Sobolev functions vanish in a ball 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. 

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 $t>1$...
