Earlier today I found myself pondering the following question for which I do not have a reasonable answer.
Suppose $f_m\to f$ and $g_m\to g$ in $L^2$. Moreover suppose that $f_m g_m\in L^2$ for all $m$ and $fg\in L^2$. Is it the case that $f_m g_m \to fg$ in $L^2$?
The naive approach of using a clever form of $0$ doesn't quite work since you get
$$\|fg - f_mg_m\| \le \|(f-f_m)g\| + \|f_m(g-g_m)\|.$$
The simplest way to ensure that this goes to zero is if you have an $L^{\infty}$ uniform boundedness on the $f_m$ and an $L^{\infty}$ boundedness on $g$. The latter condition is not overly strong, but the $L^{\infty}$ uniform boundedness of the $f_m$ is a very strong condition. Moreover, implicit in this approach is the slightly hidden assumption that $f_mg \in L^2$ for all $m$. I also thought about extracting pointwise almost everywhere convergent subsequences but I got my wires crossed.
If this is not true in general, what conditions can be placed to ensure this holds? Conditions than those stated above, of course, since those conditions are very strong.
Let $f_n$ be a square function with width $1/n$ and height $\sqrt{n}$. Then $f_n\rightarrow 0$ in $L^2$. Let $g_n=f_n$...