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The map $f:\bar{B}\to \bar{B}, $ continuous such that $ \bar{B}\subseteq C[0,1]$, need not have a fixed point.

Know about the Brouwer fixed point Theorem on $\mathbb{R} ^n$ which states that if $ \bar{B}\subseteq \mathbb{R} ^n,$ closed and $f:\bar{B}\to \bar{B} $ is continuous, then there exists $x^*$ such that $f(x^*)=x^*.$

However, I sense that it's not true on $C[0,1]$ but I can't think of an example. Can anyone, please, provide an example? Thanks for your time!

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  • $\begingroup$ What do you know about your set $B$? $\endgroup$ – TM Gallagher Apr 10 at 17:03
  • $\begingroup$ Let $f(x) = x+c$ where $c$ is a non zero constant. $\endgroup$ – copper.hat Apr 10 at 17:05
  • $\begingroup$ @TM Gallagher: It is closed. $\endgroup$ – Omojola Micheal Apr 10 at 17:11
  • $\begingroup$ And bounded and convex as well, no? Those are necessary conditions for the Brouwer fixed point theorem in $\mathbb{R}^n$ too $\endgroup$ – Charles Madeline Apr 10 at 17:12
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    $\begingroup$ Oh ok $\bar{B}$ is the unit ball I see. Then considering Schauder's theorem (same conclusion as Brouwer, but on convex compact subsets of Banach spaces), what you want to know is why the unit ball in $\mathcal{C}[0,1]$ is not compact. This is known as Riesz's lemma: unit closed balls are closed in normed vector space iff the dimension is finite. Does that help? (Of course you want an explicit counterexample so this does not answer completely the question) $\endgroup$ – Charles Madeline Apr 10 at 17:18

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