Proof of the theorem

theset

{y ∈ W ^s (x) | d(x, y) < δ}

^{isa onne ted urveoflength}

2δ

^thendene

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

^{tobethestablemanifoldof}

x

^withlength

2δ

^{. 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 | ^Ω 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 | ^Ω 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

Ω _λ

Z

Ω _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 ⁴ +λ ⁵ λ ⁵

−1+2γ−2γ ⁴ +γ ⁵ 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

dλ

 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

nλ 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    

  |λ ^t − λ ^′ | < 15|I ^t | ≤ α.

Thismeansthat

ˆ x ^′

^{doesnot rosstheverti alpie eofthe}singularityandhen e stayson thesame side ofthe singularityas

x ˆ

^{. Similarlyoneshowsthat any}

iterate

f ˆ _λ ^{n ′} (ˆ x ^′ )

^{ofstaysonthesamesideofthe}singularityas

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 (γ ⁻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.}

In document The Family of Belykh Maps Persson, Tomas (Page 49-57)