# How can we prove that a (locally bounded) semigroup is strongly continuous on the closure of its generator?

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.

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.
• It's not important for the question, but your notation suggests that $T(s)\overline{\mathcal D(A)}\subseteq\overline{\mathcal D(A)}$ for all $s\ge0$. Why is that the case? Feb 4, 2019 at 12:16
• Well, $T(s)$ maps $D(A)$ into $D(A)$ (this can be proven without strong continuity), and then continuity of $T(s)$ implies that the same is true for the closure. Feb 4, 2019 at 13:36
• From your proof I guess that's sufficient if there is a small $T>0$ such that the operator norms on $[0,T)$ are bounded, right? Feb 4, 2019 at 13:41
• That is right. You can also directly show that boundedness on $[0,T)$ for some $T>0$ implies boundedness on all bounded intervals: If $I$ is bounded, then there exists $n\in\mathbb{N}$ such that $t/n<T$ for $t\in I$. Then $\|T(t)\|\leq\|T(t/n)\|^n$ by the semigroup property. Feb 4, 2019 at 13:45