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I know that for an open set $\Omega \subset \mathbb{R}^n$ and $1 \leq p \leq \infty$, a function $u \in W^{1,p}_0(\Omega)$ can be continued to a function $v \in W^{1,p}(\mathbb{R}^n)$ by setting $v=u$ in $\Omega$ and $v=0$ in $\Omega \setminus \mathbb{R}^n$.

By Meyers-Serein, we also know that for $1 \leq p < \infty$, the set $C^{\infty}_0(\mathbb{R}^n)$ is dense in $W^{1,p}(\mathbb{R}^n)$ and therefore $W^{1,p}(\mathbb{R}^n) = W^{1,p}_0(\mathbb{R}^n)$. So in this case, we also have $v \in W^{1,p}_0(\mathbb{R}^n)$.

But in the case $p = \infty$, do we also have $v \in W^{1, \infty}_0(\mathbb{R}^n)$ or just $v \in W^{1, \infty}(\mathbb{R}^n)$?

Tom Bombadil
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2 Answers2

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The problem is that $C_0^\infty(\mathbb{R}^n)$ is not dense in $W^{1,\infty}(\mathbb{R}^n)$. Particularly, a function in $W^{1,p}(\mathbb{R}^n)$ for $1\leq p<\infty$ must decay "far outside".

As an example, choose $v=1$ which is clearly in $W^{1,\infty}(\mathbb{R}^n)$, but does not have a zero trace or is in the completion of $C_c^\infty(\mathbb{R}^n)$ w.r.t. to the $W^{1,\infty}$- norm

Quickbeam2k1
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The space $W_0^{1,\infty}(\Omega)$ consists of functions that tend to $0$ at the boundary of $\Omega$; the limit is understood in the classical way because $W^{1,\infty}$ functions have a continuous representative. See the discussion here. When $\Omega=\mathbb R^n$, the role of boundary is played by the point at infinity.

Therefore, the zero extension of a function in $W^{1,\infty}_0(\Omega)$ belongs to $W^{1,\infty}_0(\mathbb R^n)$.

Community
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  • How can one see that $W^{1, \infty}_0(\Omega)$ consists of (all?) functions that tend to $0$ at the boundary of $\Omega$? In my lecture, $W^{1, \infty}_0(\Omega)$ was defined as the closure of $C^{\infty}_0(\Omega)$ with respect to the $W^{1, \infty}$-norm. – Tom Bombadil Aug 25 '14 at 05:19
  • The link provided in the answer, contains the answer for your question. – Tomás Aug 25 '14 at 12:42