Quenched Linear Response for Smooth Expanding on Average Cocycles

Abstract

We establish an abstract quenched linear response result for random dynamical systems, which we then apply to the case of smooth expanding on average cocycles on the unit circle. In sharp contrast to the existing results in the literature, we deal with the class of random dynamics that does not necessarily exhibit uniform decay of correlations. Our techniques rely on the infinite-dimensional ergodic theory and in particular, on the study of the top Oseledets space of a parametrized transfer operator cocycle. Finally, we exhibit a surprising phenomenon: a random system and a smooth observable for which quenched linear response holds, but annealed response fails.

An Example Where Quenched Response Holds and Annealed Response Fails

Let us remind the reader that whenever quenched linear response holds (see e.g. Theorem 20), one has, for any smooth observable $\phi$, that

$$\begin{aligned} \int _{{\mathbb {S}}^1}\phi {\hat{h}}_\omega dm&= \sum _{n=0}^{\infty }\int _{{\mathbb {S}}^1}\phi \cdot {\mathcal {L}}^{n}_{\sigma ^{-n}\omega }\hat{{\mathcal {L}}}_{\sigma ^{-n-1}\omega } h_{\sigma ^{-n-1}\omega } dm\\&= \sum _{n=0}^{\infty }\int _{{\mathbb {S}}^1}\phi \circ T^n_{\sigma ^{-n}\omega }\cdot \hat{{\mathcal {L}}}_{\sigma ^{-n-1}\omega } h_{\sigma ^{-n-1}\omega } dm. \end{aligned}$$

In particular, the integral on the L.H.S is well-defined.

If one is interested in annealed linear response, the object of interest becomes the (double) integral $\int _{\Omega }\int _{{\mathbb {S}}^1}\phi {\hat{h}}_\omega dm~d{\mathbb {P}}$: when it holds this integral is well-defined, and one has

$$\begin{aligned} \int _\Omega \int _{{\mathbb {S}}^1}\phi {\hat{h}}_\omega dm~d{{\mathbb {P}}} = \sum _{n=0}^{\infty }\int _{\Omega }\int _{{\mathbb {S}}^1}\phi \circ T_\omega ^n\cdot \hat{{\mathcal {L}}}_{\sigma ^{-1}\omega } h_{\sigma ^{-1}\omega } dm~d{{\mathbb {P}}}. \end{aligned}$$

Given that quenched results concerns a.e trajectory, and that annealed one are on average, in general one expects that a quenched result implies its annealed counterpart. This intuition is validated in the context of response theory by the existing results [18, 22, 34], where stochasticity is uniform (i.e. assumptions are uniform w.r.t $\omega$). However, when there is no stochastic uniformity, this intuition can fail, as the next example shows:

Consider $({\tilde{\Omega }},{\mathcal {B}},\mathbb {Q}, S)$ the full-shift over $\{1,2,\dots \}$, with probability vector

$(Z,Z/2^{2+\delta },\dots ,Z/n^{2+\delta },\dots )$, for some $0\le \delta \le 1$, Z being the normalization constant.

Let $h:{\tilde{\Omega }}\rightarrow {\mathbb {R}}$ be the (positive) observable defined by $h(\omega )=\omega _0$ if $\omega \,{:=}\,(\omega _n)_{n\in {\mathbb {Z}}} \in {\tilde{\Omega }}$. Note that

$$\begin{aligned} \int _{{\tilde{\Omega }}}h~d\mathbb {Q}=\sum _{i\ge 1}i\cdot \frac{Z}{i^{2+\delta }}=Z\sum _{i=1}^\infty \frac{1}{i^{1+\delta }}<+\infty , \end{aligned}$$

when $0<\delta \le 1$.

Define $(\Omega ,{\mathcal {F}},{\mathbb {P}},\sigma )$ to be the suspension over S with roof function h, i.e. $\Omega \,{:=}\,\{(\omega ,i)\in {\tilde{\Omega }}\times {\mathbb {N}},~ 0\le i<h(\omega )\}$, $\sigma :\Omega \circlearrowleft$ is given by

$$\begin{aligned} \sigma (\omega ,i)\,{:=}\,\left\{ \begin{aligned}&(\omega ,i+1)~&\text {if}~i<h(\omega )-1\\&(S\omega ,0)~&\text {if}~i=h(\omega )-1 \end{aligned} \right. \end{aligned}$$

and ${\mathbb {P}}(A)\,{:=}\,\left( \int _{{\tilde{\Omega }}} h~ d\mathbb {Q}\right) ^{-1}\sum _{i\ge 0}\mathbb {Q}\left( A\cap ({\tilde{\Omega }}\times \{i\})\right)$.

We can now define our random system: take $T_0:{\mathbb {S}}^1\circlearrowleft$ to be the doubling map, i.e. $T_0(x)=2x \ (\text {mod} \ 1)$, and let $T_1$ be the identity map on ${\mathbb {S}}^1$. We consider the random circle map $T_{(\omega ,i)}$, $(\omega ,i)\in \Omega$ defined by

$$\begin{aligned} T_{(\omega ,i)}\,{:=}\,{\left\{ \begin{array}{ll} T_1 &{} \hbox {if} \ i<h(\omega )-1 \\ T_0 &{} \text {if} i=h(\omega )-1. \end{array}\right. } \end{aligned}$$

Clearly, $(T_{(\omega ,i)})_{(\omega , i)\in \Omega }$ is an expanding on average cocycle, i.e.

$$\begin{aligned} \int _{\Omega }\log \lambda _{(\omega , i)}~d{\mathbb {P}} ((\omega , i))>0. \end{aligned}$$

Indeed, observe that $(T_{(\omega ,i)})_{(\omega , i)\in \Omega }$ is a particular case of cocycles $(T_\omega )_{\omega }$ introduced in Example 17. Moreover, $\mu _{(\omega , i)}=m$ for $(\omega , i) \in \Omega$, where m denotes the Lebesgue measure on ${\mathbb {S}}^1$.

Let $n_c$ be the (random) covering time for the interval [0, 1/2]:

$$\begin{aligned} n_c (\omega , i)\,{:=}\,\min \{ k\in {\mathbb {N}}: T_{(\omega ,i)}^k([0,1/2])={\mathbb {S}}^1 \}, \quad (\omega , i) \in \Omega . \end{aligned}$$

Then, it is easy to see that $n_c(\omega , i)=h(\omega )-i$. Observe that $n_c$ is not integrable. Indeed, for each $N\in {\mathbb {N}}$ we have that

$$\begin{aligned} {\mathbb {P}}(n_c(\omega ,i)=N)&=\left( \int _{{\tilde{\Omega }}} h~ d\mathbb {Q}\right) ^{-1}\sum _{i\ge 0}\mathbb {Q}(h(\omega )-i=N)\\&=\left( \int _{{\tilde{\Omega }}} h~ d\mathbb {Q}\right) ^{-1}\sum _{i\ge N}\mathbb {Q}(\omega _0=i)\\&=\left( \int _{{\tilde{\Omega }}} h~ d\mathbb {Q}\right) ^{-1}\sum _{i\ge N}\frac{Z}{i^{2+\delta }}\sim \frac{C}{N^{1+\delta }}, \end{aligned}$$

for some constant $C>0$, which easily implies that $n_c$ is not integrable.

We now introduce the observable: consider a $\psi \in C^\infty ({\mathbb {S}}^1)$, such that:

• (i) $\text {supp}~\psi \subset I$, where $I\subset {\mathbb {S}}^1$ is an interval such that $I\cap \dfrac{1}{2}I=\emptyset$ (in particular, we can take any small enough $I\not \ni 0$).

• (ii) $\int _{{\mathbb {S}}^1}\psi ~dm=0$ and $\int _{{\mathbb {S}}^1}\psi ^2~dm=1$.

It is easy to see that $\int _{{\mathbb {S}}^1}\psi \cdot \psi \circ T_0~dm=0$. Therefore, we have

$$\begin{aligned} \int _{{\mathbb {S}}^1} \psi \cdot \psi \circ T_{(\omega ,i)}^n~dm= \left\{ \begin{aligned}&1~\text {if}~n<n_c(\omega ,i)\\&0~\text {otherwise.} \end{aligned} \right. \end{aligned}$$

In particular, it follows from the non-integrability of $n_c$³ that

$$\begin{aligned} \int _{\Omega }\sum _{n=0}^\infty \int _{{\mathbb {S}}^1} \psi \cdot \psi \circ T_{(\omega ,i)}^n~dm~d{\mathbb {P}} =+\infty . \end{aligned}$$

(61)

Let us now introduce the perturbed cocycle $T_{(\omega ,i),\varepsilon }:= D_{\varepsilon }\circ T_{(\omega ,i)}$, where $D_{\varepsilon }:{\mathbb {S}}^1\circlearrowleft$ is given by

$$\begin{aligned} D_{\varepsilon }(x)\,{:=}\,x-\varepsilon \int _0^x\psi ~dm \ (mod \ 1). \end{aligned}$$

Denote by $S(x)\,{:=}\,-\int _0^x\psi ~dm$. Since this perturbation is smooth and deterministic ($D_\varepsilon$ does not depend on $\omega$), we may apply Theorem 20, and obtain that for a.e. $(\omega ,i)\in \Omega$, the integral $\int _{{\mathbb {S}}^1}\psi {\hat{h}}_{(\omega , i)} dm$ is well defined, and satisfies

$$\begin{aligned} \int _{{\mathbb {S}}^1}\psi {\hat{h}}_{(\omega ,i)} dm&= \sum _{n=0}^{\infty }\int _{{\mathbb {S}}^1}\psi {\mathcal {L}}_{\sigma ^{-n} (\omega ,i)}^n\hat{{\mathcal {L}}}_{\sigma ^{-n-1}(\omega ,i)}h_{\sigma ^{-n-1}(\omega ,i)} dm\\&= -\sum _{n=0}^\infty \int _{{\mathbb {S}}^1}\psi \circ T_{\sigma ^{-n}(\omega ,i)}^{n} S' dm\\&=\sum _{n=0}^\infty \int _{{\mathbb {S}}^1}\psi \circ T_{\sigma ^{-n}(\omega ,i)}^{n} \psi dm, \end{aligned}$$

where we used that $\hat{{\mathcal {L}}}_{(\omega ,i)}f = -({\mathcal {L}}_{(\omega ,i)}(f)\cdot S)'$ (which can be proved by arguing as in [26, Sec 6.2]) and $h_{(\omega ,i)}=\mathbbm {1}$ for our particular example.

Hence, from (61)

$$\begin{aligned} \int _{\Omega }\int _{{\mathbb {S}}^1}\psi {\hat{h}}_{\omega ,i} dm d{\mathbb {P}}=+\infty . \end{aligned}$$

(62)

In particular, annealed response cannot hold.