When I read PDE books as a novice, $\phi \in C^\infty_0(\Omega)$ often appears.
Accidentally, given that condition I saw that function values in $\partial\Omega$ are zero.
For example, a note from my class is as follow.
My question is,
(1) Is it true that given compact support condition, the function values on boundary are zero?
(2) It is very hard for me to understand compact support. Could you explain that concept in terms of the following proof or in general?
Let $u\in L^1_{loc}(\Omega)$. u is ``weakly'' harmonic if $$\int_\Omega u\Delta \phi \, dx=0, \quad\forall \phi \in > C^\infty_0(\Omega).$$
$u\in C^2(\Omega)$ a weakly harmonic $\implies$ u is harmonic.
(proof) $\nabla \cdot (u\nabla\phi) = \nabla u \cdot \nabla \phi + u \Delta \phi$ $$U = \int_\Omega u\Delta \phi \,dx = \int_\Omega \nabla \cdot (u\nabla \phi )dx - \int_\Omega \nabla u\cdot \nabla \phi\,dx $$ By divergence theorem, $$ = \int_{\partial \Omega} \nu \cdot (u\nabla\phi)ds - \int_{\partial\Omega}\nu\cdot(\phi\nabla u)ds + \int_\Omega \phi \Delta u dx $$
$$ \therefore \int_\Omega \phi \Delta u dx=0, \quad\forall \phi \in > C^\infty_0(\Omega)\\ \text{note: I think that the fact in question is used.}$$
Therefore, $\Delta u=0$ on $\Omega$.
$\because$ if $\exists x_0 \in \Omega \mid \Delta u(x_0) \neq 0$, we may assume $\Delta u(x_0) > 0$. By continuity of $\Delta u$, $$\exists B_\delta(x_0)\subset\Omega \mid \Delta u(x)>0 \quad\forall x\in B_\delta(x_0)$$ We choose $\phi\in C^\infty_0$ so that $$\phi(x) = \begin{cases}1 & |x-x_0| < \frac \delta 2 \\ 0 & otherwise\end{cases}$$ Then $$\underbrace{0=\int_\Omega\phi\Delta u dx}_{assumption} = \int_{B_\delta (x_0)} \phi\Delta u dx \geq \int_{B_{\delta/2}(x_0)}\phi\Delta u\,dx$$ $$=\int_{B_{\delta/2}(x_0)}\Delta u\,dx > 0$$ contradiction.