Let $(X,\mu)$ be a standard probability space and let $T:X \to X$ be a measure-class preserving transformation such that there is no $T$-invariant measure absolutely continuous with respect to $\mu$. (Such transformations are called type $\mathrm{III}$.) For $1 \leq p < \infty$ let $U\_{T,p}$ be the isometry of $L^p(X,\mu)$ given by \begin{equation} (U\_{T,p} \cdot f)(x) = \left(\frac{\mathrm{d}T^{-1}\mu}{\mathrm{d}\mu}(x) \right)^{1/p} f(T^{-1}x). \end{equation} A nonzero function $f \in L^1(X,\mu)$ is invariant under $U\_{T,1}$ if and only if $\frac{|f|}{||f||\_1}$ is the density of a $T$-invariant measure, so by assumption such functions do not exist. My question is whether it is possible that there is a nonzero function $f \in L^2(X,\mu)$ which is invariant under the Koopman operator $U\_{T,2}$.


I think this is interesting because if such functions do not exist if and only if the averages \begin{equation} \frac{1}{n} \sum\_{k=0}^{n-1} U\_{T,2}^k \end{equation} always converge to zero in the strong operator topology.



Invariant function for Koopman operator of measure-class preserving tranformation
No, because if $f$ is invariant under $U\_{T,2}$ then $|f|^2$ is invariant under $U\_{T,1}$.


More generally the Mazur map $f \in L^p(X,\mu) \mapsto sgn(f) |f|^{p/q} \in L^q(X,\mu)$ is a homeomorphism intertwining the Koopman operators $U\_{T,p}$ and $U\_{T,q}$, so most properties of the Koopman operators do not depend on $p$.


