The Koopman operator

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Let $T$ be a measure-preserving transformation on a probability space $(X,\mathcal{B},\mu )$. In particular $T$ is a map from $X$ to itself, and we can use composition to induce a map on spaces of functions on $X$. For example, given $f:X\to \mathbb{C}$ we can define a new function $Tf$ by $$(Tf)(x)=f(T(x))$$ for all $x\in X$. Thus $Tf$ is just $f\circ T$. One then has the option of studying $T$ not by its direct effect on points in $X$ but by its indirect effect on functions. For any two functions $f,g$ and any $\alpha$ in $\mathbb{C}$ we have $$\begin{array}{c}(T(f+g))(x)=(Tf)(x)+(Tg)(x)\\ (T(\alpha f))(x)=(\alpha (Tf))(x)\end{array}$$ for all $x$ in $X$. Thus $$\begin{array}{c}T(f+g)=Tf+Tg\\ T(\alpha f)=\alpha (Tf)\end{array}$$ and $T$ acts linearly on functions. We can therefore study $T$ using the tools of linear algebra.

The action on Lebesgue space

In our context, where $T$ is measurable and measure-preserving, it makes sense to consider the effect $T$ has on measurable and integrable functions. The relevant spaces - Lebesgue spaces - are not just vector spaces. They also have topological structure coming from the norm we have put on them. We can therefore use functional analysis - which is essentially the study of topological vectors spaces and the continuous, linear maps between then - to study our transformation.

In this course we will mainly look at how $T$ affects functions in the Lebesgue space ${\mathsf{L}}^{\mathsf{2}}(X,\mathcal{B},\mu )$. Our first duty is to check that composition with $T$ gives us a well-defined map from ${\mathsf{L}}^{\mathsf{2}}(X,\mathcal{B},\mu )$ to ${\mathsf{L}}^{\mathsf{2}}(X,\mathcal{B},\mu )$.

Lemma

Fix a measurable space $(X,\mathcal{B})$ and a measurable map $T:X\to X$. If $f:X\to \mathbb{R}$ is $(\mathcal{B},\mathsf{Bor}(\mathbb{R}))$ measurable then $f\circ T$ is also $(\mathcal{B},\mathsf{Bor}(\mathbb{R}))$ measurable.

Proof:

We check the definition of measurability. Fix a set $B$ in $\mathsf{Bor}(\mathbb{R})$. From $$(f\circ T{)}^{-1}(B)=T^{-1}(f^{-1}(B))$$ and the fact that $f^{-1}(B)\in \mathcal{B}$ we get $(f\circ T{)}^{-1}(B)\in \mathcal{B}$ as desired. ▮

Lemma

Fix a probability space $(X,\mathcal{B},\mu )$ and a measure-preserving map $T:X\to X$. If $f$ belongs to ${\mathcal{L}}^{\mathsf{2}}(X,\mathcal{B},\mu )$ then $Tf$ does as well.

Proof:

Fix $f:X\to \mathbb{R}$ measurable such that $$\int |f{|}^2<\mathrm{\infty }$$ holds. By the previous lemma the function $Tf$ is measurable and therefore $$\int |Tf{|}^2\mathsf{d}\mu$$ exists in $[0,\mathrm{\infty }]$. Let $n\mapsto {\varphi }_n$ be a non-decreasing sequence of non-negative and simple functions with ${\varphi }_n\to |f{|}^2$ pointwise. It is immediate that $$T{\varphi }_n\to T(|f{|}^2)=|Tf{|}^2$$ and therefore $$\begin{array}{rl}\int |Tf{|}^2\mathsf{d}\mu & =\underset{n\to \mathrm{\infty }}{lim}\int T{\varphi }_n\mathsf{d}\mu \\ \int |f{|}^2\mathsf{d}\mu & =\underset{n\to \mathrm{\infty }}{lim}\int {\varphi }_n\mathsf{d}\mu \end{array}$$ by the monotone convergence theorem. It therefore suffices to check that $$\int T\psi \mathsf{d}\mu =\int \psi \mathsf{d}\mu$$ for all simple functions $\psi :X\to \mathbb{R}$.

Fix a simple and measurable function $$\psi =a(1)1_{B(1)}+\cdots +a(s)1_{B(s)}$$ from $X$ to $\mathbb{R}$ with $B(1),\dots ,B(s)$ all in $\mathcal{B}$. One calculates $$\begin{array}{rl}T\psi & =a(1)1_{B(1)}\circ T+\cdots +a(s)1_{B(s)}\circ T\\ & =a(1)1_{T^{-1}B(1)}+\cdots +a(s)1_{T^{-1}B(s)}\end{array}$$ so that $$\begin{array}{rl}\int T\psi \mathsf{d}\mu & =a(1)\mu (T^{-1}B(1))+\cdots +a(s)\mu (T^{-1}B(s))\\ & =a(1)\mu (B(1))+\cdots +a(s)\mu (B(s))=\int \psi \mathsf{d}\mu \end{array}$$ because $\mu$ is $T$ invariant. ▮

Theorem

Fix a measure-preserving transformation $T$ of a probability space $(X,\mathcal{B},\mu )$. For every $f$ in ${\mathsf{L}}^{\mathsf{2}}(X,\mathcal{B},\mu )$ the composition $Tf$ also belongs to ${\mathsf{L}}^{\mathsf{2}}(X,\mathcal{B},\mu )$.

Proof:

Fix $f$ in ${\mathsf{L}}^{\mathsf{2}}(X,\mathcal{B},\mu )$. Recall that $f$ is in fact an equivalence class of measurable functions $$f=F+{\mathcal{Z}}^{\mathsf{2}}(X,\mathcal{B},\mu )=\{g\in {\mathcal{L}}^{\mathsf{2}}(X,\mathcal{B},\mu ):||g-F|{|}_{\mathsf{2}}=0\}$$ where $F:X\to \mathbb{R}$ is a fixed representative. We would like to define $$Tf=TF+{\mathcal{Z}}^{\mathsf{2}}(X,\mathcal{B},\mu )$$ and must check this is well-defined. Thus fix $g:X\to \mathbb{R}$ measurable with $||F-g|{|}_{\mathsf{2}}=0$. By the two lemmas above $|T(F-g){|}^2$ is measurable and $$\int |T(F-g){|}^2\mathsf{d}\mu =\int |F-g{|}^2\mathsf{d}\mu$$ so $||(TF)-(Tg)|{|}_{\mathsf{2}}=0$ and $Tf$ is well-defined. ▮

The above theorem gives us a well-defined map $$T:{\mathsf{L}}^{\mathsf{2}}(X,\mathcal{B},\mu )\to {\mathsf{L}}^{\mathsf{2}}(X,\mathcal{B},\mu )$$ whenever $T$ is a measure-preserving transformation of the probability space $(X,\mathcal{B},\mu )$. This map on Lebesgue space is called the Koopman operator associated with $T$.

Dynamical properties

Our goal is to understand dynamical properties of $T$ in terms of its Koopman operator, and to use the Koopman operator to deduce dynamical properties of $T$. Generally, this involves replacing statements about sets with statements about functions, the connection being that every set $E\subset X$ gives rise to a function $1_E$. We begin by rephrasing ergodicity of $T$ in terms of its Koopman operator.

We defined $T$ to be ergodic if and only if all invariant sets have measure 0 or measure 1. Since $$T{1}_E=1_{T^{-1}E}=1_E$$ whenever $E$ is invariant, it seems reasonable to look at those functions in ${\mathsf{L}}^{\mathsf{2}}(X,\mathcal{B},\mu )$ that are fixed by the Koopman operator.

Theorem

A measure-preserving transformation $T$ on $(X,\mathcal{B},\mu )$ is ergodic if and only if all $T$ invariant functions in ${\mathsf{L}}^{\mathsf{2}}(X,\mathcal{B},\mu )$ are equal almost-everywhere to a constant function.

Proof:

First suppose that every invariant function $f$ in ${\mathsf{L}}^{\mathsf{2}}(X,\mathcal{B},\mu )$ is equal almost-everywhere to a constant function. This means the following: if $f$ is invariant then there is $c\in \mathbb{C}$ such that $$\mu (\{x\in X:f(x)=c\})=1$$ holds. Fix $E\subset X$ with $T^{-1}E=E$. Then $T{1}_E=1_E$ and $1_E$ as a member of ${\mathsf{L}}^{\mathsf{2}}(X,\mathcal{B},\mu )$ is fixed by the Koopman operator. By hypothesis there is then a constant $c\in \mathbb{C}$ with $1_E=c$ almost everywhere As $1_E$ only takes the values $0$ and $1$ it must be the case that $c\in \{0,1\}$ and therefore $\mu (E)\in \{0,1\}$.

Conversely, suppose that $T$ is ergodic and that $f$ from ${\mathsf{L}}^{\mathsf{2}}(X,\mathcal{B},\mu )$ is fixed by the Koopman operator. Suppose first that $f$ is real-valued. Fix $k\in \mathbb{N}$. Put $$E(k,N)=\{x\in X:\frac{N}{2^k}\le f(x)<\frac{N+1}{2^k}\}$$ for each $N\in \mathbb{Z}$. We have $$\mu (E(k,N)\mathrm{△}{T}^{-1}E(k,N))=0$$ and therefore $\mu (E(k,N))\in \{0,1\}$. Exactly one of these sets can have full measure. As we increase $k$ we converge to a specific value $c\in \mathbb{R}$ with $$\mu (\{x\in X:f(x)=c\})=1$$ and therefore $f$ is constant almost-everywhere. ▮

In analogy with linear algebra, we next look for eigenfunctions of $T$. Eigenfunctions of the Koopman operator indicate that $T$ has some component of regularity.

Definition

By an eigenfunction of a measure-preserving transformation $T$ we mean a member $f$ of ${\mathsf{L}}^{\mathsf{2}}(X,\mathcal{B},\mu )$ such that $Tf=\eta f$ for some $\eta \in \mathbb{C}$.

Example

For the the irrational rotation $T(x)=x+\alpha mod1$ there are lots of eigenfunctions. Indeed $${\psi }_k(x+\alpha )=e^{2\pi ik(x+\alpha )}=e^{2\pi ik\alpha }{e}^{2\pi ikx}=e^{2\pi ik\alpha }{\psi }_k(x)$$ so $$T{\psi }_k=e^{2\pi ik\alpha }{\psi }_k$$ and ${\psi }_k$ is an eigenfunction of $T$ with eigenvalue $e^{2\pi ik\alpha }$.

In contrast with the existence of eigenfunctions, we have the property of mixing.

Definition

A measure-preserving transformation $T$ on a probability space $(X,\mathcal{B},\mu )$ is mixing if $$\underset{n\to \mathrm{\infty }}{lim}⟨f,T^{n}g⟩=⟨f,1⟩⟨1,g⟩$$ for all $f,g$ in ${\mathsf{L}}^{\mathsf{2}}(X,\mathcal{B},\mu )$.