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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.

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    @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

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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.

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    @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
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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.