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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?

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    Just to clarify: Which notion of differentialbility are you using? – 2017-01-19
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    @Daniel *Differentiable* has to be understood in the sense that the difference quotiens have a limit in the sense of norm convergence here. $(2)$ shows that the right difference quotient has the limit $T(t)Ax$; I just don't know how I can show that the left-difference quotient $(3)$ has the same limit (or a limit at all). – 2017-01-19

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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 $$ \lim_{h\rightarrow 0}\frac{1}{h}\{T(t+h)-T(t)x\}=Ax,\;\; t > 0. $$

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    I've just figured it out. In fact, there is a $(C,\omega)\in[1,\infty)\times\mathbb R$ with $$\left\|T(t+h)\right\|\le Ce^{\omega (t+h)}\le Ce^{\omega t}\;\;\;\text{for all }t>0\text{ and }h\in[-t,0)\;.$$ – 2017-01-19
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    Can we use the same argument in [my other question](http://math.stackexchange.com/questions/2104514/if-s-is-a-strongly-continuous-semigroup-on-a-hilbert-space-h-and-u-in-c1)? – 2017-01-19