# Logarithm of a bounded operator

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?

• 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 $0$). Nov 30, 2023 at 13:20
• @terceira Doesn't the spectrum of your operator contain $0$, and therefore violate the condition about not meeting a half-line issued from $0$? Nov 30, 2023 at 13:22
• @Nik Weaver. Yes, you're right,of course. Sorry. Nov 30, 2023 at 15:23

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).