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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.

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  • $\begingroup$ 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 $t\geq 0$. $\endgroup$ – MaoWao Mar 16 at 21:04
  • $\begingroup$ @MaoWao In which case $(T(t))_{t\ge0}$ is conservative? $\endgroup$ – 0xbadf00d Mar 17 at 7:25
  • $\begingroup$ @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. $\endgroup$ – 0xbadf00d Mar 17 at 7:27
  • $\begingroup$ @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? $\endgroup$ – 0xbadf00d Mar 17 at 7:32
  • $\begingroup$ @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.) $\endgroup$ – 0xbadf00d Mar 17 at 7:47

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