This paper investigates the decay of correlations in slowly mixing random dynamical systems using random towers, providing bounds and asymptotics that depend on the fastest mixing map in the family.
Contribution
It introduces a general framework for analyzing quenched correlation decay in slowly mixing systems and applies it to Liverani-Saussol-Vaienti maps with various distributions.
Findings
01
Upper bounds on quenched correlation decay rates.
02
Decay governed by the fastest mixing map in the family.
03
Sharp asymptotics for return-time intervals for different distributions.
Abstract
We study random towers that are suitable to analyse the statistics of slowly mixing random systems. We obtain upper bounds on the rate of quenched correlation decay in a general setting. We apply our results to the random family of Liverani-Saussol-Vaienti maps with parameters in [Ξ±0β,Ξ±1β]β(0,1) chosen independently with respect to a distribution Ξ½ on [Ξ±0β,Ξ±1β] and show that the quenched decay of correlation is governed by the fastest mixing map in the family. In particular, we prove that for every Ξ΄>0, for almost every Οβ[Ξ±0β,Ξ±1β]Z, the upper bound n1βΞ±0β1β+Ξ΄ holds on the rate of decay of correlation for H\"older observables on the fibre over Ο. For three different distributions Ξ½ on [Ξ±0β,Ξ±1β] (discrete, uniform, quadratic), we also derive sharp asymptotics on theβ¦
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Full text
\setremarkmarkup
(#2)
Quenched decay of correlations for slowly mixing systems
Wael Bahsoun*β *
Department of Mathematical Sciences, Loughborough University,
Loughborough, Leicestershire, LE11 3TU, UK
We study random towers that are suitable to analyse the statistics of slowly mixing random systems. We obtain upper bounds on the rate of quenched correlation decay in a general setting. We apply our results to the random family of Liverani-Saussol-Vaienti maps with parameters in [Ξ±0β,Ξ±1β]β(0,1) chosen independently with respect to a distribution Ξ½ on [Ξ±0β,Ξ±1β] and show that the quenched decay of correlation is governed by the fastest mixing map in the family.
In particular, we prove that for every Ξ΄>0, for almost every Οβ[Ξ±0β,Ξ±1β]Z, the upper bound n1βΞ±0β1β+Ξ΄ holds on the rate of decay of correlation for HΓΆlder observables on the fibre over Ο.
For three different distributions Ξ½ on [Ξ±0β,Ξ±1β] (discrete, uniform, quadratic), we also derive sharp asymptotics on the measure of return-time intervals for the quenched dynamics, ranging from nβΞ±0β1β to (logn)Ξ±0β1ββ nβΞ±0β1β to (logn)Ξ±0β2ββ nβΞ±0β1β respectively.
Key words and phrases:
Random dynamical systems, slowly mixing systems, quenched decay of correlations.
1991 Mathematics Subject Classification:
Primary 37A05, 37E05
WB and MR would like to thank The Leverhulme Trust for supporting their research through the research grant RPG-2015-346. CBβs research is supported by a research grant from the National Sciences and Engineering Research Council of Canada. The authors would like to thank V.Β Baladi for useful communications and helpful comments.
In Section 2 we recall standard definitions and notation from random dynamical systems and present various natural notions of correlation decay in this setting. In Section 3 we build random towers for our system and detail the dynamical hypotheses that are in force throughout the paper. Our main general results are contained in Section 4 (Theorems 4.1 and 4.2) where we prove existence and correlation decay estimates respectively for the dynamics on the random towers. In Section 5 we present detailed computations applying our general results to the case of random LSV maps on the interval. Three different randomising distributions are
investigated: discrete, uniform and quadratic. At the end of Section 5 we also compute exact asymptotics for the measure of the return sets on the base of the
random towers. In Section 6 we prove Theorem 4.1. The expansion and distortion conditions and related estimates are the main tools used in this section.
In the next section, Section 7, we introduce random stopping times and derive asymptotics on their distributions in preparation for a coupling argument. In Section 8 we obtain decay of correlation estimates (upper bounds) for observables on our random towers. Both future and past decay estimates are derived. We conclude with Section 9 where we present some technical results that are used repeatedly in the paper.
Notation: We use aβ²b if there exists universal constant C such that aβ€Cb; βΌ, o, O will have their usual meaning.
It is easy to check that
ΞΌ is T-invariant if and only if (fΟβ)ββΞΌΟβ=ΞΌΟΟβ for P- a.e. Ο, a property
we naturally refer to as fΟββequivariance (or simply equivarariance, when the random map is understood) of the family {ΞΌΟβ}.
Notice that {Ξjβ(Ο)}jβ induces a countable partition PΟβ on each ΞΟβ:
[TABLE]
For (x,β)βΞΟβ, let R^Οβ denote the first return time to the base of the tower ΞΟR^ΟβΟβ i.e.
[TABLE]
The reference measure m and Ο-algebra on Ξ naturally lifts to ΞΟβ and by abuse of notation we call it m. The lifted Ο-algebra will be denoted by BΟβ.
Next we define the separation time s:ΞΓΞβZ+ββͺ{β} for almost every Ο by setting s(z1β,z2β)=0 if z1β and z2β lie in different towers ΞΟβ and if z1β,z2ββΞΟβ then
[TABLE]
Below we refer to Ξ as the zeroth level of the tower.
We assume that the random tower satisfies the following properties.
(P1)
Markov: for each Ξjβ(Ο) the map FΟRΟβββ£Ξjβ(Ο):Ξjβ(Ο)βΞ is a bijection;
(P2)
Bounded distortion: There are constants D>0 and 0<Ξ³<1 such that for all Ο and each Ξjβ(Ο) the map FΟRΟβββ£Ξjβ(Ο) and its inverse are non-singular with respect to m with corresponding Jacobian JFΟRΟβββ£Ξjβ(Ο) which is positive and for each x,yβΞjβ(Ο) satisfies the following
[TABLE]
(P3)
Weak expansion: PΟβ is a generating partition for FΟβ i.e. diameters of the partitions β¨j=0nβFΟβjβPΟjΟβ converge to zero as n tends to infinity;
In this section we state general theorems concerning quenched correlation decay for slowly mixing systems. We start this section by introducing some function spaces on Ξ, which are necessary to state the theorems. These spaces appeared in the present form in [7].
Below we let constants u>0,v>0, a>1, bβ₯0, Ξ³<1 be as in (P2) and (P4) above and set
Moreover, there exist constants C>0, uβ²>0, vβ²β(0,1) such that
[TABLE]
Remark 2*.*
A quenched correlation decay rate of the form naβ1(logn)bβ, which is analogous to what one expects in the deterministic setting, cannot be achieved since we want to get information on the integrability of the CΟβ in Theorem 4.2. The shift of the Lipschitz constant KΟβ, and hence the dependence of that constant on n, in equation (8.7) and the non-uniformity of the tail in (P4) are the main reasons for getting a rate at the order naβ1+Ξ΄1β, for any Ξ΄>0. See Footnote 7 for more details.
5. Applications to random LSV maps
In this section we illustrate our results with applications to the family of intermittent LSV maps as
described in [17]. Let 0<Ξ±<1 and consider fΞ±β:IβI defined as
Compositions of fΟβ are given by fΟnβ=fΟnβ1Οβββ―βfΟΟββfΟβ.
For each Ο we define a sequence of pre-images of 21β as follows.
Let x1β(Ο)=21β, and
[TABLE]
Further let
[TABLE]
The sequences {xnβ(Ο)} and {xnβ²β(Ο)} will allow us to define the
random tower structure. First of all notice that from the definition of xnβ²β(Ο) we have fΟβ(xnβ²β(Ο))=xnβ(ΟΟ), fΟ2β(xnβ²β(Ο))=xnβ1β(Ο2Ο),...,fΟnβ(xnβ²β(Ο))=x1β(ΟnΟ)=21β.
Let Ξ:=(21β,1]. The sequence {xnβ²β(Ο)}nβ₯0β generates a partition PΟβ={(xnβ²β(Ο),xnβ1β²β(Ο)]β£nβ₯0} on each ΞΓ{Ο}=ΞΟ,0β. Define the return time RΟβ:ΞΟ,0ββN by setting
The fibered map F:(Ο,ΞΟβ)β(ΟΟ,ΞΟΟβ) from equation (3.1) can be expressed in this notation as follows:
Let (x,β)βΞΟβ over Ο, with xβ(xiβ²β(ΟββΟ),xiβ1β²β(ΟββΟ)]. There are two cases. If i>β+1, then F(x,β)=(x,β+1)βΞΟΟβ since xβ(xiβ²β(ΟββΟ),xiβ1β²β(ΟββΟ)]=(xiβ²β(Οβββ1ΟΟ),xiβ1β²β(Οβββ1ΟΟ)]. On the other hand, if i=β+1 then
xβ(xβ+1β²β(ΟββΟ),xββ²β(ΟββΟ)], where
RΟββΟββ‘β+1, so the interval is
mapped bijectively to (21β,1] by fΟββΟβ+1β. Therefore in this case we have F(x,β)=(fΟββΟβ+1β(x),0).
Since every map in the family fΞ±β expands by at least a factor of 2 on return to the base interval (21β,1], with full returns, we see that the Markov and weak expansion properties are satisfied. Furthermore, for each Ο,
{xβΞΒ β£Β RΟβ(x)=1}=(43β,1], which implies that the aperiodicity condition (P5) is satisfied. Since every fΞ±β has negative Schwarzian derivative and this property is preserved under
composition, we obtain the bounded distortion condition (P3) using the Koebe principle. See Lemmas 4.8 and 4.10 in [6] for computations related to Schwarzian derivatives and [20] for more details about the use of the Koebe principle. This completes the proof.
β
It remains to establish appropriate estimates on the return time asymptotics as in (P4) and (P7) and the uniform bound in (P6).
Observe, in view of the return-time formula (5.4), that
[TABLE]
We will estimate terms on the right hand side of this expression.
Let Ξ±β0β (respectively Ξ±β1β) denote the special sequences of all Οkβ=Ξ±0β (respectively, all Οkβ=Ξ±1β).
Following Lemma 4.4 in [6], we obtain coarse estimates on the location of
the xnβ(Ο). Translated into our setting these estimates imply,
for every β,nβN,
[TABLE]
It is well known (see [27] section 6, for example) that
xββ(Ξ±β0β)βΌ21βΞ±0βΞ±0β1ββββΞ±0β1β so if we define
cββ(Ξ±β0β):=xββ(Ξ±β0β)βΞ±0β1β and write
xββ(Ξ±β0β)=βΞ±0β1βcββ(Ξ±β0β)β then
limββcββ(Ξ±β0β)=21βΞ±0βΞ±0β1ββ:=c(Ξ±β0β).
We define cββ(Ξ±β1β) and c(Ξ±β1β) analogously using xββ(Ξ±β1β).
Since, m{RΟβ>β}=21βxββ(ΟΟ)β€21βxββ(Ξ±β1β)β€CββΞ±1β1β this implies m(ΞΟβ)β€M for some M>0 independent of Ο. This proves (P6).
We now check that assumption (P4) is satisfied.
Using definition (5.1) and
the estimate (1+x)βΞ±β€1βΞ±x+2Ξ±(1+Ξ±)βx2, valid for
Ξ±>0,Β xβ₯0, and by substituting x=[2xnβ(Ο)]Ξ±(Ο) we obtain
[TABLE]
Iterating this one-step estimate along the sequence xkβ(ΟββkΟ), k=1,2,β¦β we obtain
[TABLE]
Combining equations (5.8) and (5.6) implies that, for any
parameter qβ₯0
[TABLE]
where we have introduced notation Akβ(Ο) for the sequence of independent random variables
A1ββ‘0 and for k=2,3,β¦β
[TABLE]
From now on we write Akβ:=Akβ(Ο). Note that βΞ±0β2Ξ±0ββ€Akββ€Ξ±0β2Ξ±0β for every k.
**Assumption (A1) (Asymptotics on expectations)
**Assume444Below in the examples, we will show that the assumption is satisfied for different types of distributions. In particular, we will show how different measures Ξ½ lead to different tail asymptotics. there are
constants q=q(Ξ½)β₯0 and a constant c(Ξ½)>0 such that the following holds:
[TABLE]
Fix any 0<c<c(Ξ½). Pick N1β so that
for all β>N1β,
[TABLE]
Note that, given expression (5.11), N1β depends only on the choice of c, and in
particular is independent of Ο. Set r0β=Ξ±0β2Ξ±0β.
Lemma 5.2**.**
For each t>0 we have
[TABLE]
Proof.
We may apply the classical result of Hoeffding (see [14] Theorem 1, or [18] Lemma 1.2) to the sequence of independent random variables Akβ, noting that instead of the bound
0β€Akββ€1 we have βr0ββ€Akββ€r0β, accounting for the extra factor in the exponential.
β
Next, we apply the previous lemma to obtain a large deviation estimate on the normalized
sums of Akβ. For each β>N1β:
[TABLE]
where in the last line we have used equation (5.12). Now define
provided 2Ξ±0βc(Ξ½)β<c<c(Ξ½). We may take b:=Ξ±0βqβ and a:=Ξ±0β1β and C=c(Ξ½)βΞ±0β1β in (P4). Note that the constant C is independent of n and Ο. Finally, we estimate, for n>N1β and fixed 0<v<1
[TABLE]
for suitable constants u>0 and C<β. We can remove the restriction n>N1β by substitution of a larger constant C in the final expression
[TABLE]
completing the second condition in (P4).
Finally we verify (P7).
First, we have
[TABLE]
where we have used the fact that P is Ο-invariant to write EΞ½β[xnβ1β(Ο)]=EΞ½β[xnβ1β(ΟΟ)].
Now recall that fΟβ(xnβ(Ο))=xnβ1β(ΟΟ) and fΟβ(xnβ(Ο))=xnβ(Ο)+2Ξ±(Ο)(xnβ(Ο))Ξ±(Ο)+1. Using this fact, (5.17), (5.15) and 0β€2xnβ(Ο)β€1, we obtain
[TABLE]
Choosing b^=Ξ±0βq(Ξ±0β+1)β and a=Ξ±0β1β completes the verification of (P7).
Finally, there exist constants C>0, uβ²>0 and 0<vβ²<1 such that the random variable CΟβ satisifies the following
tail estimates: for all nβN
[TABLE]
In particular, CΟβ is integrable. Every Ξ·β(0,1] can be used by choosing555Recall that Ξ³ is the regularity parameter in the distortion condition (P2). Ξ³β[21β,1) so that 2Ξ·Ξ³β₯1.
Proof.
Proposition 5.1 establishes conditions (P1) - (P3) and (P5).
Since we are assuming (A1), condition (P4) follows from equations (5.5) and (5.15)
with constants b=Ξ±0βqβ,Β a=Ξ±0β1β,Β C=c(Ξ½)βΞ±0β1β. The second condition in (P4) holds because of equation (5.16). (P7) is verified in (5.18).
Theorem 4.1 therefore applies and gives existence and mixing of the sample stationary measures ΞΌΟβ. Finally we apply Theorem 4.2 to obtain the decay of correlations.
For ΟβLβ([0,1]) and ΟβCΞ·([0,1]) define ΟΛβ=ΟβΟΟβ,ΟΛβ=ΟβΟΟβ:ΞΟββR. Then we have β«(ΟβfΟnβ)ΟdΞΌΟβ=β«(ΟΛββFΟnβ)ΟΛβhΟβdm. Now, to apply Theorem 4.2 it is sufficient to show ΟΛβhΟββFΞ³KΟββ and ΟΛββLβKΟββ. The latter is obvious, since
the projection ΟΟβ is
nonsingular and hence, for Ξ½Οββ a.e.Β (x,β) we have β£ΟΛβ(x,β)β£β€β₯Οβ₯Lββ.
For the former one we first note that since β£(fΟRΟββ)β²β£β₯2 we have β£xβyβ£β€(21β)s(x,y). Hence, for any (x,β),(y,β)βΞΟβ we have the inequality
[TABLE]
Now since s((x,β),(y,β))=s(x,y) the inequality (5.19) implies
[TABLE]
β
5.1. Sharp asymptotics on the measure of return-time intervals
Although we will not need lower bound estimates on the xnβ(Ο) to prove the main results in this
paper, it is not difficult to identify conditions (see Assumption (A2) below) under which the upperbounds from the previous section are sharp. This condition will hold for all of the examples discussed in this paper.
Notice that from equation (5.15) and the summability derived in
equation (5.16), an application of Borel-Cantelli yields, for almost every Ο,
Now we concentrate on deriving lower bounds. Using definition (5.1) and
the estimate (1+x)βΞ±β₯1βΞ±x, valid for
Ξ±>0,Β xβ₯0, and by substituting x=[2xnβ(Ο)]Ξ±(Ο) we obtain
Clearly (I)=o(1) and the same is true of (II) since
[TABLE]
In order to estimate (III) note that from equation (5.20), for all β sufficiently large (depending on Ο), for ββββ+1β€kβ€β,
[TABLE]
From now on we write Akβ²β:=Akβ²β(Ο). In addition to Assumption (A1) we now assume666We will see that for many examples, including the ones presented in the next section, Assumption (A2) will hold. the following asymptotics on the EΞ½β(Akβ²β):
Assumption (A2) (Asymptotics on expectations revisited)
[TABLE]
Another large deviations estimate as in the preceding section will give, for each c(Ξ½)<c an
integer N2β=N2β(c) so that for all β>N2β,
ββββββlog(ββββββ)qββk=ββββ+1ββEΞ½β(Akβ²β)β€2c+c(Ξ½)β and
[TABLE]
Once again, application of Borel-Cantelli implies there exists a random variable n2β(Ο), finite almost everywhere, such that
for all β>n2β(Ο)
[TABLE]
and for each 0<v<1
there are constants u>0 and C<β so that
5.2. Natural probability distributions on the parameter space
In this subsection we verify assumptions (A1) and (A2) for some natural probability distributions Ξ½ on
[Ξ±0β,Ξ±1β].
5.2.1. Example: Discrete distribution.
Here we assume Ξ½ is a discrete probability distribution; for concreteness,
Ξ½=p1βδα0ββ+p2βδα1ββ with piβ>0 and
p1β+p2β=1.
Lemma 5.4**.**
β1ββk=1ββEΞ½βAkββΞ±0β2Ξ±0βp1β.
Therefore, in condition (5.11) we can take q:=0 and
c(Ξ½):=Ξ±0β2Ξ±0βp1β>0.
Proof.
Write Akβ(Ο)=Xkβ(Ο)βYkβ(Ο) where
[TABLE]
Using Ο-invariance of P, a direct calculation shows
[TABLE]
while
[TABLE]
Therefore,
[TABLE]
whereas
[TABLE]
for Ξ΄=min{Ξ±0β/Ξ±1β,2βΞ±0β/Ξ±1β}>0.
It follows that
[TABLE]
where ΞΆ:=min{Ξ΄,1βΞ±0βΞ±1ββ}>0. Therefore
Assumption (5.11) is satisfied by taking q:=0 and
c(Ξ½):=Ξ±0β2Ξ±0βp1β>0.
β
We now show the upperbound on the xββ(Ο) obtained above is sharp.
Proposition 5.5** (Sharp asymptotics for the discrete probability distribution).**
For almost every Ο
[TABLE]
where c(Ξ½)=Ξ±0β2Ξ±0βp1β.
Proof.
We only need to verify Assumption (A2) and apply Proposition 5.3.
[TABLE]
Since Ξ±0β2Ξ±0βp1β=c(Ξ½) we have verified Assumption (A2).
β
5.2.2. Example: Uniform distribution.
Here we assume Ξ½ is the uniform probability distribution on
[Ξ±0β,Ξ±1β], that is, for Ξ±0β<t<Ξ±1β,
[TABLE]
We start with a lemma that will allow us to compute the appropriate expectations
in condition (5.11).
Lemma 5.6**.**
Let cβ₯1. Then, as uββ
[TABLE]
Proof.
We have
[TABLE]
β
As in the previous section, we decompose Akβ(Ο):=Xkβ(Ο)βYkβ(Ο) according to equation (5.26).
Using Ο invariance of P and Lemma 5.6
with u=βlog(kΞ±0β1β2ckβ(Ξ±0β)β),Β c=1 and
u=βlog(kΞ±1β1β2ckβ(Ξ±1β)β),Β c=2, respectively, we obtain
[TABLE]
It follows that logkβ EΞ½βAkββΌΞ±1ββΞ±0βΞ±02β2Ξ±0ββ.
Now apply Lemma 9.2 of the Appendix to compute the asymptotics for the sum:
[TABLE]
Therefore we can take q=1 and c(Ξ½)=Ξ±1ββΞ±0βΞ±02β2Ξ±0ββ
in Assumption (A1)
Proposition 5.7** (Sharp asymptotics for the uniform probability distribution).**
For almost every Ο
[TABLE]
where c(Ξ½)=Ξ±1ββΞ±0βΞ±02β2Ξ±0ββ.
Proof.
We verify Assumption (A2) and apply Proposition 5.3.
[TABLE]
We can evaluate the individual expectations using Lemma 5.6 to obtain:
[TABLE]
Now, two applications of Lemma 9.2 from the Appendix shows
[TABLE]
Applying this to the first estimate gives
[TABLE]
Since Ξ±1ββΞ±0βΞ±02β2Ξ±0ββ=c(Ξ½) we have verified Assumption (A2).
β
5.2.3. Example: Quadratic distribution.
Here we assume Ξ½ is the quadratic probability distribution on
[Ξ±0β,Ξ±1β], given by
[TABLE]
Again, we begin with a simple lemma that will allow us to estimate the
expectations.
Lemma 5.8**.**
Let cβ₯1. Then, as uββ
[TABLE]
Proof.
We have
[TABLE]
β
Writing Akβ=XkββYkβ as in equation (5.26), using invariance of P and Lemma 5.8 with we obtain
[TABLE]
It follows that (logk)2β EΞ½βAkββΌ2(Ξ±1ββΞ±0βΞ±0ββ)2
after which an application of Lemma 9.2 of the Appendix implies
[TABLE]
We take q=2 and c(Ξ½)=2(Ξ±1ββΞ±0βΞ±0ββ)2 in
the assumption (5.11).
For this example, lower bounds can also be obtained by essentially following the steps in the previous
example and using Lemma 5.8 in place of Lemma 5.6.
Proposition 5.9** (Sharp asymptotics for the quadratic probability distribution).**
For almost every Ο
[TABLE]
where c(Ξ½)=2(Ξ±1ββΞ±0βΞ±0ββ)2.
Proof.
We verify Assumption (A2) and apply Proposition 5.3. The key estimates are
[TABLE]
and
[TABLE]
where we have again used Lemma 9.2 in the Appendix to estimate the sum and verify
Assumption (A2) for this example.
β
6. Proof of existence of absolutely continuous mixing sample measures
The collection AΟ(n)β is a partition of FΟβnβΞΟnΟ,0β and for any xβΞΟnΟ,0β
each AβAΟ(n)β contains a single element of {FΟβnβx}.
For xβA let j(x) be the number of visits of its orbit to ΞΟkΟ,0β up to time n. Since the images of A
before time n will remain in an element of PΟkΟβ, all the points in A have the same itineraries, up to time n and so j(x) is constant on A.
Therefore JFΟnβ(x)=(JFΟRΟββ)j(x~), for the projection x~ of x onto
ΞΟnΟ,0β (i.e. if x=(z,β) then x~=(z,0)).
Thus for any x,yβA from (3.4) we obtain
[TABLE]
β
Corollary 6.3**.**
For any AβAΟ(n)β, FΟnβ:AβΞΟnΟ,0β is a
bijection and for each yβA we have
[TABLE]
Proof.
Lemma 6.2 implies JFΟnβ(x)β€(1+D)JFΟnβ(y). Integrating both sides of this inequality with
respect to x over A gives
To prove the item (i) we estimate
the density d(FΟnβ)ββm/dm at an arbitrary point xβΞΟnΟβ and consider three different cases according to the position of x. First of all, for any xβΞΟnΟ,0β, from Corollary 6.3 we have
[TABLE]
Since m(ΞΟnΟ,0β)=m(Ξ)=1 choosing M0β=D+1 finishes the proof
for the case xβΞΟnΟ,0β.
For xβΞΟnΟ,ββ with ββ₯n we have FΟβnβ(x)=yβΞΟ,ββnβ.
Since JFΟβ(y)=1 for any yβΞΟββΞΟ,0β,
[TABLE]
Finally,
let xβΞΟnΟ,ββ, for 0<β<n. Then for any yβFΟβnβx the
equality FΟnβββy=FΟnββΟβββxβΞΟnββΟ,0β holds. Hence,
JFΟnββ+jΟβ(FΟjβy)=1 for all j=0,β¦,ββ1. Therefore, by the chain
rule we obtain JFΟnβ(y)=JFΟnβββ(y). Hence the
problem is reduced to the first case since
[TABLE]
This finishes the proof of item (i).
To prove the item (ii) we first note that FΟnβ:AβΞΟnΟ,0β is invertible.
So for any xβΞΟnΟ,0β there is a unique x0ββA such that FΟnβ(x0β)=x and
[TABLE]
Let ΟΟβ=dmdΞ½ΟnΟββ
then for x,yβΞΟnΟ,0β, using Lemma 6.2 and assumption on dmdΞ»Οββ we obtain
[TABLE]
β
Remark 3*.*
It is important to note that the constant CΞ»β does not depend on Ο.
Moreover, if A and n are such that
the orbits Fj(x0β),Fj(y0β),j=1,2,β¦n see sufficiently many returns to the base, so that CΞ»βΞ³s(x0β,y0β)β€log2, then the upperbound in (ii) becomes 2D+1. The elements
A for which this holds are independent of the starting measure Ξ».
Notice that the constant in equation (6.4) is independent of j, A and Ο.
Hence for all x,yβΞΟβ letting Dβ²=e1βΞ³Dβ we have
[TABLE]
Integrating both sides of the latter inequality over ΞΟ,0β with respect to x implies
[TABLE]
On the other hand, if AβPΟβjΟ(j)ββ£ΞΟβjΟ,0β such that FΟβjΟjβ(A)βΞΟ,ββ for β>0 then Οj,AΟβ(x)=Οjββ,AΟββΟβ(FΟββΟβββ(x)). Hence we can apply (6.5). Futher define
[TABLE]
As above ΟnΟββ£ΞΟ,βββ‘0 for β>j. For xβΞΟ,ββ, ββ€j we write Οnβ as a convex combination of Οj,AΟβ and obtain
[TABLE]
Hence, {ΟnΟβ}nβNβ is a uniformly bounded family. Notice that if ΟnΟβ(x)=0 and ΟnΟβ(y)ξ =0 then s(x,y)=0. Taking this into account for all yβΞΟβ such that ΟnΟβ(y)ξ =0 we have
[TABLE]
where we have used equation (6.4) in the last step.
Exactness of the system can now be verified using the same method as detailed in
[7].
β
7. Random coupling
7.1. Estimates on the random recurrence times for the base
For a single map, the recurrence time of the base with the base gives a key construction parameter for coupling arguments. In the setting of random maps, this recurrence time is Ο dependent. Our first task is to obtain a suitable version of the recurrence time (see β0β below, and its use in the following Lemma 7.2). At this stage, it is useful for the reader to recall from section 3 our assumptions (P1)-P(7); in particular that a>1 in (P4). Moreover, recall the regularity class of the equivariant densities defined by the random variable KΟβ from Theorem 4.1.
Lemma 7.1**.**
Let N and tiβ be from the aperiodicity condition (P5). There is an β0ββN so that
for every β>β0β there are nonnegative integers ciβ such that
The result follows from the aperiodicity condition (P5) and bounded distortion (P2).
First, suppose F1β=FΟj1ββ:ΞβΞΟj1βΟβ
and F2β=FΟj1βΟj2ββ:ΞβΞΟj1β+j2βΟβ, satisfying
[TABLE]
Then, since Fiβ1β are bijections when restricted to Ξ, using bounded distortion,
we get
[TABLE]
Now, for β>β0β, Lemma 7.1 implies FΟββ can be written as a composition of
FΟiβtiββ (with at most β terms). Iterating the above estimate and using the lower bounds given in condition (P5) implies the existence of the required
lower bound V(β)>0.
β
Remark 4*.*
From the proof of the previous lemma, it is clear that one should not expect
a lower bound on the V(β),
uniform over all β.
With respect to this map, we define auxiliary stopping timesΟ1Οβ<Ο2Οβ<... to the base as follows:
Let β0β be the constant given in Lemma 7.1. For (Ο,x,xβ²)βΞβΟβΞ set
[TABLE]
and so on, with the action alternating between x and xβ². Notice that for odd iβs the first (resp. for even iβs the second) coordinate of (FΟβΓFΟβ)ΟiΟβ(x,xβ²) makes a return to ΞΟΟiΟβΟ,0β.
Let iβ₯2 be the smallest integer such that (FΟβΓFΟβ)ΟiΟβ(x,xβ²)βΞΟΟiΟβΟ,0βΓΞΟΟiΟβΟ,0β. Then we define the stopping timeTΟβ by
[TABLE]
Next define a sequence of partitions ΞΎ1ΟββΊΞΎ2ΟββΊΞΎ3ΟββΊ... of ΞΟβΓΞΟβ so that ΟiΟβ is constant on the elements of ΞΎjΟβ for all iβ€j,i,jβN. Given a partition Q of ΞΟβ we write Q(x) to denote the element of Q containing
x. With this convention, we let
[TABLE]
Letting Ο:ΞΟβΓΞΟββΞΟβ be the projection to the first coordinate, we define
[TABLE]
Let Οβ² be the projection onto the second coordinate. We define ΞΎ3Οβ by refining the partition on the first coordinate, and so on. If ΞΎ2iΟβ is defined then we define ΞΎ2i+1Οβ by refining each element of ΞΎ2iΟβ in the first coordinate so that Ο2i+1Οβ is constant on each element of ΞΎ2i+1Οβ. Similarly ΞΎ2i+2Οβ is defined by refining each element of ΞΎ2i+1Οβ in the second coordinate so that Ο2i+1Οβ is constant on each new partition element.
Now we define a partition P^Οβ of ΞΟβΓΞΟβ such that TΟβ is constant on its element. For definiteness suppose that i is even and choose ΞβΞΎiΟβ such that TΟββ£Ξβ>Οiβ1Οβ. By construction Ξ=AΓB such that FΟiΟβ(B)=ΞΟΟiΟβ,0β and FΟiΟβA is spread around ΞΟΟiΟβΟβ. We refine A into countably many pieces and choose those parts which are mapped onto the corresponding base at time ΟiΟβ. Note
that {TΟβ=ΟiΟβ} may not be measurable with respect to ΞΎiΟβ. However, since Οi+1Οββ₯β0β+ΟiΟβ and ΞΎi+1Οβ is defined by dividing A into pieces where Οi+1Οβ is constant, {TΟβ=ΟiΟβ} is measurable with respect to ΞΎi+1Οβ.
7.3. Tail of the simultaneous return times
In this section we estimate the tail of the simultaneous return time TΟβ. We start this section with the lower bound on the measure of the set that made return at time ΟiΟβ.
Finally, item (ii) of Lemma 6.4 applied to Ξ½ΟΟiβ1ΟβΟβ implies that
[TABLE]
Now, the lemma holds with CΞ»~β=min{(1+CΞ»β)(D+1)1β,(1+CΞ»β²β)(D+1)1β}. In view of Remark 3 we can use
CΞ»~β=2(D+1)22D+1β for all i sufficiently large.
β
The next lemma estimates the distribution of ΟiΟββs on ΞΟβΓΞΟβ by the measure of the tail of
the random tower.
Lemma 7.4**.**
Let CΞ»~β be as in Lemma 7.3.
For each Ο, for each i and ΞβΞΎiΟβ
[TABLE]
Proof.
Suppose that i is even. Since ΟiΟβ is
constant on the elements of ΞΎiΟβ for every ΞβΞΎiΟβ we have
[TABLE]
Letting Ξ½ΟΟiβ1ΟβΟβ=(FΟΟiβ1Οββ)ββ(Ξ»β£ΟΞ), we have
Finally, since the density of (FΟΟiβ1ΟβΟΟiΟββΟiβ1Οβ+β0ββ)ββm is bounded above by the first item of Lemma 6.4 we have
[TABLE]
For i odd the calculation is analogous and we obtain for all i
[TABLE]
β
Suppose we are given a sequence of positive
integers β0ββ€Ο1β<Ο2β<β―<Οnβ<β¦ with ΟiββΟiβ1ββ₯β0β for all iβ₯2, denoted Ο, and a positive integer
q>0. Define associated subsets of ΞβΟβΞ
[TABLE]
This is a partition into sets where a specified sequence of hitting times up to q is attained:
There exists a C>0 such that for each fixed Ο, q>0
[TABLE]
where
[TABLE]
Proof.
Assume first that q is even and let GqΟβ be as above.
Let kjβ=ΟjββΟjβ1βββ0β, j=1,β¦,q. We first show that
[TABLE]
Indeed, for any Ξqβ1ΟββΞΎqβ1Οβ with ΟjΟββ£Ξqβ1Οββ=Οjβ we have
[TABLE]
Hence we have
[TABLE]
Therefore,
[TABLE]
By induction, for any q>2, we have
[TABLE]
A similar argument applies to obtain the same formula when q is odd. Now by (i) of Lemma 6.4, we get
[TABLE]
Notice that, m{R^ΟΟjβ1β+β0βΟβ=kjβ} depends only on ΟΟjβ1β+β0ββ,β¦,ΟΟjβ1β+β0β+kjββ1β while m{R^ΟΟjβ+β0βΟβ=kj+1β} depends only on ΟΟj+β0βββ,β¦,ΟΟjβ+β0β+kj+1ββ1β. By definition of kjβ, we have Οjβ1β+β0β+kjββ1=Οjββ1<Οjβ+β0β. Therefore, the product on the right hand side of (7.1) is formed of independent random variables. Moreover, observe that mΓm(G1Οβ) depends only on Οβ0ββ,β¦ΟΟ1ββ1β and Ο1ββ1<Ο1β+β0β. Thus,
[TABLE]
Taking Cq:=(DM0βM)qβ2 gives the desired estimate.
β
We now present two lemmas that will be invoked in the proof of Proposition 7.8 below.
Lemma 7.6**.**
We have βΟj+1ββΟjβ=KββΞ³(Οj+1ββΟjβ)β€C(K)<β. Moreover, C(K)β0 as Kββ.
Proof.
Using assumption (P7) and Lemma 9.1 in the appendix, we have
[TABLE]
Moreover; C(K)β0 as Kββ.
β
Lemma 7.7**.**
We have βkβ₯β0βββP{Ο1Οβ(x,xβ²)=k}<β.
Proof.
Recall that Ο1Οβ(x,xβ²)=β0β+R^Οβ0βΟββFΟβ0ββ(x); i.e., Ο1Οβ(x,xβ²) does not depend on xβ². Therefore,
[TABLE]
where we have used the first item of Lemma 6.4 and (P7).
β
We can now present the main result of this section.
We will show that the term Y1β decays at the indicated log-polynomial rate (in n) while the term Y2β decays as stretched exponential, which implies the result. First, for the term Y1β we have:
[TABLE]
For each term in the sum (7.4), using Lemma
7.4 we obtain,
We claim that n4β has a stretched exponential tail.
[TABLE]
for an appropriate choice of uβ²β²>0, 0<vβ²β²<1.
Now, for n>n4β using the fact that Οjβ1Οββ€n and Lemma 9.1 we can further upper bound the sum in the
equation (7.5) by
[TABLE]
Now, inserting the estimate (7.6) back into equation (7.5) and
substituting that result into (7.4) we obtain the final estimate on
Y1β:
[TABLE]
Now we tackle the term Y2β by decomposing ΞβΟβΞ into two pieces.
First for parameters K>0 and 0<Ο<1, and integer q>0 define
[TABLE]
We are going to pick the parameters K, and ΟβΌ1 later, but the idea is that for points in Bqβ(K,Ο) the first q return times have many (at least Οq) large (bigger than K) gaps.
Our decomposition will be according to this Bqβ(K,Ο) for q=βncβ:
[TABLE]
In order to estimate the first term in this expression,
fix a sequence of integers
2β€t1β<t2β<β¦tsβ for Οqβ€sβ€qβ1 and define
[TABLE]
Then Bqβ(K,c)=βͺs=Οqqβ1ββͺt1β<β―<tsββBqβ(K,{tiβ}) and by Lemma 7.5 we can estimate measures of the terms on the right by
[TABLE]
Now applying Lemmas 7.7 and 7.6 we obtain, assuming Ο>21β,
We move now to estimate the individual terms in the sum over Ο,qβbad terms. Note that each GqΟβ(Ο) is ΞΎqΟβ measurable. Therefore we can write
Now, since each good Ξqβ in the above sum is a subset of GqΟβ(Ο)
that is Ο,qβbad
we know that #{iβ£2β€iβ€q,ΟiββΟiβ1ββ€K}>(1βΟ)q. Therefore, in the above product,
considering only those factors in the product, and keeping in mind the lower bound
given by Lemma 7.2 we get
[TABLE]
Finally, summing first over all the good Ξqβ and then over all GqΟβ(Ο) for
Ο,qβbadΒ Ο we obtain
[TABLE]
Set ΞΊ=max{ΞΊ1β,(1βCΞ»~βV(K))(1βΟ)}<1 and obtain
[TABLE]
for all n>n5β(Ο), giving the claimed stretched exponential decay. This completes the proof of the lemma.
β
7.4. Coupling
Here we consider F^Οβ=(FΟβΓFΟβ)TΟβ which is a mapping from Ξ^Οβ=ΞΟβΓΞΟβ into Ξ^ΟTΟβΟβ.
Let ΞΎ^β1Οβ be the partition of Ξ^Οβ on which TΟβ is constant.
Let T1,Οβ<T2,Οββ¦ be stopping times on Ξ^Οβ defined as
[TABLE]
For u,zβΞ^Οβ we define a separation time s^(u,z) associated with F^Οβ as the smallest nβ₯0 such that F^Οnβ(u) and F^Οnβ(z) lie in distinct elements of ΞΎ^β1ΟTn,ΟβΟβ 888Notice that, for any u=(x,xβ²),z=(y,yβ²)βΞΟβ if s^(u,z)>n then s(x,xβ²)>n and s(y,yβ²)>n..
Let u=(x,xβ²), z=(y,yβ²). For n>0 choose k so that F^Οnβ(u)=(FΟβΓFΟβ)k(u). Then
[TABLE]
where we have used s^(u,z)β€min{s(x,xβ²),s(y,yβ²)}. Similarly for the second item we have
[TABLE]
β
Let ΞΎ^βiΟβ be the partition of Ξ^Οβ on which T1,Οβ,β¦,Ti,Οβ are constant. For zβΞ^Οβ let ΞΎ^βiΟβ(z) be the element containing z. Given Ξ¦(x,xβ²)=Ο(x)Ο(xβ²) let i1β(Ξ¦) be such that CΞ¦βΞ³i1β<D^. For i<i1β let Ξ¦^iββ‘Ξ¦. For iβ₯i1β, let
[TABLE]
where Ξ΅ is a small number that will be defined below.
Since (Ξ¦^iββΞ¦^iβ1β)/JF^Οiβ is constant on every ΞβΞΎ^βiΟβ, we have
[TABLE]
Note that, Ξ¦^iβ is the density of the part of Ξ»~ which has not been matched up to time Ti,Οβ.
Lemma 7.10**.**
For all sufficiently small Ξ΅>0 in (7.11), there exists 0<Ξ΅1β<1 independent of
Ξ¦ such that for almost every Ο and for all iβ₯i1β
[TABLE]
We will introduce the following densities in order to prove Lemma 7.10.
For zβΞ^Οβ let
There exists C^ such that for all sufficiently small Ξ΅ the following holds: for any zβΞ^Οβ with uβΞΎ^βiΟβ(z) and iβ₯i1β
[TABLE]
Proof.
By definition of Ξ¨iβ and item (1) Lemma 7.9 of we have
[TABLE]
Since Ξ΅i,zβ is constant on ΞΎ^βiΟβ(z) we let Ξ΅iβ=Ξ΅i,zβ. We have
[TABLE]
Notice that C in the latter inequality increases as Ξ¨iβ(z)Ξ¨iβ(u)β increases. Allowing Ξ΅ to depend on i, z,u and Ο,999Notice that when i=i1β we can choose Ξ΅ uniform for all u and z, allowing dependence only on D^ and CΞ¦β. for a given 0<Ξ΅β²<Ξ³β1β1 we can choose Ξ΅ small enough so that
Note that in the last inequality we have used CΞ¦βΞ³s^(u,z)β€D^Ξ³s^(F^Οi1ββu,F^Οi1ββz).
Finally, using the relation s^(F^Οiβjβu,F^Οiβjβz)=s^(F^Οiβu,F^Οiβz)+j we have
[TABLE]
where C^=2(1+Ξ΅β²)D^βj=0ββ[(1+Ξ΅β²)Ξ³]j.
Now, we show by an inductive argument that Ξ΅ in (7.13) can be chosen independent of i, u, z, Ο.
First of all notice that we can choose Ξ΅ independent of u,z for i=i1β because
[TABLE]
Let j>i1β and suppose that Ξ΅ is small enough so that (7.16) holds for all i<j and uβΞΎ^βiΟβ(z).
Then by (7.12) we have
[TABLE]
which implies that Ξ¨iβ(z)Ξ¨iβ(u)ββ[eβ(C^+D^),eC^+D^]. Therefore C in (7.13)
is bounded by eC^+D^. Hence by choosing Ξ΅<Ξ΅β²eβ(C^+D^) we conclude that the estimate in (7.15) holds for i=j.
β
Lemma 7.12**.**
Let 0<Ξ΅1β<1 be as in Lemma 7.10. For almost every Ο and all nβN
[TABLE]
where Ξ»~=Ξ»ΟβΓΞ»Οβ²β.
Proof.
In Lemma 7.10 the estimates for the mass of Ξ»~ after the ith iterate matching was given. Now we will relate that estimate to the iterates of FΟβ. Define Ξ¦0β, Ξ¦1β,β¦ as follows: for zβΞΟβΓΞΟβ let
[TABLE]
where Ξ¦^iβ(z) is as in (7.11).
We first prove that β£(FΟnβ)ββ(Ξ»)β(FΟnβ)ββ(Ξ»β²)β£β€2β«Ξ¦nβd(mΓm).
Below we use the notation Ξ¦(mΓm) to denote a measure whose density with respect to mΓm is Ξ¦.
First of all recall that Ξ¦0β=Ξ¦ and write Ξ¦=Ξ¦nβ+βk=1nβ(Ξ¦kβ1ββΞ¦kβ). We have
Before proving the proposition, we prove the following auxiliary lemma.
Lemma 8.2**.**
There exists CΞ»~β such that for all iβ₯1 and for any ΞβΞΎ^βiΟβ
[TABLE]
Proof.
By definition we have
[TABLE]
Therefore, it remains to bound the density d(F^Οiβ)ββΞ»~/d(mΓm).
Let ΞβΞΎ^βiΟβ. Any zβ²,uβ²βΞ^0,ΟTi,ΟβΟβ have unique pre-images u,zβΞ. By definition we have
[TABLE]
Since D^ is independent of Ξ this implies d(F^Οiβ)ββΞ»~/d(mΓm)<CΞ»~β.
β
Note that, by taking T0,Οββ‘0 Lemma 7.12 implies
[TABLE]
By choosing A(n)βN so that (1βΞ΅1β)A(n)β€nβ2a, for any iβ₯A(n) we have
[TABLE]
Now we estimate Ξ»~{Tiβ1,Οββ€n<Ti,Οβ} for iβ€A(n). Let m~=m(ΞΟβ)2mΓmβ. We proceed as in equation (7.4) in the estimate of Y1β. For every i we write
For the correlations we have the following relation
[TABLE]
Let Ξ» be a probability with density dmdΞ»β=Ο~β.
Then by Proposition 8.1 for every n>n6β we have
[TABLE]
Similarly, for the probability measure Ξ»β² with the constant density m(ΞΟβ)β1 we have
[TABLE]
Define CΞ»,Ξ»β²β:=max{CΞ»,Ξ½β,CΞ»β²,Ξ½β}. Substituting (8.7) and (8.8) into (8.6), and using the inequality AΟβ1ββ€m(ΞΟβ)[1+(2CΟβ²β+1))KΟβ] we have
[TABLE]
Let n7β(Ο)=inf{kβ₯n6β(Ο)β£ββ>k,Β KΟβΟββ€βΞ΄}. Then
[TABLE]
Now, if n>n7β then
[TABLE]
If nβ€n7β then we let
[TABLE]
Hence, for all nβN we have obtained
[TABLE]
It remains to show P{CΟβ>n} has the desired decay rate. We write
[TABLE]
Notice that by the definition of n7β we have
[TABLE]
Hence, we have
[TABLE]
Then the conclusion of the theorem holds with uβ² and vβ²:=3avβ²β.
Moreover, there exist C>0, uβ²>0 and 0<vβ²<1 such that
[TABLE]
Using the above statements and following the same strategy as in the proof of future correlations we conclude decay of past correlations.
9. Appendix
9.1. Sub-polynomial tail estimates
Lemma 9.1**.**
Let a>1 and b>0. Then
[TABLE]
Proof.
The proof is based on integration by parts. Since xa(logx)bβ is monotonically decreasing on (C,+β) for C big enough, we have
[TABLE]
Let K=[b]+1. Then first making change of variables y=logx and the integrating by parts K times we obtain
[TABLE]
where Ikβ(a,b)=(aβ1)βKβj=0Kβ1β(bβj)β«lognββybβKe(1βa)ydy.
Since bβK<0 we conclude that Ikβ(a)β€(aβ1)βKβ1βj=0Kβ1β(aβj)n1βa,
This shows that the dominant term in (9.1) is aβ11β(logn)bn1βa.
β
Lemma 9.2**.**
Suppose a>0 and akββΌ(logk)a1β. Then
βk=2nβakββΌ(logn)anβ.
Proof.
A straightforward estimate, using the fact that βk=2nβ(logk)a1βββ shows that βk=2nβakβββ and
βk=2nβakββΌβk=2nβ(logk)a1β.
We work with the latter sum. An elementary estimate shows
[TABLE]
Therefore βk=2nβ(logk)a1ββΌβ«2nβ(logx)adxβ. We now
estimate the integral.
[TABLE]
using integration by parts.
The first term above is the claimed rate, and the second term is clearly
o((logn)anβ). We will show the same is true for the third, integral term.
We first upper-bound as follows:
[TABLE]
Now simply estimate the right hand side by
[TABLE]
Since both terms are o(1) in n we are done.
β
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