Convegence of regularized sequence in $L^2$

Question

Let $(\rho_n)_{n \geq 0}$ be a standard regularizing sequence on $\mathbb R$. Let $P$ be a probability measure on $\mathbb R$ such that the sequence $(P*\rho_n)_{n \geq 0}$ is bounded in $L^2$. Then, does the probability measure $P$ admit a density in $ L^2$? i.e., is there a $p \in L^2(\mathbb R)$ satisfying \begin{align} P(dx) = p (x) dx? \end{align} My understanding is that, due to the boundedness in $L^2$, the sequence $(P*\rho_n)_{n \geq 0}$ contains a subsequence, which converges weakly to a limit in $L^2$. But, how do we know that the density $p$ is the limit?

Answer 1

On the one hand, we have $\lim\limits_{n\to\infty} P \ast \rho_n \to P$ in the sense of distributions (Laurent Schwartz, not probability distributions). On the other hand, by the assumed boundedness in $L^2$, we have a subsequence such that $P\ast \rho_{n_k}$ converges weakly to $p$ in $L^2$. Weak convergence in $L^2$ implies convergence in the sense of distributions, and $\mathscr{D}'(\mathbb{R})$ is Hausdorff, so limits are unique. Hence we have "$p = P$" (more precisely $P$ is the measure with density $p$ with respect to the Lebesgue measure). Since $p$ is therefore the only possible weak limit of a subsequence, in fact the entire sequence $P\ast \rho_n$ converges weakly to $p$.