How to prove the limit of minimizing sequence of measures is again absolutely continuous(w.r.t. Lebesgue) in the minimizing movement scheme?

Question

I am considering the minimizing movement scheme related to the gradient of entropy functional in 2-Wasserstein space. The problem is to minimize the following functional for each fixed $\eta$ which is a probability density w.r.t. $(\mathbb{R}^d,Leb)$ with finite second moments: $$\int\rho\log\rho dx+W_2^2(\rho,\eta),$$ among all probability densities $\rho$(so $\rho dx\ll Leb$) with finite second moments. Now I need to show the existence of a minimizer to this problem.

So first we choose a minimizing sequence $\rho_n$, which gives that $W_2^2(\rho_n,\eta)$ are uniformly bounded. Since the second moments can be bounded by the 2-Wasserstein distance, we know the second moments of $\rho_n$ are uniformly bounded, so they are tight(and also uniformly integrable). This gives a subsequence $\rho_{n_k}$ converging weakly to some probability measure $\mu$. Now we need to show $\mu\ll Leb$ and has finite second moment.

For the second part I used Skorokhod's theorem to find $X_n\sim\rho_n$ and $X\sim\mu$ with $X_n\overset{a.s.}{\rightarrow}X$. Then Fatou's lemma gives $\mathbb{E}X^2\leq\liminf_{n\rightarrow\infty}\mathbb{E}X_n^2<\infty$.

But I have no idea how to show $\mu\ll Leb$: we can find counterexamples if we only have $X_n$ converges a.s. and in $L^1$. We might need other observations; or it is possible that the limit of the minimizing sequence of this problem is not absolutely continuous w.r.t. Lebesgue measure?

Answer 1

$\newcommand{\ep}{\varepsilon}\newcommand\R{\mathbb R}$Yes, the minimizer $\mu$ is absolutely continuous (w.r. to the Lebesgue measure $|\cdot|$).

Indeed, you showed that \begin{equation*} F(\rho_n)\to m:=\inf_\rho F(\rho) \tag{-1} \end{equation*} and \begin{equation*} \mu_{\rho_n}\to\mu \tag{-0.5} \end{equation*} weakly for some sequence $(\rho_n)$ of probability densities and some probability measure $\mu$, where \begin{equation*} F(\rho):=\int\rho\ln\rho\,dx+W_2^2(\rho,\eta) \tag{0} \end{equation*} and $\mu_\rho(dx):=\rho(x)\,dx$.

Take any set $E\subseteq\R$ with $|E|=0$. We have to show that then $\mu(E)=0$.

Take any real $\ep>0$. By the regularity of the Lebesgue measure, there is an open set $G_\ep\subset\R$ such that \begin{equation*} \text{$E\subseteq G_\ep$ and $|G_\ep|<\ep$.} \tag{0.5} \end{equation*} By (-0.5) and the Portmanteau theorem, \begin{equation*} \mu(G_\ep)\le\liminf_n\mu_{\rho_n}(G_\ep). \tag{1} \end{equation*} Next, for each real $a>1$, \begin{equation*} \mu_{\rho_n}(G_\ep)=K_n+L_n, \tag{2} \end{equation*} where \begin{equation*} K_n:=\int_{G_\ep\cap[\rho_n\le a]}\rho_n\,dx,\quad L_n:=\int_{G_\ep\cap[\rho_n>a]}\rho_n\,dx, \end{equation*} $[\rho_n\le a]:=\rho_n^{-1}((-\infty,a])$, $[\rho_n>a]:=\rho_n^{-1}((a,\infty))$. Further, \begin{equation*} K_n\le a|G_\ep|<a\ep \tag{3} \end{equation*} by (0.5), and \begin{equation*} L_n\le \int_{[\rho_n>a]}\rho_n\,dx \le\frac1{\ln a}\int \rho_n\ln\rho_n\,dx\le \frac{m+1}{\ln a} \tag{4} \end{equation*} for all large enough $n$, by (-1) and (0).

By (0.5), (1), (2), (3), (4), \begin{equation*} \mu(E)\le\mu(G_\ep)\le a\ep+\frac{m+1}{\ln a}, \end{equation*} for all real $\ep>0$ and all real $a>1$. Letting now $\ep\downarrow0$ and then $a\to\infty$, we get $\mu(E)=0$, as desired.