What is the closure of $ C^\infty_c(\mathbb{R}^n\setminus\{0\})$ in Sobolev $ W^{1,p} $ norm?

Question

For $1 \leq p < \infty, n\geq 1 $ my guess of the answer was $ W^{1,p}(\mathbb{R}^n)$ but I can't prove the inclusion $ \overline{C^\infty_c(\mathbb{R}^n\setminus\{0\})} \subseteq W^{1,p}(\mathbb{R}^n) $.

Any hints or partial solutions( like for $ p = 2$) would be extremely helpful. Thanks.

Answer 1

In the case $n = 1$ or ($n \ge 2$ and $p> n$), you have the embedding from $W^{1,p}$ to $L^\infty$. In this case $\overline{C_c^\infty}\subsetneq W^{1,p}(\mathbb R^n)$. Indeed, every limit point of $\overline{C_c^\infty}$ is continuous and zero at $0$.

In the other case, you get $\overline{C_c^\infty} = W^{1,p}(\mathbb R^n)$. This can be seen by taking an arbitrary function in $W^{1,p}(\mathbb R^n)$, mollifying it and truncating it (smoothly) at $0$ and $\infty$.

All steps but the truncation at $0$ are standard. For this truncation, recall, that there is a function $\varphi \in C_c^\infty(\mathbb R^n)$, which is equal to one in a neighbourhood of $0$ and has an arbitrary small $W^{1,p}$-seminorm. Multiplication with $1-\varphi$ will do the job.

Answer 2

I'm assuming that the closure is with respect to the $W^{1,p}(\mathbb R^n \setminus \{0\})$ norm so what you are asking is whether $W^{1,p}_0(\mathbb R^n \setminus \{0\}) \subseteq W^{1,p}(\mathbb R^n)$.

Not exactly, since the two spaces consist of functions defined on different domains. But if $\Omega$ is any open set and you extend a function $f \in W^{1,p}_0(\Omega)$ to be zero outside $\Omega$, the extension belongs to $W^{1,p}(\mathbb R^n)$.