# Can we conclude that $v_{n}\rightarrow v$ in $L^{\infty}\left(\Omega\right)$ if $p>N$

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 $p>N$ ?

• Yes, that's the other version of Sobolov embedding when $p >N$. (Theorem 7.17 in Gilbarg and Trudinger) – user99914 May 21 '15 at 1:54
• Which edition of Gilbarg and Trudinger do you mean? The first edition or the second? – user109584 May 21 '15 at 1:58
• (reprint of) 1998 edition. – user99914 May 21 '15 at 1:59
• 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 2:16
• I have found some useful discussions here math.stackexchange.com/questions/708139/… and here math.stackexchange.com/questions/709842/… – user109584 May 21 '15 at 2:43

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

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

• My concerning is the case $s=\infty$ – user109584 May 21 '15 at 3:22
• For a sequence of continuous function $v_n$, the $C^0$ norm is the $L^\infty$ norm. @user109584 – user99914 May 21 '15 at 3:24
• Ehm... one more thing, you mean $\left\Vert u\right\Vert _{L^{\infty}}\leq\left\Vert u\right\Vert _{C^{o}}$ right? @JohnMa – user109584 Jul 2 '15 at 9:23
• @user109584 : As $u$ is continuous, $\|u\|_\infty = \|u\|_{C^0}$. – user99914 Jul 2 '15 at 9:26
• Could you give me some sketch of proof? @JohnMa – user109584 Jul 2 '15 at 9:27