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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.

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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 $p$.

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A first remark is that in your argument for $U_{T,1}$, you used the fact that $U_{T,1}$ is the dual operator of the Koopman operator. This is true when $T$ is invertible, but the formula in the noninvertible case is different.

Secondly, I tried to write a comment to your comment on Mikael's answer but couldn't. Expanding on his answer, if $U_{T,2}f=f$ then $\overline{U_{T,2}f}=U_{T,2}\bar{f}=\bar{f}$ and thus $$U_{T,1}\left(|f|^2\right)=\left|U_{T,2}f\right|^2=|f|^2.$$ Therefore $|f|^2$ is a density of a $T$ invariant probability measure.

Lastly, in the case of type $\rm{III}$, a (slightly) different approach can be done using the fact $T$ is type $\rm{III}$ if and only if $\frac{d\mu\circ T}{d\mu}$ is not a coboundary in $L_0(X,\mathcal{B},\mu)$. First, expand the definition of $U_{T,1}$ to all functions in $L_0$. As $T$ is type $\mathrm{III}$, there is no $f:X\to \mathbb{R}\setminus\{0\}$ which satisfies $$\frac{d\mu\circ T^{-1}}{d\mu}(x)f\circ T^{-1}(x)=f(x).$$From here it is obvious.

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