I have a doubt on the proof of the theorem. $f \in C_0(\Omega)$, and $K = \{ w \in \Omega \ | \ |f(w)| \ge \epsilon\}$ which is compact. Then by Urysohn lemma, there is a continuous function $g: \Omega \to [0, 1]$ which has a compact support and $g(w) = 1$ for all $w \in K$. $\delta$ is chosen such that $\delta > 0$, $\delta < 1$, and $1-\delta < \epsilon$ where $\epsilon$ is a given error. Then he claims that $\|f-\delta gf\| \le \epsilon$. And this is where I have an issue.
I get that if $w \notin K$, then $|f(w)| < \epsilon$ by the definition of $K$. Hence $|f(w) -\delta g(w) f(w)| \le |f(w)| < \epsilon$ since $1-\delta g(w) \le 1$ as $g(w) \in [0, 1]$ and by the choice of $\delta$.
I have a hard time trying to show $|f(w) - \delta g(w) f(w)| \le \epsilon$ when $w \in K$.
My approach is if $w \in K$, $g(w) = 1$ and then, $|f(w) - \delta f(w)| = (1-\delta) |f(w)| \le$ $(1-\delta) \|f\|_\infty$
Should'nt the choice of $\delta$ be made such that $(1-\delta)\|f\|_\infty \le \epsilon$