Let
- $E$ be a $\mathbb R$-Banach space
- $(T(t))_{t\ge0}$ be a semigroup on $E$
- $x\in E$ with $$\left\|T(h)x-x\right\|_E\xrightarrow{h\to0+}0\tag1$$ (i.e. $(T(t))_{t\ge0}$ is strongly continuous at $x$)
In many books, we can find that $$\frac1h\int_t^{t+h}T(s)x\:{\rm d}s=T(t)x\tag2\;\;\;\text{for all }t\ge0,$$ but why does the Riemann integral on the left-hand side of $(2)$ exist at all?
By $(1)$, the only thing we shoulbe be able to conclude is that $$\operatorname{orb}x:[0,\infty)\to E\;,\;\;\;t\mapsto T(t)x$$ is right-continuous, but in order to show that $(2)$ holds, it seems like that we need that $\operatorname{orb}x$ is continuous. As far as I'm not mistaken, this cannot generally be shown.
What am I missing?
EDIT: By $(1)$, we easily see that $\operatorname{orb}x$ is right-continuous, since if $t\ge0$, then $$\|(\operatorname{orb}x)(t+h)-(\operatorname{orb}x)(t)\|_E\le\|T(t)\|_{\mathfrak L(E)}\|T(h)x-x\|_E\tag4$$ for all $h\ge0$.
But I don't think we can show left-continuity (without a local boundedness assumption). If $t\ge0$, then $$T(t)=T(t-h)T(h)\tag6$$ and hence $$\|(\operatorname{orb}x)(t-h)-(\operatorname{orb}x)(t)\|\le\|T(t-h)\|_{\mathfrak L(E)}\|x-T(h)x\|_E\tag7$$ for all $h\in[0,t]$. Now I don't see how we can conclude, since there is generally no bound for $\|T(t-h)\|_{\mathfrak L(E)}$.
The continuity at $s=0$ implies the continuity on $[t,t+h]$ (by the semigroup property and the exponential bound). Hence, the Bochner integrability.
Edit: For $t\ge h\ge 0$: \begin{align*} \|(\operatorname{orb}x)(t-h)-(\operatorname{orb}x)(t)\| &\le\|T(t-h)\|_{\mathfrak L(E)}\|x-T(h)x\|_E \\ & \le M e^{\omega (t-h)} \|x-T(h)x\|_E \; \to 0 \text{ as } h \to 0^+. \end{align*}