Let $X$ be a Banach space and $A: X \to X$ a bounded linear operator. So, $A$ is the infinitesimal generator of an uniformly continuous semigroup $\{T(t)\}_{t\geq 0}$ on $X$. The proof, as presented in Pazy's book, is to define $T(t)=e^{At}$ and verify the conditions.
After, in the same book, in order to prove that there exists an unique semigroup whose generator is $A$, is proved the following:
Theorem: Let $T(t)$ and $S(t)$ be uniformly continuous semigrops of bounded linear operators. If
$$\lim_{t\to 0^+}\left\|\frac{T(t)-I}{t}-A\right\|=0=\lim_{t\to 0^+}\left\|\frac{S(t)-I}{t}-A\right\|\tag{#}$$
then $T(t)=S(t)$ for $t\geq 0$.
Since the infinitesimal generator of a semigroup is an operator $A$ that satisfies
$$Ax=\lim_{t\to 0^+}\frac{T(t)x-x}{t},$$
why shouldn't we replace $(\#)$ by $(*)$?
$$\lim_{t\to 0^+}\left\|\frac{T(t)x-x}{t}-Ax\right\|_X=0=\lim_{t\to 0^+}\left\|\frac{S(t)x-x}{t}-Ax\right\|_X\;\forall x\in X\tag{*}$$
Are in this case $(\#)$ and $(*)$ equivalent?
Thanks.
Best Answer
The conditions are equivalent in the case that $A$ is bounded. Normally, though, where $A$ may be unbounded with a domain which is not all of $X$, you don't get uniform operator norm convergence because that would imply $A$ is bounded. The strong (vector) convergence is definitely implied by the uniform because $\|Ax\|\le \|A\|\|x\|$ for all $x\in X$ if $A$ is bounded. It's less obvious--but true--that strong convergence implies uniform convergence if the generator $A$ is bounded (this has to do with uniqueness of $C^{0}$ semigroups given the generator, and with the fact that you know how to construct one for a bounded $A$.)
This edit is to flesh out my comment that I made to you. Here's what I was saying you could prove with only minor modification to Pazy's argument:
Theorem: Let $X$ be a Banach space. Let $T : [0,\infty)\rightarrow\mathscr{L}(X)$ be a semigroup. Suppose that there exists $A\in\mathscr{L}(X)$ such that the following limits exist for all $x \in X$: $$ \lim_{t\downarrow 0}\left\|\frac{1}{t}\{T(t)x-x\}-Ax\right\|_{X}=0. $$ Then $T(t)=e^{tA}$. Therefore, for such $T$, one has $$ \lim_{t\downarrow 0}\left\|\frac{1}{t}\{T(t)x-x\}-Ax\right\|_{\mathscr{L}(X)}=0, $$ because the above holds for $e^{tA}$.
Proof: For each $x$, show that $S(t)x=e^{-tA}T(t)x$ has right derivative 0 for all $t \ge 0$. Conclude that $S(t)x=x$ for all $t \ge 0$, which means $e^{-tA}T(t)=I$ for all $t \ge 0$. Multiply on the left by $e^{tA}$ to conclude that $T(t)=e^{tA}$. $\Box$