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

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    $\begingroup$ That's not working: $S=-1$ $\endgroup$
    – Christian Remling
    Commented May 2, 2014 at 16:25
  • $\begingroup$ Can you give me more clarification $\endgroup$
    – user45340
    Commented May 2, 2014 at 18:12
  • $\begingroup$ As I just realize, you probably meant to compare $\sigma(A+\lambda S)$ with $\sigma(A)$. Our expectation here clearly has to depend on what $S$ is (consider $S=1$, $S=-1$). $\endgroup$
    – Christian Remling
    Commented May 2, 2014 at 18:53
  • $\begingroup$ I want compare $\sigma_{e,S}(\lambda S- A)$ to $\sigma_{e,S}(A)$ $\endgroup$
    – user45340
    Commented May 2, 2014 at 19:50
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    $\begingroup$ The title of this question is horrible!!! :-) $\endgroup$
    – Włodzimierz Holsztyński
    Commented Oct 18, 2014 at 11:17

1 Answer 1

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Your conjecture $\sigma_{e,S}(\lambda S- A)=\sigma_{e,S}(A) + \lambda $ has to be wrong since you can choose for $A$ and $S$ the identity operator $I$ and get:

$$ \sigma_{e,I}(I) = \sigma(I) = \{1\} $$ $$ \sigma_{e,I}(\lambda I - I) = \sigma((\lambda -1)I) = \{\lambda -1\} $$

Maybe you conjecture should be $\sigma_{e,S}(\lambda S- A)= \lambda -\sigma_{e,S}(A)$, but I really don't think this would hold in the general case.

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