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Let $\mathbb H$ be a Hilbert space and let $A\in \mathcal B(\mathbb H)$ such that the spectrum of $A$ does not meet a closed half-line issued from 0 in the complex plane. Then I guess that $ A=\exp L $ where $L\in \mathcal B(\mathbb H)$.

Question 1. Is there a reference for this result?

Question 2. Is there a Banach space version?

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  • $\begingroup$ A simple counterexample is the sequence $(\frac 1n)$ (regarded as a multiplication operator on $\ell^2$). What you require is that the spectrum be contained in a set on which the logarithm function is bounded (i.e. bounded away from $0$). $\endgroup$
    – terceira
    Commented Nov 30, 2023 at 13:20
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    $\begingroup$ @terceira Doesn't the spectrum of your operator contain $0$, and therefore violate the condition about not meeting a half-line issued from $0$? $\endgroup$
    – Nik Weaver
    Commented Nov 30, 2023 at 13:22
  • $\begingroup$ @Nik Weaver. Yes, you're right,of course. Sorry. $\endgroup$
    – terceira
    Commented Nov 30, 2023 at 15:23

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The other answer is not correct. If $L$ is the half-line issuing from the origin, then we can find a branch of the logarithm that is holomorphic on $\mathbb{C}\setminus L \supset{\rm spec}(A)$. Then applying the holomorphic functional calculus to $A$ yields the desired operator $B$ with $e^B = A$. And yes, this holds for Banach space operators too (one has holomorphic functional calculus in every Banach algebra).

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    $\begingroup$ Essentially, I aggree. I was fooled by the ambiguity of the question. It mentions a half-line, which could be open (my answer) or closed (yours). $\endgroup$
    – Denis Serre
    Commented Nov 30, 2023 at 14:40
  • $\begingroup$ (The other answer I referred to has been deleted.) $\endgroup$
    – Nik Weaver
    Commented Dec 1, 2023 at 12:35
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There is a Banach algebra version of the result (including the Banach space version) in Theorem 10.30 (page 264) of W. Rudin's book Functional analysis, 2nd ed. McGraw-Hill 1991.

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  • $\begingroup$ One of my all time favorite books. $\endgroup$
    – Nik Weaver
    Commented Nov 30, 2023 at 19:28

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