Let $H$ be a complex infinite dimensional Hilbert space, and let $T \in B(H)$ be nonscalar. An operator is Fredholm iff $\dim \ker T, \dim \ker T^* < \infty$ and the range is closed, where $T^*$ denotes the adjoint of $T$; $\Phi$ is the set of Fredholm operators in $B(H)$. The Weyl spectrum of $T$ is defined by:
$$ \sigma_{w}(T):= \lbrace \lambda \in \mathbb{C} : T- \lambda Id \not \in \Phi_0 \rbrace $$
where $\Phi_0$ is the set of Fredholm operators with index $0$, i.e. $\dim \ker T = \dim \ker T^* < \infty$. It can be seen that:
$$ \sigma_{w}(T) \subseteq \sigma(T) $$
It is known that $ \sigma_{w}(T) \neq \emptyset$. In (8.50), V. Rakočević, E. Malkowsky. Advanced Functional Analysis, 2019, it is stated that:
$$ \sigma(T)=\sigma_{w}(T) \cup \sigma_p(T) $$
(where $\sigma_p(T)$ denotes the point spectrum, i.e. the eigenvalues of $T$). How can we prove this? I started reasoning as follows: if $\lambda \in \sigma(T)$ is an eigenvalue, we have nothing to prove, so suppose not. We have to prove that it belongs to the Weyl spectrum. $T- \lambda Id$ is not invertible, and it is injective by assumption. Thus, it is not surjective. If the range is dense, by non-surjectivity we know that the range is not closed, so it is not Fredholm. Consequently, the continuous spectrum is contained in the Fredholm spectrum (the set of $\lambda$'s for which $T- \lambda Id$ is not Fredholm), which is contained in the Weyl spectrum. If it is not dense, $\lambda$ is in the residual spectrum. How can I proceed? Any suggestion is greatly appreciated.
Best Answer
I have found a way to prove it. First, if $\lambda \in \sigma_{w}(T) \subseteq \sigma(T)$, there is nothing to prove. So suppose it is not in the Weyl spectrum. Then, $T- \lambda Id$ is Fredholm with index $0$. If $T-\lambda Id$ is not injective, $\lambda \in \sigma_p(T)$ because it is an eigenvalue, so suppose instead that it is injective. Since $T- \lambda Id$ is not invertible, it follows that this operator cannot be surjective. However, since it is Fredholm, $Im(T-\lambda Id)=Im(T-\lambda Id)^-$ (it is equal to its closure, because the range is closed). Thus, the range is not dense. Consequently, since $T- \lambda Id$ is injective but does not have dense range, $\lambda \in \sigma_R(T)$ (the residual spectrum). It is well known that:
$$ \sigma^* _R(T) \subseteq \sigma_p(T^*) $$
where $A^*:= \lbrace \overline{a}, a \in A \rbrace$ (the set of complex conjugates of the elements in $A$). Thus, since $\lambda \in \sigma_R(T)$, $\overline{\lambda} \in \sigma_p(T^*)$. This means that $T^* - \overline{\lambda}Id=(T-\lambda Id)^*$ is not injective, so we have (recalling that $T-\lambda Id$ is Fredholm with index $0$):
$$ \dim \ker (T- \lambda Id)= \dim \ker ((T - \lambda Id)^*) >0 $$
So $\lambda \in \sigma_p(T)$, against our assumption. So this last case never happens, and we can conclude that if $\lambda \not \in \sigma_w(T)$, then it belongs to $\sigma_p(T)$, concluding the proof.