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Today I encountered such a problem:

Prove that the function $u(x):=|x|, x \in(-1,1),$ belongs to $W^{1, \infty}((-1,1))$ but not to the closure of $C^{\infty}((-1,1)) \cap W^{1, \infty}((-1,1))$.

The author of this textbook seems to use this exercise to illustrate conclusion in title.

Here is what he said:

The previous exercise shows that the Meyers-Serrin theorem is false for $p=\infty .$ This is intuitively clear, since if $\Omega \subseteq \mathbb{R}^{N}$ is an open set and $\left\{u_{n}\right\}_{n}$ is a sequence in $C^{\infty}(\Omega) \cap W^{1, \infty}(\Omega)$ such that $\left\|u_{n}-u\right\|_{W^{1, \infty}(\Omega)} \rightarrow 0,$ then $u \in C^{1}(\Omega)(\text { why? }) .$ However, the following weaker version of the MeyersSerrin theorem holds.

Then he gave the conclusion of the weaker version as exercise:

Let $\Omega \subset \mathbb{R}^{N}$ be an open set with finite measure and let $u \in$ $W^{1, \infty}(\Omega) .$ Modify the proof of the Meyers-Serrin theorem to show that there exists a sequence $\left\{u_{n}\right\}_{n}$ in $C^{\infty}(\Omega) \cap W^{1, \infty}(\Omega)$ such that $\left\|u_{n}-u\right\|_{L^{\infty}(\Omega)} \rightarrow 0$ $\left\|\nabla u_{n}\right\|_{L^{\infty}(\Omega)} \rightarrow\|\nabla u\|_{L^{\infty}(\Omega)},$ and $\nabla u_{n}(x) \rightarrow \nabla u(x)$ for $\mathcal{L}^{N}$ -a.e. $x \in \Omega$ as $n \rightarrow \infty$.

My first question is about the correctness of his comment.Must $u \in C^{1}(\Omega)?$.According to the definition of $L^{\infty}$,$\left\|u_{n}-u\right\|_{L^{\infty}(\Omega)} \rightarrow 0$ doesn't mean $\sup\left\|u_{n}-u\right\| \rightarrow 0$. They are equal if and only if $u_{n}-u$ is a continuous function.So this conclusion seems not to be right.

My second question is about how to solve those exercise?For (i) I cannot follow the auther's thought. For (ii) I have no idea about it...

Can someone give an opinion on my judgment or some hints for those exercise? Thanks in advance!

Johnstein
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  • Just a quick comment on your first question, if $||u_{n}-u||{W^{1,\infty}}\rightarrow 0$, then $u \in C^{1}$ since $(C^{1},||.||{W^{1,\infty}})$ is a Banach space (complete). – Winger 14 Jun 16 '20 at 12:13
  • @MF14 Thanks for your answer first.I will check $(C^{1},||.||{W^{1,\infty}})$ later.However,If $v=u $ a.e in domain,$||u{n}-v||_{W^{1,\infty}}\rightarrow 0$,so we can change accountable point of $u$ to make it not belong to $ C^{1}$. – Johnstein Jun 16 '20 at 12:25
  • In this case (as also for all Sobolev embeddings), $u \in C^{1}$ means that there exists a representative that is $C^{1}$. And that is all we need here: Choose a function that has no representative in $C^{1}$ (i.e. even if we are allowed to change it on a nullset it is never $C^{1}$) and we have a contradiction to $||u_{n}-u||_{W^{1,\infty}} \rightarrow 0$. Such a function would be $u(x)=|x|$ – Winger 14 Jun 16 '20 at 13:18
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    @ MF14 Oh,I see,thank you!!!Then how to prove that contradiction with $u(x)=|x|$.Can give me a hint? – Johnstein Jun 16 '20 at 14:35
  • Well, $u$ isn´t $C^{1}$ in the origin, and no matter what you do on a nullset, you will never make it have a continuous derivative... so there can´t exist $u_{n} \in C^{1}$ converging to $u$ in $W^{1,\infty}$ – Winger 14 Jun 16 '20 at 14:41
  • @MF14 Excellent answer! – Johnstein Jun 16 '20 at 15:04

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