The canonical Gibbs measure is stationary for underdamped Langevin dynamics

Write $z=(q,p)$ and $b(q,p)=(M^{-1}p,-\mathrm{\nabla }V(q)-\gamma p)$. The bounded Hessian makes $b$ globally Lipschitz with linear growth. By [F5], the phase-space Lebesgue measure in the definition of $Z$ is ${\lambda }_d\times {\lambda }_d$ on Borel sets, and Tonelli applied to the nonnegative Borel Gibbs factor gives $Z=Z_q{Z}_p,Z_p:={\int }_{{\mathbb{R}}^d}{e}^{-\beta p^T{M}^{-1}p\mathrm{/}2}dp\mathrm{.}$ Applying [F6] with $C=\beta M^{-1}$ yields $Z_p=\frac{(2\pi {)}^{d\mathrm{/}2}}{\sqrt{\det (\beta M^{-1})}}=(2\pi \mathrm{/}\beta {)}^{d\mathrm{/}2}\sqrt{\det M},$ so $Z$ is finite and positive. To verify the momentum second moment, let ${\lambda }_{\ast }>0$ be the smallest eigenvalue of $M^{-1}$. Then $p^T{M}^{-1}p\ge {\lambda }_{\ast }\mathrm{\mid }p{\mathrm{\mid }}^2$. The function $r\mapsto r{e}^{-\beta {\lambda }_{\ast }r\mathrm{/}4}$ is bounded on $[0,\mathrm{\infty })$, so for a finite $C_{\ast }$, $\mathrm{\mid }p{\mathrm{\mid }}^2{e}^{-\beta p^T{M}^{-1}p\mathrm{/}2}\le C_{\ast }{e}^{-\beta {\lambda }_{\ast }\mathrm{\mid }p{\mathrm{\mid }}^2\mathrm{/}4}\mathrm{.}$ The right side has finite integral by [F6] with $C=(\beta {\lambda }_{\ast }\mathrm{/}2)I$. Tonelli now gives ${\mathbb{E}}_{\mu }\mathrm{\mid }{P}_0{\mathrm{\mid }}^2<\mathrm{\infty }$, while the stated $q$-moment assumption gives ${\mathbb{E}}_{\mu }\mathrm{\mid }{Q}_0{\mathrm{\mid }}^2<\mathrm{\infty }$. Thus the Gibbs initial state is square-integrable. With the ${\mathcal{F}}_0$-measurability and future-increment independence in the Statement, [F1] gives a unique global strong solution. Since the coefficients are time-independent, the law of the state after duration $t$ depends only on the starting state and elapsed time; write its transition operator as $P_t\varphi (z)={\mathbb{E}}_z\varphi (Q_t,P_t)$. For $\varphi (z)=\cos (u\cdot z)$ or $\sin (u\cdot z)$, the solution map is twice differentiable in its initial state: its first derivative $J_t$ solves ${\dot{J}}_t=Db(Z_t)J_t$, $J_0=I$, hence $\mathrm{\parallel }{J}_t\mathrm{\parallel }\le e^{t\mathrm{\parallel }Db{\mathrm{\parallel }}_{\mathrm{\infty }}}$; its second derivative solves the corresponding variation equation with forcing $D^{2}b(Z_t)[J_t,J_t]$ and is bounded on every finite interval because $D^{2}b$ is bounded. Difference quotients in the integral equation and these bounds justify differentiating under expectation, so $P_t\varphi$ and its first two spatial derivatives are bounded and continuous.

Uniqueness of the strong solution and independence of future Brownian increments give $P_{t+s}=P_s{P}_t$. Apply [F2] to $g=P_t\varphi$ over a short interval $s$: $(P_{s}g-g)\mathrm{/}s\to Lg$ as $s\downarrow 0$, where [A1, F1, F2, F4, given, step 1.1] $Lg=(M^{-1}p)\cdot {\mathrm{\nabla }}_{q}g+(-\mathrm{\nabla }V-\gamma p)\cdot {\mathrm{\nabla }}_{p}g+\gamma {\beta }^{-1}\mathrm{tr}(M{D}_{pp}^{2}g)\mathrm{.}$ Continuity of $Lg$ and the finite second-moment bound in [F1] give this limit. Hence ${\mathrm{\partial }}_t{P}_t\varphi =L{P}_t\varphi$; the right side is integrable against $\mu$ because the derivatives of $P_t\varphi$ are bounded, $p$ has Gaussian tails, and the stated $q$ integrability controls $\mathrm{\nabla }V$.

For the generator $L$ in step 2.1, the Gibbs density $\rho =Z^{-1}{e}^{-\beta H}$ has derivatives ${\mathrm{\nabla }}_q\rho =-\beta \mathrm{\nabla }V\rho$ and ${\mathrm{\nabla }}_p\rho =-\beta M^{-1}p\rho$. Its formal adjoint is [given, step 2.1, algebra] $L^{\ast }\rho =-{\mathrm{\nabla }}_q\cdot (M^{-1}p\rho )-{\mathrm{\nabla }}_p\cdot ((-\mathrm{\nabla }V-\gamma p)\rho )+\gamma {\beta }^{-1}\mathrm{tr}(M{D}_{pp}^2\rho )\mathrm{.}$ The $q$-transport term equals $\beta (M^{-1}p\cdot \mathrm{\nabla }V)\rho$ and the force part of the $p$-divergence equals $-\beta (\mathrm{\nabla }V\cdot M^{-1}p)\rho$, so they cancel. The friction part is $\gamma d\rho -\gamma \beta (p^T{M}^{-1}p)\rho$. Direct differentiation gives $\mathrm{tr}(M{D}_{pp}^2\rho )=({\beta }^2{p}^T{M}^{-1}p-\beta d)\rho$, so diffusion cancels friction and $L^{\ast }\rho =0$.

Define $s(x)=0$ for $x\le 0$ and $s(x)=e^{-1\mathrm{/}x}$ for $x>0$, then $\eta (r)=s(2-r)\mathrm{/}(s(2-r)+s(r-1))$. This smooth function equals one for $r\le 1$, zero for $r\ge 2$, and has bounded first and second derivatives; every right derivative of $s$ tends to zero at the origin. The product cutoffs ${\chi }_R(q,p)=\eta (\mathrm{\mid }q\mathrm{\mid }\mathrm{/}R)\eta (\mathrm{\mid }p\mathrm{\mid }\mathrm{/}R)$ are smooth because each factor is constant near its origin; they equal one when $\mathrm{\mid }q\mathrm{\mid },\mathrm{\mid }p\mathrm{\mid }\le R$, vanish when either norm is at least $2R$, and satisfy $\mathrm{\mid }\mathrm{\nabla }{\chi }_R\mathrm{\mid }\le C\mathrm{/}R$, $\mathrm{\mid }{D}^2{\chi }_R\mathrm{\mid }\le C\mathrm{/}{R}^2$. For $g=P_t\varphi$, multiply the divergence terms by ${\chi }_R$ and integrate by parts coordinatewise using [F7]; product-measure integration is justified by [F5]. The $q$-cutoff boundary term is bounded by $C\mathrm{\mid }p\mathrm{\mid }\mathrm{/}R$ times the Gibbs density; the $p$-cutoff terms are bounded by $C(1+\mathrm{\mid }\mathrm{\nabla }V(q)\mathrm{\mid }+\mathrm{\mid }p\mathrm{\mid })\mathrm{/}R$ times that density, with second-derivative cutoff terms bounded by $C\mathrm{/}{R}^2$. These bounds are integrable by Gaussian $p$ tails and the hypotheses on $q$ and $\mathrm{\nabla }V$. Dominated convergence and step 3.1 therefore give $\int Lgd\mu =\int g{L}^{\ast }\rho dqdp=0$. By step 2.1, $t\mapsto \int P_t\varphi d\mu$ has derivative zero and is constant.

Apply step 4.1 to $\varphi (z)=\cos (u\cdot z)$ and $\sin (u\cdot z)$ for arbitrary $u\in {\mathbb{R}}^{2d}$. The resulting scalar characteristic functions of projections of $\mu P_t$ and $\mu$ agree; [F3] and the Cramér–Wold device identify the phase-space laws. Thus $\mu P_t=\mu$ for every $t\ge 0$. AC is inherited through [A1].