If $v_n \rightharpoonup v$ and $\phi \in C_c^{\infty}(\mathbb{R}^N)$, do we have $v_n \phi \rightharpoonup v \phi$?

Question

I'm stuck trying to solve this question. If I have a sequence in the Sobolev space $D^{1,\vec{p}}(\mathbb{R}^N)$ (or $W_0^{1,p}(\mathbb{R}^N)$ for simplicity) which converges weakly to $v$ and $\phi \in C_c^{\infty}(\mathbb{R}^N)$, do we necessarily have that $v_n \phi$ converges weakly to $v \phi$? It seems intuitive but I can't figure out a way to prove it.

Answer 1

EDIT: This answer is in fact very incomplete. It is not at all clear that $u \mapsto \phi u$ is a bounded operator: even though we get a bound on $|\phi u|$, we do NOT have a bound on its partial derivatives, which means we don't have a bound on its norm in $D^{1,\vec{p}}(\mathbb{R}^N)$.

It's in fact highly nontrivial to even show that $\phi u$ is an element of $D^{1,\vec{p}}(\mathbb{R}^N)$. See my other question: Does multiplication by a test function stay in a Sobolev space?

$\newcommand{\Dpr}{D^{1,\vec{p}}(\mathbb{R}^N)}$ An insightful comment by user357151 solves the question easily. Let $X$ be $\Dpr$, $W^{1,p}(\mathbb{R}^N)$, or even $L^p(\mathbb{R}^N)$.

Remark that the $S: X \rightarrow X : u \mapsto \phi u$ is a bounded linear operator, since $|\phi u| \leq C |u|$. Then for any $T \in X^*$, $T \circ S \in X^*$, and so: $$ \lim \limits_{n \rightarrow \infty} T(S(v_n)) = T(S(v)) $$ And hence $S(v_n)=\phi v_n$ converges weakly to $S(v) = \phi v$.