Convergence of a sequence in $L^2(\mathbb{R}^N)$ that is also bounded in $H^1(\mathbb{R}^N)$

Question

Let $(u_n)_n\subset H^1(\mathbb{R}^N)$ be a sequence and $u\in L^2(\mathbb{R}^N)$ verifying:

1. $(u_n)_n$ is bounded in $H^1(\mathbb{R}^N)$
2. $(u_n)_n$ converges to $u$ in $L^2(\mathbb{R}^N)$

Then, does $(u_n)_n$ converge to $u$ in $H^1(\mathbb{R}^N)$?

What I have thought: Since $(u_n)_n$ is bounded in $H^1(\mathbb{R}^N)$, there exists a subsequence $(u_{n_k})_k$ weakly convergent in $H^1(\mathbb{R}^N)$ to some $v \in H^1(\mathbb{R}^N)$. Now, keeping in mind that the embedding $H^1(\mathbb{R}^N) \hookrightarrow L^2(\mathbb{R}^N)$ is bounded, $(u_{n_k})_k$ also converges weakly to $v$ in $L^2(\mathbb{R}^N)$. But $(u_{n_k})_k$ also converges weakly to $u$ in $L^2(\mathbb{R}^N)$ so $v=u$ almost everywhere and we deduce $u\in H^1(\mathbb{R})$.

So it really makes sense to consider the norm $\Vert u_n -u \Vert_{H^1(\mathbb{R}^N)}$ but I don't see any reason why it should converge to $0$. I can't think of any counter example either.

Answer 1

No, here is a counterexample. Take any $\varphi\in H^1(\Bbb R^N)\setminus\{0\}$ and define $\varphi_n(x) = n^{d/2-1} \,\varphi(nx)$. Then $$ \int_{\Bbb R^N} |\varphi_n|^2 = n^{-2}\int_{\Bbb R^N} |\varphi|^2 $$ but $$ \int_{\Bbb R^N} |\nabla\varphi_n|^2 = \int_{\Bbb R^N} |\nabla\varphi|^2. $$