3
$\begingroup$

Let $B$ be a Banach space and $A:B\rightarrow B$ a bounded operator such that $A\left( B\right) $ is closed and there is some closed subspace $E\subset B$ such that $B=A\left( B\right) \oplus E$. Is there some neighborhood $V$ of $0$ such that $B=\left( A-\lambda \right) \left( B\right) +E$ for all $\lambda \in V$ ?

I know that $\ker \left( A\right) $ may not have a closed complementary subspace, but I still believe that this statement is true. So, any help please ?

Improve this question
$\endgroup$
8
  • 1
    $\begingroup$ recall that the set of surjective (linear, bounded) operators on B is open in L(B). So you can take every $A’$ in some nbd of $A$ (more generally than $A’:=A-\lambda$) $\endgroup$ Commented Dec 20, 2024 at 18:51
  • 1
    $\begingroup$ In your conditions you do not need to assume that $A(B)$ is closed. This is a consequence of $B=A(B)\oplus E$ with $E$ closed. $\endgroup$ Commented Dec 20, 2024 at 19:07
  • 1
    $\begingroup$ In fact it would be sufficient to assume $B=A(B)+E$, that is the operator $B\times E\ni (x,y)\mapsto Ax+y\in B$ is surjective. $\endgroup$ Commented Dec 20, 2024 at 19:40
  • $\begingroup$ Thank you so much @PietroMajer $\endgroup$ Commented Dec 20, 2024 at 19:54
  • 1
    $\begingroup$ Nice question . I think the answer in general is no, but there could be special spaces for which is still true —You may want to post it as a new question. $\endgroup$ Commented Dec 20, 2024 at 23:02

1 Answer 1

Reset to default
3
$\begingroup$

Denoting $Q_E:B\to B/E$ the quotient map, $Q_EA$ is surjective. Then there exists $r>0$ such that $Q_E(A-\lambda)$ is surjective, hence $B=(A-\lambda)(B)+E$, for $|\lambda|<r$.

Improve this answer
$\endgroup$
0

You must log in to answer this question.

Start asking to get answers

Find the answer to your question by asking.

Ask question

Explore related questions

See similar questions with these tags.