Analytic semigroups are operator-norm differentiable away from zero

$C^{\mathrm{\infty }}$ regularity and the first derivative. By [L1] the map $t\mapsto T(t)$ has norm derivatives of every order on $(0,\mathrm{\infty })$ and $\frac{d^m}{d{t}^m}T(t)=A^{m}T(t)$ for every $m\ge 1$; taking $m=1$ gives $\frac{d}{dt}T(t)=AT(t)$ as bounded operators, and taking all $m$ gives the $C^{\mathrm{\infty }}$ statement in the operator norm.

The vertex and bounded generators. If $\mathrm{\parallel }T(t)-I\mathrm{\parallel }\to 0$ as $t\downarrow 0$, choose $h>0$ with ${\sup }_{0\le s\le h}\mathrm{\parallel }T(s)-I\mathrm{\parallel }<1\mathrm{/}2$. By [L2], the bounded operator $K_h:=J_h\mathrm{/}h$ satisfies $\mathrm{\parallel }{K}_h-I\mathrm{\parallel }\le {\sup }_{0\le s\le h}\mathrm{\parallel }T(s)-I\mathrm{\parallel }<1\mathrm{/}2$ and is invertible. Since its range lies in $D(G)$, this forces $D(G)=X$, and $G=(T(h)-I)K_h^{-1}\mathrm{/}h$ is bounded. Thus an unbounded generator cannot have norm continuity at the vertex. Step 1.1 gives the asserted positive-time regularity, and this argument proves the general exclusion at zero rather than inferring it from one heat example.