If $f_n\in L^2$, $x f_n \in L^2$, $f_n \to 0$, then $xf_n \to 0$?

Question

Consider the Hilbert space $L^2 = L^2(\mathbb R)$. Consider a sequence $f_n \in L^2$ that satisfies $xf_n\in L^2$. Here, $xf_n$ represents the function $x\mapsto xf_n(x)$. If $f_n \to 0$, then $xf_n \to 0$?

I think this should be a simple problem, but I cannot find a proof.

The question is motivated from my trial to prove that the position operator $x$ is closed in the domain $D(x) = \{ f\in L^2: xf\in L^2\}$. I have to show that if $f_n\to f$ and $xf_n \to g$ for some $g\in L^2$, then $f\in D(x)$ and $g= xf$. Considering the special case $f=0$ reduces the above problem, but even in this simple case I cannot find a proof.

Answer 1

No, this does not always hold, consider the counter-example: $$f_n(x) = \begin{cases} 0,\quad x\notin [n, n+1)\\ \frac 1{\sqrt{n}},\quad x \in [n, n+1). \end{cases}$$

This sequence converges to zero, but $xf_n(x)$ will diverge in $L^2$.

Note that in order to show the closedness of the operator, you already assume that $xf_n$ does converge. Otherwise, this would already imply continuity of the operator, which clearly is not given.

Answer 2

Hint: Convergence in $L^2$ implies pointwise convergence a.e. along a subsequence.