3
$\begingroup$

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 :)

$\endgroup$
  • $\begingroup$ Is there supposed to be a condition on $m$? Compact embedding doesn't hold unless unless you have some conditions on $m,q$. $\endgroup$ – Christopher A. Wong Dec 19 '12 at 22:31
  • $\begingroup$ We only have the condition $1 < q < \infty$... Uh and I mingled Rellich-Kondrachov theorem with another one, but whatever, I know what I needed now :) $\endgroup$ – Nicolas Dec 20 '12 at 22:06
  • $\begingroup$ Incidentally, where is this in Evans' book? $\endgroup$ – Christopher A. Wong Dec 21 '12 at 5:48
3
$\begingroup$

You have that $(u_n,\nabla u_n)\rightharpoonup (u,F)$ in $L^q(U)\times L^q(U,\mathbb{R}^m)$. Mazur lemma implies that there is a sequence of convex combinations $$\tilde{u}_n=\sum_{j=1}^{n}\lambda_{n,j}(u_j,\nabla u_j),\ \lambda_{n,j}\geq 0,\ \sum_{j=1}^n\lambda_{n,j}=1$$

which converges to $(u,F)$ strongly in $L^q(U)\times L^q(U,\mathbb{R}^m)$. In particular, $$\tilde{v}_n=\sum_{j=1}^n\lambda_{n,j}u_j$$

is a Cauchy sequence in $W^{1,q}(U)$, and hence there exists a limit function $$\tilde{u}=\lim \tilde{v}_n$$

It follows that $\tilde{u}=u$ and $F=\nabla\tilde{u}=\nabla u$.

$\endgroup$
4
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.