Prove that if $\Delta u = 0$ in $B_1$ and $u = \frac{\partial u}{ \partial ν} = 0$ on an open portion of $\partial B_1$, then $u ≡ 0$ in $B_1$.
I'm stuck on this problem.
I tried to apply the same proof for the mean value theorem for Laplace's equation, but I couldn't get anything helpful. Any hint is appreciated.
Since you ask for a hint, I will give that and not full details. You know $u=\partial u/\partial \nu =0$ on a relatively open set $V \subseteq \partial B_1$. You don't state it, but I suppose we're allowed to assume that $V\neq \varnothing$ as otherwise we have nothing to work with. If $V= \partial B_1$ then the result follows from the maximum principle from the $u=0$ on $V$ assumption alone, and in this case the information that $\partial u /\partial \nu =0$ on $V$ is overkill. This tells us that something interesting only happens when $V \neq \partial B_1$, and since we have no information about the size of $V$ it also suggests that the result should somehow depend only on local information. In turn this suggests that we should maybe use analyticity. Harmonic functions are analytic, so perhaps you can try computing the power series for $u$ around any point $x \in V$ using the fact that $u = \partial u/\partial \nu =0$ on $V$.