I feel like this should be obvious to me but I'm blanking.
Let $\Omega$ be an open bounded subset of $\mathbb R^n$. Let $f_n\in C^\infty_c(\Omega)$ and $f\in L^\infty(\Omega)$ be such that
$$ f_n\overset*\rightharpoonup f \text{ in }L^\infty
.$$
Now let $\rho\in C^\infty_c(\mathbb R^n)$. Extending all functions by $0$ to functions defined on $\mathbb R^n$, is it true that (up to a subsequence, if you must)
$$ \rho*f_n \to \rho *f \text{ in } L^2?$$
I suspect this is indeed true. Although it is not quite my setting, if I use the Fourier basis $e_n(\theta) := e^{in\theta}$ of $\mathbb T = \mathbb R/2\pi \mathbb Z$ which converges weak* to $0$ in $L^\infty$ (which is just Riemann-Lebesgue lemma), then for $\rho\in C^\infty(\mathbb T)$,
$$ \rho* e_n (\theta) = \int_{\mathbb T}\rho(\alpha)e^{in(\theta-\alpha)} d\alpha = e^{in\theta} \hat \rho (n) \to 0 \text{ uniformly in $\theta$,} $$
and therefore in $L^2$ (and other $L^p$ too.) But I do not know how to prove a general result. By routine symbol pushing, $\rho*f_n$ converges weakly* in $L^\infty$ and weakly in $L^2$. (Sketches or counterexamples are welcome)
Best Answer
Let $f=0$. Then we want to prove $$ \int_{\mathbb R^n} \left(\int_\Omega \rho(x-y)f_n(y)dy\right)^2 dx \to0. $$ Since $\Omega$ is bounded, $f_n\rightharpoonup0$ in $L^2(\Omega)$, so $\int_\Omega \rho(x-y)f_n(y)dy \to 0$ for all $x$. That is, the integrand in the integral above converges pointwise to zero. In addition, $$ |\int_\Omega \rho(x-y)f_n(y)dy| \le |\Omega|\|\rho\|_{L^\infty} \sup_n \|f\|_{L^\infty} $$ for all $x$, which is a square-integrable pointwise upper bound. Then convergence follows by dominated convergence.
This is some kind of compactness result for the simple integral operator $f\mapsto \rho *f$ from $L^2(\Omega)$ to $L^2(\Omega)$.