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Jose27 · #1 June 8th, 2009, 06:41 PM

If $\Omega \subset \mathbb{R} ^n$ open and connected , $u \in L_{loc}^1 ( \Omega) = \{ u: \Omega \longrightarrow \mathbb{R}$ : $u$ integrable in $\Omega _1$ for every open $\Omega _1 \hspace{2 mm} such \hspace{2 mm} that \hspace{2 mm} \overline{ \Omega_1} \subset \Omega \hspace{2 mm} is \hspace{2 mm} compact \hspace{2 mm} in \hspace{2 mm} \Omega \}$ is weakly differentiable in $\Omega$ and $D_{i} u =0$ $\forall i=1, ... , n$ then $u$ is constant a.e. in $\Omega$

I don't even know where to begin with this one, the usual proof for when $u$ is differentiable does not apply since we don't really have a mean value theorem for this derivatives, so... I'm stuck. Any help is appreciated.