4
$\begingroup$

Suppose that $(\Omega,\mathcal{E},P)$ is a probability space and suppose that we have a measurable operator $T:\Omega\to\Omega$.

Recall that $T$ is said to be egodic if:

  1. $T$ is measure preserving: i.e. $P(T^{-1}E)=P(E)$;
  2. $T$ is zero-one: $E=T^{-1}(E)$ implies that $P(E)\in\{0,1\}$;

Now consider the operator $S_T:L^1(\Omega,\mathcal{E},P)\to L^1(\Omega,\mathcal{E},P)$ given by $S_T(f):=f\circ T$.

Then the Birkhoff theorem tells us that, if $T$ is ergodic, then $$ \lim_t \frac{1}{t} \sum_{j=0}^{t-1} S_T^j(f)=E[f]\quad\text{ a.s..} $$

It is known that $T$ is ergodic if and only if $S_T$ satisfies

  1. $E[S_T(f)]=E[f]$;

  2. $S_T(f)=f$ implies that $f$ is a constant a.s..

My question is: Is the claim in the Birkhoff theorem still true when an operator $S$ satisfies 1 and 2 above, but we do not necessarily have that $S=S_T$ for some ergodic transformation?

That is, suppose that $S_T:L^1(\Omega,\mathcal{E},P)\to L^1(\Omega,\mathcal{E},P)$ is any linear function that satisfies 1 and 2 above, is it true that $$ \lim_t \frac{1}{t} \sum_{j=0}^{t-1} S^j(f)=E[f]\quad\text{ a.s..} $$ ?

If true, can you give me some reference?

$\endgroup$
  • 2
    $\begingroup$ That $S$ be "any linear function" is an extremely weak condition. If you assume, however, that $S$ if positive (in the sense that $Sf \ge 0$ whenever $f \ge 0$), then the answer is "yes". Is $S$ positive in the situation you have in mind? $\endgroup$ – Jochen Glueck Jun 25 '19 at 18:37
  • 1
    $\begingroup$ Yes @JochenGlueck , $S$ could be also positive. Why is the answer yes? Could you give me a reference? $\endgroup$ – Littlefield Jun 26 '19 at 7:18
  • $\begingroup$ Probably we also need to assume that $S(1)=1$. $\endgroup$ – Littlefield Jun 26 '19 at 15:35
  • $\begingroup$ Yes, you are right that the assertion is false without an assumption of the type $S1=1$ (apparently, I was not reading carefully enough; I thought you're first assumption was $Sf=f$ if and only if $f$ is constant). $\endgroup$ – Jochen Glueck Jun 26 '19 at 18:54
2
$\begingroup$

This answer consists of two parts:

Part I. The answer is no, in general. Here is a counterexample:

Example. Let $\Omega = \{1,2\}$ and let $P$ by $1/2$ times the counting measure. The matrix \begin{align*} S = 1/2 \cdot \begin{pmatrix} 1 & 1 \\ 1 & 1 \end{pmatrix} + \begin{pmatrix} 1 & -1 \\ -1 & 1 \end{pmatrix} \end{align*} fulfils both assumptions 1. and 2., but $\frac{1}{n} \sum_{k=0}^{n-1} S^k f = \frac{1}{n} \sum_{k=0}^{n-1} 2^k f = \frac{2^n-1}{n}$ if $f$ equals the eigenvector $(1,-1)^T$ of $S$ (for the eigenvalue $2$).

Part II. The answer is yes if the operator $S$ is positive in the sense that $S f \ge 0$ whenever $f \ge 0$ and if, in addition, $S1 = 1$.

This follows, for instance, from Theorem 11.4 in [Eisner, Farkas, Haase, Nagel: Operator Theoretic Aspects of Ergodic Theory (2015)].

EDIT in response to a comment. The reference quoted above only shows almost everywhere convergence of the Cesàro means, without giving detailed information about the limit function. The fact that the limit function is given by $E[f] \cdot 1 = \langle 1, f\rangle \cdot 1$ can for instance be seen as follows (maybe there is also a more direct argument, but I find the following argument rather natural from an operator theoretic point of view):

  • Order intervals in $L^1$-spaces are weakly compact (see for instance this post). In particular, the S-invariant order interval $[-c \cdot 1, c \cdot 1]$ is weakly compact for each number $c \ge 0$.

  • Hence, it follows for instance from Corollary~A.5 in [Engel, Nagel: One-Parameter Semigroups for Linear Evolution Equations (2000)] that all orbits of $S$ are relatively weakly compact.

  • This implies in turn that the Cesàro means of $S$ converge strongly to a bounded linear operator $Q$, i.e. $S$ is mean ergodic (see Eberlein's ergodic theorem).

  • By a standard result from operator theory, this implies that the fixed space of $S$ separates the fixed space of the dual operator $S'$. Hence, as the fixed space of $S$ is one dimensional by assumption, the fixed space of the dual operator is one-dimensional, too, i.e. it is spanned by the function $1$.

  • It is easy to see that $Q$ is a projection whose range consists of the fixed points of $S$ and that the range of the dual projection $Q'$ consists of the fixed points of $S'$. Hence, we have $Q = 1 \otimes 1$, i.e. $Qf = \langle 1, f\rangle \cdot 1$ for each $f \in L^1(\Omega,P)$.

  • Hence, $(\frac{1}{n}\sum_{k=0}^{n-1} S^kf)$ converges in norm to $\langle 1, f \rangle \cdot 1$, and as quoted above the sequence converges almost everywhere to a function $g \in L^1(\Omega,P)$. This implies that actually $g = \langle 1,f\rangle \cdot 1$.

$\endgroup$
  • 1
    $\begingroup$ Thanks a lot. This makes things much more clear. However, I have another question. I read Theorem 11.4 in Eisner et al.'s book. This theorem tells us that the limit exists, but how can we know that this limit is exactly $E[f]$? Sorry, I do not know in deep ergodic theory. $\endgroup$ – Littlefield Jun 27 '19 at 7:03
  • 1
    $\begingroup$ @Littlefield: I added an argument that shows why the limit is of the claimed form. Unfortunately, I don't have much time right now, so I was a bit sketchy. Please leave another comment if you need more details or references; I'll then add them next week. $\endgroup$ – Jochen Glueck Jun 27 '19 at 16:47
  • 1
    $\begingroup$ Thanks a lot for taking your time for this outline of the proof. I will read carefully. By the way, I have realized that in that post on weak compactness of order intervals, we both interchanged also comments there already. That is a funny coincidence. $\endgroup$ – Littlefield Jun 27 '19 at 18:00
  • $\begingroup$ Theorem 8.24 in [Eisner, Farkas, Haase, Nagel: Operator Theoretic Aspects of Ergodic Theory (2015)] that the limit is in L1. So you get directly the last item in last part of your answer. $\endgroup$ – Littlefield Jul 2 '19 at 20:54

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.