Brouwer's fixed point theorem states that any continuous function $f$ mapping a compact convex set into itself has a fixed point, i.e. a point such that $f(x)=x$.

I am somewhat unclear on the restriction on the domain and range. Suppose the domain of the function is a compact and convex set $A$. I understand the theorem to require $f$ to be a mapping $A \to A$. Must the function $f$, however, be able to take on all values in $A$?

Suppose we certainly know that the range of the continuous function $f(x)$ is within the set $B \subset A$. Does Brouwer's fixed point theorem still apply? I would answer yes, beacause we can still state that the range of $f$ is the set $A$, but am not sure.

In two dimensions, I can show using the Intermediate Value Theorem, that in such a case there is a fixed point. For example if $x \in [0,a]$ and I know that $f(x) \in [1,b]$ (specifically, $f(0)=1$ and$f(a)=b$), where $a>1$ and $b<a$, then by the intermediate value theorem, it follows that a fixed point exists.

I would also conclude, that if insteaed I know that in this case, if $f(x)\in[-1,b]$, that the Brouwer fixed point theorem does not apply. Is that correct?

  • $\begingroup$ Compact convex and non-empty interior. Such sets are homeomorphic to $D^n$. $\endgroup$ – Henno Brandsma Aug 30 '17 at 18:32

The function need not be onto to apply Brouwer. A constant map $f: D^n \to D^n$ will certainly have a fixed point but not be surjective.

The "best" version of BFT requires continuity on a space homeo to a disk/ball/whatever in space.

  • $\begingroup$ Thank you very much, So does that mean, for example, that for a continuous function $f$ with domain [0,5] and range [-1,4], BWT applies and a fixed point exists? $\endgroup$ – user470533 Aug 30 '17 at 16:23
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    $\begingroup$ Ah, good question. The answer is no. Brouwer assumes $f: X \rightarrow X$, which isn't your situation in your comment. For example, what if $f: [0,1] \to [4,5]$? Having a fixed point is impossible since the domain is disjoint from the codomain. The same problem happens if they overlap as in your question: they may not overlap "correctly." $\endgroup$ – Randall Aug 30 '17 at 16:54
  • $\begingroup$ Thank you so much. So just to make sure I understood. For a mapping $f: [0,5] \to [1,4]$ BFT applies. It also applies to a mapping $f: [1,4] \to [0,5]$. Are both these statements correct? Sorry for being so confused and thanks so much again! $\endgroup$ – user470533 Aug 31 '17 at 10:10
  • $\begingroup$ For the first, yes: you simply alter the map by including $[1,4]$ into $[0,5]$ which you can do because the former is a subset of the latter. For the second, no: there is no way to similarly enlarge the domain. Where are you going to send those extra values? And even if you could, what if the resulting fixed point was not in the domain of the original problem? $\endgroup$ – Randall Aug 31 '17 at 10:49

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