Proving linear operators are Markov Generators

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I am trying to do the following question from Liggett.

Let $A$ be a linear operator defined on the space of continuous functions $C(E)$ for a compact set $E$. Define $A$ by $A=T-I$ where $T$ is a positive operator and $T 1=1$. Then show $A$ is a Markov generator.

In particular, I am struggling to prove that the maximum condition holds (the hint on the link page 13 should help, but I do not see how)

https://books.google.co.uk/books?id=7JbqBwAAQBAJ&pg=PA13&lpg=PA13&dq=markov+pregenerator+positive+operator&source=bl&ots=b3mLncAMIX&sig=_byHHPxLPUnqQNq6FiSkD4zKtqI&hl=en&sa=X&ved=0ahUKEwjdtY-W9_DKAhXhBZoKHYE1DpEQ6AEIHzAA#v=onepage&q=markov%20pregenerator%20positive%20operator&f=false

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Fix $f \in C(E)$ and $\eta \in E$ such that $$f(\eta) = \min_{\zeta \in E} f(\zeta). $$ Then

$$g(\zeta) := f(\zeta)-f(\eta) \in C(E)$$

defines a non-negative function and therefore, since $T$ is a positive operator,

$$Tg \geq 0. \tag{1}$$

On the other hand, it follows from the linearity of $T$ and $T1 = 1$ that

$$Tg = Tf - f(\eta). \tag{2}$$

Combining $(1)$ and $(2)$ yields

$$Tf(\zeta) \geq f(\eta) \qquad \text{for all $\zeta \in E$}.$$

Hence,

$$Af(\eta) = Tf(\eta)-f (\eta) \geq f(\eta)-f(\eta) = 0$$


Remark: Note that this is equivalent to $$f \in \mathcal{D}(A), f(\eta) = \max_{\zeta \in E} f(\zeta) \implies Af(\eta) \leq 0.$$ If this condition holds for an operator $A$, we say that $A$ satifies the positive maximum principle.