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Let $A$ be a closable operator in a Hilbert space $\mathcal{H}$. Do we have the following ? $$\sigma (\bar{A})=\bar{\sigma}(A)$$ thanks in advance.

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Note that operators have a bounded inverse if and only if they are closed and bijective. This is a consequence of the open mapping theorem. Hence, if $A$ is closable but not closed, then its spectrum is all of $\mathbb{C}$. It follows that the relationship you above have is not true, as $\overline{\sigma(A)}$ would always be $\mathbb{C}$.

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  • $\begingroup$ Very clear. Thank you. $\endgroup$
    – user536450
    Commented Aug 24, 2022 at 15:52
  • $\begingroup$ @user536450 If you accept this as an answer--and you should--then press the checkmark to the left of the answer, and you can optionally +1 the answer as well. $\endgroup$ Commented Aug 25, 2022 at 19:14

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