# self-adjoint bounded generates analytic semigroup

Engel Nagel A Short Course on Operator Semigroups Corollary II.4.8 states: (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.?

• There is a typo (or, more precisely, $\mathrm{arg}(0)$ is not well-defined). It should be $\sigma(A)\subseteq \{z\in\mathbb{C}: |\mathrm{arg}(-z)|<\delta\}\cup \{0\}$. – MaoWao Jan 18 at 12:50
• 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! – user41467 Jan 18 at 13:29
• 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 $\delta'\in (\delta,\pi/2)$. – MaoWao Jan 18 at 13:33
• 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. – user41467 Jan 18 at 13:40