# If $(T_t)_t$ is a strong continuous semigroup, why $T_sAf_n\to T_sg$ uniformly?

Let $$H$$ a Hilbert space and $$(T_t)_{t>0}$$ a strongly continuous semigroup. Let $$(A,\mathcal D(A))$$ its generator. Let $$(f_n)$$ a sequence of $$\mathcal D(A)$$ that converges to $$f$$ s.t. $$(Af_n)_n$$ converges to a function $$g$$. I want to show that $$g=Af$$. And the proof goes as follow :

$$T_tf_n-f_n=\int_0^t T_sAf_nds.$$ Using the fact that $$T_tAf_n\to Tg$$ uniformly, we can conclude.

Question : Why $$T_tAf_n\to Tg$$ uniformly ? My try :

$$\|T_sAf_n-T_sf_n\|_H=\|T_s(Af_n-f_n)\|_H\underset{(*)}{\leq} M_s\|Af_n-f_n\|_H,$$ Where $$(*)$$ comes from the continuity of $$T_s$$. But this constant depend on $$s$$, and then I cannot bound it uniformly. Do I ?

• In your question, it should be $\|T_s Af_n-T_sg\|$ instead of $\|T_sAf_n-T_sf_n\|.$
– Surb
Jul 7 '19 at 12:30

## 1 Answer

I would say it works : let first prove that there is $$\tau>0$$ s.t. $$p:=\sup_{0\leq s\leq \tau}\|T_s\|<\infty .$$ Suppose it's not true. Then there is a sequence $$(t_n)$$ s.t. $$t_n\to 0$$ and $$\|T_{t_n}\|\to \infty$$. In particular, by Banach-Steinhaus, there is $$f\in H$$ s.t. $$\|T_{t_n}f\|\to +\infty$$ which contradict the strong continuity. Now, let $$u=k\tau+\theta$$ where $$k\in \mathbb N$$ and $$\theta \in [0,\tau)$$. Then $$\|T_{u}\|\leq \|T_t\|^k\|T_\theta \|\leq p^{k+1}\leq p\cdot p^{\frac{u}{\tau}}.$$

At the end, if $$s\in [0,t]$$, $$\|T_s\|\leq p\cdot p^{\frac{s}{\tau}}\leq p\cdot p^{\frac{t}{\tau}},$$ and thus it's uniformly bounded. So at then end, $$\sup_{0\leq s\leq t}\|T_s(Af_n-g)\|\leq M_t\|Af_n-g\|_H\underset{n\to \infty }{\longrightarrow }0,$$ what prove the claim.