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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)?

user315279
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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}

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