Given a Banach space $E$.

Consider a C0-semigroup: $$T:\mathbb{R}_+\to\mathcal{B}(E)$$

Define its generator by: $$Ax:=\lim_{h\downarrow0}\frac{1}{h}(T(h)x-x)\in E$$ (It is a densely-defined closed operator.)

Denote the convergence radius by: $$\rho_x:=\left(\limsup_{k\to\infty}\sqrt[k]{\frac{1}{k!}\|A^kx\|}\right)^{-1}$$

Generate a semigroup via Taylor series: $$\rho_x=\infty:\quad e^{tA}x:=\sum_{k=0}^\infty\frac{1}{k!}t^kA^kx$$

Can it happen that it has no entire vectors at all?


This is the start-up for: Semigroups: Entire Vectors (II)

It is taken from: Engel & Nagel, Exercise 3.12, Page 81

  • $\begingroup$ Could you elaborate on the domain of definition of (the generated semi-group) $T(t)$ and the definition of a locally analytic element? $\endgroup$ – Jonas Dahlbæk Dec 15 '14 at 16:06
  • $\begingroup$ @user161825: Yes sure, any element in the domain of all powers give rise to a radius of convergence: $x\in\mathcal{D}^\infty(A):\quad|z|<R_x\iff\sum_{k=0}^\infty\frac{1}{k!}z^kA^kx \in E$ Call an element $x\in\mathcal{D}^\infty(A)$ locally analytic if $R_x>0$ and globally analytic if $R_x=\infty$. $\endgroup$ – C-Star-W-Star Dec 15 '14 at 16:16

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