How do we need to apply the uniform boundedness principle to show that a strongly continuous semigroup is locally bounded?

Question

Let $E$ be a $\mathbb R$-Banach space and $(T(t))_{t\ge0}\subseteq\mathfrak L(E)$ be a semigroup. Note that $$E_0:=\{x\in E:\|T(h)x-x\|_E\xrightarrow{h\to0+}0\}$$ is a closed subspace of $E$. I would like to show that there is a $t_0>0$ with $$\sup_{t\in[0,\:t_0]}\sup_{\substack{x\in E_0\\\|x\|_E=1}}\|T(t)x\|_E<\infty\tag1.$$ I've read that this can be shown using the uniform boundedness principle, but how exactly do we need to apply it? Moreover, since $E_0$ is closed (or do we need $\sup_{t\in[0,\:\delta]}\|T(t)\|_{\mathfrak L(E)}<\infty$ for some $\delta>0$ to show that?) and $T(t)E_0\subseteq E_0$ for all $t\ge0$, isn't $(\left.T(t)\right|_{E_0})$ again a semigroup? Since that semigroup would be strongly continuous by the very definition of $E_0$, we could reduce the claim to the claim that for every strongly continuous semigroup $(S(t))_{t\ge0}$ on $E$, there is a $t_0>0$ with $$\sup_{t\in[0,\:t_9]}\|S(t)\|_{\mathfrak L(E)}<\infty\tag2.$$ (While that doesn't simply the proof, it does at least simplify the notation.)

Answer 1

Here is a proof in the basic situation where $(S(t))_{t\geq 0}$ is a strongly continuous semigroup on $E$.

Suppose on the contrary that, for every $t>0$, $\sup_{s\in[0,t]}\|S(s)\|=\infty$. Then, for every $n\in\mathbb{N}$ there exists $t_{n}\in[0,1/n]$ such that $\|S(t_{n})\|\geq n$. Now, consider the family of operators \begin{equation} \{S(t_{n}) \mid n\in\mathbb{N}\} \end{equation} and note that, for every $x\in E$, there is $\sup_{n\in\mathbb{N}}\|S(t_{n})x\|<\infty$ because $(S(t))_{t\geq 0}$ is strongly continuous. It then follows that $\sup_{n\in\mathbb{N}}\|S(t_{n})\|<\infty$ by the principle of uniform boundedness but this is a contradiction because $n\leq\|S(t_{n})\|$ for every $n$. Consequently, there exists $t=t_{0}>0$ so that $\sup_{s\in[0,t_{0}]}\|S(s)\|<\infty$ as required.