Continuation of $W^{1,p}_0$-functions

Question

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 $v \in W^{1, \infty}(\mathbb{R}^n)$?

Answer 1

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

Answer 2

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 $W^{1,\infty}_0(\mathbb R^n)$.