The ultimate goal is to show that every uniformly continuous semigroup on a Banach space must be of the form $(e^{tA})_{t\geq 0}$ for some bounded linear operator $A$. There are many books discussing this but in my view they are not rigorous enough and avoid the problems. For instance I am looking at Pazy, Engel&Nagel and also Vrabie. The problem I have is that in order to show uniqueness one needs to bound differential quotiens appropriately, but in order to do so one needs already the exponential representation. A devils loop.
Let $X$ be a Banach space and denote by $\mathfrak{L}(X)$ the unital Banach algebra of bounded linear operators with the identity $I$ equipped with the operatornorm $||T||=\sup_{||u||\leq1}||Tu||$.
Definition. A family of bounded linear operators $(T_t)_{t\geq0}\subseteq\mathfrak{L}(X)$ indexed by $\mathbb{R}_+=[0;\infty)$ is said to be a semigroup of linear operators or short semigroup if $T_0=I$ and $T_{s+t}=T_s T_t$ for all $s,t\geq0$. A semigroup $(T_t)_{t\geq0}$ is said to be uniformly continuous if $||T_t-I||\rightarrow 0$ if $t\searrow0$ or merely continuous if for all $u\in X$ we have $||T_tu-u||\rightarrow 0$ if $t\searrow0$.
Now I would like to show one part of Exercise 1.15(1) of Engel&Nagel, that is
Show that for every uniformly continuous semigroup $(T_t)_{t\geq0}$ there is $c\geq0$ so that for every $t\in[0;1]$ we have
$$||T_t-I||\leq ct.$$
Of course this would be easy if we have $T_t=e^{tA}$ for some $A\in\mathfrak{L}(X)$, but this uniform bound for the differential quotient above is needed to show this exponential representation. So we need another approach but I have been trying for hours and I couldn't find anything smart in the literature nor in my head.
Using the notion of generator doesn't really help that much.
Definition. To every continuous semigroup $(T_t)_{t\geq0}$ we can associate it's generator $A:D_A\subseteq X\rightarrow X$ defined by
$$D_A=\{u\in X\mid \lim_{t\searrow0}\frac{T_tu-u}{t}\in X\}\text{ and for }u\in D_A\text{ define }Au=\lim_{t\searrow0}\frac{T_tu-u}{t}.$$
We can thus write for $t\in[0;1]$
$$\frac{||T_t-I||}{t}\leq\sup_{||u||\leq1}||\frac{T_tu-u}{t}-Au||+||A||.$$
It's possible to show that $A$ is bounded without having any exponential representaion, so this makes sense to write. However it's not possible to interchange limit and supremum here so there is no hope to bound the first term in the righthand sum above. Since in general the unit disk in a Banach space is not compact there is also not really a hope here, however one could go in the dual and use Banach-Alaoglu, but I don't really see a clear path here. Any better ideas not involving the exponential representation?
Best Answer
For the Exercise 1.15(1) of Engel&Nagel consider the Banach-Steinhaus Theorem.
For showing uniqueness read the proof of Theorem 1.2 of Pazy again.