Let $E$ be a $\mathbb R$-Banach space and $(T(t))_{t\ge0}$ be a semigroup on $E$, i.e. $T(t)$ is a bounded linear operator on $E$ for all $t\ge0$, $T(0)=\operatorname{id}_E$ and $$T(s+t)=T(s)T(t)\;\;\;\text{for all }s,t\ge0.\tag1$$ Let $$\operatorname{orb}x:[0,\infty)\to E\;,\;\;\;t\mapsto T(t)x$$ for $x\in E$, $$\mathcal D(A):=\left\{x\in E:\operatorname{orb}x\text{ is right-differentiable at }0\right\}$$ and $$Ax:=(\operatorname{orb}x)'(0)\;\;\;\text{for }x\in\mathcal D(A).$$
How can we show that $(T(t))_{t\ge0}$ is strongly continuous on $\overline{\mathcal D(A)}$?
By the semigroup property, it should suffice to show strong continuity at $0$. Moreover, by density it should suffice to consider $x\in\mathcal D(A)$. Now, my usual reflex would be to obtain the claim from the identity $$T(t)x-x=\int_0^tT(s)Ax\:{\rm d}s\;\;\;\text{for all }t\ge0\tag2$$ which is valid for any strongly continuous semigroup and its generator. However, with strong continuity being the property we're asked to prove, I don't see why $(2)$ should hold (actually, I don't see why the Riemann integral should exist in that case).
So, what do we need to do?
I think that we need to assume that $(T(t))_{t\ge0}$ is locally bounded (e.g. quasicontractive), i.e. $$\sup_{s\in[0,\:t]}\left\|T(s)\right\|_{\mathfrak L(E)}<\infty\;\;\;\text{for all }t\ge0\tag3.$$ Under that assumption we obtain $$\sup_{s\in[0,\:t]}\left\|\frac{T(s+h)x-T(s)x}h-T(s)Ax\right\|_E\le\sup_{s\in[0,\:t]}\left\|T(s)\right\|_{\mathfrak L(E)}\left\|\frac{T(h)x-x}h-Ax\right\|_E\xrightarrow{h\to0+}0\tag4$$ for all $t\ge0$ and hence locally uniform right-differentiability of $\operatorname{orb}x$. Maybe we can build up on that.
Best Answer
You are right that it suffices to show strong continuity at $0$ (by the semigroup property), but it is not true that it is enough to check strong continuity at $0$ for $x\in D(A)$. You would need some uniform bound here. On the other hand, right-differentiability at $0$ automatically implies right-continuity at $0$: If $T_t x-x$ does not tend to zero, then there is no chance for the limit $\frac 1 t(T_t x-x)$ to exist.
In your situation, strong continuity on $\overline{D(A)}$ is equivalent to local boundedness on $\overline{D(A)}$. One implication follows directly from the uniform boundedness principle and the semigroup property. For the other implication (the one you ask about), let $x\in\overline{D(A)}$ and $(x_n)$ a sequence in $D(A)$ such that $x_n\to x$. Then $$ \|T(t)x-x\|\leq \sup_{s\in[0,T]}\|T(s)\|_{\mathcal{L}(\overline{D(A)})}\|x-x_n\|+\|T_t x_n-x_n\|+\|x-x_n\|. $$ Letting first $t\to 0$ and then $n\to\infty$ yields the desired convergence.