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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 :)

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Is there supposed to be a condition on $m$? Compact embedding doesn't hold unless unless you have some conditions on $m,q$. –  Christopher A. Wong Dec 19 '12 at 22:31
    
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 :) –  Nicolas Dec 20 '12 at 22:06
    
Incidentally, where is this in Evans' book? –  Christopher A. Wong Dec 21 '12 at 5:48

2 Answers 2

up vote 2 down vote accepted

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 $F=\nabla\tilde{u}=\nabla u$.

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Thanks for the idea :) –  Nicolas Dec 20 '12 at 22:04

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.

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