2
$\begingroup$

Let $E$ be a locally compact separable metric space, $$C_0(E):=\left\{f\in C(E):\left\{|f|\ge\varepsilon\right\}\text{ is compact for all }\varepsilon>0\right\},$$ $\kappa$ be a Markov kernel on $(E,\mathcal B(E))$, $$\kappa f:=\int\kappa(\;\cdot\;,{\rm d}y)f(y)\;\;\;\text{for bounded Borel measurable}f:E\to\mathbb R,$$ $\lambda\ge0$ and $$Af:=\lambda(\kappa f-f)\;\;\;\text{for }f\in C_0(E).$$

Are we able to show that $A$ is the generator of a Feller semigroup (as defined in my other question) on $C_0(E)$?

Note that $(\mathcal D(A),A)$ is the generator of the uniformly continuous semigroup $$S(t)f:=e^{tA}f\;\;\;\text{for }f\in C_0(E)\text{ and }t\ge0.$$

EDIT: For whatever it may be useful, we may note that since $\kappa$ is contractive, $$\lambda\left\|\kappa f\right\|_\infty\le\lambda\left\|f\right\|_\infty\le(\lambda+\mu)\left\|f\right\|_\infty\tag2$$ and hence $$\left\|A_\mu f\right\|_\infty\ge\left|(\lambda+\mu)\left\|f\right\|_\infty-\lambda\left\|\kappa f\right\|_\infty\right|\ge(\lambda+\mu)\left\|f\right\|_\infty-\lambda\left\|f\right\|_\infty=\mu\left\|f\right\|_\infty\tag3$$ by the reverse triangle inequality for all bounded Borel measurable $f:E\to\mathbb R$ (in particular, for $f\in C_0(E)$) and $\mu\ge0$. So, $(\mathcal D(A),A)$ is dissipative.

EDIT2: It's trivial to see that $A$ satisfies the nonnegative maximum principle: If $f\in C_0(E)$ and $x_0\in E$ with $$f(x_0)=\sup_{x\in E}f(x)$$ (it's not necessary to assume $f(x_0)\ge0$ for the following conclusion), then $$(\kappa f)(x_0)-f(x_0)=\int\kappa(x_0,{\rm d}y)\underbrace{\left(f(y)-f(x_0)\right)}_{\le\:0}\le0\tag4.$$

$\endgroup$
  • $\begingroup$ Basically it seems like you are asking about the semigroup $S'(t) = e^{-\lambda t} S(\lambda t)$. This is going to be strongly continuous since the original semigroup $S(t)$ is, and because $|e^{-\lambda t}-1| \leq \lambda t$. $\endgroup$ – Shalop Mar 24 at 5:45
  • $\begingroup$ @Shalop Thank you for your comment. I guess this shows indeed strong continuity. However, it's still not clear to me why $S(t)C_0(E)\subseteq C_0(E)$. (I thought it might be easier to show that $(S(t))_{t\ge0}$ is Feller by showing the mentioned properties of the generator, but if you know how we directly show that this semigroup is Felelr, this would be fine as well for me.) $\endgroup$ – 0xbadf00d Mar 26 at 19:02
  • $\begingroup$ Of course if $\kappa$ maps $C_0$ to $C_0$, then clearly $e^{t\kappa}$ maps $C_0$ to $C_0$. Because you assumed “\kappa is a contractive operator on $C_0$”. $\endgroup$ – Shalop Mar 27 at 1:43
  • $\begingroup$ @Shalop Sorry, I guess the title is misleading. I only assume that $\kappa$ is a Markov kernel on $(E,\mathcal B(E))$ and hence induces a contractive operator on the Banach space of bounded Borel measurable $f:E\to\mathbb R$ (by the given definition of $\kappa f$) equipped with the sup-norm. So, in particular, $\left\|\kappa f\right\|_\infty\le\left\|f\right\|_\infty$ for all $f\in C_0(E)$, but we don't know that $\kappa f\in C_0(E)$. $\endgroup$ – 0xbadf00d Apr 9 at 11:37
  • $\begingroup$ Then the answer is no. Basically set $E=[0,1]$. Then define $\kappa$ by $(\kappa f)(x) = f(0)$ for $x \in [0,1/2]$ and define $(\kappa f)(x) = f(1)$ for $x \in (1/2,1]$. It is clear that $\kappa$ acts contractively on bounded measurable functions. However, $\kappa$ clearly fails to send $C(E)$ to itself. Since $\kappa$ is idempotent, it then follows that $e^{t\kappa} = I + (e^t-1)\kappa$ also fails to send $C(E)$ to itself. $\endgroup$ – Shalop Apr 10 at 5:15

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.