Liouville probability transport and constant fine entropy

By [F2], ${\mathrm{\Phi }}_t$ preserves $m_0{\mathrm{\mid }}_R$. Applying [F3] to ${\rho }_0$ gives ${\int }_R{\rho }_{t}d{m}_0={\int }_R{\rho }_{0}d{m}_0=1$. Applying [F3] to the nonnegative and negative parts of the bounded observable times the probability density gives ${\int }_{R}A{\rho }_{t}d{m}_0={\int }_R(A\circ {\mathrm{\Phi }}_t){\rho }_{0}d{m}_0$. All these integrals are finite because $A$ is bounded and ${\rho }_0\in L^1$.

For $L^1$ data and a test function $\phi$ as in the statement, use the measure-preserving change $z={\mathrm{\Phi }}_t(w)$ from [F2]. The integrand becomes ${\rho }_0(w)\frac{d}{dt}\phi (t,{\mathrm{\Phi }}_{t}w)$. Its time support is compact; the union of the inverse images of the compact spatial support over that compact time interval is compact by continuity of the complete flow. The derivative of $\phi$ is bounded there, so ${\rho }_0$ times a constant times the indicator of this compact set is integrable by [F5]. [F4] permits Fubini, and the inner time integral is the integral of a total derivative with zero endpoint values. It is therefore zero, proving the distributional identity.

If ${\rho }_0$ is smooth, differentiate ${\rho }_t(z)={\rho }_0({\mathrm{\Phi }}_{-t}z)$ by [F6] and the flow composition law. The derivative of ${\mathrm{\Phi }}_{-t}$ along the flow is $-X_H$, yielding ${\mathrm{\partial }}_t{\rho }_t+X_H{\rho }_t=0$.

Pointwise, ${\rho }_t\log {\rho }_t=({\rho }_0\log {\rho }_0)\circ {\mathrm{\Phi }}_{-t}$, including zeros under the convention $0\log 0=0$. By [F3], the integral of this integrable function is independent of $t$. Its absolute integrability and the $L^1$-integrability of ${\rho }_0$ make $S_k(t)$ finite, and the displayed identity makes it constant. In particular, when the finite canonical density of [F7] is used, it is stationary.