I'm studying problems from some qualifiers and I came across a problem I couldn't solve:
Let $\Omega\subseteq \mathbb{R}^3$ be a domain and suppose that $\Sigma\subseteq \Omega$ is a smooth (nontrivial) surface. If $u\in C^2(\Omega)$ is harmonic (i.e. $\Delta u = 0$ in $\omega$) and $u = \partial_\nu u = 0$ on $\Sigma$ then u vanishes identically.
Here, $\partial_\nu u = \nabla u\cdot \nu$ where $\nu$ is the normal vector.
Below is the solution I came up with and my questions about said solution.
Let $x_0\in \Sigma$ and $r>0$ be such that $B(x_0, r) \subseteq \Omega$ is devided by $\Sigma$ into two "side". We call these sides $\Omega_L$ and $\Omega_R$.
While it feels intuitively clear that I can do this, how to I justify the existence of such a ball?
Now, define $\tilde{u} : B(x_0, r) \to \mathbb{R}$ by $$ \tilde{u}(x)=\begin{cases} 0 & \text{if } x\in \Omega_L\\ u(x) & \text{otherwise}. \end{cases} $$ We claim that $\tilde{u}$ is harmonic. It would then follow from analiticity that $u \equiv \tilde{u} \equiv 0$.
Observe first that $\tilde{u}\in C\left(\overline{B(x_0, r)}\right)$. It therefore suffices to show that for all $x\in B(x_0, r)$ $$ \int\limits_{\partial B(x, \delta)}\partial_\nu \tilde{u} = 0 $$ for all sufficiently small $\delta > 0$.
I am looking for a reference on this last result (if this is true!).
Since $u$ is harmonic, this is trivial unless $x\in\Sigma$. In this case, observe that $$ \begin{split} \int\limits_{\partial B(x, \delta)}\partial_\nu \tilde{u} &= \int\limits_{\partial \left[B(x, \delta)\cap \Omega_L\right]}\partial_\nu \tilde{u} + \int\limits_{\partial \left[B(x, \delta)\cap \Omega_R\right]}\partial_\nu \tilde{u} \\ &=\int\limits_{\partial \left[B(x, \delta)\cap \Omega_R\right]}\partial_\nu u\\ & = 0 \end{split} $$ since $u$ is harmonic and the boundary of $B(x, \delta)\cap \Omega_R$ is piecewise smooth.