Some claims about spectra

Question

I see the following equalities used sometimes, but couldn't find proofs. How is it done?

$$\overline{Ran(L-\lambda\mathbb{1})}^{\perp}=ker(L^*-\overline{\lambda}\mathbb{1})$$ and $$\lambda\in\sigma_p(L)\implies\overline{\sigma_p(L^*)\cup\sigma_r(L^*)}$$

I couldn't find a rigorous set of assumptions either, so I guess we are talking densely defined bounded operators on Hilbert spaces, but I could be wrong.

Answer 1

The first part is a standard result (easy to prove also!) available in most books on operator theory for eg: Proposition 4.6 in Banach Algebra Techniques in Operator Theory by R.G. Douglas, 2nd edition states it as follows

If $T$ is an operator on the Hilbert space $H$, then $ker T = (ran T^*)^\perp $ and $(ran T)^\perp = ker T^*$

This proposition along with the result

If $M$ is a subset of a Hilbert Space then $M^\perp = {\overline M}^\perp $

which is again easy to prove gives answer to your first part (Take $T = L-\lambda 1$).

For the second part, let $H$ be the underlying Hilbert space. The correct statement is $\lambda \in \sigma_p (L) \Rightarrow \overline \lambda \in \sigma_r(L^*) \cup \sigma_p(L^*)$

$Proof: $ First of all note that

$\sigma (L)= \sigma_p(L) \cup\sigma_c(L) \cup \sigma_r (L)$

for reference https://en.wikipedia.org/wiki/Decomposition_of_spectrum_(functional_analysis)

Observe that if $\lambda \in \sigma (L)$ then $ \overline \lambda \in \sigma(L^*)$. Also if $\lambda \in \sigma_p (L)$ then there exists a non zero $x\in H $ such that $ (L-\lambda) x = 0 $ then

$\forall y \in H, $

$ \langle(L-\lambda) x , y \rangle = 0 \Rightarrow \langle x ,(L^*-\overline\lambda) y \rangle = 0 $

If $Ran (L^*-\overline \lambda) $ is dense then $x$ must be zero which is a contradiction. Therefore, $\overline \lambda \notin \sigma_c(L^*) \Rightarrow \overline \lambda \in \sigma_r(L^*) \cup \sigma_p(L^*)$ (since we already know that $\overline \lambda \in \sigma(L^*)$)