Let $X$ be a complex infinite dimensional Banach space, and let $T \in B(X)$ be nonscalar. The Fredholm spectrum of $T$ is defined by:
$$ \sigma_{\Phi} (T) := \lbrace \lambda \in \mathbb{C} : T- \lambda Id \not \in \Phi \rbrace $$
(an operator is Fredholm iff $\dim \ker T, \dim \ker T^* < \infty$, where $T^*$ denotes the adjoint of $T$; $\Phi$ is the set of Fredholm operators in $B(X)$). Moreover, 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:
$$ \sigma_{\Phi}(T) \subseteq \sigma_{w}(T) \subseteq \sigma(T) $$ While it is known that $\sigma(T)$ is always nonempty, is it possible that $\sigma_{\Phi}(T) = \emptyset$? According to formula (8.50) in V. Rakočević, E. Malkowsky. Advanced Functional Analysis, 2019, we have (at least when $X$ is Hilbert):
$$ \sigma(T)=\sigma_{w}(T) \cup \sigma_p(T) $$
(where $\sigma_p(T)$ denotes the point spectrum, i.e. the eigenvalues of $T$), so if $T$ does not have any eigenvalue the Weyl spectrum is certainly nonempty. More generally, what about $\sigma_{\Phi}$? Is it always nonempty or can it be empty in some cases?
