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Let $T(t)$, $t\in [0,\tau]$, be a $C_0$ semigroup on an Banach space $X$. Also, let $T_n(t)$ be a sequence of semigroups that satisfies for all $x\in X$

$$\lim_{n\to \infty}\sup_{t\in [0,\tau]}\|T_n(t)x-T(t)x\|\to 0.$$

Let $y(t)\in L^2(0,\tau;X)$, which of the following properties holds

  1. $$\lim_{n\to \infty}\sup_{t\in [0,\tau]}\|T_n(t)y(t)-T(t)y(t)\|= 0$$

  2. $$\lim_{n\to \infty}\|T_n(t)y(t)-T(t)y(t)\|=0, \quad a.e. \; t\in [0,\tau]$$

  3. $$\lim_{n\to \infty}\int_0^\tau\|T_n(t)y(t)-T(t)y(t)\|^2 dt= 0$$

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  • $\begingroup$ Isn't 2. a direct consequence of the assumption? $\endgroup$
    – Mateusz Kwaśnicki
    Commented Nov 13, 2019 at 22:53
  • $\begingroup$ Also, if $X = \ell^2$, $T(t)x = x$, $(T_n(t)x)_k = e^{-kt/n}x_k$, and $(y(t))_k = \delta_{k,\lfloor 1/t^2 \rfloor}$, then $T_n(1/n) y(1/n) = e^{-1} y(1/n)$, and so 1. fails. $\endgroup$
    – Mateusz Kwaśnicki
    Commented Nov 13, 2019 at 23:00
  • $\begingroup$ Finally, 3. follows from 2. if all $T_n(t)$ are uniformly bounded. If I am not mistaken, this is the case, by the uniform boundedness principle, right? $\endgroup$
    – Mateusz Kwaśnicki
    Commented Nov 13, 2019 at 23:02

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