This is a question on the proof of Lemma 7.18 in the book of Schilling and Partzsch (Brownian motion, An introduction to stochastic processes).
The setup is that there is a generator associated with a strongly continuous contraction semigroup and the generator satisfies the positive maximum principle. The claim is that the corresponding resolvent, $U$, is positivity preserving.
I understand all the steps in the proof except for the first one, which goes as follows.
Assume that $v \geq 0$ and let $x_0$ be the minimum point of $U_{\lambda}v$, i.e. $$U_{\lambda}v(x_0) = \inf_{x \in \mathbb{R}^d}U_{\lambda}v(x)$$ Since $U_{\lambda}v$ vanishes at infinity, it is clear that $U_{\lambda}v(x_0)\leq 0$.
My question is why this is true. I don't see why a vanishing function needs to take a nonpositive value at its lowest point. Clearly, there is some other consideration here that I am not seeing.
The passage defines Uλv(x0) as the infimum, but also refers to it as the "minimum". "Minimum" and "infimum" aren't quite the same thing; the infimum of a set need not be a member of the set, while the term "minimum" does imply membership. Using both of these terms apparently interchangeably implies that this is a set in which the infimum is indeed a member of the set, and is thus both the infimum and the minimum.
As for your question as to why the minimum can't be positive, suppose it were positive. So there's some positive number m such that m is the minimum. It follows that all other values are greater than or equal to m (that's the definition of minimum). But then how can it vanish? "Vanish at infinity" means "The limit as x goes to infinity is zero", which in turn means that for all $\epsilon$, there is some X for which if x>X, then the function is within $\epsilon$ of 0. But if we take $\epsilon$ as, say, m/2, then being within $\epsilon$ of 0 would mean being less than m/2, which would mean that it's less than the minimum, which is by definition of "minimum" impossible.