# weak* convergence definition in Sobolev space

I have a question which might quite trivial but I would appreciate any assistance.

Why does it follow that for Sobolev spaces, say $W^{1,p}(\Omega)$, where $\Omega \subset \mathbb{R}^n$, it follows that $$u_n \rightharpoonup^{*} u \in W^{1,p}(\Omega)~~~\Leftrightarrow~~~ u_n \rightharpoonup^{*} u \in L^p(\Omega) \text{ and } \nabla u_n \rightharpoonup^{*} \in L^p(U;\mathbb{R}^n).$$

Thanks for any help.

## 1 Answer

First of all, we only use weak star convergence in the case of $p=\infty$, i.e., weak star convergence stands for convergence test against the pre-dual, not dual which is the case of weak convergence used for $p<\infty$.

Next, for your question, you can think $W^{1,p}$ space as $N+1$ fold of $L^p$ space, i.e. you have $(u,\partial_1u,\partial_2u,\ldots,\partial_Nu)\in L^p(\Omega)\times L^p(\Omega)\times\cdots\times L^p(\Omega)$ for $N+1$ times. Hence, weak compactness in $W^{1,p}(\Omega)$ is equivalent to weak compactness in $N+1$ times $L^p$ space and hence it gives you what you want.

For the reason why we are using weak star convergence in the case of $L^\infty$ is that $L^\infty$ is not a reflexive space, the dual of $L^\infty$ is not so clear, we only know it is strictly greater then $L^1$. However, the pre-dual of $L^\infty$ is $L^1$ and hence we adapted to use weak star convergence as $p=\infty$. In another word, we say $u_n\to u$ weak star in $L^\infty$ by means of $$\int u_nv\to\int uv$$ for all $v\in L^1$.

• Thanks for your answer. With regards to your first point, I am actually interested in the case $p=\infty$ (I should have stated this). But I don't understand why weak* convergence applies for this case ( where $p=\infty$), since as you state weak* convergence is usually associated with convergence in a dual space? I'm not quite getting your reason for why weak* convergence applies to $p=\infty$. – user116403 Jan 14 '15 at 14:39
• I updated my answer. – spatially Jan 14 '15 at 14:57
• I think I understand. Since $L^{p}$ is reflexive for $1 < p < \infty$ you can use Riesz representation theorem to get that $L^{q}$ is the dual space, where $q:= \frac{p-1}{p}$, and then weak convergence follows. But as you say the lack of reflexivity means that it is more common to consider the weak* convergence on $L^{\infty}$ since the case of its dual is not as simple. Very nice responses, thanks. – user116403 Jan 14 '15 at 15:04