Spectrum of the laplacian on a Banach space

Question

Is the spectrum of the laplacian on $L^1(0,1)$ with Neumann boundary conditions known?

Answer 1

Show that the resolvent is $$ R(\lambda)f = \frac{\cos(\sqrt{\lambda}(1-x))}{\sqrt{\lambda}\sin(\sqrt{\lambda})}\int_{0}^{x}\cos(\sqrt{\lambda}t)f(t)dt+\frac{\cos(\sqrt{\lambda}x)}{{\sqrt{\lambda}\sin(\sqrt{\lambda})}}\int_{x}^{1}\cos(\sqrt{\lambda}(1-t))dt. $$ That is, $g=R(\lambda)f$ solves $$ -g''-\lambda g = f \\ g'(0)=0,\;\;\; g'(1)=0, $$ provided that $\lambda \ne n^2\pi^2$ for $n=0,1,2,3,\cdots$. The same resolvent expression works for $L^1$ and for $L^2$ if $L =-\frac{d^2}{dx^2}$. So, in either case, the spectrum is a subset of $\{ n^2\pi^2 : n =0,1,2,3\cdots\}$. Conversely, every such point is in the spectrum because $L\cos(n\pi x)=n^2\pi^2\cos(n\pi x)$, which proves that $n^2\pi^2$ is an eigenvalue.