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Brouwer fixed point theorem states that for any compact convex set $X$, a continuous mapping from $X$ to $X$ has at least one fixed point.

Brouwer fixed point theorem applies in particular on the close unit ball of a finite dimensional space.

Do you have counterexample(s) where Brouwer fixed point theorem does not hold on the close unit ball of a Banach space (of infinite dimension)?

Note: this question is a refinement of this one which was put on hold.

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  • $\begingroup$ Does the fact that in infinite dimension the close unit ball is not compact ensure the fact that Brouwer fixed point theorem cannot be applied ? In that case there will be a lot of counterexample $\endgroup$ – Maman Oct 25 '16 at 15:07
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A comment in the other question mentioned that the wiki page for the Brouwer Fixed Point Theorem had a counterexample for the Hilbert space $\ell^2$. I'll adapt the statement there to $\ell^p$ where $p < \infty$ so you can compare the two:

In the Banach space $\ell^p$ of $p$-norm summable real (or complex) sequences, consider the map $f : \ell^p → \ell^p$ which sends a sequence $x_n$ from the closed unit ball of $\ell^p$ to the sequence $y_n$ defined by

$$y_0=\sqrt[p]{1-\|x\|_p^p} \,\,\textrm{ and } \,\, y_n=x_{n-1} \textrm{ for } n\geq 1$$

It is not difficult to check that this map is continuous, has its image in the unit sphere of $\ell^p$, but does not have a fixed point.

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    $\begingroup$ Thanks. I imagine that you meant $y_0=\sqrt[p]{1-\|x\|_p^p}$ $\endgroup$ – mathcounterexamples.net Jun 26 '15 at 12:07
  • $\begingroup$ Yes, I did. Thanks. $\endgroup$ – muaddib Jun 26 '15 at 12:10

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