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

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  • $\begingroup$ Your example follows from the regular EVT. $\endgroup$ – zhw. Jun 28 '17 at 18:31
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    $\begingroup$ $\lVert f(x)\rVert$ 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
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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.

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  • $\begingroup$ 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
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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 $m>1$.

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  • $\begingroup$ 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

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