I want to prove that the Koopman operator $U_f : L^2 (\mu) \rightarrow L^2 (\mu)$ such that $U_f(\phi) =\phi \circ f$ is not surjective. Where $ \mu $ is a measure preserving mapping $f$.
I was testing with $f(x)= 2x\ mod \ 1$ in $[0,1]$ and $\mu =m$ then $f(0.25)=f(0.75)$ and $U_f(L^2 (\mu) )=\{U_f (\phi)(0.25)=U_f (\phi)(0.75) \}\subset L^2 (\mu)$ strict inclusion but this does not prove the required, this idea could be used to build a $ f $ and $ \mu $ appropriate?
thanks for any suggestions
If $\phi\in U_f(L^2)$, then there is $\phi'$ such that for almost every $x\in [0,1]$, $\phi'(f(x))=\phi(x)$. This gives $\phi(x)=\phi(x+1/2)$ for almost every $x\in [0,1/2]$. There are elements of $L^2[0,1]$ which doesn't satisfy this property. (take the characteristic function of $A\subset [0,1/2]$ of positive Lebesgue measure).