Why is the spectrum of the Laplace operator positive?

Question

I took a look at the Lax-Milgram method and using Rayleigh quotient and the max-min principle it is clear that the Laplace Operator should have a positive specturm. However, if I take $L : [0,1] \to \mathbb C$, $Lf(x) = -f''(x)$, and $f(x) = \sin(ix)$, I have

$$Lf(x) = -\sin''(ix) = -(i\cos'(ix)) = - (-i^2\sin(ix)) = -\sin(ix),$$ so $-1$ is an eigenvalue. Where am I wrong?

Answer 1

As Ian said, the spectrum of the Laplacian operator is normally considered on a function space with some boundary condition, e.g. Dirichlet or Neumann. Either of those conditions would exclude $\sin ix$ (which is simply a multiple of $e^x-e^{-x}$) from consideration.

We need boundary conditions to show that Laplacian is a symmetric operator, and to promote it to self-adjoint by suitably extending the domain of definition. Only then does its spectrum become a useful tool.

Answer 2

(This is more a commentary to @zaq answer than an answer in itself). Consider the function $f(x)=e^{kx}$, with $k\in\mathbb C$. You have the formula, analogous to your example, $$ -\frac{d^2}{dx^2} f= -k^2 f.$$ So, any complex number can be an eigenvalue of $-\frac{d^2}{dx^2}$, if you do not specify some additional condition. See Example 5 here.