For a semigroup $\{T(t)\}_{t \geq 0}$, $x=0$ is the only entire element if $T(s)=0$ for some $s>0$

Question

This question is asked in a PDE course, but is primarily focused on semigroup theory.

On Banach space $X$, let $\{T(t)\}_{t \geq 0}$ be a strongly continuous semigroup with generator $A:D(A) \subset X \rightarrow X$.

We call an element $x \in \cap_{k=1}^{\infty} D(A^k)$ entire if $\sum_{n=0}^{\infty}\frac{|t|^n}{n!}\|A^n x\|_X < \infty$ for all $t \in \mathbb{R}$.

($D(A^k)=\{x \in D^{k-1} \ | \ A^{k-1}x\in D(A)\}$, inductively defined for integers $k\geq2$.)

Show that $x=0$ is the only entire element if $T(s)=0$ for some $s>0$.

Questions which I've already proven before this question:

• I've already proven that $T(t)x=\sum_{k=0}^{\infty}\frac{t^k}{k!}A^kx$ for all $t \geq 0$ and $x\in X$ entire.

• Also I've proven that if $x \in D(A^k)$, then $T(t)x \in D(A^k)$ for all $t\geq 0$.

I don't know whether these are relevant, but I posted it anyway. I keep trying to use continuity, or linearity or the semigroup properties to derive a contradiction, but it won't work. Can someone please help me? It bothers me that I can't seem to solve it when it looks kinda simple.

Answer 1

Let me just expand the answer by Tim.

Given a Banach space $E$.

Consider a semigroup: $$T:\mathbb{R}_+\to\mathcal{B}(E):\quad T(t+t')=T(t)T(t')$$

By the assumption: $$T(t_0)=0\quad(t_0>0)\implies T(t)=0\quad(t\geq t_0)$$

Regard the entire functions: $$F_\varphi(\zeta):=\sum_{k=0}^\infty\frac{1}{n!}\varphi\left(A^nx\right)\zeta^n\quad(\varphi\in E')$$

By the identity theorem: $$F_\varphi(t)=0\quad(t\geq t_0)\implies F_\varphi=0$$

By Cauchy's theorem: $$F_\varphi=0\implies\varphi(A^kx)=0\quad(k\in\mathbb{N}_0)$$

By Hahn-Banach: $$\varphi(1x)\quad(\varphi\in E')\implies 1x=0$$

Concluding the assertion.

Answer 2

Okay, so I found a answer to my own question. I however don't know whether this is the right way to post it. Correct me if I'm wrong.

Anyway, here is the answer:

$x=0$ is entire, which is easy to prove.

Assume $x \neq0$ is entire.

$T(s)=0$ for some $s>0$. Hence for all $t \geq s$: $T(t)=T(t-s+s)=T(t-s)T(s)=T(t-s)0=0$.

So for all $t \geq s: \, T(t)x=0=\sum_{n=0}^{\infty}\frac{t^n}{n!}A^n x $ (1).

Note $x \neq 0$ and $t>0$. Now we look at (1) as a 'polynomial of degree $\infty$' which has uncountable many zeros. This is either a contradiction with the number of zeros a polynomial can have (main theorem of algebra), or we have that for all $n \in \mathbb{N}$: $A^n x=0$. Which is to say that $y=0= \cap_{k=1}^{\infty} D(A^k)$. Hence 0 is the only entire element.