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.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.