Spectrum of a differential operator

Question

I'm dealing with this exercise: let $p \ge 2$. On $X=L^p(0,\pi)$ we consider the operator defined by

\begin{cases} D(A) = W^{2,p}(0,\pi) \cap W_0^{1,p}(0,\pi)\\ Af(x)=f''(x) \hspace{5mm}x \in(0,\pi) \hspace{3mm} a.e. \end{cases}

Where $W_0^{1,p}(0,\pi) = \{f \in W^{1,p}(0,\pi) : f(0)=0=f(\pi)\}$

it should result that $\sigma(A) = \{-n^2 : n \in \mathbb{N}\}$.

Well I know that $\lambda \in \sigma(A) \Leftrightarrow \exists f \in D(A) : Af = \lambda f$.

So it results that \begin{equation} f''-\lambda f = 0 \end{equation}

then I suppose that one should solve the differential equation and find $\lambda$ values which satisfy it, but how to show that $\sigma(A)$ is exactly that set? thank you

Answer 1

I think you have to show that the general solution to $f''-\lambda f=0$ is $f(x)=\sin(kx)$ and $f(x)=\cos(kx)$ for $k \in \mathbb{R}$. Then you check which of these solutions lie in $D(A)$.