# Show that the semigroup given by the product of two semigroups is generated by the product of the generators

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 $$$$\begin{split}\left\|\frac{\tilde U(t)(x,y)-(x,y)}t-(x',y')\right\|_Z&=\left\|\left(\frac{(S_n(t)x_n)_{n\in\mathbb N}-x}t-x',\frac{T(t)y-y}t-y'\right)\right\|_Z\\&=\sup\left(\sup_{n\in\mathbb N}\left\|\frac{S_n(t)x_n-x_n}t-x_n'\right\|_{X_n},\left\|\frac{T(t)y-y}t-y'\right\|_Y\right)\end{split}\tag4$$$$ for all $$((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?