I am confused about a proof.
Let $E$ be a locally compact and separable metric space, denote $C_0(E)$ the space of continuous functions vanishing at infinity, and $E^\Delta$ the one-point compactification of $E$. We let $T(t)$ be a strongly continuous contraction (i.e., $||T(t)|| \le 1$) semi-group on $E$ which is positive (i.e., maps non-negative functions into non-negative functions). We define the operator $T^\Delta(t)f = f(\Delta) + T(t)(f - f(\Delta))$ and wish to show that it is also positive.
The purported proof claims that "we must show that if $\alpha \in R, f \in C_0(E), \alpha + f \ge 0$ then $\alpha + T(t)f \ge 0$". I don't see why this would constitute proof that $T^\Delta$ is positive, what am I missing?
As far as I understand, $T^\Delta$ is supposed to be an extension of $T$ to $C(E^\Delta)$, correct?
Since $$ C(E^\Delta)= \mathbb R\cdot 1 \oplus C_0(E), $$ any such extension is determined by the value of $T^\Delta(t)$ at $1$, and the given formula seems to indicate a desire to set $T^\Delta(t)(1)=1$.
This said, an equivalent way to define $T^\Delta(t)$ is by setting $$ T^\Delta(t)(\alpha 1+f) = \alpha 1+T(t)(f),\quad \forall \alpha\in \mathbb R,\quad \forall f\in C_0(E). $$ Seeing things from this angle positivity of $T^\Delta(t)$ is clearly expressed by the given condition.