Operator version of Birkhoff ergodic theorem

Question

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?

Answer 1

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 $g = \langle 1,f\rangle \cdot 1$.