# Proof of (8.50) in the book Advanced Functional Analysis

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.

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.