Let

- $E$ be a $\mathbb R$-Banach space
- $(T_n(t))_{t\ge0}$ and $(T(t))_{t\ge0}$ be strongly continuous contraction semigroups on $E$ with generators $(\mathcal D(A_n),A_n)$ and $(\mathcal D(A),A)$, respectively
- $D$ be a core of $(\mathcal D(A),A)$

Consider the following assertions:

- $D\subseteq\mathcal D(A_n)$ for all $n\in\mathbb N$ and $$\left\|A_nx-Ax\right\|_E\xrightarrow{n\to\infty}0\;\;\;\text{for all }x\in E\tag1$$
- For each $x\in D$ there is a sequence $x_n\in\mathcal D(A_n)$, n$\in\mathbb N$, with $$\left\|x_n-x\right\|_E+\left\|A_nx_n-Ax\right\|_E\xrightarrow{n\to\infty}0\tag2$$
- For each bounded interval $I\subseteq[0,\infty)$ and $x\in E$, $$\sup_{t\in I}\left\|T_n(t)x-T(t)x\right\|_E\xrightarrow{n\to\infty}0\tag3$$

By the Trotter-Kato Approximation Theorem, the following implications hold: $$\text{1.}\Rightarrow\text{2.}\Leftrightarrow\text{3.}\tag4$$

Now assume $(T_n(t))_{t\ge0}$ is not a semigroup on $E$, but on another $\mathbb R$-Banach space $E_n$. Moreover, assume there is a bounded linear operator $\iota_n:E\to E_n$ such that $$\sup_{n\in\mathbb N}\left\|\iota_n\right\|<\infty\tag5$$ (if necessary, assume that $\left\|\iota_n\right\|\le1$). Now consider the following assertions:

- For each $x\in D$ there is a sequence $x_n\in\mathcal D(A_n)$, n$\in\mathbb N$, with $$\left\|x_n-\iota_nx\right\|_{E_n}+\left\|A_nx_n-\iota_nAx\right\|_{E_n}\xrightarrow{n\to\infty}0\tag5$$
- For each bounded interval $I\subseteq[0,\infty)$ and $x\in E$, $$\sup_{t\in I}\left\|T_n(t)\iota_nx-\iota_nT(t)x\right\|_{E_n}\xrightarrow{n\to\infty}0\tag6$$

Question 1: Are we able to infer the equivalence $$\text{4.}\Leftrightarrow{5.}\tag7$$ by the known result $(4)$ or do we need to mimic its proof from scratch?

Question 2: Is there an easier proof of $\text{2.}\Leftrightarrow{3.}$ (in the former setting) available, if we assume that each $(T_n(t))_{t\ge0}$ is even uniformly continuous (and hence each $A_n$ is bounded)?

In the context of question 2 I've got the Hille-Yosida approximation theorem in mind where something similar is shown. The crucial fact therein seems that $A_n$ and $T(t)$ commute. While this should be wrong in our general setting (is it?), there might still be an easier proof available.