Let $E$ be a locally compact Hausdorff space, $E^\ast=E\cup\left\{\infty\right\}$ denote the Alexandroff one-point compactification of $E$, $(T(t))_{t\ge0}$ be a strongly continuous contraction semigroup on $C_0(E)$ and $$T^\ast(t)f:=f(\infty)+T(t)(\left.f\right|E-f(\infty))\;\;\;\text{for }f\in C(E^\ast)\text{ and }t\ge0.\tag1$$
It's easy to see that $(T^\ast(t))_{t\ge0}$ is a strongly continuous semigroup on $C(E^\ast)$, but is it contractive as well?
We identify $g\in C_0(E)$ with $g^\ast\in C(E^\ast)$, $$g^\ast:E^\ast\to\mathbb R\;,\;\;\;x\mapsto\begin{cases}g(x)&\text{, if }x\in E\\0&\text{, if }x=\infty.\end{cases}$$ This is done in $(1)$ with $T(t)(\left.f\right|E-f(\infty))\in C_0(E)$, noting that $\left.f\right|E-f(\infty)\in C_0(E)$, for $f\in C(E^\ast)$.
I know how the desired claim can be proved if we assume that $(T^\ast(t))_{t\ge0}$ is nonnegativity-preserving (e.g. by assuming that $(T(t))_{t\ge0}$ is nonnegativity-preserving), but I wonder whether we really need this.
So, let $t\ge0$, $f\in C(E^\ast)$ and $g:=\left.f\right|_E-f(\infty)\in C_0(E)$. By contractivity of $T(t)$, $$\left\|T(t)g\right\|_\infty\le\left\|g\right\|_\infty=\left\|f-f(\infty)\right\|_\infty\tag2$$ (I hope it's clear the the supremum in the definition of $\left\|;\cdot\;\right\|_\infty$ is taken over $E$ in $\left\|g\right\|_\infty$, while it is taken over $E^\ast$ in $\left\|f-f(\infty)\right\|_\infty$). Now $$\left\|T^\ast(t)f\right\|_\infty=\sup\left(\sup_{x\in E}\left|f(\infty)+\left(T(t)\left(\left.f\right|_E-f(\infty)\right)\right)(x)\right|,|f(\infty)|\right)\tag3$$ and $$\left\|f\right\|_\infty=\sup\left(\left\|\left.f\right|_E\right\|_{\infty},|f(\infty)|\right).\tag4$$