the ba kward orbit of

S v

ontains nitely many pie es, ea h bounded away

from

f (S)

, and there areuniform numbers

α > 0

and

N > 0

su h that

U = ([ −α, α] × [−|κ|, |κ|])

satises

f −N (U ) = ∅

. This implies that

U ∩ Λ γ,λ,κ = ∅

forany

(γ, λ, κ) ∈ P

.

Re allthat wehavetheparameters

γ

and

κ

xedand

I

isanintervalsu h

that

{γ} × I × {κ} ⊂ P

. Partition

I

into sub intervals

{I t } p t=1

, su h that

|I t | < 15 1 α

. This anbedonesothat

p < 15|I| α + 1

. Forea h

t = 1, . . . , p

xa

iterate oordinates

numeri al

values ontainedin

0

0

0

0

0 f (Q 1 ) ∩ f(Q −1 )

−1

−1+λ

−1+γ λ γ

−0.8519

0.4618 f (Q −1 ) \ f(Q 1 )

−2

−1+2λ−λ 2 λ 2

−1+2γ−γ 2 γ 2

−0.7257

−0.2132 f (Q −1 ) \ f(Q 1 )

−3

−1+2λ−λ 3 λ 3

−1+2γ−γ 3 γ 3

−0.4919

−0.5766 f (Q −1 ) \ f(Q 1 )

−4

−1+2λ−λ 4 λ 4

−1+2γ−γ 4 γ 4

−0.05916

−0.7721 f (Q 1 ) ∩ f(Q −1 )

−5

−1+2λ−2λ 45 λ 5

−1+2γ−2γ 45 γ 5

−0.9614

0.04623 f (Q −1 ) \ f(Q 1 )

−6

−1+2λ−2λ 4 +2λ 5 −λ 6 λ 6

−1+2γ−γ 4 +2γ 5 −γ 6 γ 6

−0.9285

−0.4369 f (Q −1 ) \ f(Q 1 )

−7

−1+2λ−2λ 4 +2λ 5 −λ 7 λ 7

−1+2γ−γ 4 +2γ 5 −γ 7 γ 7

−0.8677

−0.6969 f (Q −1 ) \ f(Q 1 )

−8

−1+2λ−2λ 4 +2λ 5 −λ 8 λ 8

−1+2γ−γ 4 +2γ 5 −γ 8 γ 8

−0.7549

−0.8369 f (Q −1 ) \ f(Q 1 )

−9

−1+2λ−2λ 4 +2λ 5 −λ 9 λ 9

−1+2γ−γ 4 +2γ 5 −γ 9 γ 9

−0.5462

−0.9122 f (Q −1 ) \ f(Q 1 )

−10

−1+2λ−2λ 4 +2λ 5 −λ 10 λ 10

−1+2γ−γ 4 +2γ 5 −γ 10 γ 10

−0.1596

−0.9528 f (Q −1 ) \ f(Q 1 )

−11

−1+2λ−2λ 4 +2λ 5 −λ 11 λ 11

−1+2γ−γ 4 +2γ 5 −γ 11 γ 11

0.5563

−0.9746 Q \ (f(Q 1 ) ∪ f(Q −1 ))

Table3.1: ThepointsinFigure 3.3.

iterate oordinates

numeri al

values ontainedin

0

0

0

0

0 f (Q 1 ) ∩ f(Q −1 )

−1

−1+λ

−1+γ λ γ

−0.8519

0.4618 f (Q −1 ) \ f(Q 1 )

−2

−1+2λ−λ 2 λ 2

−1+2γ−γ 2 γ 2

−0.7257

−0.2132 f (Q −1 ) \ f(Q 1 )

−3

−1+2λ−λ 3 λ 3

−1+2γ−γ 3 γ 3

−0.4919

−0.5766 f (Q −1 ) \ f(Q 1 )

−4

−1+2λ−λ 4 λ 4

−1+2γ−γ 4 γ 4

−0.05916

−0.7721 f (Q 1 ) ∩ f(Q −1 )

−5

−1+2λ−λ 5 λ 5

−1+2γ−γ 5 γ 5

0.7423

−0.8773 Q \ (f(Q 1 ) ∪ f(Q −1 ))

Table3.2: ThepointsinFigure3.4.

λ t ∈ I t

. Thefollowinglemmawillprovidesu ient ontrolwhen hangingthe

parameter

λ

.

Lemma3.4.0.2. Forany

t

andany

λ, λ ∈ I t

the symboli spa es

Σ λ,γ,κ

and

Σ λ ,γ,κ

oin ide.

Proof. Apoint

x = (x ˆ 1 , x 2 , x 3 ) ∈ ˆ Q

liesin

Λ ˆ λ t

ifandonlyifthereisasequen e

{i n } n∈Z

su hthat

x 1 = 1 − λ t

λ t

X ∞ n=1

i −n λ n t , x 2 = γ − 1 γ

X ∞ n=0

i n γ −n , x 3 = 1 − τ τ

X ∞ n=1

i −n τ n

and

f ˆ λ n t (ˆ x) ∈ ˆ Q i n

forall

n ∈ Z

. Let

x = (x ˆ 1 , x 2 , x 3 ) ∈ ˆ Λ λ t

. Weshowthat for

any

λ ∈ I t

thereisapoint

x ˆ = (x 1 , x 2 , x 3 ) ∈ ˆ Λ λ

su hthatthe orresponding sequen e

{i n } n∈Z

satises

i n = i n

for all

n ∈ Z

. Inthiswaywedeneamap

Ξ λ t ,λ : ˆ Λ λ t → ˆ Λ λ

by

Ξ λ t ,λ : ˆ x 7→ ˆx

.

Itsu estoshowthatthepoint

x ˆ = (x 1 , x 2 , x 3 )

dened by

x 1 = 1 − λ

λ X ∞ n=1

i −n (λ ) n , x 2 = γ − 1 γ

X ∞ n=0

i n γ n , x 3 = 1 − τ τ

X ∞ n=1

i −n τ n

satises

f ˆ λ n (ˆ x ) ∈ ˆ Q i n

forall

n ∈ Z

. Thenthisimpliesthat

x ˆ ∈ ˆ Λ λ

. A hange

of

λ

hasonlyinuen eonthese ond oordinateof

f (ˆ ˆ x)

. Sin ethelo alstable

manifoldsare parallel and oriented in the dire tion of the se ond oordinate

it su esto he k that when hanging

λ

, these ond oordinate nevermove

overtheverti aldis ontinuity

S v

. Wedothisbyanestimateofthederivative

d

 1−λ λ

P ∞

n=0 i −n λ n 

. Asimple al ulationgivesthat

d dλ

 1 − λ λ

X ∞ n=0

i −n λ n  ≤

1 λ 2

X ∞ n=0

i n λ n +

1 − λ λ

X ∞ n=1

ni n λ n−1

≤ 1 λ 2

X ∞ n=0

λ n + 1 − λ λ

X ∞ n=1

n−1 = 1 + λ

λ 2 (1 − λ) < 15,

if

1

2 < λ < 3 4

. Thisimpliesthat

|x 1 − x 1 | ≤ sup

λ

d dλ

 1 − λ λ

X ∞ n=0

i −n λ n 

t − λ | < 15|I t | ≤ α.

Thismeansthat

ˆ x

doesnot rosstheverti alpie eofthesingularityandhen e stayson thesame side ofthe singularityas

x ˆ

. Similarlyoneshowsthat any

iterate

f ˆ λ n (ˆ x )

ofstaysonthesamesideofthesingularityas

f ˆ λ n t (ˆ x)

.

Remark. The partition of

I

intosubintervalsisarbitrary soinfa t the sym-boli spa es oin ide for any

λ, λ ∈ I

.

We have shown that the symboli spa e

Σ λ,γ,κ

does not hange when

λ

varies. Wealsoneedtoestimatehowthemeasure

µ ˆ λ SBR

hanges.

Lemma 3.4.0.3. There is a onstant

c 1 > 0

su h that for any

t

and any

λ, λ ∈ I t

c −1 1 µ ˆ λ SBR t (C λ t ) ≤ ˆµ λ SBR (C λ ) ≤ c 1 µ ˆ λ SBR t (C λ t ),

(3.5)

forany ylinder setoftheform

C λ = k [i k i k+1 · · · i n ] n = T n

j=k f ˆ λ −j ( ˆ Q i j )

,

k < n

.

Proof. Sin ethereareonlynitelymanyinverseimagesoftheset

S v

,allother

inverseimagesofthesingularitywill onsistofahorizontalline. As

λ

variesover

I

thesehorizontallinesarenot hanged,onlytheinverseimagesof

S v

hanges.

Thereisthusa onstant

c 1

,independentof

t

,su hthatforany

λ ∈ I t

andany

ylindersetoftheform

C λ = k [i k i k+1 · · · i n ] n = T n

j=k f ˆ λ −j ( ˆ Q i j )

,

0 ≤ k < n

we

have

c −1 1 ˆ ν(C λ t ) ≤ ˆν(C λ ) ≤ c 1 ˆ ν(C λ t ).

Espe ially

c −1 1 ν( ˆ ˆ f λ −m t (C λ t )) ≤ ˆν( ˆ f λ −m (C λ )) ≤ c 1 ν( ˆ ˆ f λ −m t (C λ t ))

for any

m ∈ N

andso

c −1 1 µ ˆ λ SBR t (C λ t ) ≤ ˆµ λ SBR (C λ ) ≤ c 1 µ ˆ λ SBR t (C λ t ),

for any ylinder set of the form

C λ = k [i k i k+1 · · · i n ] n = T n

j=k f ˆ λ −j ( ˆ Q i j )

with

k < n

.

Remark. Lemma 3.4.0.3 is also valid for the onditional measures on the

stable manifold.

Sin e the entropy of the onditional measures are a.s. equal that of the

measure, the Shannon-M Millan-Breiman theorem implies that given

ε > 0

,

thereexists anumber

n 0 (λ, ˆ x, ε)

su hthat

ˆ

µ λ,s,ˆ SBR x n ˆ

y | ˆµ λ,s,ˆ SBR x ( ˆ R q λ (ˆ y)) < e −q log(γ−ε) , q > n 0 (λ, ˆ x, ε) o

> 1 − ε.

Lusin'sTheoremimpliesthatthereisanumber

n 1 (ε)

andaset

Ω ˆ 0,λ t

su hthat

n 1 (ε) ≥ n 0 (λ t , ˆ x, ε)

when

x ˆ ∈ ˆ Ω 0,λ t

and

µ ˆ λ SBR t ( ˆ Ω 0,λ t ) > 1 − ε

. Weput

Ω ˆ λ t = n

ˆ

y ∈ ˆ Ω 0,λ t ∩ ˆ W s (ˆ x) | ˆx ∈ Ω 0,λ t

and

ˆ

µ λ,s,ˆ SBR x ( ˆ R q λ (ˆ y)) < e −q log(γ−ε) , q > n 1 (ε) o .

Then

µ ˆ λ SBR t ( ˆ Ω λ t ) > 1 − 2ε

.

Put

Ω ˆ λ = Ξ λ t ,λ ( ˆ Ω t )

for

λ ∈ I t

. ByLemma3.4.0.3itfollowsthat if

y ˆ ∈ ˆ Ω λ

then

ˆ

µ λ SBR ( ˆ R q λ (ˆ y)) < c 1 e −q log(γ−ε) , q > n 1 (ε)

and

ˆ

µ λ SBR ( ˆ Ω λ ) > c −1 1 µ ˆ λ SBR t (Ω λ t ) > c −1 1 (1 − 2ε).

Takea

λ 0 ∈ I

and

x ˆ λ 0 ∈ ˆ Ω λ 0

. Dene

x(λ ˆ 0 ) = ˆ x λ 0

andfor

λ ∈ I

dene

x(λ) ˆ

sothat

ρ −1 λ,γ,κ (ˆ x(λ)) = ρ −1 λ 0 ,γ,κ (ˆ x(λ 0 ))

. Then

x ˆ

is ontinuous. ByLemma3.4.0.1

T i,t,k = Z

I t

Z

Ω ˆ λ ∩ ˆ R k i1 (λ)

Z

Ω ˆ λ ∩ ˆ R k i−1 (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 2 1

Z

I t

Z

Ω ˆ λt ∩ ˆ R k i1 (λ t )

Z

Ω ˆ λt ∩ ˆ R k i−1 (λ t )

χ {|Ξ λt,λ (ˆ y) 1 −Ξ λt,λ (ˆ z) 1 |<r}

dˆ µ λ SBR t ,s,ˆ x(λ t ) (ˆ z)dˆ µ λ SBR t ,s,ˆ x(λ t ) (ˆ y)dλ

= c 2 1 Z

Ω ˆ λt ∩ ˆ R k i1 (λ t )

Z

Ω ˆ λt ∩ ˆ R k i−1 (λ t )

Z

I p

χ {|Ξ λt,λ (ˆ y) 1 −Ξ λt,λ (ˆ z) 1 |<r}

dλdˆ µ λ SBR t ,s,ˆ x(λ t ) (ˆ z)dˆ µ λ SBR t ,s,ˆ x(λ t ) (ˆ y)

≤ c 2−1 + ε) −k r ˆ µ λ SBR t ,s,ˆ x(λ t ) ( ˆ R k i1 (λ t ))ˆ µ λ SBR t ,s,ˆ x(λ t ) ( ˆ R k i−1 (λ t ))

≤ c 2 r(γ −1 + ε) −k (γ − ε) −k µ ˆ λ SBR t ,s,ˆ x(λ t ) ( ˆ R k i1 (λ t )).

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.

I dokument The Family of Belykh Maps Persson, Tomas (sidor 49-57)

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