A space is *linearly Lindelöf* iff every open cover $C$ has a subcover $S$ with $\operatorname{cf} (|S|)= \aleph_{0}$.

Question.Is there a linearly Lindelöf space $X$ with $\operatorname{cf} (L(X))> \aleph_{0}$ (where $L$ is the cardinal function Lindelöf degree)?

$L(X)$ must be a limit cardinal, like $\aleph_{\omega_{1}}$ or an inaccessible cardinal, because if $L(X)$ is a successor cardinal, then there is an open cover $C$ of size $L(X)$ with no subcover of smaller size, and such space cannot be linearly Lindelöf.

But consider the following situation for a space $X$: let $\kappa$ be a limit cardinal of uncountable cofinality (like $\aleph_{\omega_{1}}$), and $\Gamma = \{ \lambda_{\alpha} : \alpha < \operatorname{cf} (\kappa) \}$ a set of infinite cardinals, with $\operatorname{cf} (\lambda_{\alpha}) = \aleph_{0}$ for every $\alpha < \operatorname{cf} (\kappa)$, and $\sup \Gamma = \kappa$; and for every $\alpha < \operatorname{cf} (\kappa)$ there is an open cover $C_{\alpha}$ with $|C_{\alpha}| = \lambda_{\alpha}$ and no subcover of smaller size. This space $X$ may be linearly Lindelöf with $L(X) = \sup \Gamma = \kappa$.

In particular, is there a linearly Lindelöf space $X$ such that $L(X)$ is a weakly inaccessible cardinal?