May I ask something to the community:
If $A: H \rightarrow H$ is a normal, compact operator on a complex Hilbert space H,
then $\sigma(A)=\sigma_p(A)\cup \lbrace 0\rbrace$, since if $z-A$ $(z\neq 0)$ is injective it is continuously invertible (because $A$ is compact) and $0$ does not have to be an eigenvalue.
If $\sigma_{ap}(A)$ is the approximate point spectrum, that means it contains all complex numbers $z$, satisfying $\|(A-z)x_n\|\rightarrow 0$, for some normed sequence $(x_n)$.
Then apperantly $\sigma_p(A)\subseteq \sigma_{ap}(A)$.
By the spectral theorem there is a sequence of eigenvalues converging to $0$, therefore $0 \in \sigma_{ap}(A)$.
What also holds is $\sigma_{ap}(A)\subseteq \sigma(A)$, since $z-A$ can't be bounded from bellow for any $z\in \sigma_{ap}(A)$.
Overall this would mean $\sigma_{ap}(A)=\sigma(A)$ for compact, normal operators on a Hilbert space.
Is the converse maybe also true? Can we deduce some of the properties like, normal, selfadjoint, compactness to $A$, if $\sigma_{ap}(A)=\sigma(A)$ holds? Or are there any counter examples.