If $A$ is local, and $x$ is a point of relative maximum for $f$, then $Af(x)\le 0$,(Markov transition functions)

Question

I have to set up the quesiton with some definitions, so before I get to the question there will be some definitions.

Markov transition function(definition)

Let $E$ be a topological space and let $\mathcal{E}$ be the Borel sigma-algebra on $E$. A markov transition function on $(E,\mathcal{E})$ is a function $p(s,t,x,A)$ where $s,t \in [0,\infty), x \in E, A \in \mathcal{E}$ such that:

1. for fixed $s,t,A$ we have $x \rightarrow p(s,t,x,A)$ is $\mathcal{E}$-measurable.

2. for fixed $s,t,x$ we have $p(s,t,x,A)$ is a probability on $(E,\mathcal{E})$.

3. p satisfies the Chapman-Kolmogorov equation

$$p(s,t,x,A)=\int_E p(u,t,y,A)p(s,u,x,dy),$$

for every $s\le u\le t.$

$T_tf, Af, \mathcal{D}(A)$ definition

Assume that $f$ is bounded Borel function from $E$ to $\mathbb{R}$ we then define

$$T_{s,t}f(x)=\int_e f(y)p(s,t,x,dy).$$

For $f$ a bounded Borel function from $E$ to $\mathbb{R}$ define

$$A_sf(x)= \lim\limits_{t \rightarrow 0^+}\frac{1}{t}[T_{s,s+t}f(x)-f(x)],$$

if the limit exists.

Define $f \in \mathcal{D}(A)$ if $A_sf(x)$ exist for every $x \in E$.

We say that $A_s$ is local if $A_sf(x)$ only depends on the behaviour of $f$ in a neighborhood of $x$, this means that if $f \in \mathcal{D}(A)$, $g$ is Borel from $E$ to $\mathbb{R}$, and $f=g$ in a neighborhood around $x$ then $A_s g(x)$ is defined, and $A_s f(x)=A_s g(x)$.

The question

It is stated:

If $A_s$ is local, then if $f \in \mathcal{D}(A_s)$ and $x$ is a point of relative maximum for $f$, then $A_sf(x)\le 0.$

How do we show this?

Attempt

Since $x$ is a relative maximum, there exists an open set $E_1$ such that $f(y)\le f(x)$ for all $y \in E_1$. We have

$$T_{s,s+t}f(x)=\int_E f(y) p(s,s+t,x,dy)=\int_{E_1} f(y) p(s,s+t,x,dy)+\int_{E\backslash E_1} f(y) p(s,s+t,x,dy).$$

We know that the first integral is less than $\int_{E_1} f(x) p(s,s+t,x,dy)$, but I do not know what do do with the second integral. The idea must be that the second integral does not matter because of the fact that $A_s$ is local.

My idea was too look at a function which is equal to $f$ on $E_1$ and something else on $E\backslash E_1$, but I have to do this in such a way that the new function is also in $\mathcal{D}(A_s)$ and I am not sure how to do this. Any idea on how to solve this problem?

Answer 1

Hint: The error term is $T_{s,s+t}(f \chi_{E \setminus E_{1}})$ and $f \chi_{E \setminus E_{1}} = 0$ in a neighborhood of $x$.