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 $\|f(x_0)\|\le \|f(x)\|, \forall x\in K$?

  • $\begingroup$ Your example follows from the regular EVT. $\endgroup$ – zhw. Jun 28 '17 at 18:31
  • 2
    $\begingroup$ $\lVert f(x)\rVert$ is a function from $\mathbf R^n$ to $\mathbf R$! $\endgroup$ – Bernard Jun 28 '17 at 18:33
  • $\begingroup$ en.wikipedia.org/wiki/Vector_optimization $\endgroup$ – Red shoes Jun 29 '17 at 3:57

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.

  • $\begingroup$ The thing is that I'm wondering whether or not I should have been deducted marks for stating my theorem in the way I did. I think that my statement was correct. $\endgroup$ – sequence Jun 28 '17 at 18:32
  • $\begingroup$ @sequence the broad strokes of your statement are correct, but you're not being precise. See my latest edit. $\endgroup$ – Omnomnomnom Jun 28 '17 at 18:37
  • $\begingroup$ If $f:\mathbb{R}^m\to \mathbb{R}^m$ is bounded on $K$ then why can't we use the same notion of boundedness in EVT without the additional step of introducing the norm function? $\endgroup$ – sequence Jun 28 '17 at 18:44
  • $\begingroup$ Because that's not what people other than you mean by the "EVT". $\endgroup$ – Omnomnomnom Jun 28 '17 at 19:05

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 $m>1$.

  • $\begingroup$ But isn't this just the way in which one writes the definition? It looks like both definitions are actually equivalent. For if we consider your definition, then a bounded function on a compact set also has no order if $n>1$, but we do take norms there! $\endgroup$ – sequence Jun 28 '17 at 18:36
  • $\begingroup$ You are mixing the notion of boundedness with Extreme values. They are related ONLY in the one-dimensional case. $\endgroup$ – uniquesolution Jun 28 '17 at 18:37

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.