# Generation theorem for Feller semigroups

Let $$E$$ be a locally compact Hausdorff space.

I want to show that a linear operator $$(\mathcal D(A),A)$$ on $$C_0(E)^1$$ is closable and the closure $$(\mathcal D(\overline A),\overline A)$$ is the generator of a Feller$$^2$$ semigroup on $$C_0(E)$$ if and only if

1. $$\mathcal D(A)$$ is dense;
2. $$(\mathcal D(A),A)$$ satisfies the nonnegative maximum principle, i.e. $$\forall f\in\mathcal D(A):\forall x_0\in E:f(x_0)=\sup_{x\in E}f(x)\ge0\Rightarrow (Af)(x_0)\le0;\tag1$$
3. $$A_\lambda:=\lambda\operatorname{id}_{\mathcal D(A)}-A$$ has dense range for some $$\lambda>0$$; and

Now, I know that if $$F$$ is a $$\mathbb R$$-Banach space, a linear operator $$(\mathcal D(B),B)$$ on $$F$$ is closable and $$(\mathcal D(\overline B),\overline B)$$ is the generator of a strongly continuous contraction semigroup on $$F$$ if and only if

1. $$\mathcal D(B)$$ is dense;
2. $$\mathcal D(B)$$ is dissipative; and
3. $$B_\lambda\mathcal D(B)$$ is dense for some (and hence all) $$\lambda>0$$.

This is the Lumer-Phillips theorem. So, the desired claim seems to be a simple reformulation of this equivalence. I know that a linear operator on $$C_0(E)$$ satisfying the nonnegative maximum principle $$(1)$$ is dissipative. On the other hand, the generator of a contractive nonnegativity preserving semigroup on $$C_0(E)$$ satisfies the nonnegative maximum principle.

So, it seems like (please tell me if anything is wrong) the only missing piece is the sub-Markovity. How can we embed this into the equivalence?

$$^1$$ $$C_0(E)$$ denotes the space of continuous functions $$E\to\mathbb R$$ vanishing at infinity equipped with the supremum norm.

$$^2$$ A semigroup $$(T(t))_{t\ge0}$$ is called Feller if it is contractive (i.e. $$\left\|T(t)\right\|_{C_0(E)}\le1$$ for all $$t\ge0$$), sub-Markov (i.e. $$0\le T(t)f\le 1$$ for all $$f\in C_0(E)$$ with $$0\le f\le1$$) and $$(T(t)f)(x)\xrightarrow{t\to0}f(x)\;\;\;\text{for all }x\in\mathbb R\text{ and }f\in C_0(\mathbb R)\tag4.$$

$$^3$$ $$1$$ stands for the function $$E\to\mathbb R$$ which is constantly $$1$$ here.

• If $E$ is non-compact, a sequence as in (4) cannot exist, since $1\notin C_0(E)$. Even if $E$ is compact, (4) is not necessarily true. In this case this property means that $(T_t)$ is conservative, i.e. $T_t1=1$ for all $t\geq 0$. – MaoWao Mar 16 at 21:04
• @MaoWao In which case $(T(t))_{t\ge0}$ is conservative? – 0xbadf00d Mar 17 at 7:25
• @MaoWao Did you really meant $(4)$? This is the usual definition of a Feller semigroup as it can be found, for example, in the book of Kallenberg on page 369. – 0xbadf00d Mar 17 at 7:27
• @MaoWao And in this question $(\kappa_t)_{t\ge0}$ is a strongly continuous semigroup on $C_0(\mathbb R)$. In particular, it holds $(4)$. Am I missing something? – 0xbadf00d Mar 17 at 7:32
• @MaoWao It seems to be me that you've actually meant $(2)$ instead. And I guess you're right in that case. Clearly, $1\not\in C_0(E)$. I guess one needs to consider the one-point compactification of $E$ in that case. However, meanwhile I think we don't need $4.$ (oh, I guess you've meant the fourth bullet point, right?) at all, unless we want to force the conservativeness. (I've removed the fourth bullet point from the question.) – 0xbadf00d Mar 17 at 7:47