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

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?

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, $$\begin{equation*} K_n\le a|G_\ep| by (0.5), and $$\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.