Koopman operator

Remarks

The operator is well defined and isometric: since $\mu(T^{-1}A) = \mu(A)$ for all $A \in \mathcal{B}$, one gets $\int \abs{f \circ T}^2 \, d\mu = \int \abs{f}^2 \, d\mu$. If $T$ is invertible with measurable measure-preserving inverse, then $U_T$ is a unitary operator with inverse $U_{T^{-1}}$.

Example: for the rotation $T(x) = x + \alpha \bmod 1$ with Lebesgue measure, the functions $e_k(x) = e^{2\pi i k x}$ satisfy $U_T e_k = e^{2\pi i k \alpha} e_k$. They form an orthonormal basis of $L^2$, so $U_T$ is diagonal in this basis.

Non-example: if $T$ does not preserve $\mu$, the formula need not define a bounded operator on $L^2(\mu)$. For $T(x) = x^2$ on $[0,1]$ with Lebesgue measure and $f(x) = x^{-1/3}$, the function $f$ is square-integrable but $f \circ T = x^{-2/3}$ is not.

The same formula defines an isometry of $L^p(\mu)$ for every $1 \le p \le \infty$. The constants are always fixed by $U_T$, and $T$ is ergodic if and only if they are the only fixed vectors.