Weak convergence in $L^2$ of a sequence bounded in $H^1$ implies weak convergence in $H^1$ I have a very specific problem, that probably has a more general solution. Let $(v_n)_n$ be a sequence in $H^1(\mathbb{R}^3)$ such that


*

*$\forall n \in \mathbb{N}: ||v_n||_{L^2} = 1$;

*$v_n \rightharpoonup 0$ in $L^2(\mathbb{R}^3)$ for $n \to \infty$;

*$(v_n)_n$ bounded with respect to the norm of $H^1(\mathbb{R}^3)$.


Is it true then that $v_n \rightharpoonup 0$ in $H^1(\mathbb{R}^3)$?
 A: It is true, even assuming only 2. and 3.
First note, and that is a very useful fact for these kinds of questions, that it suffices to show that every subsequence of $(v_n)$ has a subsequence that converges to $0$ weakly in $H^1$. This can be done as follows:
Since $H^1(\mathbb{R}^3)$ is a separable Hilbert space, property 3 implies that every subsequence of $(v_n)$ has a subsequence converging weakly in $H^1$ (not necessarily to $0$). Moreover, as the embedding $H^1\hookrightarrow L^2$ is bounded, it is also continuous with respect to the weak topologies on both spaces. Thus, if a subsequence of $(v_n)$ converges weakly in $H^1$, it also converges weakly in $L^2$ to the same limit. But according to property 2, the only possible limit is $0$. Hence every subsequence of $(v_n)$ has a subsequence converging weakly to $0$ in $H^1$, which is what we wanted to prove.
A: After the elegant answer of MaoWao, let me add some dirty version as well :)
Let's start with $f\in H^2$. Then we have (using integration by parts)
$$ \langle v_n, f \rangle_{H^1} 
= \langle v_n, f \rangle_{L^2} + \langle \nabla v_n, \nabla f \rangle_{L^2} 
= \langle v_n, f \rangle_{L^2} - \langle v_n, \Delta f \rangle_{L^2}.  $$
As $v_n \rightharpoonup 0$ in $L^2$, we get $\langle v_n, f \rangle_{H^1} \rightarrow 0$. However, for a general $f\in H^1$ and $\tilde{f}\in H^2$ we find by Cauchy-Schwarz
$$ \vert \langle v_n, f \rangle_{H^1} \vert 
\leq \vert \langle v_n, \tilde{f} \rangle_{H^1}\vert + \vert \langle v_n, f-\tilde{f} \rangle_{H^1} \vert 
\leq \vert \langle v_n, \tilde{f} \rangle_{H^1}\vert + \left(\sup_{n\geq 1} \Vert v_n \Vert_{H^1} \right) \Vert f - \tilde{f} \Vert_{H^1}. $$
Hence, by density of $H^2$ in $H^1$ we get that $\langle v_n, f\rangle_{H^1} \rightarrow 0$ for any $f\in H^1$ and thus $v_n \rightharpoonup 0$ in $H^1$.
