Let
- $E$ be a locally compact separable metric space
- $(\mathcal D(A),A)$ be the generator of a strongly continuous contraction semigroup on $C_0(E)$
- $E_n$ be a metric space for $n\in\mathbb N$
- $(\mathcal D(A_n),A_n)$ be the generator of a strongly continuous contraction semigroup on$^1$ $B(E_n)$
- $\pi_n:E_n\to E$ be continuous and $$\iota_nf:=f\circ\pi_n\;\;\;\text{for }f\in C_0(E)$$ for $n\in\mathbb N$
Let $\lambda>0$ and $f\in C_0(E)$. Assume$^2$ $$\left|\left(R_\lambda(A_n)\iota_nf\right)(x_n)-\left(R_\lambda(A)f\right)(x)\right|\xrightarrow{n\to\infty}0\tag1$$ for all $x_n\in E_n$, $n\in\mathbb N$, and $x\in E$ with $\pi_n(x_n)\xrightarrow{n\to\infty}x$. Are we able to conclude $$\left\|R_\lambda(A_n)\iota_nf-\iota_nR_\lambda(A)f\right\|_\infty\xrightarrow{n\to\infty}0;\tag2$$ at least under suitable further assumptions (e.g. compactness of $E$)?
Note that the result holds if $E_n=E$ for all $n\in\mathbb N$, $E$ is compact and $\iota_n$ is the identity for all $n\in\mathbb N$: https://math.stackexchange.com/q/3139957/47771.
We may note that by contractivity, $(0,\infty)$ is contained in the resolvent sets of $(\mathcal D(A_n),A_n)$, $n\in\mathbb N$, and $(\mathcal D(A),A)$. Moreover, $$\left\|R_\lambda(A_n)\right\|,\left\|R_\lambda(A)\right\|\le\frac1\lambda\;\;\;\text{for all }n\in\mathbb N.\tag3$$ This might be crucial.
$^1$ If $S$ is a set, let $B(S)$ denote the space of bounded functions from $S$ to $\mathbb R$ equipped with the supremum norm.
$^2$ If $(\mathcal D(B),B)$ is a bounded linear operator on a Banach space and $\lambda$ is a regular value of $(\mathcal D(B),B)$, let $R_\lambda(B)$ denote the resolvent operator of $(\mathcal D(B),B)$.
