8
$\begingroup$

Let $B$ denote the closed unit ball in $\mathbf{R}^n$. Brouwer's fixed point theorem states that every continuous map $f:B\to B$ has a fixed point. There is a simple proof using Stokes's theorem, at least for the special case in which $f$ is smooth, as presented on Wikipedia here.

The page also states that this case contains the full generality of the theorem, because if $f:B\to B$ is continuous without fixed points then $\epsilon = \inf_{x\in B} |x-f(x)| > 0$, so we can just convolve (each component of) $f$ with a smooth bump $\psi:\mathbf{R}^n\to\mathbf{R}$ supported on $\epsilon B$ to get a smooth counterexample to the theorem.

Unfortunately, as it stands the proof doesn't work, because the distance of $f(B)$ to $\partial B$ could well be zero, in which case $\tilde{f} = \psi\ast f$ might not satisfy $\tilde{f}(B)\subset B$. Does anybody see a resolution to this difficulty?

EDIT, following Willy's answer. I've just realised that I was confused when I asked this question. $\tilde{f}(B)\subset B$ was never really an issue; the issue was rather that convolution isn't fully defined near the boundary. The most immediate interpretation is to extend $f:B\to B$ by $0$ to $\mathbf{R}^n\to B$, but then mollifying $f$ doesn't give you a uniformly nearby $\tilde{f}$. The interpretation that works is to extend $f:B\to B$ to any uniformly continuous $F:\mathbf{R}^n\to B$, such as

$$F(x) = \begin{cases} f(x) & \text{if $|x|\leq 1$,}\\ f(x/|x|) & \text{if $|x|\geq 1$,}\end{cases}$$

and then mollify.

  • 2
    Why would you want to convolve? Isn't the point just that the smooth functions $f\colon B \to B$ are uniformly dense?2012-09-12
  • 1
    Yes, Wikipedia often contains irresponsible statements. In this case, see [CS's comment](http://en.wikipedia.org/wiki/Talk:Brouwer_fixed-point_theorem#elemantary_proof_with_stokes.27_theorem) which I have not verified myself.2012-09-12
  • 0
    @t.b. I think Sean is quoting Wikipedia there...2012-09-12
  • 0
    @Willie: Okay, then I ask the responsible Wikipedist :)2012-09-12
  • 0
    @WillieWong I saw his comment, but it wasn't clear to me how one homotopes a continuous map to a nearby smooth one. Is it obvious?2012-09-12
  • 0
    @t.b. Good point. Then my question is: What's the easiest way of approximating continuous $f$ by smooth $f$ without leaving $B$?2012-09-12
  • 1
    I would just convolve in charts but the details are somewhat fiddly (I don't have the time to write this up now). Concerning the homotopy question, see [this thread](http://math.stackexchange.com/q/176399/5363) (I'm not sure if the manifolds with boundary case is included in those references, but Hirsch *should* do it).2012-09-12
  • 1
    Hmm, I suppose we could just convolve with a bump supported on $\epsilon B$, and then scale by $1/(1+\epsilon)$. Points will move by at most $2\epsilon$.2012-09-12
  • 0
    @Sean: that won't work exactly on the boundary; if at a boundary point $F(x)$ is nonzero, $\eta * F$ at that point can be in principle only $F(x)/2$ (Think the one dimensional case.) And after rescaling, the boundary will certainly change by $F(x)$, since the value at the new boundary is 0.2012-09-12
  • 0
    To not let the boundary all go wonky, you have to take a partition of unity. The required proof that @t.b. has in mind is probably a modification of Theorem 13 in [my notes on Sobolev spaces](https://documents.epfl.ch/users/w/ww/wwywong/www/lecturenotes/NotesOnSobolevSpaces.pdf).2012-09-12
  • 0
    @WillieWong What if I first extend $f:B\to B$ to a continuous function $f:\mathbf{R}^n\to\mathbf{R}^n$?2012-09-12
  • 0
    @Willie, yes essentially that argument. I just checked in my copy of Hirsch and the required theorem is proved as Theorem 3.3 in Chapter 2 on page 57: $C^s(M,N)$ is dense in $C^r(M,N)$ whenever $0 \leq r \lt s \leq \infty$ (same with $(M,\partial M; N, \partial N)$ in place of $(M,N)$) for compact $C^s$-manifolds with boundary. Google doesn't let me check but maybe this [direct link](http://books.google.com/books?id=iSvnvOodWl8C&pg=PA57) works for you.2012-09-12
  • 0
    Not sure where to put my comment, but since I'm referenced above... It's apparent there's a bunch of analysts here :-) But to a topologist, the theorem is quite easy to show. Since we know the smooth maps are not fixed-point free and the continuous map can be approximated by smooth maps, just show the set of continuous maps with a fixed point is closed.2013-05-08
  • 0
    @Chan-HoSuh Thanks for your comment. I think this discussion is exactly about showing (in the fewest words) that continuous maps $B\to B$ can be uniformly approximated by smooth such maps. The resolution was that the Wikipedia party-line "to prove that a map has fixed points, one can assume that it is smooth, because if a map has no fixed points then convolving it with a smooth function of sufficiently small support produces a smooth function with no fixed points" is true for an appropriate meaning of "convolve" (one must first extend to a uniformly continuous function $\mathbf{R}^n\to B$).2013-05-09

2 Answers 2

6

Sean's last comment inspired the following answer:

Let $100\epsilon < \inf |x - f(x)|$. Let $g(x) = \frac{1}{1 + 10\epsilon} f(x)$. Then by triangle inequality we have that $|x - g(x)| > \epsilon/2$.

Let $h: (1+10\epsilon)^{-1}B \to (1+10\epsilon)^{-1}B$ be the smooth map formed by $$ h(x) = \eta* g(x) $$ where $\eta$ is a mollifier supported in $\epsilon B$. We have that $h(x)$ is smooth and has no fixed points etc.

  • 0
    Neat trick!${}{}{}$2012-09-12
  • 0
    @t.b. Thanks. It occurred to me that [uniform approximation](http://math.stackexchange.com/questions/194599/continuous-brouwers-fixed-point-theorem-via-stokess-theorem/194614#comment447745_194599) is a bit stronger than we actually need (since we are arguing by contradiction).2012-09-12
  • 1
    Nice. If I might paraphrase, this argument can be described as follows. If convolving $f$ with a mollifier is saying "replace $f(x)$ with the its average (of sorts) on $x + \epsilon B$", then what you're saying is "replace $f(x)$ with its average on $x/(1+10\epsilon) + \epsilon B$". This is no longer a simple convolution, but it still replaces $f$ with a uniformly nearby smooth $\tilde{f}:B\to B$.2012-09-12
  • 1
    A similar idea, more closely related to my comment: Given $f:B\to B$ define $F:\mathbf{R}^n\to B$ by $F(x) = f(x)$ if $x\in B$ and $F(x) = f(x/|x|)$ otherwise. Now mollify $F$ and restrict the result to $B$. The result is a smooth $\tilde{f}:B\to B$ such that $\|f-\tilde{f}\|_\infty < \epsilon$.2012-09-12
  • 1
    @Sean: indeed. Furthermore, if $f$ has no fixed points, then clearly $F$ that you constructed also has no fixed points on $rB$ for any $r > 0$. In both of our constructions the trick is to get a map from a ball to a strictly smaller subset of itself. After which we can mollify and restrict.2012-09-12
0

hmm yes fix $\varepsilon$ greater than $0$ and let $x_n$ be a convergent subsequence of a convergent sequence that converges to the same limit $l$. Then fix $\delta$ greater than, but not equal to $0$, and let $|x_n-l|<\delta$ iff $f(x_n)\rightarrow f(l)$.

Furthermore I would suggest fixing $\varepsilon$ greater than $2\delta/5$ and then let the convergent bijective map $f$ converge to a fixed limit $cl$, where $c$ is a non-linear constant.