# Sectorial operator: $0\in \rho(A)$ or $0\not\in \rho(A)$?

I am confused with a characterisation of the infinitesimal generators that generates analytic semigroups.

In the following characterisation do we or do we not need the origin (or if the sector is shifted then the 'origin' of the sector) to be in the resolvent set? Why is this happening?

The question is that's it, and below I will spell out my confusion.

Pazy's book Theorem 2.5.2 states,

Let $$T(t)$$ be a uniformly bounded $$C_0$$ semigroup. Let $$A$$ be the infinitesimal generator of $$T(t)$$ and assume $$0\in\rho(A)$$. The following statements are equivalent:

(a) $$T( t)$$ can be extended to an analvtic semigroup in a sector $$\Delta_\delta =\{ z : | \arg z | < \delta \}$$ and $$\|T(z)\|$$ is uniformzy bounded in every closed sub-sector $$\overline{\Delta_{\delta'}}$$, $$\delta'<\delta$$ of $$\Delta_\delta$$.

(c) There exist $$0<\delta<\pi/2$$ and $$M>0$$ such that $$\rho ( A ) \supset \Sigma = \left\{ \lambda : | \arg \lambda | < \frac { \pi } { 2 } + \delta \right\} \cup \{ 0 \}$$ $$\| R ( \lambda : A ) \| \leq \frac { M } { | \lambda | } \quad \text { for } \quad \lambda \in \Sigma ,\ \lambda \neq 0$$

He explicitly states $$0$$ is in the resolvent set of $$A$$. But also note (c) does not require an estimate of $$A^{-1}$$.

Also his notation indicates that $$0\not\in\Sigma$$.

On the other hand, Renardy--Rogers book Theorem 12.31 states

A closed, densely defined operator $$A$$ in $$X$$ is the generator of an analytic semigroup if and only if there exists $$\omega\in\mathbb{R}$$ such that the half-plane $$\mathrm{Re}(\lambda)>\omega$$ is contained in the resolvent set of $$A$$ and, moreover, there is a constant $$C$$ such that $$\left\| R _ { \lambda } ( A ) \right\| \leq C / | \lambda - \omega |\qquad (12.65)$$ for $$\mathrm{Re}(\lambda)>\omega$$. If this is the case, then actually the resolvent set contains a sector $$|\arg (\lambda -\omega)|<\pi/2+\delta$$ for some $$\delta > 0$$, and an analogous resolvent estimate holds in this sector.

In particular for $$\omega=0$$, $$0$$ does not have to be in the resolvent set (I am assuming "contains" does not mean as a proper subset).

Kato's book is even more confusing. Theorem IX.1.24 states

Let $$T$$ be an $$m$$-sectorial operator in a Hilbert space $$H$$ with a vertex $$0$$ (so that its numerical range $$\Theta(T)$$ is a subset of a sector $$|\arg \zeta | \leqq \frac { \pi } { 2 } - \omega , 0 < \omega \leqq \frac { \pi } { 2 } )$$. Then, $$e^{-tT}$$ is analytic.

But since the numerical range is $$\Theta(T)=\{(Tu,u)\mid \|u\|=1,\, u\in D(T)\},$$ this seems to exclude the case where $$T$$ is non-negative but not positive definite, i.e., the case where $$(Tv,v)=0$$ can happen for $$v\neq 0$$.

But apparently it does not exclude this case, as Example IX.1.25. says if $$H$$ nonnegative (not "positive definite") self-adjoint operator then $$-H$$ generates an analytic semigroup.

Then there is the shifted version, which says consider the situation where the spectrum of $$A$$ satisfies $$\sigma ( A ) \subset \beta + \Sigma _ { \omega } , \quad - \infty < \beta < \infty , 0 < \omega < \frac { \pi } { 2 },$$ then consider the estimate $$\left\| ( \lambda - A ) ^ { - 1 } \right\| \leq \frac { M _ { \omega } } { | \lambda - \beta | } , \quad \lambda \notin \beta + \Sigma _ { \omega }$$ In the assumption, implicitly $$\beta$$ is in the resolvent set so basically it is the same as Pazy.

The generator of a bounded analytic semigroup (if you accept that as definition of sectorial operator) can have $$0$$ in the spectrum. For example, the Laplacian on $$L^2(\mathbb{R}^n)$$ generates a bounded analytic semigroup and has spectrum $$[0,\infty)$$.
Now to the different versions. Pazy makes the "global" assumption that $$0\in \rho(A)$$. I don't know why, but the equivalence is certainly true under this additional assumption. Renardy-Rogers give the definition without the condition $$0\in \rho(A)$$. Kato's statement is a little trickier since $$\mathrm{arg}(0)$$ is not well-defined (maybe it's defined somewhere in the book, but I could not find it). You just have to interpret it correctly, i.e., in such a way that it allows for $$0\in\Theta(T)$$. I could not find the shifted version you mention (nor the notation $$\Sigma_\omega$$) in Kato's book, but one should have $$0\in \Sigma_\omega$$ to make this correct.
• I don't have access to Yagi's book, but the preview shows that he does define sectorial operators in such a way that $0\in \rho(A)$. – MaoWao Jan 18 at 13:29
• Actually on page 55 he says "The condition (2.1) '$\sigma ( A ) \subset \Sigma _ { \omega } = \{ \lambda \in \mathbb { C } ; | \arg \lambda | < \omega \} , \quad 0 < \omega \leq \pi$' implicitly means that the origin is not in $\sigma(A)$, namely, $A$ has a bounded inverse $A^{−1}$ on $X$" – user41467 Jan 18 at 13:33