I am trying to understand a proof in Evan's book "Partial Differential Equations".
We have a sequence $(u_n)_{n\in\mathbb{N}}$ in $L^q(U)$ where $U$ is a bounded open set of $\mathbb{R}^m$. We know that $\sup_n||u_n||_{L^q(U)}<+\infty$ and $\sup_n||\nabla u_n||_{L^q(U,\mathbb{R}^m)}<+\infty$ (so $\sup_n||u_n||_{W^{1,q}(U)}<+\infty$).
It is then said that there exists a subsequence $(u_{n_k})_{k\in\mathbb{N}}$ and a function $u \in W^{1,q}(U)$ such that
$(u_{n_k})_k$ converges weakly to $u$ in $L^q(U)$ and $(\nabla u_{n_k})_k$ converges weakly to $\nabla u$ in $L^q(U,\mathbb{R}^m)$.
The author abbreviate it in "$(u_n)_n$ converges weakly to $u$ in $W^{1,q}(U)$".
I understand that from Rellich-Kondrachov theorem, we can extract a subsequence such that $(u_{n_k})_k$ converges to a function $u$ and then another subsequence such that $(\nabla u_{n_{k}})_k$ converges to a function $F$, but I don't understand why we would have $u \in W^{1,q}(U)$ and $\nabla u = F$.
Perhaps we could directly use Rellich-Kondrachov with the weak convergence in $W^{1,q}(U)$ since $(u_n)_n$ is bounded in $W^{1,q}(U)$, but I thought this weak convergence only was an abbreviation and not really a weak convergence in a Banach Space…
Thank you in advance 🙂
Best Answer
The space $W^{1,q}$ is reflexive (as a closed subspace of $L^q \times L^q$), so every norm-bounded set is relatively weakly compact. And weakly compact is equivalent to weak sequentially compact in Banach spaces by the Eberlein-Smulian Theorem.