$-\Delta u=\lambda u$ in $\Omega$, $u=0$ in some ball, then $u\equiv 0$ Let $-\Delta u=\lambda u$ in $\Omega\subset\mathbb{R}^n$, $\lambda>0$. Suppose $u=0$ in a ball $B\subset\Omega$. I want to prove that $u\equiv0$ in $\Omega$.
If $\lambda\leq 0$, integrating by part solves this problem. But if $\lambda>0$, things become more subtle.
Besides, I think $\Omega$ must be connected. (am I right?)
Any hints will be appreciated a lot!
 A: This follows from regularity theory. Indeed, elliptic regularity theory tells us that solutions to $-\Delta u = \lambda u $ must be analytic in $\Omega$ (since all the coefficients in the elliptic operator $Lv=-\Delta v -\lambda v$ are analytic). Hence, by unique continuation, if $u=0$ in a ball $B\subset \Omega$ then $u =0$ in the connected component of $\Omega$ containing $B$ . If $\Omega$ is connected then there is only one connected component, so $u=0$ in $\Omega$.

Inspired by @Guiseppe Negro's comment, here is a more 'elementary' solution. I put elementary in air quotes because it is certainly not as easy or as elegant as the above solution, but it is elementary in the sense that it only need the mean-value formula for solutions to $-\Delta u = \lambda u$.
Here is the idea: If $u\in C^2(\Omega)$ and $B_\rho(x_0) \subset \Omega$ then it can be shown that $$ \tag{$\ast$}\frac 1 {\vert B_\rho \vert}\int_{B_\rho(x_0)} u(x) \, d x = \alpha_n u(x_0) \lambda^{-n/4} \rho^{-n/2} J_{n/2}(\rho)  $$ where $\alpha_n = n \Gamma(n/2) \cdot 2^{n/2-1}$ and $J_\beta$ is the Bessel function of order $\beta$. This is the analogue of the classical mean-value formula for harmonic functions. You can obtain also obtain it in a similar way: you let $\phi(\rho)$ be the average $u$ over $B_\rho(x_0)$ (as above) and $\psi(\rho)$ be the average of $u$ over $\partial B_\rho(x_0)$. Following the proof of the classical mean-value formula you can find an ODE in terms of $\phi$ and $\psi$ which, paired with a second ODE obtained from polar coordinates, you can solve to get the above formula.
For the sake of contradiction, suppose that $\Omega \setminus \{u =0\} \neq \varnothing$. Then either $\{u>0\}$ or $\{u<0\}$ is non-empty - assume without loss of generality that it is $\{u>0\} $. Moreover, we may assume without loss of generality that $\{u>0\}$ must touch {u=0} (i.e. that $\partial \{ u>0\} \cap \partial \{u=0\} \neq \varnothing$).
Now let $x_0\in \{u=0\}$ and $\rho>0$ be such that $B_\rho(x_0) \subset \{u\geqslant 0\}$ with $B_\rho(x_0) \cap \{u>0\} \neq \varnothing$. I believe this is possible since we know that $B\subset \{u=0\}$, so $\{u=0\}$ is not $(n-1)$-dimensional (or lower dimension). Now we use ($\ast$) to obtain that$$0<\frac 1 {\vert B_\rho \vert}\int_{B_\rho(x_0)} u(x) \, d x = \alpha_n u(x_0) \rho^{-n/2} J_{n/2}(\rho) =0 $$ which is a contradiction. Hence, our initial assumption that $\Omega \setminus \{u =0\} \neq \varnothing$ was false.
