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Engel Nagel A Short Course on Operator Semigroups Corollary II.4.8 states: enter image description here (There should be a typo. If $\delta=0$ then the spectum is empty, but normal operator has a non-empty spectrum? Anyways,)

Then they proceed,

In particular, Corollary 4.8 shows that the semigroup generated by a self-adjoint operator A that is bounded above, which means that there exists $w\in\mathbb{R}$ such that $$ ( A x | x ) \leq w \| x \| ^ { 2 } \quad \text { for all } x \in D ( A ) $$ is analytic of angle $\pi/2$. Moreover, this semigroup is bounded if and only if $w\leq 0$.

I do not see how Corollary 4.8 implies this. If $w=0$, then $(Ax|x)\leq 0$, so it is still possible that $0\in \sigma(A)$. But $0\not\in \{z\in\mathbb{C}\mid |\arg(-z)|<\delta\}$.

Are we meant to consider something like $A+w$, consider ${e}^{t(A+w)}$ and then say ${e}^{tA}$ is well-defined further ${e}^{t(A+w)}={e}^{tA}e^{tw}$ etc.?

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  • $\begingroup$ There is a typo (or, more precisely, $\mathrm{arg}(0)$ is not well-defined). It should be $\sigma(A)\subseteq \{z\in\mathbb{C}: |\mathrm{arg}(-z)|<\delta\}\cup \{0\}$. $\endgroup$ – MaoWao Jan 18 at 12:50
  • $\begingroup$ Ah, I thought it should be $|\arg(-z)|\leq \delta$ so that for $\delta=0$ we have self-adjoint negative definite. Your corrected assumption makes the result stronger than my version, which I guess is related to an answer of yours (Pazy's $0\in \rho(A)$ is not needed). Thanks! $\endgroup$ – user41467 Jan 18 at 13:29
  • $\begingroup$ You have to clarify whether $0\in \sigma(A)$ is allowed or not. For every $z\neq 0$ it does not matter if there is $<$ or $\leq$, you can always replace $\delta$ by $\delta'\in (\delta,\pi/2)$. $\endgroup$ – MaoWao Jan 18 at 13:33
  • $\begingroup$ I was looking at Yagi before this and thought $\sigma(A)\subseteq {|\arg(−z)|\leq \delta}$ implicitly excludes $0$. But now I see $0\in\sigma(A)$ is fine. Thanks for the discussion on this typo. $\endgroup$ – user41467 Jan 18 at 13:40

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