# Weak* convergence in $L^\infty$ and the strong convergence in $L^2$ of a mollification?

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)

• @squid thanks for the bounty; is there something you want answered or improved? Nov 22 at 2:16
• "Extending all functions by 0" so that one has $f_n\in C_c^\infty(\mathbb{R}^n)$ and $f\in L^\infty(\mathbb{R}^n)$, correct? I am curious what the motivation is for these two modes of convergence.
– user995162
Nov 22 at 15:17
• I have just earned back 50 pts to be able to leave a comment:-) I thought this well-written question interesting and hence the bounty.
– user995162
Nov 22 at 15:24
• I can answer that. I was reading a paper (specifically De Lellis and Székelyhidi’s “the Euler equations as a differential inclusion”) and this is part of getting a limit which solves the equation in a distributional sense. PS this is the first time I can guess where the “-10: user was removed” is from :) Nov 23 at 1:40

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 $$L^2(\Omega)$$.