Let $X_n,Y$ be $\mathbb R$-Banach spaces for $n\in\mathbb N$, $A_n\subseteq X_n\times X_n$ and $B\subseteq Y\times Y$ be linear and dissipative with $\mathcal R(\lambda-A_n)=X_n$ and $\overline{\mathcal R(\lambda-B)}=Y$ for all $\lambda>0$. It can be shown that $$A_{n,\:0}:=\left\{(x,x')\in\overline{A_n}:x'\in\overline{\mathcal D(A_n)}\right\}$$ and $$B_0:=\left\{(y,y')\in\overline B:y'\in\overline{\mathcal D(B)}\right\}$$ are single-valued and generate strongly continuous contraction semigroups $(S_n(t))_{t\ge0}$ and $(T(t))_{t\ge0}$ on $\overline{\mathcal D(A_n)}$ and $\overline{\mathcal D(B)}$, respectively. Let $X:=\left\{x\in\prod_{n\in\mathbb N}X_n:\sup_{n\in\mathbb N}\left\|x_n\right\|_{X_n}<\infty\right\}$ be equipped with $\left\|x\right\|_X:=\sup_{n\in\mathbb N}\left\|x_n\right\|_{X_n}$ for $x\in X$ and $Z:=X\times Y$ be equipped with $\left\|(x,y)\right\|_Z:=\max(\left\|x\right\|_X,\left\|y\right\|_Y)$ for $(x,y)\in Z$. Now let $$C:=\left\{((x,y),(x',y'))\in Z\times Z:(x,x')\in\prod_{n\in\mathbb N}A_n\text{ and }(y,y')\in B\right\}.$$ It's easy to see that $C$ is a (multi-valued) dissipative linear operator on $Z$.

It's not clear to me why $\mathcal R(\lambda-C)=Z$ for all $\lambda>0$ and I've asked for that here: Show that this multi-valued operator is surjective (Theorem 1.6.9 of Ethier and Kurtz).

In this question, assume we know that $\mathcal R(\lambda-C)=Z$ and hence $$C_0:=\left\{(z,z')\in\overline C:z'\in\overline{\mathcal D(C)}\right\}$$ is single-valued and the generator of a strongly continuous contraction semigroup $(U(t))_{t\ge0}$ on $\overline{\mathcal D(C)}$.

How can we show that $$U(t)(x,y)=((S_n(t)x_n)_{n\in\mathbb N},T(t)y)\tag1$$ for all $t\ge0$ and $(x,y)\in\overline{\mathcal D(C)}$?

The claim can be found in the proof of Theorem 6.9 of Chapter 1 in the book of Ethier and Kurtz.

We now that $$\left\|\frac{S_n(t)x-x}t-x'\right\|_X\xrightarrow{t\to0+}\;\;\;\text{for all }(x,x')\in A_{n,\:0}\tag2$$ for all $n\in\mathbb N$ and $$\left\|\frac{T(t)y-y}t-y'\right\|_Y\xrightarrow{t\to0+}\;\;\;\text{for all }(y,y')\in B_0\tag3.$$ It's not hard to see that if $((x,y),(x',y'))\in C_0$, then $(x,x')\in\prod_{n\in\mathbb N}A_{n,\:0}$ and $(y,y')\in B_0$. Letting $$\tilde U(t)(x,y):=((S_n(t)x_n)_{n\in\mathbb N},T(t)y)\;\;\;\text{for }(x,y)\in\overline{\mathcal D(C)}\text{ and }t\ge0,$$ we obtain \begin{equation}\begin{split}\left\|\frac{\tilde U(t)(x,y)-(x,y)}t-(x',y')\right\|_Z&=\left\|\left(\frac{(S_n(t)x_n)_{n\in\mathbb N}-x}t-x',\frac{T(t)y-y}t-y'\right)\right\|_Z\\&=\sup\left(\sup_{n\in\mathbb N}\left\|\frac{S_n(t)x_n-x_n}t-x_n'\right\|_{X_n},\left\|\frac{T(t)y-y}t-y'\right\|_Y\right)\end{split}\tag4\end{equation} for all $((x,y),(x',y'))\in\overline{\mathcal D(C)}$. So, if $((x,y),(x',y'))\in C_0$, then each of the terms we are taking the supremum of tends to $0$ as $t\to0+$. However, that doesn't mean that the supremum tends to $0$. What am I missing?


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