is the pullback operator associated to a flow bounded in L^2?

Question

Let $M$ be a smooth compact manifold with a finite Borel measure $m$. Let $\{f_t\}_{t\in\mathbb R}$ be a $C^1$ flow on $M$. That is, a $C^1$ function $$ \mathbb R\times M\ni(t,x)\mapsto f_t(x)\in M $$ such that $f_0(x)=x$ for all $x\in M$, and $f_{t_1}\circ f_{t_2}=f_{t_1+t_2}$ for all $t_1,t_2\in\mathbb R$. Then, the pullback operator $f_t^*$ on $C(M)$ (with the sup-norm), $$ f_t^*\;\!\psi=\psi\circ f_t,\quad t\in\mathbb R,~\psi\in C(M), $$ is a bounded operator.

I am wondering if $\;\!f_t^*$ could be extended to a bounded operator in $L^2(M,m)$ (with the $L^2$-norm), for instance when $|t|$ is small$\;\!$?

The problem is we cannot use the tools of differential calculus such as the Jacobian, integration by parts, and so on, to estimate the norm of $f_t^*$ in $L^2(M,m)$ because (1) the measure $m$ is not given by a volume form on $M$ and (2) the flow $\{f_t\}_{t\in\mathbb R}$ does not preserve the measure $m$.

Answer 1

Absent any relation between the measure $m$ and the manifold/flow structure, you can't do this. For example, let $M=S^1$, and let $f_t$ be the rotation of $S^1$ by angle $t$, and choose $m$ to be the restriction of the arclength measure to the upper semi-circle. For any $t\in (0,\pi)$ there exists a function supported in the lower semi-circle such that after rotation by $t$, its support goes to the upper semi-circle. So its $L^2(m)$ norm goes from $0$ to positive under the action of $f_t$.