Skip to main content
MathOverflow

Return to Question

Post Closed as "Needs details or clarity" by Yemon Choi, Stefan Kohl♦, Ryan Budney, Ricardo Andrade, S. Carnahan♦
deleted 1 character in body
Source Link
user45340
user45340
  • 1
  • 3

Let $X$ be a Banach space. $A$ is a linear closed and densely defined operator and $S$ is a bounded invertible operator. What' s the relation between $\sigma_{e,S}(\lambda S - A)$ and $\sigma_{e,S}(A)$? Is it true that $\sigma_{e,S}(\lambda S- A)=\sigma_{e,S}(A) + \lambda S?$$\sigma_{e,S}(\lambda S- A)=\sigma_{e,S}(A) + \lambda ?$

Here $\sigma_{e,S}(A): = \{\lambda \in\mathbb{C}\,\,\hbox{such that} \ \lambda S-A \,\,\hbox{isn't a Fredholm operator on}\, X \}$.

Let $X$ be a Banach space. $A$ is a linear closed and densely defined operator and $S$ is a bounded invertible operator. What' s the relation between $\sigma_{e,S}(\lambda S - A)$ and $\sigma_{e,S}(A)$? Is it true that $\sigma_{e,S}(\lambda S- A)=\sigma_{e,S}(A) + \lambda S?$

Here $\sigma_{e,S}(A): = \{\lambda \in\mathbb{C}\,\,\hbox{such that} \ \lambda S-A \,\,\hbox{isn't a Fredholm operator on}\, X \}$.

Let $X$ be a Banach space. $A$ is a linear closed and densely defined operator and $S$ is a bounded invertible operator. What' s the relation between $\sigma_{e,S}(\lambda S - A)$ and $\sigma_{e,S}(A)$? Is it true that $\sigma_{e,S}(\lambda S- A)=\sigma_{e,S}(A) + \lambda ?$

Here $\sigma_{e,S}(A): = \{\lambda \in\mathbb{C}\,\,\hbox{such that} \ \lambda S-A \,\,\hbox{isn't a Fredholm operator on}\, X \}$.

Source Link
user45340
user45340
  • 1
  • 3

On a characterization of some subsets

Let $X$ be a Banach space. $A$ is a linear closed and densely defined operator and $S$ is a bounded invertible operator. What' s the relation between $\sigma_{e,S}(\lambda S - A)$ and $\sigma_{e,S}(A)$? Is it true that $\sigma_{e,S}(\lambda S- A)=\sigma_{e,S}(A) + \lambda S?$

Here $\sigma_{e,S}(A): = \{\lambda \in\mathbb{C}\,\,\hbox{such that} \ \lambda S-A \,\,\hbox{isn't a Fredholm operator on}\, X \}$.