I'm trying to understand the proof of Theorem 2.5.6 (chapter 2.5) in Amnon Pazy, *Semigroups of Linear Operators and Applications to Partial Differential Equations*, Springer 1983.


For the direction (a) $\implies$ (b) it would be helpful to have a proposition along the lines of:


An analytic semigroup of bounded linear operators on a sector $\Delta\_\delta$ (see definition below) that is uniformly bounded on the non-negative real axis is uniformly bounded on every closed subsector of $\Delta\_\delta$.


Does anyone know of a proof of this proposition? Or do you know a counterexample?


Thank you very much in advance!


For an analytic semigroup, I use the following definition:


Let $\Delta\_\delta := \{ z \in \mathbb{C} : | \arg(z) | < \delta \}$. For $z \in \Delta\_\delta$ let $T(z)$ be a bounded linear operator. The family $\{T(z)\}\_{z \in \Delta\_\delta}$ is an analytic semigroup on $\Delta\_\delta$, if


* $z \to T(z)$ is analytic on $\Delta\_\delta$.
* $T(0) = I$ and $T(z\_1 + z\_2) = T(z\_1) T(z\_2)$ for all $z\_1, z\_2 \in \Delta\_\delta$.
* $\lim\_{\substack{z \to 0 \\z \in \Delta\_{\delta'}}} T(z)x = x$ for all $x \in X$ and $0< \delta' < \delta$.



Does uniform boundedness carry over from the non-negative real axis to closed sectors of $\mathbb{C}$ for analytic semigroups?
There is a one-dimensional counterexample: Consider the analytic semigroup $z \mapsto e^{iz}$. This semigroup is bounded on the non-negative real line, but it is not bounded on any sector $\Delta\_\delta$.


**EDIT** in response to the comments: One can "modify" each analytic semigroup to obtain the following boundedness property: 


Let $0 < \delta' < \delta \le \frac{\pi}{2}$ and let $(T(z))\_{z \in \Delta\_\delta}$ be an analytic semigroup. Then there exists a real number $\omega \ge 0$ such that the rescaled semigroup $(e^{-\omega z} T(z))\_{z \in \Delta\_\delta}$ is bounded on the sector $\Delta\_{\delta'}$. 


This can be seen as follows: Let $\lambda$ be the complex number with modulus $1$ and argument $\delta'$. Then $(T(t\lambda))\_{t \in [0,\infty)}$ and $(T(t\overline{\lambda}))\_{t \in [0,\infty)}$ are $C\_0$-semigroups, so there exists a number $\eta \ge 0$ and a number $M \ge 1$ such that $\|T(t\lambda)\| \le Me^{\eta t}$ and $T(t\overline{\lambda}) \le M e^{t\eta}$ for all $t \in [0,\infty)$.


Choose $\omega = \frac{\eta}{\operatorname{Re} \lambda}$. Let $z$ be a number on the boundary of the sector $\Delta\_{\delta'}$. Then $z$ is of the form $z = t \lambda$ or $z = t\overline{\lambda}$ for a real number $t \ge 0$. Hence,
$\|e^{-\omega z} T(z)\| \le e^{-\omega t \operatorname{Re} \lambda} M e^{\eta t} = M$. Since the sector $\Delta\_{\delta'}$ is contained in the convex hull its boundary, we conclude that $\|e^{-\omega z} T(z)\| \le M^2$ for all $z \in \Delta\_{\delta'}$. Hence, the semigroup $(e^{-\omega z}T(z))\_{z \in \Delta\_\delta}$ is bounded on the sector $\Delta\_{\delta'}$.


Note however that $M$ and $\omega$ might both depend on $\delta'$. The one-dimensional counterexample above shows that we cannot choose $\omega$ independently of $\delta'$ in general.


