Cfrgtkky

Let $(X_t)_{tge0}$ be a solution of $${rm d}X_t=-h'(X_t){rm d}t+sqrt 2W_t,tag1$$ where $(W_t)_{tge0}$ is a Brownian motion and $h$ is such that $X$ is the unique strong solution of $(1)$. Assume $c:=int e^{-h}:{rm d}lambdain(0,infty)$, 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 $mu:=glambda$ with density $g$ with respect to $lambda$.

Are we able to show that the distribution $mathcal L(X_t)$ converges to $mu$ as $ttoinfty$? 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 $X_0$, $$kappa_t(x,B)=operatorname Pleft[X_tin Bmid X_0=xright];;;text{for }operatorname Pcirc:X_0^{-1}text{-almost all }xinmathbb Rtext{ and }Binmathcal B(mathbb R)tag2.$$ Assume $nu:=operatorname Pcirc:X_0^{-1}$ has a density $f$ with respect to $lambda$ and let $left|mu-nukappa_tright|$ denote the total variation distance of $mu$ and $$(nukappa_t)(B):=intnu({rm d}x)kappa_t(x,B);;;text{for }Binmathcal B(mathbb R).$$ If we could show that $mathcal L(X_t)=nukappa_t$ has a density $h_t$ with respect to $lambda$, it's well-known that $$left|mu-nukappa_tright|=frac12left|g-h_tright|_{L^1(lambda)}tag3$$ and we could conclude if we would been able to show that this converges to $0$ as $ttoinfty$.

EDIT 2:
Some thoughts: Let $Lvarphi:=-h'varphi'+varphi''$ and $L^astvarphi:=(h'varphi)'+varphi''$ for $varphiin C^2(mathbb R)$. Note that $$L^ast g=0tag4$$ and $$mu(Lvarphi):=int Lvarphi:{rm d}mu=0;;;text{for all }varphiin C_c^infty(mathbb R).tag4$$ Moreover, $${rm d}varphi(X_t)=(Lvarphi)(X_t){rm d}t+varphi'(X_t){rm d}W_ttag5$$ for all $varphiin C^2(mathbb R)$. In particular, $$mathcal L(X_t)varphi=(nukappa_t)varphi=underbrace{lambda(fvarphi)}_{=:nuvarphi}+int_0^toperatorname Eleft[(Lvarphi)(X_s)right]:{rm d}stag6$$ for all $varphiin C_c^2(mathbb R)$.

EDIT 3:
Let $(mathcal D(A),A)$ denote the generator of $(kappa_t)_{tge0}$. We know that $C_c^infty(mathbb R)$ is a core of $(mathcal D(A),A)$, $$mathcal D(A)=left{varphiin C_0(mathbb R)cap C^2(mathbb R):Lvarphiin C_0(mathbb R)right}tag7$$ and $$A=left.Lright|_{mathcal D(A)}.tag8$$ Now, let $$mathcal E(varphi,psi):=-langlevarphi,Apsirangle_{L^2(mu)};;;text{for }varphi,psiinmathcal D(A).$$ $mathcal E$ 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 $$operatorname{Var}_muleft[varphiright]lefrac1rhomathcal E(varphi,varphi);;;text{for all }varphiinmathcal D(A)tag9$$ holds, then $$left|nukappa_t-muright|^2lefrac14e^{-2rho t}chi^2(nu,mu)tag{10}$$ for all probability measures $nu$ (not only the special one above) on $(mathbb R,mathcal B(mathbb R))$, where $$chi^2(operatorname P,operatorname Q):=begin{cases}operatorname Pleft|frac{{rm d}operatorname P}{{rm d}operatorname Q}-1right|^2&text{, if }operatorname Plloperatorname Q\infty&text{, otherwise}end{cases}$$ 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 $nu=flambda$, we have $mulllambda$ and $nulllambdallmu$ (and hence $nullmu$). As a last note $$mathcal E(varphi,psi)=frac12left(langlevarphi',psi'rangle_{L^2(mu)}+langlevarphi,h'psi'rangle_{L^2(mu)}right);;;text{for all }varphi,psiin C_c^infty(mathbb R)tag{11}$$ (as can be seen by partial integration). Oh, and note that since $C_c^infty(mathbb R)$ is dense in $L^2(mu)$, $(kappa_t)_{tge0}$ 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 $fin C^2(mathbb R)$.

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^hpartial_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 as 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 $L^2(mathbb R, e^h)$.