The analytic semigroup generated by a bounded operator

The exponential and its generator. By [L1] the series $E(z)$ converges for every complex $z$ (the scalar series $\sum \mathrm{\mid }z{\mathrm{\mid }}^n\mathrm{\parallel }A{\mathrm{\parallel }}^n\mathrm{/}n!$ bounds it), its derivative series converges uniformly on $\mathrm{\mid }z\mathrm{\mid }\le R$ since it is bounded by $\mathrm{\parallel }A\mathrm{\parallel }{e}^{R\mathrm{\parallel }A\mathrm{\parallel }}$. For the difference quotients the remainder after the linear term is bounded by $\mathrm{\mid }h\mathrm{\mid }\mathrm{\parallel }A{\mathrm{\parallel }}^2{e}^{(R+\mathrm{\mid }h\mathrm{\mid })\mathrm{\parallel }A\mathrm{\parallel }}\mathrm{/}2$, so $E^{\mathrm{\prime }}(z)=AE(z)$ and $E$ is entire. Absolute convergence permits regrouping the double product series; the binomial identity then gives $E(z+w)=E(z)E(w)$, and its restriction $T$ is a uniformly continuous strongly continuous semigroup with $(T(t)-I)\mathrm{/}t\to A$, so its generator is the bounded operator $A$ with $D(A)=X$; [L6] identifies $T$ with any other strongly continuous semigroup having generator $A$.

Unboundedness examples. For $A=\left(\begin{array}{cc}0& 1\\ 0& 0\end{array}\right)$ on ${\mathbb{C}}^2$ the series terminates: $A^0=I$, $A^1=A$ and $A^n=0$ for $n\ge 2$, so $E(z)=I+zA$; on the unit vector $e_2$ one has $E(z)e_2=(z,1)$ with Euclidean norm $\sqrt{1+\mathrm{\mid }z{\mathrm{\mid }}^2}$, so $\mathrm{\parallel }E(z)\mathrm{\parallel }\ge \sqrt{1+\mathrm{\mid }z{\mathrm{\mid }}^2}$ by [L3] and the operator norm is unbounded along every ray; for $A=1$ on $\mathbb{C}$ the series gives $E(z)=e^z$ with $\mathrm{\mid }E(z)\mathrm{\mid }=e^{\mathrm{Re}z}$, which is unbounded on every sector that meets the open right half-plane.

The resolvent estimate remains a separate condition. For bounded $A$, the Neumann-series estimate of [L3] controls $R(\lambda ,A)$ when $\mathrm{\mid }\lambda \mathrm{\mid }>2\mathrm{\parallel }A\mathrm{\parallel }$, but it does not control behavior as $\lambda \to 0$. The sectorial resolvent condition in [L2] therefore retains the separate $O(\mathrm{\mid }\lambda {\mathrm{\mid }}^{-1})$ bound on every smaller sector; spectrum avoidance alone is not used to infer it.

The sectorial extension criterion. By [L2], for bounded $A$ the sectorial resolvent condition with exponent $\delta$ is exactly the conjunction of ${\mathrm{\Sigma }}_{\pi \mathrm{/}2+\delta }\subseteq \rho (A)$ and the resolvent bounds on the smaller sectors; step 1.3 explains why the bound at the vertex must remain explicit. The same characterization makes this condition equivalent to generation of a bounded analytic semigroup of angle $\delta$. By [step 1.1], $T$ is the semigroup generated by $A$, so it has such an extension exactly under that sectorial condition; the spectrum formulation in (1) is just the equivalent resolvent-set clause together with the bound.

The numerical range case. Assume $X$ is a complex Hilbert space and $W(A)$ lies in the closed sector $C_{\theta }:=\{\mu :\mathrm{\mid }\arg (-\mu )\mathrm{\mid }\le \pi \mathrm{/}2-\theta \}\cup \{0\}$ with $\theta \in (0,\pi \mathrm{/}2)$; for $\lambda \in {\mathrm{\Sigma }}_{\pi \mathrm{/}2+\theta -\epsilon }$ and $u\mathrm{\ne }0$ the normalised value $z_u:=⟨Au,u⟩\mathrm{/}\mathrm{\parallel }u{\mathrm{\parallel }}^2$ lies in $C_{\theta }$ by [L4] and the angular distance from $\lambda$ to $C_{\theta }$ is at least $\epsilon$, so $\mathrm{\mid }\lambda -z_u\mathrm{\mid }\ge \mathrm{\mid }\lambda \mathrm{\mid }\sin \epsilon$ and Cauchy-Schwarz gives $\mathrm{\parallel }(\lambda I-A)u\mathrm{\parallel }\ge \mathrm{\mid }\lambda \mathrm{\mid }\sin \epsilon \mathrm{\parallel }u\mathrm{\parallel }$; hence $\lambda I-A$ is injective with closed range and, since for $\mathrm{\mid }\lambda \mathrm{\mid }>\mathrm{\parallel }A\mathrm{\parallel }$ it is invertible by [L3] and the set of surjectivity points is open (Neumann series) and closed in the connected complement of $C_{\theta }$: if ${\lambda }_n\to \lambda$ there, the positive distance of $\lambda$ to $C_{\theta }$ bounds $\mathrm{\parallel }R({\lambda }_n,A)\mathrm{\parallel }$ uniformly for large $n$, and the resolvent identity makes these inverses Cauchy in operator norm. Their limit $R$ satisfies $(\lambda I-A)R=R(\lambda I-A)=I$ by boundedness of $A$. Thus surjectivity is closed, every such $\lambda$ is in $\rho (A)$ with $\mathrm{\parallel }R(\lambda ,A)\mathrm{\parallel }\le 1\mathrm{/}(\mathrm{\mid }\lambda \mathrm{\mid }\sin \epsilon )$; by [step 2.1] the sectorial condition with exponent $\theta$ holds and $T$ extends to a bounded analytic semigroup on ${\mathrm{\Sigma }}_{\theta }$, so its maximal analytic angle is at least $\theta$.