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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 ?

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

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