# Operator version of Birkhoff ergodic theorem

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?

• 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? – Jochen Glueck Jun 25 '19 at 18:37
• Yes @JochenGlueck , $S$ could be also positive. Why is the answer yes? Could you give me a reference? – Littlefield Jun 26 '19 at 7:18
• Probably we also need to assume that $S(1)=1$. – Littlefield Jun 26 '19 at 15:35
• 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). – Jochen Glueck Jun 26 '19 at 18:54

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

• 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. – Littlefield Jun 27 '19 at 7:03
• @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. – Jochen Glueck Jun 27 '19 at 16:47
• 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. – Littlefield Jun 27 '19 at 18:00
• 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. – Littlefield Jul 2 '19 at 20:54