Show that the $\parallel supremum\: of\: a \: function \parallel$is finite and that is a norm on a compact set

Question

Let $K$ be a compact subset of $\mathbb{R^n}$ and let $C(K,\mathbb{R^M})$ denote the vector space of all continuous function from $K$ to $\mathbb{R^n}$. Show that for $f$ in $C(K,\mathbb{R^M})$, the quantity $\parallel f \parallel_\infty = sup_{x\in K}\parallel f(X)\parallel_2$ is finite and $\parallel . \parallel_{\infty}$ is a norm on $C(K,\mathbb{R^M})$.

Im not really sure how to go about this one, i know intuitively that since he dimention of $\mathbb{R^M}$ is finite so the subset should also have a finite dimention and so the norm has to be finite but isnt that obvious or si there something i am not seeing? also im unfamilir with the notation $\parallel f(X)\parallel_2$, why is there a 2 in subscript?

Any help would be useful, thanks

Answer 1

Do you know the fact that every real-valued continuous function from a compact set obtains a maximum? Then you can use the fact that the function $x\mapsto \|f(x)\|_2$ is continuous to show that $\|f\|_{\infty}$ is equal to that maximum value (and thus finite).

As for the subscript 2, there are many choices of norm for $\mathbb{R}^M$, in particular, the one you are taking $\|v\|_2=\sum_{i=1}^{M}|v_i|^2$, where $v=(v_1,v_2,\dotsc,v_M)$. The 2 is from the fact that you're squaring.

EDIT: As commented, statement was false without compactness.