Koopman Operator is Isometry

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Theorem

Let $\struct {X, \BB, \mu, T}$ be a measure-preserving dynamical system.

Let $\map {L^2_\C} \mu$ be the complex-valued $L^2$ space of $\mu$.

Let $U_T : \map {L^2_\C} \mu \to \map {L^2_\C} \mu$ of $T$ be the Koopman operator.

Let $\innerprod \cdot \cdot$ denote the inner product on $\map {L^2_\C} \mu$, i.e.

$\ds \forall f, g \in \map {L^2_\C} \mu : \innerprod f g := \int \overline f g \rd \mu$

where $\overline f$ denotes the complex conjugate of $f$.


Then $U_T$ is isometry on the $L^2$ Hilbert space $\struct {\map {L^2_\C} \mu, \innerprod \cdot \cdot}$.

That is:

$\ds \forall f, g \in \map {L^2_\C} \mu : \innerprod {U_T f} {U_T g} = \innerprod f g$


Proof

Let $f, g \in \map {L^2_\C} \mu $.

Then:

\(\ds \innerprod {U_T f} {U_T g}\) \(=\) \(\ds \int {\overline {U_T f} } \; {U_T g} \rd \mu\)
\(\ds \) \(=\) \(\ds \int {\overline {f \circ T} } \; {g \circ T} \rd \mu\) Definition of Koopman operator
\(\ds \) \(=\) \(\ds \int {\overline f} \circ T \; {g \circ T} \rd \mu\) as $\map {\overline {f \circ T} } x = \overline {\map f {T x} } = \map {\overline f} {T x} = \map { {\overline f} \circ T} x$
\(\ds \) \(=\) \(\ds \int \paren { {\overline f} g} \circ T \rd \mu\)
\(\ds \) \(=\) \(\ds \int {\overline f} g \rd \mu\) Definition of Measure-Preserving Transformation
\(\ds \) \(=\) \(\ds \innerprod f g\)

$\blacksquare$