Regularity of Poisson equation Let $\Omega$ an open and bounded open set, and let $f\in C_0^\infty(\Omega)$. Is it true that the unique weak solution in $u\in H_0^1(\Omega)$ of the Poisson equation $$-\Delta u = f$$ is in $C_0^\infty(\Omega)$.
Thanks.
 A: It's not true if $C^\infty_0(\Omega)$ means smooth functions compactly supported in $\Omega$. 
Here is an example. Consider the ball $\Omega=B(0,1)$ in $\mathbb{R}^2$. Let $\Phi:\mathbb{R}\to\mathbb{R}$ be a smooth function satisfying $\Phi(t)=t$ for $t \le 1$ and $\Phi(t)=2$ for $t\geq 3$. Finally define
$$u(x,y) = \Phi(-\log(x^2+y^2)) \ \ \ \text{for } (x,y) \in \Omega\setminus \{(0,0)\},$$
and $u(0,0)=2$. Then $u \in C^\infty(\Omega)$ is harmonic in a neigborhood of $\partial \Omega$, hence $f:=-\Delta u$ is smooth with compact support in $\Omega$. On the other hand $u$ is smooth and vanishes on the boundary, but is not compactly supported in $\Omega$. 
If your questions is just about regularity, then yes, every weak solution of Poisson's equation with smooth right hand side belongs to $C^\infty(\Omega)$. This is called interior regularity, and does not depend on any properties of the boundary. If you want $u\in C^\infty(\bar{\Omega})$, then you need to assume the boundary is also smooth. 
EDIT: Here is an example in dimension $n=1$ to help clarify my additional remarks in the comment section. Suppose that $\Omega = (-2,2)$ and
$$f(x) = \begin{cases}
1,&\text{if } -1 < x < 1\\
0,&\text{otherwise.}\end{cases}$$
The fact that $f$ is not smooth is not really an issue for the example, and we could smooth it out without changing the main ideas. The key point is that $f$ is compactly supported. Now let $B_0=(-1,1)$ be the support of $f$. The solution of $-u_0'' = f$ in $B_0$ with $u_0(-1)=0=u_0(1)$ is
$$u_0(x) = \frac{1}{2} - \frac{1}{2}x^2.$$
The solution of $-u'' = f$ in $\Omega$ with $u(-2)=0=u(2)$ cannot be the zero extension of $u_0$, since $u$ must be $C^1$. In fact, the solution is the $C^1$ function
$$u(x) = \begin{cases}
x+2,&\text{if } -2 < x < -1\\
\frac{3}{2} - \frac{1}{2}x^2,&\text{if } -1 \leq x \leq 1\\
-x+2,&\text{if } 1 < x < 2.\end{cases}$$
Notice that $u$ is not compactly supported and does not agree with $u_0$ (it is a coincidence that $u(x)=u_0(x) + 1$ for $x \in B_0$, and this will not be true in general). 
Basically, you cannot localize solutions of elliptic equations in the way you would like. The solution at any point $x$ depends on $f$ throughout the entire domain (think Green's functions). 
