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Let $X$ be a compact topological space and $(E, |\cdot|)$ a Banach space. Let $\mathcal C$ be the space of all continuous functions from $X$ to $E$. Let $\|\cdot\|_\infty$ be the supremum norm on $\mathcal C$. Then $(\mathcal C, \|\cdot\|_\infty)$ is a Banach space. We fix a continuous map $g:X \to (0, \infty)$ and define a new norm $[\cdot]$ on $\mathcal C$ by $$ [f] := \sup_{x\in X} g(x) |f(x)| \quad \forall f \in \mathcal C. $$

Because $X$ is compact and $g$ continuous, there are $c_1, c_2 >0$ such that $c_1 \le |g(x)| \le c_2$ for all $x \in X$. As such, $$ c_1 \|\cdot\|_\infty \le [\cdot] \le c_2 \|\cdot\|_\infty. $$

It follows that $[\cdot]$ is equivalent to $\|\cdot\|_\infty$. Hence $(\mathcal C, [\cdot])$ is a Banach space.

Could you confirm if my above understanding is correct?

Akira
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    It looks fine to me. – José Carlos Santos Dec 24 '22 at 13:27
  • Since $g$ must be uniformly continuous we see that there is no problem with continuity of $\sup g(x)$ so your understanding is correct. – Ataulfo Dec 24 '22 at 13:53
  • @Piquito: the proof is totally correct as it is. Uniform continuity is off topic here. By the way, $X$ is barely a topological space, so it does not make sense to speak about uniformly continuous functions defined on $X$. – vizietto Dec 24 '22 at 14:29

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