# Show that ||f|| attains its maximum on a compact subset

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.

• 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. – Callus - Reinstate Monica Mar 1 '16 at 3:25

Prove that the norm $\|\cdot\| : \mathbb R^d\to [0,\infty)$ is continuous. Then everything follows from 1D analysis since $\|f\| = \|\cdot\|\circ f$.