I am asked to show that for the operator
$$ Af = -f'' $$
with $D(A)=\left\{f\in H^2(0,1), f(1)=f(0)=0 \right\} \subset L^2(0,1)$
is self Adjoint in $L^2(0,1)$ (This part is solved).
I cannot see why the Spectrum and set of Eigenvalues of the operator A concide.
The set of $\{\sin(\pi n x)\mid k\in\mathbb N\}$ of $A$ is an orthogonal base in $L^2(0,1).$ To show this use the fact that every function $f\in L^2(0,1)$ can be continued to an odd function in $L^2(-1,1),$ and then use Fourier series in $L^2(-1,1)$ w.r.t. $\{\sin{\pi nx},\cos{\pi mx},\ n\geq 1,m\geq 0\}.$
Since $\{\sin(\pi n x)\mid k\in\mathbb N\}$ are eigenfunctions of $A,$ $A$ is unitary equivalent to the diagonal operator $Ae_k=k^2e_k,\ k\in\mathbb N.$
The spectrum $\sigma(A)$ of such operator is the closure of the set $\{k^2\mid k\in\mathbb N\}$ of its eigenvalues. But $\{k^2\mid k\in\mathbb N\}$ is closed.