Functional Kreyszig section 7.3 exercise 8

Question

I need some help with this exercise, I do not have idea about how solve it.

Let $X=C[0,\pi]$ and define $T: \mathcal{D}\left( T\right) \longrightarrow X$ by $f \to D^2(f)$, where $$\mathcal{D}\left( T\right)=\left\lbrace f \in X \mid D(f), D^2(f) \in X, f(0)=f(\pi)=0 \right\rbrace $$ Show that $\sigma\left( T\right) $ is not compact.

PD: Sorry, but I do not speak English well.

Answer 1

If $f(x)=\sin (nx)$ then $f \in \mathbb D$and $Tf=-n^{2}f$. Hence $-n^{2}$ is an eigen value of $T$ which implies it belongs to the spectrum. Thus the spectrum is not bounded.

Answer 2

$f_n(x)=\sin(nx)$ for $n=1,2,3,\cdots$ are functions in the domain, and $Tf_n = -n^2 f_n$.