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Let $\Omega\subset\mathbb{R}^{N}$ be a smooth bounded domain, $v_{n}\rightharpoonup v$ in $W_{0}^{1,p}\left(\Omega\right)$ , $\left\Vert v_{n}\right\Vert _{W_{0}^{1,p}}=1$ $\forall n$ . So we can assume that $v_{n}\rightarrow v$ a.e, $v_{n}\rightarrow v$ in $L^{s}\left(\Omega\right)$ for $1<s<\dfrac{Np}{N-p}$ . Can we conclude that $v_{n}\rightarrow v$ in $L^{\infty}\left(\Omega\right)$ if $p>N$ ?

user109584
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  • Yes, that's the other version of Sobolov embedding when $p >N$. (Theorem 7.17 in Gilbarg and Trudinger) –  May 21 '15 at 01:54
  • Which edition of Gilbarg and Trudinger do you mean? The first edition or the second? – user109584 May 21 '15 at 01:58
  • (reprint of) 1998 edition. –  May 21 '15 at 01:59
  • What is the relationship between $L^{\infty}\left(\Omega\right)$ and $C^{1-N/p}\left(\overline{\Omega}\right)$ ? Theorem 7.17 in Gilbarg - Trudinger just states that $W_{0}^{1,p}\left(\Omega\right)\subset C^{1-N/p}\left(\overline{\Omega}\right)$ . – user109584 May 21 '15 at 02:16
  • I have found some useful discussions here http://math.stackexchange.com/questions/708139/compactness-of-sobolev-space-in-l-infinty?rq=1 and here http://math.stackexchange.com/questions/709842/sobolev-holder-lp-spaces-continuous-and-compact-embeddings-proof – user109584 May 21 '15 at 02:43

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The argument is the following:

Let $v_n$ be a sequence in $W^{1, p}_0(\Omega)$ for some $\Omega \subset \mathbb R^N$ and $p >N$. If $v_n \to v$ weakly in $W^{1,p}_0(\Omega)$, then $||v_n||_{1, p}$ is uniformly bounded. By the Sobolev Embedding (Theorem 7.17) and the fact that $$C^{0, \alpha}(\overline \Omega) \to C(\overline \Omega)$$

is compact, there is $\tilde v\in C(\overline \Omega)$ such that $v_n \to \tilde v$ in $C(\overline \Omega)$

But you know also that $v_n \to v$ strongly in $L^s$ for some $s$, so $\tilde v = v$.

(We are essentially using the fact that a compact operator sends weakly convergent sequence to strong convergent one in the image, see here)

Remark: Let $ u \in C(\overline\Omega)$. Then by definition of $\|u\|_{C^0}$,

$$\{ x\in \Omega : |u(x)| > \|u\|_{C^0}\} = \emptyset \Rightarrow \|u\|_\infty \le \|u\|_{C^0}.$$

On the other hand, we show by contradiction that we do not have strictly inequality. Assume that $\|u\|_\infty < \|u\|_{C^0}$. Then by definition, there is $x\in \Omega$ so that $u(x) > \| u\|_\infty + \epsilon$ for some small epsilon. As $u$ is continuous, there is $\delta>0$ so that $u(y) > \| u\|_\infty + \epsilon/2$ for all $y$ such that $|y-x|<\delta$. In particular, the set

$$\{ x\in \Omega: u(x) > \|u\|_\infty + \epsilon/3\}$$ has positive measure. But that contradict the definition of $\|u\|_\infty$, which is

$$\|u\|_\infty := \inf\{ r>0 : \{|u|\ge r\} \text{ has measure zero}\}\}.$$

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