Why Koopman operator is defined on $L^\infty$ space?

Question

Let $(X, \mathcal{A}, \mu)$ be a measure space, $S:X\rightarrow X$ a nonsingular transformation, and $f\in L^\infty$. The operator $U:L^\infty \rightarrow L^\infty$ defined by

$ Uf(x) = f(S(x)) $

is called the Koopman operator with respect to $S$.

This is the definition of Koopman operator according to reference [1] and similar definitions can be found also in another textbooks/papers.

But, why we need functions defined on $L^\infty$ and not on other spaces like $L^1$?

[1]: Andrzej Lasota, Michael C. Mackey-Probabilistic properties of deterministic systems-Cambridge University Press (1985)

Answer 1

One possible reason is that on $L^\infty$ you get an isometry. That's not the case on $L^1$. You can still define the operator, of course.

Say you take $L^1[0,1]$ with Lebesgue measure, and $S(x)=x^2$. Then, with $f(t)=t$, you have $$ \|f\|_1=\tfrac12,\ \ \ \|Uf\|_1=\int_0^1 t^2\,dt=\tfrac13. $$ In fact, we can push the example and show that in $L^1$ the operator $U$ is not even bounded. Take $S$ as above and $f_n(t)=n\,1_{[0,1/n]}$. Then $$ \|f_n\|_1=1, \ \ \ \|Uf_n\|_1=n^2\int_0^11_{[0,1/n]},\,dt=n $$

Answer 2

Actually there is no need to define the Koopman operator just on $L^\infty$. I found it defined over several functional spaces.