# Weak continuity in Sobolev Spaces

First consider the following two Sobolev Embedding Theorems.

Theorem 1:

The continuous embedding $W^{1,p}(\Omega) \subset L^{p^{*}}(\Omega)$ holds provided the exponent $p^{*}$ is defined as $p^{*} := \{ \frac{np}{n-p} \text{ for } p <n, \text{ an arbitrary large real for } p=n, + \infty \text { for } p>n \}$

Theorem 2(Rellich, Kondrachov Theorem):

The compact embedding $W^{1,p}(\Omega) \Subset L^{p^{*}-\epsilon}(\Omega)$ for $\epsilon \in (0,p^{*}-1]$ holds for $p^{*}$.

Consider a bounded continuous mapping $f$:

$f: L^{p^{*}-\epsilon}(\Omega) \times L^{p}(\Omega;\mathbb{R}^{n}) \rightarrow L^{p'}(\Omega;\mathbb{R}^{n})$ $\text{ where } p^{'}=\frac{1}{p-1}$

How does it follow then that we can state that: $f: W^{1,p}(\Omega)\times L^{p}(\Omega;\mathbb{R}^{n}) \rightarrow L^{p'}(\Omega;\mathbb{R}^{n})$ is (weak $\times$ norm, norm) -continuous?

Thanks for assistance. Let me know if something is unclear.

• Is $f$ a linear function? What is in between $L^{p^\star-\epsilon}$ and $L^p$ in the product above? – Tomás Jan 9 '14 at 11:06
• $f$ is not necessarily linear. There is nothing in between the product, sorry was a mistake. Editing it now. But if you think the problem resolves easier if $f$ is linear then I would be interested in how that works. Thanks. – user100431 Jan 9 '14 at 12:08
• What do you mean by weak $\times$ norm? – i like xkcd Jan 9 '14 at 14:38
• If by (weak $\times$ norm, norm) you mean the product norm $W^{1,p}(\Omega) \times L^p(\Omega; \mathbb{R}^n)$ then the result follows from Theorem 1... – i like xkcd Jan 9 '14 at 14:40
• By 'weak x norm' I mean that it is the product topology where $W^{1,p}(\Omega)$ is endowed with it's weak topology and $L^{p}$ is endowed with it's norm topology. – user100431 Jan 9 '14 at 14:46

Assume that $w_\alpha=(x_\alpha,y_\alpha)$ is a net in $W^{1,p}(\Omega)\times L^p(\Omega:\mathbb{R}^n)$ such that $w_\alpha\to w=(x,y)$. We need to prove that $$f(w_\alpha)\to f(w)\tag{1}$$

We have that $x_\alpha\to x$ in the weak topology of $W^{1,p}(\Omega)$. Once Theorem 2 is valid, we can conclude that $x_\alpha \to x$ in the strong topology of $L^{p^\star-\epsilon}(\Omega)$, hence, as $f$ is continuoous with respesct to norm$\times$norm topology, we conclude that $(1)$ is true.

Remark: there is a assumption in my proof which must be verified.

• Thanks for the response, how do you get weak convergence in $W^{1,p}(\Omega)$ implies strong convergence in $L^{p^{*}-\epsilon}(\Omega)$ from Theorem 2? And what assumption are your referring to? – user100431 Jan 10 '14 at 19:12
• This is part of the assumption I left to you verify. Try to understand it first, then if you need some help, post a comment here. – Tomás Jan 10 '14 at 19:46
• Okay I see, I will revise it and then comment if I have further queries,thanks. – user100431 Jan 11 '14 at 19:00
• There is a lemma which states 'Let $X$ be a topological space and $(x_{n})_{n}$ be a sequence in $X$ and $x \in X$. If every subsequence of $(x_{n})_{n}$ has a subsequence converging to $x$, then $x_{n} \rightarrow x$' – user100431 Jan 12 '14 at 19:28
• The same argument also work for nets. Note that every net in a compact set has a convergent subnet. – Tomás Jan 12 '14 at 22:59