Brouwer's Fixed Point Theorem states, essentially, that any continuous function on a closed disc to itself has a fixed point. I am familiar with the proof based on the impossibility of a retraction from a disc to its boundary and the proof based on Sperner's Lemma. Wikipedia lists a number of other proofs.

However, it seems there is a simpler "proof" - quoted because it could be wrong - that uses no fancy machinery and I'm wondering if it is right and if so, why it isn't well known.

We will prove that any continuous $f : [0, 1]^n \rightarrow [0, 1]^n$ has a fixed point by induction on $n$. $n = 1$ amounts to the Intermediate Value Theorem. For $n > 1$ our space is $[0, 1] \times [0, 1]^{n-1}$. By the 1-d case, for each $\mathbf{u} \in [0, 1]^{n-1}$, $x \mapsto f(x, \mathbf{u})_1$, the first component of $f(x, \mathbf{u})$, has a fixed point $x$. By continuity of $f$ we may choose this fixed point, $x(\mathbf{u})$, to vary continuously in $\mathbf{u}$. By the $(n-1)$-d case, for each $y \in [0, 1]$, $\mathbf{v} \mapsto f(y, \mathbf{v})_{2, \ldots, n}$ has a fixed point $\mathbf{v}$ and we may let $\mathbf{v}(y)$ vary continuously.

If $x(\mathbf{v}(0)) = 0$ then $(0, \mathbf{v}(0))$ is a fixed point of $f$; similarly for 1. Otherwise let $X = \{(x(\mathbf{u}), \mathbf{u}) \mid u \in [0, 1]^{n-1}\}$. $X$ is the graph of a continuous function so it is closed and $[0, 1]^n \setminus X$ is open. Furthermore, $\mathbf{v}(0)$ and $\mathbf{v}(1)$ are in different components of $[0, 1]^n \setminus X$ so by an argument like the proof of the Intermediate Value Theorem, $\mathbb{v}$ must cross $X$ at some point, which is a fixed point of $f$.

  • $\begingroup$ I think the reason the textbooks exclude this is the same reason you include it: it doesn't use any other machinery (except calculus). I'm sure some authors have thought of this because I also independently came up with something similar. $\endgroup$ – Jacob Wakem Jul 8 '16 at 15:33
  • 9
    $\begingroup$ Showing that $x(u)$ varies continuously in $u$ seems to be nontrivial. There might be more than one fixed point for each $u$. $\endgroup$ – Seewoo Lee Jul 8 '16 at 15:35
  • $\begingroup$ @See-WooLee You are right, that seems to be the catch. $\endgroup$ – Reinstate Monica Jul 8 '16 at 15:45
  • 1
    $\begingroup$ I'm satisfied that you've reduced the proof to a proof of the $[0,1]^2 \to [0,1]^2$ case (assuming that the fixed point can be continuously chosen). However, I don't quite see that you've satisfactorily proved a fixed point in this case. $\endgroup$ – Ben Grossmann Jul 8 '16 at 15:52
  • 2
    $\begingroup$ "By continuity, we can pick..." I don't see how you can be sure to do that. That's a much bigger leap than you seem to think it is. $\endgroup$ – Thomas Andrews Jul 8 '16 at 16:32

In general, $x(\mathbf{u})$ can't be chosen to be continuous. Here is a simple counter-example in $\mathbb R^2$:

$$f(x,y) = \begin{cases} 2xy, & \text{if } y \le \frac12 \\ 1-2(1-x)(1-y), & \text{if } y \ge \frac12 \end{cases}$$

If $y < \frac12$, the unique fixed $x$ is $x=0$; if $y > \frac12$, the unique fixed $x$ is $x=1$. (If $y=\frac12$, then $f(x,y)=x$ fixes all points in $[0,1]$.)


I think your proof works fine - as long as the fixed points involved are all unique. What if the map taking $x $ to $(x, \mathbf {u} )$ has multiple fixed points, so your $x (\mathbf {u}) $ is ill-defined? There might not be a continuous curve of fixed points over the whole space.

  • $\begingroup$ Note that the proof arbitrarily chooses some continuous $x(\mathbb{u})$. There may be additional fixed points not chosen. $\endgroup$ – Reinstate Monica Jul 8 '16 at 16:07
  • $\begingroup$ What I mean is there might be no such curve. For example, in the case of $[0,1]^2$, the fixed points could make two disjoint parabolas that together cover all values of each coordinate. $\endgroup$ – Reese Jul 8 '16 at 16:45
  • 1
    $\begingroup$ Not sure why this answer is downvoted, when it points out essentially the same thing as the accepted answer. $\endgroup$ – Steven Gubkin Jul 8 '16 at 20:20

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.