Understanding the definition of the generator of a semigroup of operators Given a strongly continuous semigroup $T : \mathbb{R}_+ \to L(B)$ on a Banach space $B$, its infinitesimal generator $A$ of a strongly continuous semigroup $T$ is defined as a mapping $B \to B$ as
$$
    A\,x = \lim_{t\downarrow0} \frac1t\,(T(t)- I)\,x , \forall x \in B
$$
whenever the limit exists wrt the norm on the Banach space $B$. Let $D(A)$ be the set of $x$'s in $B$, where the limit exists.


*

*Can the definition of $A$ be rewritten as the right-derivative of
$T: \mathbb{R}_+ \to L(B)$ at $t=0^+$, wrt some norm
$\|\cdot\|_{L(D(A))}$on $L(D(A))$, as $$
        \lim_{t\downarrow0} \frac1t\,\|T(t)- I - tA\|_{L(D(A))} = 0? $$ What is the  norm $\|\cdot\|_{L(D(A))}$on $L(D(A))$ then?

*Can the generator "generate" back the one-parameter semigroup of operators? I was wondering why $A$ is called a "generator"? What
can the generator generate?
Can the generator $A$  "generate" back the  one-parameter semigroup
of operators?
References are also appreciated!
Thanks and regards!
 A: Ad 1.: I don't think so. You would at least have to include $A$ into the norm, since otherwise the domain of the right-hand side is $\mathbb{R}$. But even so, that derivative with respect to the norm topology should be stronger than the one with respect to the strong operator topology, at least in the infinite-dimensional case.
Edit: Here is an example that the two are not equivalent: Consider $T_t: L^2(\mathbb{R}) \rightarrow L^2(\mathbb{R}),\, T_t f(x) = f(x-t)$. Then $t \mapsto T$ defines a strongly continuous semigroup. The limit of $(T_t f - f)/t$ for $t \to 0$ exists for instance for $C^1$-functions with bounded support, which form a dense subset of $L^2(\mathbb{R})$. Let $Af$ be the limit where it exists. If you let $f_t \neq 0$ be a $C^1_c$-function with support in $[0,t]$ then
$$\|T_t f_t - f_t\|_{L^2} = \sqrt{2} \|f_t\|_{L^2},$$
so $\|T_t - I\|_{L(D(A))} \geq \sqrt{2}$. In particular, $T_t$ cannot be differentiable with respect to the norm topology.
Ad 2.: Doesn't the next paragraph in that Wikipedia article answer that question? If and only if $A$ is an infinitesimal generator of a strongly continuous semigroup, the semigroup can be pointwisely recovered as the unique mild solution to the Cauchy problem.
