# What's exactly meant if one is saying that the closure of a multivalued operator is the generator of a $C^0$-semigroup?

Let

• $$E$$ be $$\mathbb R$$-Banach space
• $$A$$ be a subspace of $$E\times E$$ and $$\mathcal D(A):=\left\{x\in E\mid\exists y\in E:(x,y)\in A\right\}$$
• $$(T(t))_{t\ge0}$$ be a contractive $$C^0$$-semigroup on $$\overline{\mathcal D(A)}$$

$$A$$ is considered as being a multivalued operator. What's meant by "the closure $$\overline A$$ of $$A$$ is a generator of $$(T(t))_{t\ge0}$$"?

Is this related to the concept of a "full generator"? My assumption is that the meaning of the sentence is $$\overline A=\left\{(f,g)\in\overline{\mathcal D(A)}\times\overline{\mathcal D(A)}:T(t)f-f=\int_0^tT(s)g\:{\rm d}s\text{ for all }t\ge0\right\}.\tag1$$

Is that correct? Is that what it's meant?

If so, what can we say about existence and uniqueness of $$(T(t))_{t\ge0}$$ and $$A$$? And is $$\overline A$$ "single-valued" (which means that $$(0,y)\in\overline A$$ implies $$y=0$$)?

Remark: I've seen the terminology used in Theorem 8.2 of Chapter 4 in Markov Processes: Characterization and Convergence by Ethier and Kurtz: • How many questions run through your mind in a given day? – mathworker21 Nov 14 '18 at 12:00
• Can you give some more context and/or an example where this kind of terminology is used? – MaoWao Nov 14 '18 at 12:31
• @MaoWao It's used in Theorem 8.2 of Chapter 4 in Markov Processes: Characterization and Convergence by Ethier and Kurtz. – 0xbadf00d Nov 14 '18 at 12:46
• There is a notion of pre-semigroups for which multi-valued generators make sense (defined in the expected way). However, I don't see how this fits into the context of this theorem. In fact, what you wrote down is exactly the usual single-valued generator of $(T_t)$, if I'm not mistaken. – MaoWao Nov 14 '18 at 13:42
• @MaoWao No, it's not. If $E=\mathcal L^\infty(\mathbb R)$ and $$(T(t)f)(x):=f(x+t)\;\;\;\text{for }x\in\mathbb R\text{ and }f\in E.$$ Then, $$(0,g)\in\left\{(f,g)\in E\times E:T(t)f-f=\int_0^tT(s)g\:{\rm d}s\text{ for all }t\ge0\right\}$$ for all $g\in E$ with $g=0$ almost everywhere. – 0xbadf00d Nov 14 '18 at 13:53