If $E^\ast$ is the Alexandroff one-point compactification of $E$ and $f\in C(E^\ast)$, then $\left.f\right|_E-f(\infty)\in C_0(E)$

Question

Let $E$ be a locally compact Hausdorff space, $E^\ast=E\cup\{\infty\}$ denote the Alexandroff one-point compactification of $E$, $f\in C(E^\ast)$ and $g:=\left.f\right|_E-f(\infty)$.

How can we show that $g\in C_0(E)$?

The "reverse" of the claim is clear to me: If $h\in C_0(E)$, then $$h^\ast:E^\ast\to\mathbb R\;,\;\;\;x\mapsto\begin{cases}f(x)&\text{, if }x\in E\\0&\text{, if }x=\infty\end{cases}$$ is continuous.

Note that $g\in C_0(E)$ if and only if $g$ is continuous and $\left\{|g|\ge\varepsilon\right\}$ is compact for all $\varepsilon>0$. So, we might need a characterization of the compact sets in $E^\ast$.

Answer 1

$f$ is continuous at $\infty$ which means that for each $\varepsilon > 0$ there exists an open neighborhood $U$ of $\infty$ such that $\lvert f(e) - f(\infty) \rvert <\varepsilon$ for all $e \in U$. But $U$ is an open neighborhood $\infty$ if and only if $K = E^* \setminus U$ is compact. This shows $$M_\varepsilon = \{ e \in E \mid \lvert g(e) \rvert \ge \varepsilon \} \subset K .$$ Hence $M_\varepsilon$ is compact.