# Smooth function composed with sobolev function vanishes at 0

Let $\Omega$ be a bounded domain with sufficiently smooth boundary. Let $u \in W^{1, 2}_{0}(\Omega)$ and $F \in C^{\infty}(\mathbb{R} \rightarrow \mathbb{R})$ such that $F(u(x)) = 0$ for almost every $x \in \Omega$. Must $F(0) = 0$?

My thoughts are: I think the answer is yes. Suppose $u$ instead was smooth. The $F(u(x)) = 0$ for every $x \in \Omega$. By continuity, $F(u(x)) = 0$ for every $x \in \overline{\Omega}$. Since $u = 0$ on the boundary $\partial \Omega$, then there exists an $x_{0}$ such that $u(x_{0}) = 0$. Thus $F(u(x_{0})) = 0$ and hence $F(0) = 0$. However, I am not sure how to argue if $u$ was not smooth. My first thought was to approximate $u \in W^{1, 2}_{0}$ by smooth functions, but then I'll be working with a Sobolev norm.

• Ok. Go on now! Take an approximating sequence $u_n$. As you know, convergence $u_n\to u$ does not necessarily imply that $F(u_n(x))\to F(u(x))$. You need pointwise convergence of $u_n$ to $u$ for that. But there is something you can do to recover something which converges pointwise to $u$... – Giuseppe Negro May 25 '14 at 20:19
• Let $\Omega\subset\mathbb{R}^N$. I think that the case $2<N$ may be cause of trouble. – Tomás May 25 '14 at 20:52
• @GiuseppeNegro: Maybe take an approximation to the identity? Let $\phi$ be a mollifier and $\phi_{\epsilon} := \epsilon^{-d}\phi(x/\epsilon)$ where $\Omega \subset \mathbb{R}^{d}$. Then $u \ast \phi_{\epsilon} \rightarrow u$ for almost every $x \in \Omega$ as $\epsilon \rightarrow 0$. Thus $F((u \ast \phi_{\epsilon})(x)) \rightarrow F(u(x)) = 0$ for almost every $x \in \Omega$. But $F((u \ast \phi_{\epsilon})(x))$ is smooth and hence we have $F((u \ast \phi_{\epsilon})(x)) \rightarrow 0$ for every $x \in \Omega$. But I'm not sure how to relate this to an approximating sequence $u_{n}$... – user153325 May 25 '14 at 21:00
• Instead of taking a generic approximating sequence $u_n$, you could take $u_\epsilon=u\star \phi_{\epsilon}$. Of course you need to prove that $u_\epsilon\to 0$ on the boundary. (P.S. I had in mind something different, but this idea might very well work and is nice). – Giuseppe Negro May 25 '14 at 21:22

## 1 Answer

Suppose $F(0)\ne 0$. Then there is $\epsilon>0$ such that $F(x)\ne 0$ when $|x|<\epsilon$. Since $F\circ u=0$ a.e., it follows that the set $\{x\in \Omega: |u(x)|<\epsilon\}$ has zero measure.

So, the question reduces to showing that a function $u$ with $|u|\ge \epsilon$ a.e. cannot be in $W^{1,2}_0(\Omega)$. One way to do this is to recall that the zero extension of a $W^{1,p}_0$ function is in $W^{1,p}_0(\mathbb R^n)$. But if $|u|\ge \epsilon$ a.e., on $\Omega$, the zero extension of $u$ has an essential (unfixable by redefining on a set of measure zero) jump discontinuity on almost every line crossing $\Omega$. Thus, it does not have a representative with the ACL property, meaning it's not in any Sobolev class.

Note that the smoothness of the boundary is not needed.