Neumann Problem - how to prove that a weak solution is also a classic one? I think a similar problem appears in Evans' book: For a given Neumann problem, i.e. -$\nabla^2 u=f$ in $\Omega$, $\partial{u}/\partial{\nu}=0$ on $\partial \Omega$ where $\Omega$ is a bounded domain in $\mathbb{R}^N$, $\partial \Omega$ is smooth, $f\in C(\bar\Omega)$, $u\in C^2(\bar\Omega)$. I want to prove that $u$ is a classical solution if and only if it is a weak solution of the problem. One direction (assuming a classical solution and proving it is also a weak solution) is quite easy. But how do I prove the other direction (assuming a weak solution and proving that it is also a classic one)?
 A: We need a conclusion
For any smooth function $w$ defined on $\partial\Omega$, there exists a $v\in C^\infty(\overline\Omega)$ such that $v=w$ on $\partial\Omega$. 
Define 
\begin{equation}
B[u,v]=\int_\Omega Du\cdot Dv\;{\rm d}x\quad u,v\in H^1(\Omega)
\end{equation}
and $u\in H^1(\Omega)$ is a weak solution iff
\begin{equation}
B[u,v]=(f,v)_{L^2(\Omega)}\quad\forall v\in H^1(\Omega)
\end{equation}
When $u\in C^2(\Omega)\cap C^1(\overline\Omega)$ is a weak solution then $\forall v\in C^\infty(\overline\Omega)$
\begin{align}
&\int_\Omega(\triangle u+f)v\;{\rm d}x=\int_\Omega\triangle u\,v\;{\rm d}x+(f,v)_{L^2(\Omega)}\\
=&\int_{\partial\Omega}v\frac{\partial u}{\partial\nu}\;{\rm d}S-\int_\Omega Du\cdot Dv\;{\rm d}x+(f,v)_{L^2(\Omega)}\\
=&\int_{\partial\Omega}v\frac{\partial u}{\partial\nu}\;{\rm d}S
\end{align}
First, choose $v\in C^\infty_c(\Omega)$, then $v=0$ on $\partial\Omega$, then
\begin{equation}
\int_\Omega(\triangle u+f)v\;{\rm d}x=0\quad\forall v\in C^\infty_c(\Omega)
\end{equation}
then
\begin{equation}
\triangle u+f=0\quad {\rm in}\;U
\end{equation}
Then for any smooth function $w$ defined on $\partial\Omega$, denote $v$ to be its extension
\begin{equation}
0=\int_\Omega(\triangle u+f)v\;{\rm d}x=\int_{\partial\Omega}v\frac{\partial u}{\partial\nu}\;{\rm d}S=\int_{\partial\Omega}w\frac{\partial u}{\partial\nu}\;{\rm d}S
\end{equation}
$\int_{\partial\Omega}w\frac{\partial u}{\partial\nu}\;{\rm d}S=0\;\forall w\in C^\infty(\Omega)$ implies
\begin{equation}
\frac{\partial u}{\partial\nu}=0\quad{\rm on}\;\partial\Omega
\end{equation}
