Does uniform boundedness carry over from the non-negative real axis to closed sectors of $\mathbb{C}$ for analytic semigroups? 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 $0< \delta' < \delta$.

 A: 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.
