If $T$ is a strongly continuous semigroup, then $t\mapsto T(t)x$ is differentiable, for all $x$ 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?
 A: 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
$$
         \lim_{h\rightarrow 0}\frac{1}{h}\{T(t+h)-T(t)x\}=Ax,\;\; t > 0.
$$
