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Let $\Omega \subset \mathbb R^d$ a connected domain. If $u\in W^{1,p}(\Omega )$ is s.t. $\nabla u=0$ show that $u$ is constant a.e.

Proof of my course

Let $x\in \mathbb R^d$ and $\varepsilon>0$ small enough. Extend $u$ by $0$ on $\mathbb R^d\backslash \Omega $. Let $(\varphi_n)$ a sequence of standard mollifier. For $n$ large enough $$\nabla (\varphi_n* u)=\varphi_n*\nabla u=0$$ a.e. in $B(x,\varepsilon)$. In particular $\varphi_n*u=c_n$ a.e. in $B(x,\varepsilon)$ for some sequence $(c_n)$.

Letting $n\to \infty $ we derive that $u=c$ a.e. in $B(x,\varepsilon)$ for some constant $c$. Since $\Omega $ is connected, the conclusion follow.

Questions

I really don't understand this proof. My question are the following one :

1) Why do we have to extend $u$ by $0$ on $\mathbb R^d\subset \Omega $ since $\Omega $ is open and $\varepsilon>0$ small enough to have $B(x,\varepsilon)\subset \Omega $.

2) Why for $n$ large enough $\varphi_n*\nabla u=0$. Don't we have $\partial _n*\nabla u$ always ? (since $\nabla u=0$ a.e.)

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  • $\begingroup$ the extension allow you to use the convolution product and hence the mollification $\endgroup$ – Guy Fsone Jan 15 '18 at 18:45
  • $\begingroup$ Where is that proof from? $\endgroup$ – PtF Jan 15 '18 at 18:54
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I agree, it's not necessary to extend $u$ to zero. Fix $x \in \Omega$ and then choose $\epsilon > 0$ such that $\overline{B(x,\epsilon)} \subseteq \Omega$. Let $d = \text{dist}(\partial \Omega, \overline{B}(x,\epsilon))$. Pick a $N \in \mathbb{N}$ such that $\text{sppt}(\varphi_{n}) \subseteq B(0,d)$ if $n \geq N$. Now $\varphi_{n} * u$ is well-defined in $B(x,\epsilon)$ and $\nabla(\varphi_{n} * u) = \varphi_{n} * \nabla u = 0$. Thus, $\varphi_{n} * u = c_{n}$ in $B(x,\epsilon)$ and so on.

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    $\begingroup$ The convolution here is probably defined on the whole $\Bbb R^d$, hence we need to extend in order to apply the operation. $\endgroup$ – BigbearZzz Jan 15 '18 at 18:47

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