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For a topological space $X$ we define $$\mathcal{C}_{0}(X) : = \left\lbrace f \colon X \longrightarrow \mathbb{C} \ \text{continuous} \colon \forall \, \varepsilon >0 \ \exists \, K \subseteq X \ \text{kompact} \ \forall \, x \notin K \colon |f(x)| < \varepsilon \right\rbrace.$$

How can I show that for $f \in \mathcal{C}_{0}(X)$ we have that $\sup_{x \in X} |f(x)| = \max_{x \in X} |f(x)|$?

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You take $\epsilon = \sup_{x\in X} |f(x)|$, then there exist a compact K such that $|f(x)| < \frac{\sup_{x\in X} |f(x)|}{2}$ outside of $K$

And as $K$ is compact and f continuous, $\sup_{x\in K} |f(x)| = \max_{x\in K} |f(x)|$.

Hence $$\sup_{x\in X} |f(x)| = \sup_{x\in K} |f(x)| = \max_{x\in K} |f(x)| =\max_{x\in X} |f(x)| $$

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