Let $L_n,L$ be $\mathbb R$-Banach spaces for $n\in\mathbb N$, $A_n\subseteq L_n\times L_n$ and $A\subseteq L\times L$ be linear and dissipative with $\mathcal R(\lambda-A_n)=L_n$ and $\overline{\mathcal R(\lambda-A)}=L$ for some $\lambda>0$.
Now let $$\mathcal L:=\prod_{n\in\mathbb N}L_n\times L$$ and $$\mathcal A:=\left\{(\{f_n\},f),(\{g_n\},g))\in\mathcal L\times\mathcal L:(f_n,g_n)\in A_n\text{ for all }n\ge1\text{ and }(f,g)\in A\right\}.$$
How can we show that $\mathcal R(\lambda-\mathcal A)=\mathcal L$?
The claim can be found in the proof of Theorem 6.9 of Chapter 1 in the book of Ethier and Kurtz. What's worrying me is that it's quite surprising that $\mathcal R(\lambda-\mathcal A)=\mathcal L$, while we've only got $\overline{\mathcal R(\lambda-A)}=L$ (and not even $\mathcal R(\lambda-A)=L$).
Remark: The authors write that they equip $\mathcal L$ with the norm $\left\|(\{f_n\},f)\right\|:=\sup(\sup_{n\ge1}\left\|f_n\right\|,\left\|f\right\|)$. However, this doesn't yield a normed space, since the supremum could be $\infty$. So, shouldn't they define $\mathcal L$ to consist of those elements for which this supremum is finite?
EDIT: In the paragraph above the proof of item (a) of Theorem 6.9 the authors write: "Given $(\{h_n\},h)\in\mathcal L_0$ and $\lambda>0$, there exists $(f,g)\in A$ such that $\lambda f-g=h$." Either I'm missing something or this means that they actually assume $\mathcal R(\lambda-A)=L$.