In a course on Markov processes, we defined the infinitesimal generator of a Feller process as follows:
A (not necessarily bounded) linear operator $\mathcal{L}$ from a subspace $\mathcal{D}$ of $\mathcal{C}_0(S)$ to $\mathcal{C}_0(S)$ is called a probability generator if
- (G1) $\mathcal{D}$ is dense in $\mathcal{C}_0(S)$
- (G2) For every $\lambda > 0$ and $f \in \mathcal{D}$, $$\inf_{x \in S} f(x) \geq \inf_{x \in S} \left(f(x) - \lambda \mathcal{L} f (x)\right)$$
- (G3) For all $\lambda>0$ small enough, the range of $I - \lambda \mathcal{L}$ is equal to $\mathcal{C}_0(S)$.
- (G4) If $S$ is compact, then $\mathcal{L}{\mathbf 1} = 0$. If not, then for every $\lambda$ there exists a sequence $(f_n)$ in $\mathcal{D}$ such that $f_n \to {\mathbf 1}$ pointwise, $f_n - \lambda \mathcal{L} f_n \to {\mathbf 1}$ pointwise, and $\sup\{\|f_n - \lambda \mathcal{L} f_n\| : n = 1,2,\ldots\} < \infty$.
After this definition, the following lemma was stated:
If an operator $\mathcal{L}$ satisfies (G1), (G3) and (G4) of the above definition, then (G2) is equivalent to the following condition: For every $f\in\mathcal{D}$ that attains a minimum, there exists $x_0$ such that $f$ attains its minimum at $x_0$ and $\mathcal{L} f(x_0) \geq 0$.
However, this lemma was only proved for the case where $S$ is compact (see below). I would appreciate any help in proving the above lemma in the general case. Either in form of a reference to a proof or in form of some hints which give the general proof outline.
For reference, here is the proof in the compact case: assume that $S$ is compact, and set $g(x) = f(x) - \lambda \mathcal{L} f(x)$.
- Condition in lemma $\implies$ (G2): let $f \in \mathcal{D}$. Since $S$ is compact, $f$ attains a minimum. By the condition in the lemma, there is some $x_0$ such that $f$ attains its minimum at $x_0$ and $\mathcal{L} f(x_0) \geq 0$. Consquently, \begin{align*} \inf_x f(x) = f(x_0) \geq f(x_0) - \lambda\mathcal{L} f (x_0) = g(x_0 )\geq \inf_x g(x) \end{align*}
- $\neg$ Condition in lemma $\implies$ $\neg$ (G2): if $\mathcal{L}$ does not satisfy the condition in the lemma, then there exists $f\in \mathcal{D}$ (which attains its minimum) such that $\mathcal{L} f(y) < 0$ for all $y \in A$, where $A=\{y:f(y) = \min f\}$. Let $B = \{x : \mathcal{L} f(x) \geq 0\}$. The sets $A$ and $B$ are closed by continuity of $f$ and $\mathcal{L} f$ and thus compact (since $S$ is compact), and $A \cap B = \emptyset$. Let $M = \max \{|\mathcal{L} f(x)| : x\in S\}$ (the maximum exists since $S$ is compact and $\mathcal{L} f$ continuous) and $m = \min \{f(x) - \inf f : x \in B\} > 0$. Let $\lambda < m / M$, write as before $g = f - \lambda \mathcal{L} f$. Then for every $x \in B^\textrm{c}$, $\mathcal{L} f(x) < 0$, so $$ g (x) = f(x) - \lambda \mathcal{L} f(x) > f(x) \geq \inf f, $$ and for every $x \in B$, $$ g (x) = \inf f + \big(f(x) - \inf f \big) - \lambda \mathcal{L} f(x) > \inf f + m - \frac mM M = \inf f. $$ In total we get that $g(x) > \inf f$ for all $x\in S$, and thus by compactness $\inf g > \inf f$, so (G2) is not satisfied.