Let $(\Omega,\mathcal{F},P)$ be a probability space where $\Omega$ is a complete separable metric space, let $T:\Omega\to \Omega$ ` be an ergodic transformation, let $\hat{T}:L^{2}_{_P}(\Omega)\to L^{2}_{_P}(\Omega)$ be the (linear, isometric) operator $\hat{T}X:=X\circ T$, and let $\hat{T}^{*}: L^{2}_{_P}(\Omega)\to L^{2}_{_P}(\Omega)$ be the adjoint of $\hat{T}$.

Supose $\mathcal{F}_{0}\subset \mathcal{F}$ is a given sigma field and that, upon definning $$\mathcal{F}_{k}:= T^{-k}\mathcal{F}_{0}:=\{T^{-k}A: A\in \mathcal{F}_{0}\}$$ one has $\mathcal{F}_{k}\subset \mathcal{F}_{k-1}\subset \mathcal{F}$ for all $k\in \mathbb{Z}$, and let $E_{n}: L^{2}_{_P}(\Omega)\to L^{2}_{_P}(\Omega)$ be the projection

$$E_{n}(X):=E[X|\mathcal{F}_{n}]$$ onto the subspace of $\mathcal{F}_{n}-$measurable functions.

In his paper "Central Limit Theorem for Deterministic Systems", C.Liverani presents the following theorem (be aware! I made significant changes in notation)

Theorem 1 Let $X_{0}\in L^{\infty}_{_P}(\Omega)$ be an $\mathcal{F}_{0}-$measurable function such that $EX_0=0$, and define $X_{n}:=\widehat{T}^n X_{0}$. Then the Central Limit Theorem for $\{X_{n}\}_{n\geq 0}$ holds under the following assumptions:

(0) $E_{1}\hat{T}\hat{T}^{*}=E_{1}$

(1) $\sum_{n=0}^{\infty} |E[X_{0}X_{n}]|<\infty$

(2) $P[\sum_{n=0}^{\infty} |E_{0}[\hat{T}^{*n}X_{0}]|=\infty]=0$

I.e, the assumtions $(0)-(2)$ imply that $$\frac{1}{\sqrt{n}}\sum_{k=0}^{n}X_{k}\Rightarrow N(0,\sigma^{2}) $$ where

(A) $\sigma^{2}\leq EX_{0}^2+2\sum_{k=1}^{\infty}EX_{0}X_{k}$, and

(A') equality (for $\sigma^2$) occurs if, for the series in $(2)$, we strengthen to convergence in $L^{1}_P(\Omega)$.

(B) $\sigma=0$ if and only if $\hat{T}X_0= (\hat{T}-Id)Y_0$ for some $\mathcal{F}_{0}-$measurable function $Y_0$.

I have been working in this result and I have found myself stuck in the following part of his argument (I just summarize here, assuming that he who can help me will go to the original paper if she doesn't know the details. I will give more explanations upon request): let be $\lambda>1$ be given and let be $Y_{0}(\lambda)$, $D_{1}(\lambda)$, $\mathcal{F}_{0}$ measurable such that $E_{1}D_{1}(\lambda)=0$ and, if $Y_{k}(\lambda):=\hat{T}^{k}Y_{0}(\lambda)$, $D_{k}(\lambda)=\hat{T}^{k-1}D_{1}(\lambda)$ ($k\geq 1$)`, then


Indeed, this gives $Y_{0}(\lambda)=\sum_{n\geq 0}\lambda^{-n}E_{0}\hat{T}^{*n}X_{0}$, $D_{k}(\lambda)$ happens to be a reverse martingale difference with respect to $\{\mathcal{F}_{k}\}_{k}$ (this follows from $E_{0}D_{1}(\lambda)=D_{1}(\lambda)$ and $E_{1}D_{1}(\lambda)=0$), and $D_{1}:=\lim_{\lambda\to 1} D_{1}(\lambda)$ is a (well defined) measurable function which he proves has second moment.

Here is what I do not understand, and is essential to his argument: how does one know, under the theorem's hypothesis, that $E_{1}D_{1}=0$? An exchange of limits and conditional expectation seems to be involved here, and I can't justify it without passing to the hypothesis (A'). Can you give some light on this?

Unfortunately, my imagination is too short and can't see an example where (2) holds but (A') does not, so I don't even know how to start looking for possible counterexamples to the theorem as stated here.

PS: The reason of my concerns is that, with those things that I've understood, this theorem (or better: its restricted version, assuming (A')) can be extended to the case $X_{0}\in L^2$ by means of just a couple of very simple observations to Liverani's argument. As I find this fact worth communicating I decided to try writing this extension down, but I realized I was missing that step, which I can't fill neither see in the paper's proof.



(Important editions were made following the discussion in the comments. My acknowledgments to Ian Morris for his enlightment on this question.)

Indeed, Liverani proves that $$E[(D_{1}(\lambda))^2]\leq E[X_{0}^2]+2\sum_{n\geq 1}E[X_{0}X_{n}]$$ In particular, $\{D_{1}(\lambda)\}_{\lambda>1}$ is a bounded set in $L^{2}$ and there exists, by Alaoglu Theorem, a sequence $\lambda_{n}\to 1$ and a weak-$L^2$ limit for $D_{1}(\lambda_{n})$, say $D'$.

For such a limit one has in particular that, if $I_{A}$ denotes the characteristic function of a measurable set $A$ then $$E(D_{1}(\lambda_{n})I_{A})\to E(D'I_{A})$$ and in particular one has that, for all $A\in \mathcal{F}_{1}$, $E(D'I_{A})=0$.

This proves that $E_{1}D'=0$, and the result follows from the fact (see the discussion below in the comments) that a.e limits and weak limits coincide when both exist.

  • $\begingroup$ I am unsure about the details of your application of Alaoglu's theorem: in general a bounded sequence in $L^2$ will not have a convergent subsequence with respect to the $L^2$ distance, only with respect to the weak topology. However, $g \mapsto \mathbb{E}(g\chi_A)$ is a continuous linear functional on $L^2$ so Alaoglu's theorem delivers the result you need and $\mathbb{E}(D_1(\lambda_n)\chi_A)$ does converge to $\mathbb{E}(f\chi_A)$ where $f$ is the limit of the subsequence. I don't recall the details, but it should not be hard to show that the weak limit and a.e. limit agree when both exist. $\endgroup$ – Ian Morris Mar 22 '13 at 16:59
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    $\begingroup$ (I think I saw your point. I will think about it) $\endgroup$ – David Mar 22 '13 at 17:58
  • $\begingroup$ If $(e_n)$ is an orthonormal basis sequence for $L^2$ then its weak limit is zero but it does not converge in the $L^2$ distance. Your argument shows -- correctly I think, if you directly use the definition of weak convergence to justify $E(D_1(\lambda_n I_A)) \to E(D_1I_A)$ instead of attempting to use norm convergence in $L^2$ and $L^1$ -- that $D(\lambda_n)$ has a limit in the weak topology, and that the conditional expectation with respect to $\mathcal{F}_1$ of that limit is zero. It remains only to show that the weak limit really is $D_1$: math.stackexchange.com/questions/160306 $\endgroup$ – Ian Morris Mar 22 '13 at 18:20
  • $\begingroup$ Yes you are right. Here are my current findings in that direction: apparently one can not say that the weak limit is $D_{1}$ just from the fact that there is a.e convergence to $D_{1}$ and the weak limit exists. Here is the reason: you can repeat the argument above taking absolute values to conclude that $|D_{1}(\lambda_{n})|$ has a weak limit $f$. In particular $E[|D_{1}(\lambda_{n})|]\to E[f]$ (take the constant function $1=\xi_{\Omega}$). Now, clearly $|D_{1}(\lambda_{n})|\to |D_{1}|$ but, according to Billingsley's "Probability and Measure", corollary to Theorem 16.14 $\endgroup$ – David Mar 22 '13 at 18:50
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    $\begingroup$ The absolute value of the weak limit of $f_n$ does not have to equal the weak limit of the sequence of absolute values $|f_n|$. For example, in $L^2([0,1])$ take $f_n(x)=sin(nx)$ to obtain $f_n \to 0$ weakly and $|f_n| \to \frac{1}{2}$ weakly. $\endgroup$ – Ian Morris Mar 22 '13 at 19:30

Which instance of $\mathbb{E}(Y_1|\mathcal{F}_1)-0$ (or $E_1D_1=0$ in your notation) is problematic? I can find one instance shortly after (1.6) but I think that this is a typo, and the text there should read

`$\ldots$since $\mathbb{E}(\hat{T}f - \hat{T}g(\lambda) + \lambda^{-1}g(\lambda)|\mathcal{F}_1) = \mathbb{E}(Y_1(\lambda)|\mathcal{F}_1)=0$.''

This statement can be justified by some straightforward manipulations since the sums involved now converge in $L^1$. Conversely, at this stage in the argument $g$, and hence $Y_1$, is only measurable, and I can see no reason for $\mathbb{E}(Y_1|\mathcal{F}_1)$ to even be a well-defined expression.

Once one has established $\mathbb{E}(Y_1^2)<\infty$, it is then implicitly necessary to go back and show that indeed $\mathbb{E}(Y_1|\mathcal{F}_1)=0$. Is this the problematic step? (I am certainly not quite sure at this stage how to do it.)

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    $\begingroup$ Yes, that is precisely the point. As you see this is the launching point for the Martingale approximation to run, but I don't see how to justify that statement. $\endgroup$ – David Mar 19 '13 at 15:53

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