TITLE: Limits and convolution 
QUESTION [5 upvotes]: Let $f,g \in L^2(\mathbb{R^n})$, $\{ f_n \}, \{ g_m \} \subset C^\infty_0(\mathbb{R}^n)$ (infinitely differentiable functions with compact support) where $f_n \to f$ in $L^2$, and $g_n \to g$ in $L^2$.
Then $f_n \star g_m(x) \to f \star g(x)$ pointwise as $n,m \to \infty$ (where $\star$ denotes convolution). 
Is it true that $\int_{\mathbb{R}^n} f_n \star g_m dx \to \int_{\mathbb{R}^n} f \star g dx$? 
I've thought about using the dominated convergence theorem, but I couldn't figure out how, since I don't really know what I can do with the sequences of functions given only the facts at hand.

REPLY [3 votes]: For any integrable $f$ and $g$ (so in particular for functions in $C_O({\mathbb R}^n)$ one has $\int_{{\mathbb R}^n} f \ast g = \int_{{\mathbb R}^n} f \int_{{\mathbb R}^n} g$.
So your question is asking if $\int_{{\mathbb R}^n} f_n \int_{{\mathbb R}^n} g_n$ necessarily converges to $\int_{{\mathbb R}^n} f \int_{{\mathbb R}^n} g$. A counterexample will be provided by letting $f_n = g_n$ such that $f_n$ converges to $f$ in $L^2$ but such that $\int_{{\mathbb R}^n} f_n$ does not converge to $\int_{{\mathbb R}^n} f$. To do this, one can let $\phi(x)$ be any smooth function with compact support and integral one, and then let $f_n(x) = {1 \over n} \phi({x \over n})$. Each $f_n$ will have integral $1$, but $f_n(x)$ converges in $L^2$ to $f(x) = 0$ which has integral zero.