I am looking for a proof of this well known fact and I guess it has to do with integration by parts (Green's identity). Unfortunately, I only know about 1-d integration by parts( I am just 3rd semester). could anybody sketch a short proof that shows that the spectrum of the Laplacian is negative?
My problem with the hint that brom gave me is that I do not know what I can assume about this surface integral and so on that appear in green's identity, but maybe I am looking at this from the wrong direction.
If I start:
$$ \int_V u \Delta u + \underbrace{\langle \nabla u, \nabla u \rangle}_{\ge 0} = \int_{\partial V}u(\nabla u n) dS$$ I mean intuitively, the thing is, that if we assume that our function $\psi$ vanishes at infinity and we integrate over the whole space, then the surface integral is maybe zero. In this case, we had $$ \int_V u \Delta u=-\int_V\langle \nabla u, \nabla u \rangle \le 0$$ But I do not see the correct mathematical reasoning behind this!
As brom already pointed out in the comments, for a bounded domain $\Omega\subset \mathbb{R}^n$, if $\lambda $ is a Dirichlet eigenvalue for Laplacian: $$ \Delta u = \lambda u \; \text{ in }\Omega, \\ u = 0 \text{ on }\partial \Omega.\tag{1} $$ Then $$ \lambda \int_{\Omega} u^2 = \int_{\Omega} u\Delta u = - \int_{\Omega} |\nabla u|^2 + \color{blue}{\int_{\partial \Omega} u(\nabla u\cdot n)\,dS}, $$ where blue term vanishes and we get the Rayleigh quotient for Laplacian: for non-trivial $u$ solving (1) $$ \lambda = - \frac{\int_{\Omega} |\nabla u|^2}{\int_{\Omega} u^2 } <0. $$ For Neumann eigenvalues we have the first one being 0 and others being negative, in that there exists a non-trivial function $\Delta u =0 $ and $\nabla u\cdot n = 0$.
A key part here for either Dirichlet or Neumann eigenvalues for Laplacian, is to assume proper boundary conditions so that the boundary surface integral will vanish.