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Ferguson, Andrew and Pollicott, Mark. (2012) Escape rates for Gibbs measures. Ergodic
Theory and Dynamical Systems, Vol.32 (No.3). pp. 961-988. ISSN 0143-3857
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Escape rates for Gibbs measures
ANDREW FERGUSON and MARK POLLICOTT
Ergodic Theory and Dynamical Systems / Volume 32 / Issue 03 / May 2012, pp 961 - 988 DOI: 10.1017/S0143385711000058, Published online: 27 April 2011
Link to this article: http://journals.cambridge.org/abstract_S0143385711000058
How to cite this article:
ANDREW FERGUSON and MARK POLLICOTT (2012). Escape rates for Gibbs measures. Ergodic Theory and Dynamical Systems, 32, pp 961-988 doi:10.1017/S0143385711000058
doi:10.1017/S0143385711000058
Escape rates for Gibbs measures
ANDREW FERGUSON and MARK POLLICOTT
Mathematics Institute, Zeeman Building, University of Warwick, Coventry CV4 7AL, UK (e-mail: a.j.ferguson@warwick.ac.uk, m.pollicott@warwick.ac.uk)
(Received30August2010and accepted in revised form4January2011)
Abstract. In this paper we study the asymptotic behaviour of the escape rate of a Gibbs measure supported on a conformal repeller through a small hole. There are additional applications to the convergence of the Hausdorff dimension of the survivor set.
1. Introduction
Given any transformation T :X→X preserving an ergodic probability measure µ and any Borel set A⊂X, the escape rate quantifies the asymptotic behaviour of the measure of the set of points x∈X for which none of the firstn terms in the orbit intersectsU. Bunimovich and Yurchenko[7]considered the fundamental case of the doubling map and Haar measure, and whereU is a dyadic interval. Subsequently, Keller and Liverani[16]
proved a more general perturbation result which was then used to show, amongst other things, that a similar formula holds in the case thatT is an expanding interval map andµ the absolutely continuous invariant probability measure. Other papers regarding this topic include[1, 6, 9, 19]and references therein.
In this paper, we prove analogous results in the more general setting of Gibbs measures supported on conformal repellers. Much of the analysis is undertaken in the setting of subshifts of finite type; this not only allows us to prove similar results for a broad class of maps which can be modelled symbolically but also improve on the work of Lind[18]who considered the convergence of topological entropy for a topologically mixing subshift.
Another interesting aspect of our analysis is the connection with the work of Hirata[12]
on the exponential law for first return times for Axiom A diffeomorphisms. Some of the ingredients in our approach were suggested by Hirata’s paper, although we had to significantly modify the actual details.
LetMbe a Riemannian manifold and f :M→MaC1-map. Let J be a compact subset ofMsuch that f(J)=J. We say that the pair(J, f)is a conformal repeller if: (1) f|J is a conformal map;
(2) there existc>0 andλ >1 such thatkd fxnvk ≥cλnkvkfor allx∈ J,v∈TxM, and
n≥1;
(4) Jis maximal, i.e. there exists an open neighbourhoodV ⊃Jsuch that
J= {x∈V : fn(x)∈V for alln≥0}.
Letφ:J→Rbeα-Hölder and letµdenote the associated equilibrium state, i.e.
P(φ)=sup
hν(f)+
Z
φdν: f∗(ν)=ν, ν(J)=1
=hµ(f)+
Z
φdµ,
wherehν(f)denotes the Kolmogorov–Sinai entropy of the measureν(see[26]for further details).
Fixz∈J; for >0, we define the escape rate of µ through B(z, ) (i.e. the rate at which mass ‘escapes’ or ‘leaks’ through the holeB(z, )) by
rµ(B(z, ))= −lim supk→∞
1
k logµ{x∈J: f
i(x)6∈B(z, ),0≤i≤k−1}.
Our first result concerns the asymptotic behaviour ofrµ(B(z, ))for small.
THEOREM1.1. Let(J, f)be a conformal repeller,φ:J→RH¨older continuous, andµ the associated equilibrium state. Fix z∈ J ; then
lim
→0
rµ(B(z, ))
µ(B(z, )) =dφ(z)=
(
1 if z is not periodic,
1−eφp(z)−p P(φ) if z has prime period p, whereφp(z)=φ(z)+φ(f(z))+ · · · +φ(fp−1(z)).
We also obtain an asymptotic formula for the Hausdorff dimension of the survivor set:
J= {x∈J: fk(x)6∈B(z, )for allk≥0},
i.e. all points whose orbits are-bounded away fromz.
Suppose now that f ∈C1+α(J)for some α >0. Let µ denote the equilibrium state related to the potentialφ= −slog|f0|, wheres=dimH(J). For >0, we lets denote
the Hausdorff dimension of the setJ.
THEOREM1.2. Let(J, f)be a conformal repeller with f∈C1+α(J). Letφ= −slog|f0|
and letµdenote the associated equilibrium state. Fix z∈J ; then
lim
→0
s−s µ(B(z, ))=
dφ(z)
R
log|f0|dµ.
Remark 1.3. A similar formula was obtained by Hensley[11]in the setting of continued fractions.
The paper is structured as follows: in §2 we apply Theorems 1.1 and 1.2 to concrete examples. In §3 we study the spectral properties of transfer operators acting on a certain class of Banach spaces. Section 4 contains a perturbation result, while in §5 we prove the result in the analogue of Theorem 1.1 in the setting of subshifts of finite type. Finally, §§6 and 7 contain the proofs of Theorems 1.1 and 1.2, respectively.
2. Examples
To illustrate the main results, we briefly consider two simple examples.
2.1. Hyperbolic Julia sets. Let f :bC→bCbe a rational map of degreed≥2, wherebC
points of f, i.e.
J=cl({z∈bC: fp(z)=z,for somep≥1 and|(fp)0(z)|>1}).
The map f :J→J is a conformal expanding map and the results of the previous section apply. As an example, the map f(z)=z2+cfor|c|<1/4 is hyperbolic. Define φ:J→R byφ(z)= −slog|2z|, wheres denotes the Hausdorff dimension of J. Let µ denote the associated equilibrium state. Setting z=(1+
√
1−4c)/2, we see that f(z)=zand|f0(z)|>1 and accordingly Theorem 1.1 implies that
lim
→0
rµ(B(z, )) µ(B(z, )) =1−
1
|2z|s.
2.2. One-dimensional Markov maps. Assume that there exist a finite family of disjoint closed intervalsI1,I2, . . . ,Im⊂ [0,1]and aC1+α-map f :Si Ii→ [0,1]such that:
(1) for everyi, there is a subsetP=P(i)of indices with f(Ii)∩Si Ii=Sk∈P Ik;
(2) for every x∈S
iint(Ii), the derivative of f satisfies |f0(x)| ≥ρ for some fixed
ρ >0;
(3) there existλ >1 andn0>0 such that if fm(x)∈Si Iifor all 0≤m≤n0−1, then |(fn0)0(x)| ≥λ.
Let J= {x∈ [0,1] : fn(x)∈S
i Ii for alln∈N}. The set J is a repeller for the map f
and conformality follows from the domain being one dimensional.
If we take I1= [0,1/3], I2= [2/3,1], and let f(x)=3x(mod 1), the associated
repeller Jis the middle-third Cantor set. Letz=1/4; thenz∈Jand has prime period 2. Setφ(x)= −log(2)and letµdenote the associated equilibrium state. Then Theorem 1.1 implies that
lim
→0
rµ(B(1/4, )) µ(B(1/4, )) =1−
1 22 =
3 4. For >0, we set
J= {x∈J: fk(x)6∈B(1/4, )fork=0,1,2, . . .}.
Lets=dimH(J)ands=log(2)/log(3); then Theorem 1.2 implies that
lim
→0
s−s µ(B(1/4, ))=
3 4 log(3).
3. Spectral properties of the transfer operator
In this section, we study the spectral properties of the transfer operator. We first fix notation which will be used for the rest of the paper. Throughout the rest of the paper,cwill denote a positive and finite constant which may change in value with successive uses. Let Adenote an irreducible and aperiodic l×l matrix of zeroes and ones, i.e. there exists a positive integerdsuch thatAd>0. We define the subshift of finite type (associated with matrixA) to be
6= {(xn)∞n=0:A(xn,xn+1)=1 for alln}.
If we equip the set{0,1, . . . ,l−1}with the discrete topology, then6is compact in the corresponding Tychonov product topology. The shiftσ :6→6 is defined byσ (x)=y, where yn=xn+1 for all n, i.e. the sequence is shifted one place to the left and the first
Forθ∈(0,1), we define a metric on6bydθ(x,y)=θm, wheremis the least positive integer (assuming that such anmexists) with xm6=ym; otherwise, we set dθ(x,x)=0.
Equipped with the metricdθ, the space(6,dθ)is complete, and moreover the topology induced bydθ agrees with the previously mentioned Tychonov product topology. Finally, forx∈6and a positive integern≥1, we define the cylinder of lengthncentred onxto be the set
[x]n= [x0,x1, . . . ,xn−1] = {y∈6:yi=xi fori=0,1, . . . ,n−1}.
Fix adθ-Lipschitz continuous functionφ:6→Rand recall that we letµdenote its equilibrium state defined in the introduction, i.e.
P(φ):=sup
hν+
Z
φdν:σ∗(ν)=ν, ν(6)=1
=hµ+
Z
φdµ.
We let
L1(µ):=
w:6→C:wis measurable and
Z
|w|dµ <∞
,
which, equipped with the normkwk1=R |w|dµ, is a Banach space. We now describe a particular subspace of L1(µ) on which the transfer operator will act: forw∈L1(µ), x∈6, and a positive integerm, we set
osc(w,m,x)=esssup{|w(y)−w(z)| :y,z∈ [x]m}.
We introduce the semi-norm
|w|θ =sup
m≥1
θ−mkosc(w,m,·)k 1.
We let
Bθ= {w∈L1(µ): |w|θ<∞}.
It is worth noting that if we were to take the supremum normk · k∞in place of theL1 norm, then the space coincides with Lipschitz continuous functions (with respect to the metricdθ).
We equipBθ with the norm
kwkθ= |w|θ + kwk1.
This space was first introduced by Keller[14], in a more general framework, where the following result was also proved.
PROPOSITION3.1. (Keller) The space (Bθ,k · kθ) is complete. Furthermore, the set
{w∈Bθ : kwkθ ≤c}is L1-compact for any c>0.
We introduce the transfer operatorL=Lφ:Bθ →Bθ
(Lw)(x)= X σ(y)=x
eφ(y)w(y).
We leti=(i0,i1, . . . ,ik−1)denote an allowed string of lengthk; then we can write (Lkw)(x)=P
|i|=keφ
k(i x)
Another Banach space that we require is that of Lipschitz functions
Fθ=
w:6→C:sup
m≥1
θ−mkosc(w,m,·)k
∞<∞
.
The following theorem describes the spectral properties ofLacting on the spaceFθ of dθ-Lipschitz continuous functions; for a proof, see[21, Theorem 2.2].
PROPOSITION3.2. (Ruelle)Letφ∈Fθ be real valued and suppose that A is irreducible and aperiodic.
(1) There is a simple maximal positive eigenvalue λ=λφ of L with corresponding strictly positive eigenfunction g=gφ∈Fθ.
(2) The remainder of the spectrum ofL:Fθ→Fθ (excludingλ >0) is contained in a disk of radius strictly smaller thanλ.
(3) There is a unique probability measureνsuch thatL∗ν=λν.
(4) λ−kLkw→gR
wdνuniformly for allw∈Fθ, where g is as above andR g dν=1.
We remark that the equilibrium state µ is absolutely continuous with respect to the eigenmeasureν, with the Radon–Nikodym derivative being given by the eigenfunctiong. By scaling the operator L, if necessary, we may assume without loss of generality that λ=1; further, asg>0, we may assume thatL1=1.
Another useful property ofµis the Gibbs property (see[3]for further details). Namely, there exists a constantc>1 such that for anyx∈6and positive integernwe have that
c−1≤µ[x]n
eφn(x) ≤c. (1)
We now prove a result relating to the spectrum ofLacting onBθ, namely that it has a spectral gap. A crucial part in this process is proving a Lasota–Yorke inequality†.
LEMMA3.3. There exists c>0such that for anyw∈Bθ we have
|Lkw|θ≤c(θk|w|θ+ kwk1).
Proof. Letx,y∈6be such thatdθ(x,y)≤θm; then
|Lkw(x)−Lkw(y)| ≤ X
|i|=k
|eφk(i x)w(i x)−eφk(i y)w(i y)|
≤ X
|i|=k
eφk(i x)osc(w,k+m,i x)+eφk(i x)|1−eφk(i y)−φk(i x)||w(i y)|
≤cX
|i|=k
eφk(i x)osc(w,k+m,i x)+θm e φk(i x)
µ[i x]k+m
Z
[i x]k+m
|w|dµ
≤cX
|i|=k
eφk(i x)osc(w,k+m,i x)+θm c
µ[x]m
Z
[i x]k+m
|w|dµ
,
where we used the Gibbs property (1) in the final line, i.e.
eφk(i x) µ[i x]k+m ≤
c
eφm(x) ≤
c2 µ[x]m.
Thus,
osc(Lkw,m,x)≤cX
|i|=k
eφk(i x)osc(w,k+m,i x)+θm c
µ[x]m
Z
[i x]k+m
|w|dµ
.
Integrating with respect toµ, and again invoking (1), we see that
Z
osc(Lkw,m,x)dµ(x)≤c
Z
osc(w,k+m,x)dµ(x)+θmkwk1
;
dividing byθmand taking suprema yields
|Lw|θ≤c(θk|w|θ+ kwk1).
Finally, we see that
kLkwk
θ = |Lkw|
θ+ kLkwk 1
≤cθk(|w|θ+ kwk1)+ kwk1
≤c(θk|w|θ+ kwk1). 2
LEMMA3.4. The operatorL:Bθ→Bθ has a simple maximal eigenvalueλ=1, while the rest of the spectrum is contained in a ball of radius strictly less than1.
Proof. We begin by proving that for any w∈Bθ, Lkw converges toR
wdµ in L1(µ). Fix >0 and choosev∈Fθ such thatkv−wk1< /3; by Proposition 3.2, there exists a positive integerN such thatkLn(v)−R v
dµk1< /3 for alln≥N, in which case we see that
Ln(w)−
Z
wdµ
1
≤ kLn(w−v)k1+
Ln(v)−
Z
vdµ
1 + Z
vdµ−
Z
wdµ
1
≤2kv−wk1+
Ln(v)−
Z
vdµ
1< .
This in turn implies that for eachw∈B= {v∈Bθ : kvkθ≤1},
kLn(w)|
C⊥k1=cinf∈ C
kLn(w)−ck1→0 asn→ ∞,
whereC⊥= {w∈Bθ :R wdµ=0}. We claim that this convergence is uniform over B. To see this, fixδ >0 andw∈B; then there exists a positive integer N=N(w)such that
kLn(w)|
C⊥k1≤δ/2 for all n≥N. By Proposition 3.1, B is compact and so the cover {B1(w, δ/2)}w∈Bhas a finite subcover, sayB1(w1, δ/2),B1(w2, δ/2), . . . ,B1(wm, δ/2).
In which case, if n≥N:=maxi=1,2,...,m N(wi), we have kLnφ(w)|C⊥k1≤δ for
anyw∈B.
Finally, to show the existence of a spectral gap, from Proposition 3.3, we observe for w∈Bandn≥N that
kL2n(w)|
C⊥kθ ≤c(θ
n|Ln(w)|
C⊥|θ+ kL n(w)|
C⊥k1) ≤c(θ2n|w|
C⊥|θ+θ nkw|
C⊥k1+ kL n(w)|
C⊥k1) ≤c(θ2n+θn+δ).
We may choosenandδso thatkL2n(w)|
C⊥kθ<1, which proves thatLhas a spectral
3.1. Singular perturbations of the transfer operator. We introduce a perturbation of the transfer operatorL: let{Un}nbe a family of open sets; further, we require that they satisfy
the following technical conditions:
(1) {Un}nare nested withTn≥1Un= {z};
(2) eachUnconsists of a finite union of cylinder sets, with each cylinder having lengthn;
(3) there exist constantsc>0, 0< ρ <1 such thatµ(Un)≤cρnforn≥1;
(4) there are a sequence {ln}n⊂N and a constant κ >0 such that κ <ln/n≤1 and
Un⊂ [z]ln for alln≥1;
(5) if σp(z)=z has prime period p, thenσ−p(Un)∩ [z0z1· · ·zp−1] ⊆Un for large
enoughn.
Remark 3.5. We observe that (5) above is not absolutely essential for the application to conformal repellers and serves only to greatly simplify the analysis.
Forn≥1, we define the perturbed operatorLn:Bθ →Bθ by
Ln(w)(x)=L(χUc nw)(x).
For a positive integern, we let6n=Tk≥06\σ−j(Un). By choosingnlarge enough,
we can ensure that the system(6n, σ|6n)is topologically mixing, and so the results of[8]
apply, namely we have the following proposition.
PROPOSITION3.6. (Collet, Mart´ınez, Schmitt)For each n, there exist a continuous gn:
6→Rwith gn>0 and aλn>0such thatLngn=λngn; moreover, for any continuous
w:6→C, we have
kλ−nkLknw−νn(w|6
n)gnk∞→0,
whereνn denotes the unique probability measure guaranteed by Proposition 3.2, i.e.νn
satisfiessupp(νn)=6nand(L∗nνn)(w)=λnνn(w)forw∈Fθ(6n).
Moreover, we may prove a Lasota–Yorke style inequality forLn:Bθ →Bθ, which, in
conjunction with Proposition 3.6 and the methods of Lemma 3.4, we can use to show that gn∈Bθ and thatλnis a simple maximal eigenvalue forLn:Bθ→Bθ.
The perturbationLnis singular with respect to thek · kθnorm. We adopt the approach
of[15]and introduce a weak norm
kwkh:= |w|h+ kwk1=sup
j≥0
sup
m≥1 θ−mZ
σ−j(U m)
|w|dµ+ kwk1.
Throughout this section, we assume thatθ∈(ρ,1). Our first result states that the weak norm is dominated by the strong norm.
LEMMA3.7. Under the assumptions above, we have
kwkh≤ckwkθ
Proof. We first relate the strong norm with theL∞norm. Letc=maxi=0,1,...,l−1µ[i]−11;
then, forµalmost allx∈6,
|w(x)| ≤osc(w,1,x)+c
Z
[x0]1 |w|dµ
≤c
Z
[x0]1
osc(w,1,y)dµ(y)+
Z
[x0]1 |w|dµ
≤ckwkθ. (2)
Ifθ∈(ρ,1), then
|w|h≤sup
m≥1
θ−mµ(U
m)kwk∞≤ckwkθ. 2
3.2. Convergence of the spectral radii. In this section, we prove a preliminary result relating to the behaviour of the spectra of the operators Ln acting on Bθ. From
Proposition 3.6, it is easy to see that for anyu∈6we have
P6n(φ):=logλn= lim k→∞
1 k log(L
k
n1(u)). (3)
PROPOSITION3.8. Under assumptions (1)–(5), we havelimn→∞λn=λ.
Proof. As Un⊂ [z]ln, setting 6˜n=6\
T
k≥0σ
−k[z]
ln, it is easy to see that 6˜n⊂6n.
Accordingly, it suffices to show thatP6˜n(φ)→P(φ).
As(6, σ)is topologically mixing, we may find a positive integerd such thatAd>0. Fixu∈6; for integerskandn, we set
Bk= {x0x1· · ·xk−1:x0x1· · ·xk−1u∈6}, Bk,n= {x0x1· · ·xk−1∈Bk: [x0x1· · ·xk−1] ∩6n6= ∅},
Zk(φ)=
X
x0x1···xk−1∈Bk
eφk(x0x1···xk−1u), Z
k,n(φ)=
X
x0x1···xk−1∈Bk,n
eφk(x0x1···xk−1u).
It is easy to see thatLk1(u)=Z
k(φ)(respectivelyLkn1(u)=Zk,n(φ)) and so by (2)
we have that P(φ)=limk→∞(1/k)logZk(φ) (respectively P6n(φ)=limk→∞(1/k)
logZk,n(φ)).
Fix >0. By (3), there existsa>0 such that Zk(φ)≥aek(P(φ)−) for allk≥1. In
addition, ashtop(σ ) >0, there existsb>0 such that|Bk| ≥bek(htop(σ )−)for allk≥1.
Fix large integersk andn such that bothbek(htop(σ)−)>l
n−k+1 and 2(k+d) <
ln. Observe that the stringz0z1· · ·zln−1has preciselyln−k+1 subwords of lengthk;
accordingly, the first condition on k and n guarantees the existence of a finite word x∈Bk such that x does not appear as a subword of z0z1· · ·zln−1. Fix m∈N and
let y1,y2, . . . ,ym∈Bln−2k−2d; we now associate with this list a unique element
of Bm(ln−k). Choose s
1,s2, . . . ,sm,t1,t2, . . . ,tm∈B
d so that the word w:=
y1s1x t1y2s2x t2· · ·tm−1ymsmx tm∈ {1,2, . . . ,l}m(ln−k) is such that tmu∈6; this is
possible asAd>0.
It is easy to see that asxis contained in any subword of lengthn, the wordz0z1· · ·zln−1
cannot be contained as a subword of the periodic extension ofw. Hence,w∈Bm(ln−k),ln
and so
Zm(ln−k),ln(φ)≥(ae(
Taking logarithms, dividing bym, and lettingm→ ∞yields
P6˜n(φ)≥
log(a) ln−k
+(1−) ln
ln−k(
P(φ)−).
Finally, lettingn→ ∞and→0 gives the result. 2
Remark 3.9. The proof of Proposition 3.8 is modified from[4], where an analogous result for topological entropy is proved.
3.3. A uniform Lasota–Yorke inequality. We now prove that the transfer operatorsLn
satisfy a uniform Lasota–Yorke inequality. We assume that the transfer operator L is normalized, i.e.L1=1. Iterating the perturbed operatorLn, we see that
(Lk
nw)(x)=
X
σk(y)=x
hn,k(y)eφ
k(y)
w(y),
wherehn,k(x)=Qkj−=10χUc n(σ
jx)andφk(y)=Pk−1
j=0φ(σj(y)).
LEMMA3.10. For any positive integers k,n, we have
kLk nkh≤1.
Proof. Letw∈L1; then
kLnwk1=
Z
|LχUc nw|dµ
≤
Z
L|χUc nw|dµ
=
Z
|χUc
nw|dµ≤ kwk1. (4)
In addition, fixing j≥0,m≥1, we see that
θ−mZ σ−j(U
m)
|Lnw|dµ≤θ−m
Z
σ−j(U m)
L(|χUc nw|)dµ
=θ−m
Z
σ−(j+1)(Um)
χUc n|w|dµ
≤θ−m
Z
σ−(j+1)(Um)
|w|dµ≤ |w|h.
Taking the supremum over j andmyields
|Lnw|h≤ |w|h. (5)
Combining (4) and (5) and iterating completes the proof. 2
LEMMA3.11. There exists a constant c>0 such that for any positive integers n,k we have
|hn,kw|θ≤ |w|θ+cθ−kkwkh
for allw∈Bθ.
Proof. We prove this by induction, namely we prove that for anyw∈Bθ we have
|χσ−j(Uc
n)w|θ≤ |w|θ+θ −jkwk
To show this, fix a positive integerm. We consider two cases, namely: j+n≤mand m<j+n. If we suppose that j+n≤m, then osc(χσ−j(Uc
n)w,m,x)≤osc(w,m,x)for
allx∈6, and thus
θ−mZ osc(χ σ−j(Uc
n)w,m,x)dµ(x)≤θ
−mZ osc(w,m,x)dµ(x)≤ |w|
θ. (7)
On the other hand, if m<j+n it is easy to see that if [x]m⊂σ−j(Unc), then osc(χσ−j(Uc
n)w,m,x)=osc(w,m,x). If, however,[x]m∩σ −j(U
n)6= ∅then
osc(χσ−j(Uc
n)w,m,x)=max(osc(w,m,x),kχ[x]mwk∞),
in which case
osc(χσ−j(Uc
n)w,m,x)=max(osc(w,m,x),kχ[x]mwk∞)
≤osc(w,m,x)+ 1
µ[x]m
Z
[x]m
|w|dµ,
which implies that
θ−mZ osc(χ σ−j(Uc
n)w,m,x)dµ(x)≤ |w|θ +θ −mZ
{x:[x]m∩σ−j(Un)6=∅}
|w|dµ. (8)
We now analyse two further subcases. Ifm≤ j, we see that
θ−mZ
{x:[x]m∩σ−j(Un)6=∅}
|w|dµ≤θ−jkwk1. (9)
If j<m<j+n, the fact that the open sets{Un}nare nested implies that
{x: [x]m∩σ−j(Un)6= ∅} ⊂σ−j(Um−j).
In which case,
θ−m
Z
{x:[x]m∩σ−j(Un)6=∅}
|w|dµ≤θ−j|w|h. (10)
If we combine (7), (9), and (10), we obtain (6). This completes the proof. 2
LEMMA3.12. There exists a constant c>0such that
kLknwkθ ≤c(θkkwkθ + kwkh)
for allw∈Bθand n,k≥1.
Proof. Fixx,y∈6and suppose thatdθ(x,y)=θmwithm≥1; then
|(Lk
nw)(x)−(Lknw)(y)| ≤
X
|i|=k
|eφk(i x)hn,k(i x)w(i x)−eφ
k(i y)
hn,k(i y)w(i y)|
≤ X
|i|=k
eφk(i x)|hn,k(i x)w(i x)−hn,k(i y)w(i y)|
+eφk(i x)|1−eφk(i y)−φk(i x)||w(i y)| ≤ X
|i|=k
eφk(i x)[osc(hn,kw,k+m,i x)+c·osc(w,k+m,i x)]
+cθm X
|i|=k
eφk(i x) µ[i x]k+m
Z
µ[i x]k+m
Integrating and dividing byθmimplies that
|Lk
nw|θ≤cθk(|w|θ+ |hn,kw|θ)+ckwk1. (11)
And so, from (4) and (11) along with Lemma 3.11, we deduce that
kLknwkθ = |Lnkw|θ+ kLknwk1
≤cθk(|hn,kw|θ+ |w|θ)+ kwk1
≤cθk|w|θ+ckwkh≤cθkkwkθ +ckwkh.
This completes the proof. 2
Remark 3.13. The advantage of introducing the weak normk · khis that it overcomes the restrictions imposed by the usual weak normk · k1. In particular, had we considered the usualk · k1norm it would have imposed the condition that 0< θ <1 be chosen sufficiently small (leading to complications later in the proof when we also require thatρ < θ <1).
3.4. Quasi-compactness of Ln. A prerequisite for proving quasi-compactness ofLnis
that the unit ball is compact with respect to the weak norm.
PROPOSITION3.14. The set B= {w∈Bθ: kwkθ≤1}isk · kh-compact.
Proof. Let (fn)n∈B be any sequence. By Proposition 3.1, there exist a subsequence
(fnk)k and f ∈B such thatkfnk− fk1→0. It suffices to show that |fnk− f|h→0.
As f, fnk∈B, we have thatc=supk≥1kf − fnkk∞<∞. Fix >0 and choose a positive
integer M such thatθ−mµ(Um)≤/cfor allm>M. Choose a positive integerK such
thatkf − fnkk1≤θ
Mfor allk≥K. For fixedm, j, ifm>Mwe have
θ−mZ σ−j(U
m)
|f − fnk|dµ≤θ −mµ(U
m)kf − fnkk∞< . (12)
Otherwise,m≤M, in which case fork≥Kwe have
θ−m
Z
σ−j(U m)
|f − fnk|dµ≤θ −M
Z
|f − fnk|dµ < . (13)
Taking (12) and (13) together implies that|f − fnk|h< fork≥K. This completes
the proof. 2
We now prove quasi-compactness ofLnusing a criterion of Hennion.
LEMMA3.15. The essential spectral radii of the operators Ln are uniformly bounded
byθ.
Proof. To show that the essential spectral radii of the operatorsLnare bounded byθ, we
note that Lemmas 3.12 and 3.14 show that the operatorsLnsatisfy the hypotheses of[10,
Corollary 1], namely:
(1) Ln({w∈Bθ: kwkθ≤1})is conditionally compact in(Bθ,k · kh);
(2) for eachk, there exist positive real numbers Rk,rk such that lim infk→∞(rk)1/k=
r< λnfor which
In which case, we conclude thatLn is quasi-compact with essential spectral radius
bounded byr. Condition (1) can be deduced from Proposition 3.1, while condition (2) is the uniform Lasota–Yorke inequality proved in Lemma 3.12. Finally, Proposition 3.8 implies that for anyθ∈(0,1)we have thatλn> θfor largen. 2
3.5. Stability of the spectrum. We introduce a so-called ‘asymmetric operator norm’ for which the operatorsLnconverge toLasn→ ∞. For a linear operatorQ:Bθ→Bθ, we
define
k|Q|k =sup{kQwkh: kwkθ≤1}.
Recall the Gibbs property (1) ofµ, namely that there exists a constantc>1 such that for anyx∈6and a positive integernwe have that
c−1≤ µ[x]n
eφn(x) ≤c. (14)
Using the above, it is relatively easy to show the following proposition, which is stated without proof.
PROPOSITION3.16. (Gibbs property) There exists a constant c>0 such that for any positive integers n,m and j≥n we have
µ(Un∩σ−j(Um))≤cµ(Un)µ(Um).
LEMMA3.17. There exists a constant c>0such that
k|L−Ln|k ≤c(ρθ−1)n
for all n.
Proof. Letw∈Bθ be such thatkwkθ≤1; then
k(L−Ln)wk1= kLχUnwk1
≤ kχU
nwk1
≤µ(Un)kwk∞≤cµ(Un)kwkθ ≤cµ(Un). (15)
On the other hand, for fixedm, jwe have
θ−mZ σ−j(U
m)
|(L−Ln)w|dµ≤cθ−mµ(σ−(j+1)(Um)∩Un)kwkθ.
Fix positive integersm, j. We study three cases, namely: (1) n≤ j+1;
(2) j+1<n<m+ j+1; (3) m+j+1≤n.
First, we suppose thatn≤ j+1, which implies from Proposition 3.16 that
θ−mµ(σ−(j+1)(U
m)∩Un)≤cθ−mµ(Um)µ(Un)≤cρn. (16)
with Proposition 3.16 we see that θ−mµ(σ−(j+1)(U
m)∩Un)≤µ(σ−(j+1)(Um)∩Uj+1) ≤cθ−mµ(Um)µ(Uj+1) ≤cθ−mρm+j+1
≤c(θ−1ρ)m+j+1
≤c(θ−1ρ)n. (17)
Ifn≥m+ j+1, θ−mµ(σ−(j+1)(U
m)∩Un)≤θ−mµ(Un)≤θ−nµ(Un)≤c(θ−1ρ)n. (18)
Combining (16), (17), and (18) yields
|(L−Ln)w|h≤c(θ−1ρ)nkwkθ.
Combining this with (15) completes the proof. 2
We note that Lemmas 3.10, 3.12, 3.15, and 3.17 show that the operatorsLnsatisfy the
hypotheses of[15, Theorem 1]. We now cite a specific consequence of the result. Forδ >0 andr> θ, let
Vδ,r = {z∈C: |z| ≤ror dist(z,spec(L))≤δ}.
Then, by[15, Theorem 1], there existsN=N(δ,r)such that
Sδ,r=sup{k(z−Ln)−1kθ :n≥N,z∈C\Vδ,r}<∞, (19)
wherek(z−Ln)−1kθ denotes the operator norm of(z−Ln)−1:Bθ→Bθ.
We may use quasi-compactness ofLnto write
Ln=λnEn+9n,
whereEnis a projection onto the eigenspace{cgn:c∈C}andEn9n=9nEn=0.
PROPOSITION3.18. There exist a positive integer N and constants c>0and0<q<1 such that for all n≥N we have
k9nk1k∞≤cqk for any k≥1.
Proof. Fix q∈(θ,1)such that spec(L)\{1} ⊂B(0,q). Then by Proposition 3.8 there exists a positive integerN such that for alln≥N, we may write using standard operator calculus
9k n=
1 2πi
Z
|t|=q
tk(t−Ln)−1dt.
Then, from Lemma 3.7 and (19) above, we see that
k9k
n1k∞≤ck9nk1kθ
≤c
Z
|t|=q
|t|kk(t−Ln)−1kθdt
≤cqk. 2
PROPOSITION3.20. There exists a constant c>0such that for all n
kEn1k∞≤c.
Proof. Forn≥N, write
En=
1
2πi
Z
|t−1|=1−q
(t−Ln)−1dt.
Then, from Lemma 3.7 and (19) above, we see that
kEn1k∞≤ckEn1kθ ≤c
Z
|t−1|=1−q
k(t−Ln)−1kθdt
≤c. 2
4. An asymptotic formula forλn
In this section, we prove the following proposition.
PROPOSITION4.1. Fixφ∈Bθ; then
lim
n→∞
λ−λn
µ(Un) =
(
λ if z is not periodic,
λ(1−λ−peφp(z)) if z has prime period p.
We prove the proposition in the case thatLis normalized, i.e.L1=1; the more general statement above can be deduced by scaling the operator.
Letmndenote the restriction ofµtoIn, i.e.
mn= µ |U
n
µ(Un)
.
The following four lemmas were motivated by corresponding results in[12].
LEMMA4.2. If z is non-periodic, then
lim
n→∞
R
En(LχUn)dmn
1−λn =nlim→∞
Z
En1dmn=1.
Proof. For simplicity, we put
[En] =
Z
En(LχUn)dmn.
Then, by usingLχUn=1−Ln1,
[En] =(1−λn)
Z
En1dmn.
Asz is non-periodic, it follows from the fact that a countable intersection of nested compact sets is non-empty that for any integerk≥1, there exists Nk such thatUNk ∩
σ−j(U
Nk)= ∅for j=1,2, . . . ,k.
Then, for anyx∈σ−jUNk, 1≤ j≤k, we have thatx6∈UNk. So, for anyn>Nk and
anyx∈σ−kUn, we see that
χUc
n(x)χUnc(σ (x))· · ·χUnc(σ
So, forn>Nk, we see that
χUn(x)L
k
n1(x)=χUn(x)L
k1(x)=χ Un(x).
And soR Ln1dmn=1 for alln>Nk.
We now use the decompositionLk
n=λknEn+9nk to see that for anykandn>Nk we
have 1− Z
En1dmn
= (λ k n−1)
Z
En1dmn+
Z
9k n1dmn
≤ |1−λk
n|kEn1k∞+ k9nk1k∞
≤c(|1−λkn| +qk),
where Propositions 3.18 and 3.20 were used in the final line. This completes the proof. 2
LEMMA4.3. If z is non-periodic, then
lim
n→∞
R
En(LχUn)dmn
µ(Un)
=1.
Proof. We letTn(x)denote the first return time (assuming that it exists) forx∈Un, i.e.
Tn(x)=inf{i∈N:σi(x)∈Un};
then
Z
Tndmn =
∞
X
i=1
i mn(Tn=i)
=mn(Tn=1)+
∞
X
i=2 i
Z
Li−1
n (LχUn)dmn
=mn(Tn=1)+
Z
En(LχUn)dmn ∞
X
i=2 iλin−1
+
∞
X
i=2 i
Z
9i−1
n (LnχUn)dmn
=mn(Tn=1)+
Z
En(LχUn)dmn
1
(1−λn)2 −1
+
∞
X
i=2
Z
9i−1
n 1dmn. (20)
But, by Kac’s theorem,R
Tndmn=1/µ(Un)and thus
R
En(LχUn)dmn
µ(Un)
=
R E
n(LχUn)dµn
1−λn
2
| {z }
→1
+
Z
En(LχUn)dmn
| {z }
→0
×
mn(Tn=1)−
Z
En(LχUn)dmn+ ∞
X
k=1 9k
n1dmn
| {z }
=O(1)
.
LEMMA4.4. If z has prime period p, then
lim
n→∞
R
En(LχUn)dmn
1−λn =nlim→∞
Z
En1dmn=1−eφ
p(z)
.
Proof. Fix a large positive integertand setk=pt. We have for largenthat
χUn(x)−χUn(x)L
k
n1(x)=χUn(x)
X
σk(y)=x
χSk−1
j=0σ−j(Un)(y)e
φk(y)
=χUn(x) X
σk(y)=x
χσk−p[z
0z1···zp−1](y)e
φk(y)
=χUn(x) X
σpm(y)=x
eφpm(y)χσ−p(m−1)[z0z1···zp−1](y)
=χU
n(x)L
pm(χ
[z0z1···zp−1]◦σ
p(m−1))(x) =χUn(x)Lp(χ
[z0z1···zp−1])(x),
where assumption (5) on the family{Un}nwas utilized in the second line. Hence,
1−eφp(z)−
Z
Lk n1dmn
≤ Z
Lp(χ
[z0z1···zp−1])(x)−eφ
p(z)
dmn(x)
≤ sup
y∈[z]ln+p
|φp(y)−φp(z)|
≤ |φ|θ,∞
1−θ diam(Un)→0 (n→ ∞),
where| · |θ,∞denotes the usual Hölder semi-norm. Hence, for anyk=pt,
lim
n→∞
Z
Lkn1dmn=1−eφ
p(z)
.
On the other hand, by Lemma 4.2, for largen
Z
Lkn1dmn−λkn
Z
Endmn
= Z 9k n1dmn
≤ k9nkk∞≤cqk.
We fixk=ptandλn→1 asn→ ∞; hence,
lim
n→∞
Z
En1dmn=1−eφ
p(z)
. 2
LEMMA4.5. If z has prime period p, then
lim
n→∞
R
En(LχUn)dmn
µ(Un)
=(1−eφp(z))2.
Proof. The proof of this is a combination of the methods from Lemma 4.3 and the result
of Lemma 4.4. 2
5. Escape rates for Gibbs measures
In this section, we prove the analogue of Theorem 1.1 in the setting of a topologically mixing subshift of finite type, namely, we prove the following theorem.
THEOREM5.1. Suppose that the{Un}n satisfy assumptions (1)–(5). Let φ:6→Rbe
H¨older continuous and letµdenote the associated equilibrium state; then
lim
n→∞
rµ(Un)
µ(Un) =
(
1 if z is not periodic,
1−eφp(z)−p P(φ) if z has prime period p,
whereφp(z)=φ(z)+φ(σ (z))+ · · · +φ(σp−1(z)).
It is well known that the escape raterµ(Un)is related to the spectral radiusλnand we
include the proof of the following proposition only for completeness.
PROPOSITION5.2. We have the following relation:
rµ(Un)= −log(λn).
Proof. We can write
µ{x∈6:σi(x)6∈Un,0≤i≤k−1} =
Z k−1 Y
i=0 χUc
n(σ
ix)
dµ(x)
=
Z
Lk
k−1 Y
i=0 χUc
n(σ
ix)
dµ(x)
=
Z
Lkn1(x)dµ(x)
=λk n
Z
En1dµ+
Z
9k n1dµ.
Using Propositions 3.18 and 3.20, we see that
rµ(Un)= lim k→∞
−1
klogµ{x∈6:σ
i(x)6∈U
n,0≤i≤k−1} = −log(λn).
We now prove Theorem 5.1.
Proof. We assume without loss of generality that P(φ)=0. In which case, we see from Proposition 5.2 that
rµ(Un)
µ(Un)
= −log(λn)
µ(Un)
= log(λ)−log(λn)
µ(Un)
= λ−λn
µ(Un)
log(λ)−log(λn)
λ−λn .
We also can obtain results relating to the convergence of the topological pressure.
THEOREM5.3. Suppose that the{Un}n satisfy assumptions (1)–(5). Let φ:6→R be
H¨older continuous and letµdenote the associated equilibrium state; then
lim
n→∞
P(φ)−P6n(φ) µ(Un)
=
(
1 if z is not periodic,
1−eφp(z)−p P(φ) if z has prime period p. Proof. Usingλ=eP(φ), we see that
P(φ)−P6n(φ) µ(Un)
= P(φ)−P6n(φ)
eP(φ)−eP6n(φ)
λ−λn
µ(Un).
(21)
Observing that
lim
n→∞
P(φ)−P6n(φ) eP(φ)−eP6n(φ) =e
−P(φ),
and combining this, (21), and Proposition 4.1 completes the proof. 2
An immediate corollary is the following.
COROLLARY5.4. Let µ denote the measure of maximal entropy (i.e. the Parry measure[20]); then
lim
n→∞
htop(σ )−htop(σ|6n))
µ(Un)
=
(
1 if z is not periodic,
1−e−phtop(σ ) if z has prime period p.
Remark 5.5. The rate of convergence of the topological entropy of the restriction of the shift to these sets was studied by Lind[18]who proved, in the case that theUnconsisted
of a single cylinder of lengthn, i.e.Un= [z]n, the existence of a constantc>1 such that
1/c≤htop(σ)−htop(σ|6n)
µ(Un)
≤c for alln.
6. Proof of Theorem 1.1
In this section, we prove Theorem 1.1. LetMbe a Riemannian manifold and f :M→M
aC1-map. LetJbe a compact subset ofMsuch that f(J)=J. We say that the pair(J, f) is a conformal repeller if:
(1) f|J is a conformal map;
(2) there existc>0 andλ >1 such thatkd fxnvk ≥cλnkvkfor allx∈J,v∈TxM, and
n≥1;
(3) f is topologically mixing onJ;
(4) Jis maximal, i.e. there exists an open neighbourhoodV ⊃Jsuch that
J= {x∈V : fn(x)∈V for alln≥0}.
Letφ:J→Rbeα-Hölder and letµdenote the associated equilibrium state. For an open setU⊂J, we letrµ(U)denote the escape rate ofµthroughU.
It is well known that an expanding map has a finite Markov partition{R1,R2, . . . ,Rl},
We state without proof the following result of Bowen[2].
PROPOSITION6.1. (Bowen)There exists a positive integer d such that the cardinality of π−1(x)is at most d for all x∈J .
This proposition was used to prove the following corollary.
COROLLARY6.2. (Bowen)x∈6is periodic if and only ifπ(x)∈J is periodic.
We also require the following technical lemmas.
LEMMA6.3. For any periodic point z∈J , there exists a Markov partition{R1,R2, . . . , Rm}such that z∈Smi=1int(Ri).
Proof. This follows easily from the standard construction of Markov partitions (using
shadowing); for example see[27]. 2
LEMMA6.4. There exist constants s,c1>0such thatµ(B(z, ))≤c1s for all >0.
Proof. Let φ˜:6→R be defined by φ(˜ x)=φ(π(x)), and denote the associated equilibrium state byµ˜; thenµ=π∗(µ)˜ .
For >0, letU denote the Moran cover associated with the Markov partition {R1, R2, . . . ,Rm}(see[23, p. 200]). Then forz∈Jwe choose elementsU1,U2, . . . ,Uk∈U
which intersectB(z, ). A basic property of Moran covers is that: (1) Ui=π[zi0zi1· · ·zin
(zi)], wherez
i∈6;
(2) diam(Ui)≤ <diam(π[zi0zi1· · ·zin(zi)−1]);
(3) k≤K, whereK is independent of bothzand.
In which case, it suffices to show thatµ(Ui)≤cs for some constantc>0. To see this,
we observe that a basic property of Gibbs measures is that for any x∈6, there exist c>0 andγ∈(0,1)such that µ˜[x]n≤cγn for n=1,2, . . . . In addition to f ∈C1+α and conformal, we have thatcλ−n(zi)≤for any >0. In which case, we see that
µ(B(z, ))≤ k
X
i=1
µ(Ui)= k
X
i=1 ˜
µ[zi0zi1· · ·zin
(zi)] ≤K c1
+log(γ )/log(λ)−log(γ )/log(λ). 2
Next, we require the so-called ‘D-annular decay property’, that is, there exists a constant c2>0 such that for allx∈J, >0 and 0< δ <1, we have that
µ(B(x, )\B(x, (1−δ)))≤c2δDµ(B(x, )). (22)
A related condition is the ‘doubling’ or ‘Federer’ property, namely there exists a constantK>1 such that for allx∈J and >0, we have
µ(B(x,2))≤Kµ(B(x, )).
Evidently, a measure that satisfies the D-annular decay property also satisfies the doubling property. The converse was shown by Buckley in [5, Corollary 2.2]. In the context of an equilibrium stateµsupported on a conformal repeller, Pesin and Weiss[22]
PROPOSITION6.5. There exists a D such thatµsatisfies the D-annular decay property.
We now prove Theorem 1.1.
Proof. We first prove the result if z∈ J is not periodic. As a consequence of Proposition 6.1 we have thatπ−1{z} = {z1,z2, . . . ,zr}. Further, Corollary 6.2 implies that eachziis non-periodic.
Hence, to show Theorem 1.1 it suffices to show that
lim
→0
rµ˜(π−1(B(z, )))
˜
µ(π−1(B(z, ))) =1.
First, we observe that Theorem 5.1 may be modified to accommodate multiple non-periodic points appearing in the intersection; this modification is trivial and we therefore omit the proof. For non-periodic points, the new hypotheses become:
(1) let {Vn} be a family of nested sets with each Vn being a finite union of
cylinders. Suppose further thatT
n≥1Vn consists of finitely many non-periodic
points{z1,z2, . . . ,zr};
(2) there exist constantsc>0 and 0< ρ <1 such thatµ(˜ Vn)≤cρkn for alln≥1; here
kndenotes the maximum length of a cylinder inVn;
(3) There exist a sequence (ln)n and a constant κ >0 such that κ <ln/kn≥1 and
Vn⊂Sri=1[zi]ln for alln≥1.
If the sets{Vn}nsatisfy these hypotheses, then we conclude that
lim
n→∞
rµ˜(Vn) ˜
µ(Vn) =1.
For >0 and a positive integerk, we set
Uk,=
U∈ k−1
_
i=0
f−iR:U∩B(z, )6= ∅
.
We observe that due to f being uniformly expanding, there exist constantsc3>0 and
0< ρ <1 such that
diam(U)≤c3ρk
for anyU∈Uk,.
Letδk=c3ρk/(+c3ρk), in which case it is easy to see that
[
U∈Uk,
U⊂B(z, +c3ρk)=B(z, (1−δk)−1).
Fixη >0 small and choosek=k(, η)such that
ρk≤
c3((c2η−1)1/D−1)< ρ k−1,
in which case we see that
(1−η)µ
[
U∈Uk,
U
≤(1−c2δkD)µ
[
U∈Uk,
U
≤(1−c2δkD)µ(B(z, (1−δk)−1))
where the D-annular decay property was used in the final line. Now let {n}n be any monotonic sequence withn→0 and set
Un=
[
U∈Uk(n,η),n
U.
Observing that Un is a finite union of kn:=k(n, η)th level refinements of the
Markov partition, there exists Vn⊂6, a finite union of cylinders of lengthkn, such that
π(Vn)=Un.
We claim thatVnsatisfies the hypotheses of the modified Theorem 5.1. Clearly, theVn
are nested (1), so it suffices to show thatµ(˜ Vn)decays exponentially inn. To see this, we
observe that
˜
µ(Vn)=µ(Un)
≤(1−η)−1µ(B(z, n))
≤c1ns≤c1(c3((c2η−1)1/D−1))sρs(kn−1).
And, thus, we see thatµ(˜ Vn)decreases exponentially inkn, which proves (2).
As f is conformal andz∈ [zi]lfor alliandl, there exist constantsc4>0 and 0< % <1
such that for anyi∈ {1,2, . . . ,r}andl∈N, we have thatc4−1≤diam(π[zi]l)/%l. Letln
be the minimum such l such that c4−1%l≥2n. It is easy to see that for such a choice
ofl, we have thatVn⊂Sri=1[zi]ln. In addition, we have thatln>c5kn for some constant
c5>0; this proves (3). Thus, we deduce from the modified Theorem 5.1
lim
n→∞
rµ˜(Vn) ˜
µ(Vn) =1.
And, so, by monotonicity of escape rates and (23), we see that
lim sup
n→∞
rµ(B(z, n))
µ(B(z, n))
≤(1−η)−1lim sup
n→∞
rµ(Un)
µ(Un)
=(1−η)−1lim sup
n→∞
rµ˜(Vn) ˜
µ(Vn)
=(1−η)−1. (24)
Similarly, using the same method we may obtain a lower bound, which in conjunction with (24) gives
lim
n→∞
rµ(B(z, n))
µ(B(z, n)) =1.
We now turn our attention to the case where z is periodic. By Lemma 6.3, we may assume thatπ−1(z)consists of a single point of prime period p, sayπ(z0)=z.
As before, we approximateB(z, )from outside using elements ofWk−1 i=0 f
−iR, which
may be thought of as cylinders of lengthkin a subshift of finite type. Recall the hypotheses for Theorem 5.1:
(1) let{Vn}be a family of nested sets with each Vn being a finite union of cylinders.
Suppose further thatT
n≥1Vn= {z0}, wherez0has prime period p;
(2) there exist constantsc>0 and 0< ρ <1 such thatµ(˜ Vn)≤cρkn forn=1,2, . . .;
herekndenotes the maximum length of a cylinder inVn;
(3) for eachn≥1, we have thatσ−p(Vn)∩ [z00z
0
1· · ·z
0
In which case, we deduce from Theorem 5.1 that
lim
n→∞
rµ˜(Vn) ˜
µ(Vn)
=1−eφ˜p(z0).
We first approximate B(z, n) from outside using the same method employed
previously. For η >0, we obtain Un⊃B(z, n) nested, each being a finite union of
elements from Wk(n)−1
i=0 f
−iR for some k, with the property that µ(U
n)≤(1−η)−1
µ(B(z, n)). As before, we may find aVn⊂6 which is a finite union of cylinders. It
is easy to see thatVnsatisfies conditions (1) and (2). To see (3), we observe that forn
small, expansivity of f and the fact that z has prime period p yields f−p(B(z, n))∩
π[z00,z01, . . . ,z0p−1] ⊂B(z, n). A simple argument extends this to approximations of
balls centred onz. Using monotonicity of escape rates together with the conclusions of Theorem 5.1 yields
lim sup
n→∞
rµ(B(z, n))
µ(B(z, n))
≤(1−η)−1lim sup
n→∞
rµ(Un)
µ(Un)
=(1−η)−1lim sup
n→∞
rµ˜(Vn) ˜
µ(Vn) =(1−η)−1(1−eφ˜p(z0))
=(1−η)−1(1−eφp(z)). (25)
Similarly, using the same method we may obtain a lower bound, which, in conjunction with (25), yields
lim
n→∞
rµ(B(z, n))
µ(B(z, n))
=1−eφp(z). 2
7. Proof of Theorem 1.2
In this section, we study the asymptotic behaviour of the Hausdorff dimension of the non-trapped set. Let f :J→J be a conformal repeller as defined in the previous section; we make the further assumption that f ∈C1+α(J)for someα >0. Fixz∈J. For >0, we define
J= {x∈J: fk(x)6∈B(z, ),for allk≥0},
i.e. all points whose orbits are-bounded away fromz.
Let µ denote the equilibrium state related to the potential ψ= −slog|f0|, where s=dimH(J). As before, we may study the escape raterµ(B(z, ))ofµthroughB(z, )
and its associated asymptotic, i.e.
dφ(z):=lim
→0
rµ(B(z, )) µ(B(z, )).
The method of proof is as follows: in a similar vein to the proof of Theorem 1.1, we first prove the result where the hole consists of a finite union of refinements of the Markov partition and then extend it to the case of geometric balls via an approximation argument.
LetR= {R1,R2, . . . ,Rm}denote a Markov partition for the conformal repellerJ; this
repeller(J, f). LetIn∈Wni=−01 f
−jRbe a nested family such thatT
n≥0In= {z}. We let
Jndenote the set of points inJwhich do not fall down the holeIn, i.e.
Jn= {x∈J: fk(x)6∈In,for allk≥0}.
Letsdenote the Hausdorff dimension of the setJ.
PROPOSITION7.1. Under the assumptions above,
lim
n→∞
s−sn
µ(In)
=R dφ(z)
log|f0|dµ.
A crucial ingredient to the proof of Proposition 7.1 is the following result of Ruelle[24].
PROPOSITION7.2. (Ruelle) Let s≥0 be the unique real number for which P(−slog|f0|)=0; thendimH(J)=s.
Letφ(˜ x):= −log|f0(π(x))|. It is easy to see that the semi-conjugacy π being one–
one on a set of full measure for all equilibrium states for Hölder potentials implies that the Hausdorff dimension of J is the unique real number s for which P(sφ)˜ =0. A similar argument shows that dimH(Jn)=sn, wheresnis the unique real number satisfying
P6n(snφ)˜ =0. We may therefore translate the problem into the language of subshifts of
finite type. As the family{In}is nested, there exists a pointz0∈6 such thatπ[z0] =In.
Accordingly, if we set
6n= {x∈6:σk(x)6∈ [z0]nfork=0,1,2, . . .},
then π(6n)=Jn. Let φ(˜ x)= −log|f0(π(x))|; then it is easy to see that the
semi-conjugacyπ being one–one on a set of full measure for all equilibrium states for Hölder potentials implies that the Hausdorff dimension ofJis the unique real numbersfor which P(sφ)˜ =0. A similar argument shows that dimH(Jn)=sn, wheresn is the unique real
number satisfying P6n(snφ)˜ =0. We therefore may prove the result in the setting of
subshifts of finite type.
Fort≥0, we letLt:Bθ →Bθdenote the transfer operator associated with the potential
tφ˜, i.e.
(Ltw)(x)=
X
σ(y)=x
w(y)
|f0(π(y))|t;
analogously, we define the perturbed transfer operatorLt,n:Bθ→Bθ to be(Lt,nw)(x)=
(Ltχ[z0]c
nw)(x). We letgt (respectivelygt,n) andνt (respectivelyνt,n) denote the
eigen-functions and eigenmeasures guaranteed by Proposition 3.2 applied to Lt (respectively
Lt,n). We shall assume without loss of generality thatR gtdνt =R gt,ndνt,n=1 for all
t≥0 and n≥1. The associated equilibrium states will be denoted by µt and µt,n,
observing that one can show that dµt =gt dνt (respectively dµt,n=gt,ndνt,n). We
proved earlier that both Lt and Lt,n have spectral gaps; we denote their maximal
eigenvalues by λt and λt,n, respectively. As log(λt)=P(tφ)˜ (respectively log(λt,n)=
P6n(tφ)˜ ), the problem of finding the Hausdorff dimensions ofJ(respectivelyJn) reduces
to finding the values oft(respectivelytn) such thatλt =1 (respectivelyλtn,n=1).
(and converges to λt), we use Taylor’s theorem applied to λt,n about t=s to obtain
an approximation of λt,n close to λs,n, we then use Theorem 5.1 and let n→ ∞ to
prove the result. The main problem then reduces to analysing the behaviour of the first λ0
t,n=d/dt(λt,n)and secondλ00t,n=d2/dt2(λt,n)derivatives ofλt,n, which is the focus of
the following two technical lemmas.
LEMMA7.3. For any t≥0, we have thatlimn→∞λ0t,n=λ0t.
Proof. We first obtain an explicit formula for λ0
t,n; to do this, we follow an argument of
Ruelle[25, p. 96, Exercise 5]to prove that for anyt≥0 andn=1,2, . . .,
λ0
t,n= −λt,n
Z
log|f0|dµt,n. (26)
Analogously, for the unperturbed operator,
λ0
t= −λt
Z
log|f0|dµt. (27)
To see this, we take the eigenfunction equation
Lt,ngt,n=λt,ngt,n. (28)
Differentiating once yields
L0t,ngt,n+Lt,ngt0,n=λ
0
t,ngt,n+λt,ngt0,n,
and then integrating with respect toνt,nand cancelling terms yields
λ0
t,n=
Z
L0
t,n(gt,n)dνt,n=
Z
Lt,n(φgt,n)dνt,n=λt,n
Z
φdµt,n
whereφ= −log|f0|. This shows (26); the proof of (27) is analogous and is omitted. Without loss of generality, we may assume that gt =1, that is, Lt1(x)=λt. We
decompose the transfer operatorsLt,nandLtas
Lt,n=λt,nEt,n+9t,n, Lt=λtEt+9t,
whereEt,nandEt are projection operators given by
Et,nw=
Z
wdνt,ngt,n, Etw=
Z
wdνt (29)
and9t,n(respectively9t) has a spectral radius strictly less thanλt,n(respectivelyλt).
From[15], we have that limn→∞k|Et,n−Et|k =0 and so
kgt,n−gtk1≤ kgt,n−gtkh
= k(Et,n−Et)(1)kh≤ k|Et,n−Et|kk1kθ→0. (30)
Finally, to show that λ0t,n→λ0t, it suffices to show that Et,n(gt,nφ)→Et(φ). We
that by [15, Corollary 1]there exist a constantc>0 and a positive integer N such that
kEt,nwkθ≤ckEtwkhfor anyw∈Bθ andn≥N. In which case,
kgt,nφkθ = kgt,nφkθ+ kgt,nφk1
≤ kgt,nk∞|φ|θ+ |gt,n|θkφk∞+ kφk∞kgt,nk1 ≤2kφkθkgt,nkθ+ kφk∞kgt,nkh
=2kφkθkEt,n1kθ+ kφk∞kgt,nkh ≤2ckφkθkEt,n1kh+ kφk∞kgt,nkh =(2ckφkθ+ kφk∞)kgt,nkh.
We observe thatkgt,n−1k1→0 implies thatkgt,n−1kh→0 and sokgt,nkθ is bounded.
Next, we note that
kEt,n(gt,nφ)−Et(φ)k1= k|Et,n−Et|kkgt,nφkθ + kEt(φ(gt,n−1))k1
≤ck|Et,n−Et|k + kφk∞kEtk1kgt,n−1k1.
Both terms tend to zero by (30) and[15]. This completes the proof. 2
LEMMA7.4. For any s>0, there existsδ >0such thatsupn≥1supt∈(s−δ,s+δ)λ00t,n<∞.
Proof. We first obtain an expression for λ00t,n. Fix a positive integer N. Taking the eigenfunction equation LN
t,ngt,n=λNt,ngt,n and differentiating twice, integrating with
respect toνt,n, and cancelling yields
λ−1 t,nλ
00
t,n=
1 N
Z
(φN)2g
t,ndνt,n−N(N−1)(λ−t,n1λ
0
t,n) 2 +2 1 N Z
φNg0
t,ndνt,n−λ−t,n1λ
0
t,n
Z
g0t,ndνt,n
.
We observe that asdµt,n=gt,ndνt,n is strong mixing, this second term tends to zero
as N→ ∞, and thus
λ−1 t,nλ
00
t,n= lim N→∞
1 N
Z
(φN)2g
t,ndνt,n−N(N−1)(λ−t,n1λ
0
t,n)2
. (31)
We now estimate the termN−1R(φN)2gt,ndνt,n: expanding the term(φN)2and using
the dual identityL∗
t,n(νt,n)=λt,nνt,nyields fornlarge enough
N−1
Z
(φN)2g
t,ndνt,n= N−1
X
i=0 N−1
X
j=0
Z
gt,nφ◦σiφ◦σj dνt,n
= kφk22+ 2
N
N−1
X
k=0
(N−k)
Z
gt,nφφ◦σkdνt,n
= kφk2 2+
2 N
N−1
X
k=0
(N−k)λ−t,nk
Z
Lk
We apply the decompositionLt,n=λt,nEt,n+9t,nalong with Proposition 3.18 to (32) to
obtain
N−1
Z
(φN)2g
t,ndνt,n = kφk22+
2 N
N−1
X
k=0
(N−k)
×
Z
Et,n(gt,nφ)φ+λ−t,nk9tk,n(gt,nφ)φdνt,n
= kφk22+(N+1)
Z
φgt,ndνt,n
2
+ 2
N
N−1
X
k=1
(N −k)λ−t,nk
Z
9k
t,n(gt,nφ)φdνt,n. (33)
We note that(R φgt,ndνt,n)2=(λ−t,n1λ0t,n)2and so combining (31) and (33) we obtain
λ−1 t,nλ
00
t,n= kφk22+2(λ
−1 t,nλ
0
t,n)2+ lim N→∞
2 N
N−1
X
k=0
(N−k)λ−t,nk
Z
9k
t,n(gt,nφ)φdνt,n.
Finally, we observe that the perturbationt7→Lt is analytic and so for anyq>0 such
that spec(Ls)\{λs} ⊂B(0,q), there exist a positive integerMandδ >0 such thatλt,n>q
and spec(Lt,n)\{λt,n} ⊂B(0,q) for all n≥M and t∈(s−δ,s+δ). Combining this
observation with Proposition 3.18 completes the proof. 2
We now prove Proposition 7.1.
Proof. We begin by proving thats−sn=O(µ(In)); to see this, we observe that the map
t7→λt,nis analytic, and so using Taylor’s theorem we may write
λsn,n=1=λs,n+λ 0
ξn,n(sn−s) (34)
for someξn∈(sn,s). We note that Proposition 4.1 and Lemma 7.3 prove the claim. Next,
we use Taylor’s theorem once again to see that
λsn,n=1=λs,n+λ 0
s,n(sn−s)+λ00ξn,nO(µ(In)2)
forξn∈(sn,s). Rearranging yields
s−sn
µ(In) = 1
−λ0
s,n
1−λ
s,n
µ(In) +λ00ξ
n,nO(µ(In))
.
Finally, we let n→ ∞, observing that the right-hand side converges by Lemmas 7.3
and 7.4. This completes the proof. 2
We note that as in the case of escape rates Proposition 7.1 generalizes easily to the case of finite unions of symbolic holes. We now prove Theorem 1.2.
Proof. Let{n}n be any monotonic sequence withn→0. Fixη >0 and chooseUn⊂
B(z, n)⊂Vn which consist of finite unions of refinements of the Markov partition R
such that
From the proof of Theorem 1.1, it is clear that we may choose the families{Un}nand {Vn}nso that they satisfy the hypotheses of Proposition 7.1. Letsn(respectivelysn) denote
the Hausdorff dimension of the non-trapped set with respect to the holeUn (respectively
Vn). Monotonicity of the Hausdorff dimension along with (35) yields
1 1+η
s−sn µ(Un)
≤ s−sn
µ(B(z, n)) ≤ 1
1−η
s−sn
µ(Vn)
for anyn; combining this with Proposition 7.1 and lettingη→0 completes the proof. 2
Acknowledgement. We would like to thank the referee for his careful reading of the original version of this paper and his numerous comments.
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