I have a in $H^1(\mathbb{R^N})$ uniformly bounded sequence $u_n \in H^1$. I also know $u_n\to u$ in $L^p$ for every $2\leq p < 2^\ast$, where $\ast$ means the Sobolev exponent.
Can I conclude that $u\in H^1$ as well and $\liminf_{n\to \infty} \|\nabla u_n\|_2 \geq \|\nabla u \|_2$?
My attempt: Does that follow from Banach-Alaoglu?
Since $H^1$ is a Hilbert space, I could consider $u_n \in (H^1)^\ast$. By Banach-Alaoglu $u_n$ has a weak-*-convergent subsequence. In particular, I have $\langle u_n,v\rangle \to \langle u_w,v\rangle$ for every $v\in H^1$. By uniqueness of the limit, I have $u_w = u$. Hence $u_n \to u$ weakly and then the rest follows from the weak lower semicontinuity of the $H^1$ norm.
$H^1$ is reflexive, no need to go the weak-star convergence way. So one obtains weak convergence of a subsequence $u_{n_k}\to u^*$ in $H^1$. Due to the continuity of the embedding $H^1\hookrightarrow L^p$, we know $u_{n_k}\to u^*$. By uniqueness of the weak limit, we find $u^*=u$, which proves $u\in H^1$.