# The inclusion relations among Sobolev spaces: $W^{1,p}$, $W^{1,p}_0$ and $W^{1,p}_c$

Denote $C^\infty$ the set of all real-valued smooth function on $\mathbb R$

Denote $C^\infty_0$ the set of all real-valued smooth function $f$ on $\mathbb R$ such that $\lim_{x\to \pm\infty} f(x)=0$

Denote $C^\infty_c$ the set of all real-valued smooth function $f$ on $\mathbb R$such that the support of $f$ is compact.

The Sobolev norm is given by $||f||_{W^{1,p}}:=||f||_{L^p} + ||f'||_{L^p}$ and define $W^{1,p}$, $W^{1,p}_0$ and $W^{1,p}_c$ to be the completion, with respect to this norm, of $C^\infty$, $C^\infty_0$ and $C^\infty_c$ respectively.

Clearly we have inclusion relation: $W^{1,p} \supset W^{1,p}_0 \supset W^{1,p}_c$.

Question: Can we say more information than the mere inclusion relation? For example, is $W^{1,p}_0$ (or $W^{1,p}_c$) an open subset of $W^{1,p}$?

You have that $W^{1,p}_c=W^{1,p}_0$ and $W^{1,p}_0$ is closed in $W^{1,p}$. To prove the first, given $f\in W^{1,p}_0$ you first approximate $f$ with a function $g\in C^1_0$ and then you construct a smooth function $\varphi$ which is $1$ in $B(0,1)$ and $0$ outside $B(0,2)$ and then you multiply $g$ by $\varphi_n(x)=\varphi(x/n)$. You can prove that $g_n=g\varphi_n\in C^\infty_c\to g$ in $W^{1,p}$. That $W^{1,p}_0$ is closed in $W^{1,p}$ comes from its definition. It is the closure of $C^1_0$ in $W^{1,p}$.
• You also have $W^{1,p}(\mathbb{R})=W^{1,p}_c(\mathbb{R})$ by the same use of cut-off functions $\varphi_n$ and a mollification argument. So if the domain is $\mathbb{R}$ all three spaces coincide. – Jose27 Jun 11 '17 at 8:57