Extreme value theorem for $f:\mathbb{R}^n\to \mathbb{R}^m$

Question

A marker comments that the EVT only considers functions of the form $f:\mathbb{R}^n\to \mathbb{R}$. However, I don't understand why this should be the case. For there is, for example, the notion of a bounded function of the form $f:\mathbb{R}^n\to \mathbb{R}^m$ on a compact set, which assumes the existence of $M\in \mathbb{R}$ such that $\|f(x)\|\le M, \forall x\in A\subset \mathbb{R}^n$, with $A$ being the domain of $f$.

So why can't we then apply the EVT to functions of the form $f:\mathbb{R}^n\to \mathbb{R}^m$ and say that on a compact set $K$ $f$ achieves a maximum and a minimum in the sense that $\exists x_0 \in K$ such that $\|f(x_0)\|\le \|f(x)\|, \forall x\in K$?

Answer 1

You can apply the EVT to such functions, but there is no need to invent such a theorem. If $f:\Bbb R^n \to \Bbb R^m$ is continuous, then the map $$ g(x) = \|f(x)\| $$ is a just another example of a continuous map from $\Bbb R^n$ to $\Bbb R$.

---

If you said that your statement is a consequence of the EVT, then your statement is correct. If you said that your statement is a version of the EVT, then you have misstated/misunderstood the EVT and perhaps deserve a slight deduction.

Answer 2

In order to speak about an extreme value of a function $f$, in the sense of maximal value or minimal value, you need the notion of order. That is, you want to be able to determine whether $f(x)<f(y)$ , or $f(x)>f(y)$. Since $\mathbb{R}^m$ has no natural order except when $m=1$, there is no EVT for functions whose range is $\mathbb{R}^m$, in general. Also, observe that the function $x\to||f(x)||$ is a function whose range is $\mathbb{R}$, not $\mathbb{R}^m$ with $m>1$.