Langevin diffusion has a reversible density and compact minorization

Put $g=\mathrm{\nabla }U$. By [F8], $g(x_{\star })=0$, so $\mathrm{\parallel }g(x)\mathrm{\parallel }\le L\mathrm{\parallel }x-x_{\star }\mathrm{\parallel }$; the equation meets [F1]. Its additive-noise integral equation is $X_s=x-{\int }_0^{s}g(X_u)du+\sqrt{2}{B}_s$. For each continuous input path the deterministic Picard iteration converges uniformly on finite intervals: successive differences are bounded by a fixed finite constant times $L^n{s}^n\mathrm{/}n!$, by repeated integration. The limit is the unique solution; subtraction and the same iterative integral bound give ${\sup }_{s\le t}\mathrm{\parallel }{X}_s^{x,w}-X_s^{z,v}\mathrm{\parallel }\le e^{Lt}(\mathrm{\parallel }x-z\mathrm{\parallel }+\sqrt{2}{\sup }_{s\le t}\mathrm{\parallel }{w}_s-v_s\mathrm{\parallel })$. Thus the solution is a continuous function of its initial point and driving path. Its flow composition identity follows from uniqueness of the integral equation. Independent Brownian future increments [F4] now give the Markov and semigroup identities; integration of its Borel solution map against the Brownian law gives a Borel transition kernel. Solutions on integer horizons agree by uniqueness and hence define all $t\ge 0$.

Define ${\beta }_s=\sqrt{2}(B_s-s{B}_t\mathrm{/}t)$ for $0\le s\le t$. For any finite list of times its coordinates together with $B_t$ are jointly Gaussian and $\mathrm{Cov}(B_s-s{B}_t\mathrm{/}t,B_t)=0$. By [F4] that finite bridge block is independent of $B_t$. Apply [F5] to rational-time cylinders to obtain independence of the entire bridge path and $B_t$. Thus Brownian paths disintegrate explicitly as $Y_s=x+(s\mathrm{/}t)(y-x)+{\beta }_s$ when their endpoint has value $y$, with endpoint density $h_t(y-x)$ and bridge law independent of $y$. This identity holds first for cylinder functions by independence and the Gaussian endpoint density, then for all bounded Borel path functions by [F5], and for nonnegative ones by monotone approximation. The bridge has covariance $2(\min (s,u)-su\mathrm{/}t)I_d$, which is unchanged when $(s,u)$ is replaced by $(t-s,t-u)$; its mean is zero. Gaussian finite-dimensional laws and [F5] therefore also give equality in law of ${\beta }_s$ and ${\beta }_{t-s}$.

Under Brownian law put $Y_s=x+\sqrt{2}{B}_s$. The stochastic exponential with integrand $-g(Y_s)\mathrm{/}\sqrt{2}$ is, by [F3], $Z_s=\exp \text{\hspace{0.17em}⁣}(\frac{U(x)-U(Y_s)}{2}+{\int }_0^s\{\frac{\mathrm{\Delta }U(Y_u)}{2}-\frac{\mathrm{\parallel }g(Y_u){\mathrm{\parallel }}^2}{4}\}du)\mathrm{.}$ Lipschitz continuity gives $\mathrm{\mid }{\mathrm{\partial }}_{ii}U\mathrm{\mid }\le L$ by taking coordinate difference quotients. Since $U\ge 0$, for $s\le t$ this positive local martingale is bounded above by $e^{U(x)\mathrm{/}2+tdL\mathrm{/}2}$. Stop at increasing energy/local-martingale levels. Dominated convergence [F7] in the stopped martingale identities makes $Z$ a true martingale of mean one. By [F2], under $Z_t$ times Brownian law the process $B_s+{\int }_0^{s}g(Y_u)\mathrm{/}\sqrt{2}du$ is Brownian, so $Y$ solves the Langevin equation. The explicit unique solution map of step 1.1 identifies its law with $P_t(x,\cdot )$.

Put $H(z)=\mathrm{\Delta }U(z)\mathrm{/}2-\mathrm{\parallel }g(z){\mathrm{\parallel }}^2\mathrm{/}4$ and $b_s^{x,y}=x+(s\mathrm{/}t)(y-x)+{\beta }_s$. Combining steps 2.1 and 1.2 gives the density $p_t(x,y)=h_t(y-x)e^{(U(x)-U(y))\mathrm{/}2}\mathbb{E}\exp \text{\hspace{0.17em}⁣}({\int }_0^{t}H(b_s^{x,y})ds)\mathrm{.}$ The integral is a continuous function of $(x,y,\beta )$ in the uniform path topology, since $H$ is continuous and uniformly continuous on each compact set containing the convergent paths. Its exponential is strictly positive and bounded above by $e^{tdL\mathrm{/}2}$. Consequently its expectation is positive, finite, and continuous in $(x,y)$ by dominated convergence; the density is jointly Borel. Its integral in $y$ is one by step 2.1. Bridge reversal from step 1.2 interchanges $x,y$ without changing the integral of $H$, and $h_t(y-x)=h_t(x-y)$. Multiplication by $e^{-U(x)}$ proves the asserted symmetry. If $Z<\mathrm{\infty }$, Tonelli and the mass-one identity give $\int Z^{-1}{e}^{-U(x)}{p}_t(x,y)dx=Z^{-1}{e}^{-U(y)}\int p_t(y,x)dx=Z^{-1}{e}^{-U(y)},$ proving invariance.

By [F6], $\mathbb{E}{\sup }_{s\le t}\mathrm{\parallel }{B}_s{\mathrm{\parallel }}^2\le 4dt$, and $\sup \mathrm{\parallel }{\beta }_s{\mathrm{\parallel }}^2\le 8\sup \mathrm{\parallel }{B}_s{\mathrm{\parallel }}^2$. With $a_0=\sqrt{64dt}$, Markov's inequality (integrate $1_{\{Y>a_0^2\}}\le Y\mathrm{/}{a}_0^2$ for $Y=\sup \mathrm{\parallel }\beta {\mathrm{\parallel }}^2$) gives $\mathrm{Pr}(\sup \mathrm{\parallel }\beta \mathrm{\parallel }\le a_0)\ge 1\mathrm{/}2$. If both endpoints lie in the ball $D=\{\mathrm{\parallel }z-x_{\star }\mathrm{\parallel }\le R\}$, on this event the bridge lies in the radius $R+a_0$ ball. Therefore $H(b_s^{x,y})\ge -dL\mathrm{/}2-L^2(R+a_0{)}^2\mathrm{/}4$ and the bridge expectation in step 3.1 is at least one half times the exponential of $-t[dL\mathrm{/}2+L^2(R+a_0{)}^2\mathrm{/}4]$. The heat factor is at least $(4\pi t{)}^{-d\mathrm{/}2}{e}^{-R^2\mathrm{/}t}$ on $D^2$, and $e^{(U(x)-U(y))\mathrm{/}2}\ge e^{-{\sup }_{D}U\mathrm{/}2}>0$. Thus a constant $c>0$ bounds $p_t$ below on $D^2$. Let $\nu$ be normalized Lebesgue measure on $D$ and $ϵ=\min \{c\mathrm{Leb}(D),1\mathrm{/}2\}>0$. Integrating the density over $A\cap D$ proves the stated minorization. The radius is positive, so the normalizing volume is finite and nonzero.