Transfer operator

Remarks

In practice $X$ carries more structure, and $\mathcal{L}_g$ is studied on a suitable Banach space of functions, such as Hölder or bounded variation functions.

Example: let $f(x) = 2x \bmod 1$ on $[0,1)$ and $g = 1/2$. Then $(\mathcal{L}_g \varphi)(x) = \frac{1}{2}\big(\varphi(x/2) + \varphi((x+1)/2)\big)$, and $\mathcal{L}_g 1 = 1$. The substitutions $y = x/2$ and $y = (x+1)/2$ give, for integrable $\varphi$ and bounded $\psi$, $$\int_0^1 (\mathcal{L}_g \varphi) \, \psi \, dx = \int_0^1 \varphi \, (\psi \circ f) \, dx.$$ So $\mathcal{L}_g$ is the adjoint of the Koopman operator $\psi \mapsto \psi \circ f$ for Lebesgue measure, and taking $\psi$ an indicator function shows that $\varphi \, dx$ is $f$-invariant exactly when $\mathcal{L}_g \varphi = \varphi$.

Non-example of this duality: with the weight $g = 1$ for the same map, $\mathcal{L}_1 1 = 2$, and $\int \mathcal{L}_1 \varphi \, dx = 2 \int \varphi \, dx$, so $\mathcal{L}_1$ is not the adjoint of a Koopman operator for Lebesgue measure.

For a piecewise smooth expanding map of an interval, the weight $g = 1/\abs{f'}$ gives the Perron–Frobenius operator, the adjoint of the Koopman operator for Lebesgue measure.