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.
2026-04-12 09:45:39.1775987139
On the spectrum of the closure of a closable operator in a Hilbert space
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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 $\mathbb{C}$.