Let's consider Laplace eigenvalue problem \begin{eqnarray} -\Delta u = \kappa u\text{ on }\Omega,\\ \frac{\partial u}{\partial n}=0 \text{ on }\partial\Omega, \end{eqnarray} where $\Omega\subset\mathbb{R}^{n}$ is a bounded domain with Lipschitz boundary.
It is well known that the eigenfunction $e_{0}$ corresponding to the eigenvalue $\kappa_{0}=0$ is constant and any other eigenfunction $e_{j}$, $j>0$ change the sign on the domain $\Omega$.
My question: Is any linear combination of these eigenfunctions $e_{j}$ changing the sign too?
I was unable to find the answer in usual PDE and functional analysis related books and I did not find a way to prove this myself. I would be very grateful for a reference or hint how to prove this.
Since $\Delta$ is self-adjoint, the eigenfunctions $e_j$ can be taken to be orthogonal; so any linear combination $f = \sum_{j>0} a_j e_j$ satisfies $$\int_\Omega f = \sum_{j>0} a_j \int_\Omega e_j = C\sum_{j>0} a_j \langle e_j, e_0 \rangle_{L^2} = 0$$ since $e_0$ is constant. (Here $C$ is just the constant such that $Ce_0(x) = 1.$) Thus $f$ must change sign unless it is identically zero.