Convergence of the distribution of the Langevin diffusion to its invariant measure Let $(X_t)_{t\ge0}$ be a solution of $${\rm d}X_t=-h'(X_t){\rm d}t+\sqrt 2W_t,\tag1$$ where $(W_t)_{t\ge0}$ is a Brownian motion and $h$ is such that $X$ is the unique strong solution of $(1)$. Assume $c:=\int e^{-h}\:{\rm d}\lambda\in(0,\infty)$, where $\lambda$ denotes the Lebesgue measure, and let $g:=c^{-1}e^{-h}$. $X$ is a time-homogeneous Markov process whose transition semigroup is stationary with respect to the measure $\mu:=g\lambda$ with density $g$ with respect to $\lambda$.

Are we able to show that the distribution $\mathcal L(X_t)$ converges to $\mu$ as $t\to\infty$? If so, for which mode of convergence? Weak convergence? Convergence in total variation distance?

EDIT:
Let me precise the question: Let $\kappa_t$ denote a regular version of $X_t$ given $X_0$, $$\kappa_t(x,B)=\operatorname P\left[X_t\in B\mid X_0=x\right]\;\;\;\text{for }\operatorname P\circ\:X_0^{-1}\text{-almost all }x\in\mathbb R\text{ and }B\in\mathcal B(\mathbb R)\tag2.$$ Assume $\nu:=\operatorname P\circ\:X_0^{-1}$ has a density $f$ with respect to $\lambda$ and let $\left|\mu-\nu\kappa_t\right|$ denote the total variation distance of $\mu$ and $$(\nu\kappa_t)(B):=\int\nu({\rm d}x)\kappa_t(x,B)\;\;\;\text{for }B\in\mathcal B(\mathbb R).$$ If we could show that $\mathcal L(X_t)=\nu\kappa_t$ has a density $h_t$ with respect to $\lambda$, it's well-known that $$\left|\mu-\nu\kappa_t\right|=\frac12\left\|g-h_t\right\|_{L^1(\lambda)}\tag3$$ and we could conclude if we would been able to show that this converges to $0$ as $t\to\infty$.
EDIT 2:
Some thoughts: Let $L\varphi:=-h'\varphi'+\varphi''$ and $L^\ast\varphi:=(h'\varphi)'+\varphi''$ for $\varphi\in C^2(\mathbb R)$. Note that $$L^\ast g=0\tag4$$ and $$\mu(L\varphi):=\int L\varphi\:{\rm d}\mu=0\;\;\;\text{for all }\varphi\in C_c^\infty(\mathbb R).\tag4$$ Moreover, $${\rm d}\varphi(X_t)=(L\varphi)(X_t){\rm d}t+\varphi'(X_t){\rm d}W_t\tag5$$ for all $\varphi\in C^2(\mathbb R)$. In particular, $$\mathcal L(X_t)\varphi=(\nu\kappa_t)\varphi=\underbrace{\lambda(f\varphi)}_{=\:\nu\varphi}+\int_0^t\operatorname E\left[(L\varphi)(X_s)\right]\:{\rm d}s\tag6$$ for all $\varphi\in C_c^2(\mathbb R)$.
EDIT 3:
Let $(\mathcal D(A),A)$ denote the generator of $(\kappa_t)_{t\ge0}$. We know that $C_c^\infty(\mathbb R)$ is a core of $(\mathcal D(A),A)$, $$\mathcal D(A)=\left\{\varphi\in C_0(\mathbb R)\cap C^2(\mathbb R):L\varphi\in C_0(\mathbb R)\right\}\tag7$$ and $$A=\left.L\right|_{\mathcal D(A)}.\tag8$$ Now, let $$\mathcal E(\varphi,\psi):=-\langle\varphi,A\psi\rangle_{L^2(\mu)}\;\;\;\text{for }\varphi,\psi\in\mathcal D(A).$$ $\mathcal E$ is called the Dirichlet form associated to $(\mathcal D(A),A)$ on $L^2(\mu)$. It's easily seen that if $\rho>0$ and the Poincaré inequality $$\operatorname{Var}_\mu\left[\varphi\right]\le\frac1\rho\mathcal E(\varphi,\varphi)\;\;\;\text{for all }\varphi\in\mathcal D(A)\tag9$$ holds, then $$\left|\nu\kappa_t-\mu\right|^2\le\frac14e^{-2\rho t}\chi^2(\nu,\mu)\tag{10}$$ for all probability measures $\nu$ (not only the special one above) on $(\mathbb R,\mathcal B(\mathbb R))$, where $$\chi^2(\operatorname P,\operatorname Q):=\begin{cases}\operatorname P\left|\frac{{\rm d}\operatorname P}{{\rm d}\operatorname Q}-1\right|^2&\text{, if }\operatorname P\ll\operatorname Q\\\infty&\text{, otherwise}\end{cases}$$ is the $\chi^2$-distance of probability measures $\operatorname P$ and $\operatorname Q$ on any common measurable space. Note that in our special choice for $\nu=f\lambda$, we have $\mu\ll\lambda$ and $\nu\ll\lambda\ll\mu$ (and hence $\nu\ll\mu$). As a last note $$\mathcal E(\varphi,\psi)=\frac12\left(\langle\varphi',\psi'\rangle_{L^2(\mu)}+\langle\varphi,h'\psi'\rangle_{L^2(\mu)}\right)\;\;\;\text{for all }\varphi,\psi\in C_c^\infty(\mathbb R)\tag{11}$$ (as can be seen by partial integration). Oh, and note that since $C_c^\infty(\mathbb R)$ is dense in $L^2(\mu)$, $(\kappa_t)_{t\ge0}$ is a strongly continuous contraction semigroup on $L^2(\mu)$ and the corresponding generator coincides with $A$ on $\mathcal D(A)$.

So, one approach could be to show $(9)$ and somehow use $(10)$ to conclude.

If it is of any use, it would be fine for me to assume $h=-\ln f$ for some positive $f\in C^2(\mathbb R)$.
 A: The standard proof of convergence I am aware of works as follows.
Let $v(t, x)$ be the solution of the associated Fokker-Planck equation,
\begin{equation}
    \partial_t v = \partial_x(h' \, v + \partial_x v), \quad v(0, x) = v_0(x),
\end{equation}
where $v_0$ is the (normalized) initial distribution.
Let $u := e^{h} v$, and note that $v$ converging to $e^{-V}$ corresponds to $u$ converging to $1$.
Substituting, we find that $u$ satisfies
\begin{equation}
    \partial_t u = - h' \partial_x u + \partial_x^2 u = e^h\partial_x(e^{-h} \partial_x u), \quad u(0, x) = e^{h(x)} \, v_0(x).
\end{equation}
Taking the $e^{-h}$-weighted $L^2$ inner product of each side with $u - 1$ and integrating by parts,
\begin{equation}
    \frac{1}{2} \, \frac{d}{dt} \int_{\mathbb R} (u - 1)^2 \, e^{-h} \, dx = - \int_{\mathbb R} (\partial_x u)^2 \, e^{-h} \, dx, \quad u(0, x) = e^{h(x)} \, v_0(x).
\end{equation}
I omitted some technical details here about why it makes sense to integrate by parts,
so consider this only a formal argument.
Notice now that
\begin{equation}
    \overline u := \int_{\mathbb R} u \, e^{-h} \, dx = \int_{\mathbb R} v \, dx = 1.
\end{equation}
To be able to conclude the argument, you need the measure $e^{-h}$ to satisfy a sort of Poincar'e inequality.
Let us therefore assume it:
\begin{equation}
    \int_{\mathbb R} (f - \overline f)^2 \, e^{-h} \, dx \leq C \int_{\mathbb R} (f')^2 \, e^{-h} \, dx , \quad \forall f \in H^1(\mathbb R, e^{-h}),
\end{equation}
where $\overline u$ is the average of $u$:
\begin{equation}
    \overline f := \int_{\mathbb R} f \, e^{-h} \, dx.
\end{equation}
I recommend you take a look at this paper for details.
With the Poincar'e inequality,
we obtain
\begin{equation}
    \frac{d}{dt} \int_{\mathbb R} (u - 1)^2 \, e^{-h} \, dx \leq - C \int_{\mathbb R} (u - 1)^2 \, e^{-h} \, dx, \quad u(0, x) = e^{h(x)} \, v_0(x),
\end{equation}
which by Gronwall lemma implies convergence of $u$ to 1 in $L^2(\mathbb R, e^{-h})$,
and therefore convergence of $v$ to $e^{-h}$ in $L^2(\mathbb R, e^h)$.
