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Suppose that $f: C \rightarrow R^d$ is continuous on a compact set $C$. Show that $\|f\|$ attains its maximum on C.

First, I am wondering if I can just apply the Extreme Value theorem and then I'm done? I.e: By the Extreme Value Theorem, $\implies$ ... $\implies$ $\|f(a)\|$ $\leq$ $\|f(x)\|$ $\leq$ $\|f(b)\|$, $\forall x \in C$.

Alternatively, I have: Suppose $f$ is continuous on the compact set C. This imples the image set {$f(C) : c \in C$} is compact, since a compact set is compact under continuous maps. This implies the setis closed and bounded. Then, there is $M \in R^d$ s.t: $\|f \|$ $\leq M$. Thus, $\|f\|$ attains its maximum on C.

I guess I'm just sort of confused how is this proof/problem is any different from the proof of the Extreme Value Theorem(???). What is it about $\|f\|$ that changes the nature of what is to be shown? I'm not entirely clear on what is being asked to be shown.

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  • $\begingroup$ Everything is fine except that $M\in \mathbb{R}^d$. You want $M\in \mathbb{R}$, and more specifically that there is a least such $M$ and that that number is in the image set. $\endgroup$ – Callus - Reinstate Monica Mar 1 '16 at 3:25
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Prove that the norm $\|\cdot\| : \mathbb R^d\to [0,\infty)$ is continuous. Then everything follows from 1D analysis since $\|f\| = \|\cdot\|\circ f$.

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  • $\begingroup$ The question has already done this step. It is asking something else, so this isn't an answer as it stands. $\endgroup$ – Milo Brandt Mar 1 '16 at 3:26
  • $\begingroup$ He/she asked: "First, I am wondering if I can just apply the Extreme Value theorem and then I'm done?" I told him yes and why. $\endgroup$ – Friedrich Philipp Mar 1 '16 at 3:29

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