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Proposition 9.18 from Brezis' book mentions that if $u\in W_{0}^{1,p}(\Omega)$, then the extension of $u$ outside $\Omega$ by zero belongs to $W^{1,p}(\mathbb{R}^n)$ for $1<p<\infty$. I felt that this should be true for all $p\geq 1$.

Is there a counterexample where this does not hold in the case $p=1$?

The proof as given in Brezis uses the fact that $D(\mathbb{R}^n)$ is dense in $L^p(\mathbb{R}^n)$ for $1\leq p<\infty$.

Is my line of reasoning valid? We need only show that $D_{e_1}\bar{u}=\overline{(D_{e_1}u)}$

$\int_{\mathbb{R^n}}\bar{u}D_{e_1}\phi=\int_{\Omega}{u}D_{e_1}\phi=\mathrm{lim}_{m\to\infty}\int_{\Omega}{u_m}D_{e_1}\phi=\mathrm{lim}_{m\to\infty}-\int_{\Omega}{D_{e_1}u_m}\phi=-\int_{\Omega}({D_{e_1}u})\phi=-\int_{\mathbb{R^n}}(\overline{{D_{e_1}u}})\phi$

where $D_{e_i}$ represents the weak derivative on the i-th variable, $u_m$ is the sequence in $D(\Omega)$ converging to $u$ in the norm of $W^{1,p}(\Omega)$ and $\phi$ belongs to $D(\mathbb{R^n})$. The overbar denotes extension of the subject outside its domain by zero.

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up vote 4 down vote accepted

Yes, you are right. Brezis excludes $p=1$ because of part (ii) of the proposition, which indeed breaks down when $p=1$. For completeness, here is a 1-dimensional example: $u=\chi_{(0,1)}$ satisfies (ii) on $\Omega=(0,1)$ but does not belong to $W^{1,1}_0(\Omega)$.

In fact, the implication you are asking about is more or less a tautology. An element $u\in W^{1,p}_0(\Omega)$ is represented by a sequence of $C^1_c(\Omega)$ functions $\{u_m\}$ which is Cauchy in the $W^{1,p}(\Omega)$ norm. This same sequence, extended by zero, is Cauchy in the norm of $W^{1,p}(\mathbb R^n)$ and therefore represents an element of $W^{1,p}(\mathbb R^n)$.

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