Cauchy estimates for an analytic semigroup give generator power bounds

The first-order identity. Fix $t>0$ and $y\in X$. For $h>0$ small, the semigroup law [L1] with $t,t+h\in {\mathrm{\Sigma }}_{\delta }$ gives $h^{-1}(T(h)-I)T(t)y=h^{-1}(T(t+h)y-T(t)y)$, and the right-hand side converges as $h\downarrow 0$ to the complex derivative $T^{\mathrm{\prime }}(t)y$ of the holomorphic map $z\mapsto T(z)y$ at $t$, because that derivative exists in operator norm by [L1]; hence by [L2] $T(t)y\in D(A)$ and $AT(t)y=T^{\mathrm{\prime }}(t)y$.

The inductive identity. Assume $T(s)X\subseteq D(A^m)$ and $A^{m}T(s)=T^{(m)}(s)\in \mathcal{B}(X)$ for all $s>0$. Fix $t>0$ and $y\in X$, and put $x:=A^{m}T(t)y=T^{(m)}(t)y$. For $h>0$, the semigroup law [L1] gives $T(h)T(t+s)=T(t+h+s)$ for real $s$ near $0$; differentiating this identity $m$ times in operator norm, justified by [L3, L4], gives $T(h)T^{(m)}(t)y=T^{(m)}(t+h)y$. Therefore $\frac{T(h)x-x}{h}=\frac{T^{(m)}(t+h)y-T^{(m)}(t)y}{h}⟶T^{(m+1)}(t)y(h\downarrow 0)\mathrm{.}$ By the generator definition [L2], $x\in D(A)$ and $Ax=T^{(m+1)}(t)y$. Since $T(t)y\in D(A^m)$ by the induction hypothesis and $A^{m}T(t)y=x\in D(A)$, the recursive definition of powers gives $T(t)y\in D(A^{m+1})$ and $A^{m+1}T(t)y=T^{(m+1)}(t)y$. As $y$ was arbitrary, $T(t)X\subseteq D(A^{m+1})$ and $A^{m+1}T(t)=T^{(m+1)}(t)\in \mathcal{B}(X)$.

Conclusion of the identity. [step 1.1] is the case $m=1$ and [step 2.1] carries every higher $m$, so $T(t)X\subseteq D(A^m)$ and $T^{(m)}(t)=A^{m}T(t)\in \mathcal{B}(X)$ for every $t>0$ and $m\ge 1$.

The Cauchy estimate. Fix $t>0$ and $0<{\delta }^{\mathrm{\prime }\mathrm{\prime }}<{\delta }^{\mathrm{\prime }}<\delta$. Choose $r:=t\sin {\delta }^{\mathrm{\prime }\mathrm{\prime }}$ and $R$ with $r<R<t\sin {\delta }^{\mathrm{\prime }}$. The closed disc $\overline{D(t,R)}$ lies in ${\mathrm{\Sigma }}_{{\delta }^{\mathrm{\prime }}}$: its radius is smaller than the distance $t\sin {\delta }^{\mathrm{\prime }}$ from $t$ to either boundary ray, and $R<t$ keeps it away from the vertex. In particular the circle $\mathrm{\mid }w-t\mathrm{\mid }=r$ lies in ${\mathrm{\Sigma }}_{{\delta }^{\mathrm{\prime }}}$, where [L1] bounds $\mathrm{\parallel }T(w)\mathrm{\parallel }$ by $M_{{\delta }^{\mathrm{\prime }}}$. Applying the Cauchy estimate [L3] to the $\mathcal{B}(X)$-valued holomorphic map $F=T$ at $z_0=t$ gives $\mathrm{\parallel }{T}^{(m)}(t)\mathrm{\parallel }\le m!{\sup }_{\mathrm{\mid }w-t\mathrm{\mid }=r}\mathrm{\parallel }T(w)\mathrm{\parallel }\mathrm{/}{r}^m\le m!M_{{\delta }^{\mathrm{\prime }}}\mathrm{/}(t\sin {\delta }^{\mathrm{\prime }\mathrm{\prime }}{)}^m$, and with [step 3.1] this is $\mathrm{\parallel }{A}^{m}T(t)\mathrm{\parallel }\le m!M_{{\delta }^{\mathrm{\prime }}}\mathrm{/}(t\sin {\delta }^{\mathrm{\prime }\mathrm{\prime }}{)}^m$.