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.
Best Answer
No it is not true.
First note that we can improve pointwise convergence to uniform convergence of $f_{n} \ast g_{m} \to f \ast g$. This is an easy consequence of Hölder's inequality (markup doesn't like my subscripts, so I'll use $\|\cdot\|$ for the $L^{2}$-norm): \[ |(f \ast g)(x)| \leq \int |f(x - y) g(y)|\,dy \leq \|f\| \|g\|. \] As $f_{n} \ast g_{m} - f \ast g = f_{n} \ast (g_{m} - g) + (f_{n} - f) \ast g$ we get \[ |(f_{n} \ast g_{m})(x) - (f \ast g)(x)| \leq \|f_{n}\| \|g_{m} - g\| + \|f_{n} - f\| \|g\|. \] Since $\|f_{n}\| \to \|f\|$ and both $\|f_{n} - f\| \to 0$ and $\|g_{m} - g\| \to 0$, the right hand side can be made arbitrarily small independently of $x$. It follows in particular that $f \ast g$ is continuous and vanishes at infinity (that's one reason why I prefer writing $C_{c}$ for functions with compact support and $C_{0}$ for functions vanishing at infinity).
But we cannot do better, i.e., get $L^1$-convergence without further hypotheses. To see what happens, I prefer to ignore the condition that $f_{n} \in C^{\infty}$ but rather look at the function $f_{n} = \frac{1}{n}[-n,n]$. Then $f_{n} \to 0$ in $L^{2}$ but not in $L^{1}$. By the above argument we have $f_{n} \ast f_{n} \to 0$ uniformly on $\mathbb{R}$. On the other hand it is easy to see that for $x \in [-\frac{n}{2},\frac{n}{2}]$ \[ (f_{n} \ast f_{n})(x) = \int f_{n}(y) f_{n}(x-y)\,dy \geq \frac{1}{n^{2}} \frac{n}{2} = \frac{1}{2n} \] (the intervals $[-n,n]$ and $[-n-x,n-x]$ have an overlap of length at least $\frac{n}{2}$) and therefore we have the estimate $\int (f_{n} \ast f_{n}) \geq \frac{1}{2}$. Thus we cannot have $\int f_{n} \ast f_{n} \to \int f \ast f = 0$. It is now easy to cook up an example along these lines with smooth functions.