Let
- $E$ be a $\mathbb R$-Banach space
- $T:[0,\infty)\to E$ be a strongly continuous semigroup
- $A$ be the infinitesimal generator of $T$
- $x\in\mathcal D(A)$
How can we show that $$[0,\infty)\to E\;,\;\;\;t\mapsto T(t)x\tag1$$ is differentiable? It's clear that we have $$\frac{T(t+h)x-T(t)x}h=T(t)\frac{T(h)x-x}h\xrightarrow{h\to0+}T(t)Ax\;,\tag2$$ but I don't know what I need to do with $$\frac{T(t)x-T(t-h)x}h=T(t-h)\frac{T(h)x-x}h\tag3\;.$$ The convergence of the right-hand side of $(3)$ would be clear, if $T$ would be uniformly continuous. How can we proceed in general?
The strong continuity of $t\mapsto T(t)$ from the right implies the uniform operator norm boundedness of $T(t)$ near $0$ by the uniform boundedness principle. This can be used to show that $T(t)$ is also strongly continuous from the left at any $t > 0$ by writing $$ T(t)x-T(t-h)x=T(t-h)\{T(t+h)x-T(t)x\}, $$ and using the uniform operator norm boundedness of $T(t-h)$ for $h$ near $0$.
The domain of $A$ consists of all $x\in X$ such that $t\mapsto T(t)x$ has a right derivative at $0$, and that right derivative is $Ax$. Suppose $x\in\mathcal{D}(A)$. Then, for small positive $h$, \begin{align} &\frac{1}{h}\{T(t)x-T(t-h)x\}-T(t)Ax \\ =& T(t-h)\frac{1}{h}\{T(h)x-x\}-T(t)Ax \\ =& T(t-h)\left[\frac{1}{h}\{T(h)x-x\}-Ax\right] \\ &+\{T(t-h)-T(t)\}Ax. \end{align} The first term on the far right tends to $0$ because $T(t-h)$ is uniformly bounded for all $h > 0$ near $0$, and the term inside square brackets tends strongly to $0$ as $h\downarrow 0$. The second term on the far right tends to $0$ by the strong left-continuity of $T$. Therefore, the left derivative of $T(t)x$ exists for all $t > 0$ and $x\in\mathcal{D}(A)$, and the left derivative equals the right derivative, giving the two-sided limit $$ \lim_{h\rightarrow 0}\frac{1}{h}\{T(t+h)-T(t)x\}=Ax,\;\; t > 0. $$