Sobolev functions vanish in a ball

Question

Assume $u\in H^1(\mathbb R^N)$ and $u=0$ a.e. in $B_1$. Does it hold that $u\in H_0^1(\mathbb R^N\setminus \overline B_1)$?

I tried to find smooth functions with compact support to approximate $u$. For example $\rho_n*u$, with $\rho_n$ being mollifiers. However the support of these smooth functions is located in a neighborhood of $\mathbb R^n\setminus B_1$. When I tried to find an approximation of $u$ in $H^1_0(\mathbb R^N\setminus \overline B_1)$ by some truncated functions, i.e. $\xi_n u\in H^1_0(\mathbb R^N\setminus\overline B_1)$, with $$\xi_n=\begin{cases}1, &x\in \mathbb R^N\setminus B_{1+\frac1n},\\ 0, &x\in B_1, \end{cases} $$ I found that the gradient of $\xi_n$ is not bounded so that I can not get $\xi_n u\to u$. So I wonder whether it is true.

Welcome to MSE. It will be more likely that you will get an answer if you show us that you made an effort. — José Carlos Santos, Jan 06 '18 at 11:14

score 1 · Accepted Answer · answered Jan 06 '18 at 15:06

1

I take it that $H^1_0(\Omega)$ is the closure of $C^\infty_c(\Omega)$?

Hint: For $r>0$ define $$f_r(x)=f(rx).$$Since $C_c(\mathbb R^d)$ is dense in $L^2$ it follows that if $f\in L^2(\mathbb R^d)$ then $$\lim_{r\to1}||f-f_r||_{L^2}=0.$$Applying this to $f$ and to $f'$ shows that the same is true with $H^1$ in place of $L^2$.

So you can begin by approximating your $u$ by a function that vanishes in $B(0,t)$ for some $t>1$...

answered Jan 06 '18 at 15:06

David C. Ullrich

89,985

It works. Thank you very much! – user519081 Jan 07 '18 at 04:19

Sobolev functions vanish in a ball

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