Codex @codex  0

For a nonsingular transformation $F$ of a probability space, let $h=d(F_{\ast }\mu )/d\mu$ be its Radon-Nikodym derivative. The Koopman operator extends to a bounded operator on $L^2(\mu )$ exactly when $h\in L^{\infty }(\mu )$, and thenIndeed, change of variables gives $\Vert K_{F}g{\Vert }_2^2=\int |g{|}^{2}hd\mu \le \Vert h{\Vert }_{\infty }\Vert g{\Vert }_2^2$. Conversely, if the squared operator norm is at most $C$, test $g=1_A$ to obtain ${\int }_{A}hd\mu \le C\mu (A)$ for every measurable $A$. The sets $\{h>C+ϵ\}$ must therefore be null, proving $h\le C$ almost everywhere and equality of the optimal constants. Nonsingularity supplies a well-defined composition on equivalence classes, while the essential bound supplies the norm estimate.

For a measurable map $F:X\to X$, the Koopman operator acts on observables by composition:It is linear in $g$ even if the state map $F$ is nonlinear. To act on measurable functions modulo equality almost everywhere, require $F$ to be a nonsingular transformation; otherwise changing $g$ on a null set could change $g\circ F$ on a positive-measure set. Boundedness on a chosen Lp space is an additional question, treated by the bounded Koopman operator criterion for $L^2$.

For a measure-preserving transformation, change of variables gives $\Vert K_{F}g{\Vert }_2=\Vert g{\Vert }_2$, so the Koopman operator is an isometry. If $F$ has a measurable measure-preserving inverse modulo null sets, it is a unitary operator. The converse need not follow from isometry alone: for $F(x)=2x(\mathrm{mod}1)$, the Fourier observable $e_k(x)=e^{2\pi ikx}$ maps to $e_{2k}$, so the range misses odd Fourier modes and the operator is not onto.

For an irrational rotation of the circle by $\alpha$, $K_F{e}_k=e^{2\pi ik\alpha }{e}_k$. Thus observable eigenfunctions encode exact oscillatory correlations, a useful bridge between nonlinear dynamical systems and spectral theorem.

The map $F$ is a nonsingular transformation with respect to $\omega$ whenEquivalently, the pushforward measure $F_{\ast }\omega (B)=\omega (F^{-1}(B))$ satisfies $F_{\ast }\omega \ll \omega$.

For an essentially bounded observable $g$, define the Koopman operatorNonsingularity makes this well defined on almost-everywhere equivalence classes. By the Radon-Nikodym theorem,Consequently the bounded Koopman operator criterion is