The Brouwer fixed point theorem states for the continuous case that,

Every continuous application $G: D^n \rightarrow D^n$ has a fixed point

I don't fully understand the J. Milnor proof.

The proof is done by contradiction. We assume $G$ has no fixed points.

Then $\forall x \in D^n$, $||x - G(x)|| > 0$. We know that $D^n \subset R^n$. Then $D^n$ is compact means that $D^n$ is closed and bounded.

Therefore, the application that compute the distance between $x$ and $G(x)$, $$ D^n \rightarrow [0, +\infty[, \quad x \rightarrow ||x - G(x)|| $$

reach its minimal value for $||x - G(x)|| = \delta > 0$

By the Stone-Weierstrass approximation theorem, $\exists$ a polynomial, $$ P: R^n \rightarrow R^n $$

of $n$ variables and $n$ components such that $G$ be the continuous uniform limit of a sequence of polynomial $P_k$ and then we can assume that there exist a polynomial $P$ "close enough" to $G$,

$$ \forall x \in D^n: \quad 0 < ||P(x) - G(x)|| < \delta / 2 $$

We know by the reverse triangular inequality, $$ \delta / 2 > ||P(x) - G(x)|| \geq \Bigl| ||P(x)|| - ||G(x)|| \Bigr| \geq ||P(x)|| - 1 $$

Then, $$ ||P(x)|| \leq \delta / 2 + 1 $$

Let $\hat{P}(x) \equiv \frac{P(x)}{||P(x)||} \geq \frac{P(x)}{\delta / 2 + 1}$ wich is smooth.

We then compute that,

$$ ||\hat{P}(x) - G(x)|| \geq \left\Vert \frac{P(x)}{\delta / 2 + 1} - G(x) \right\Vert $$

I should conclude that this last expression is $< \delta$ and conclude that $\hat{P}$ has no fixed points which contradict the Brouwer fixed point theorem for the smooth case. I don't know how to do that. How can I finish this proof ?

  • $\begingroup$ To conclude $\hat P$ doesn't have a fixed point, you'd rather need to approximate $\|\hat P(x)-x\|$. Also, I guess $\hat P(x)=\frac{P(x)}{\delta/2+1}$ because $\|P(x)\|$ might be zero. $\endgroup$
    – Berci
    Dec 22, 2021 at 15:59

1 Answer 1


Let $\overline{P} = \frac{P}{\delta/2+1}$. $\overline{P}$ takes its values in $D^n$ as for $x \in D^n$: $\left\Vert P(x) \right\Vert \le \delta/2+1$.

You have for $x \in D^n$

$$\begin{aligned} \left\Vert \overline{P}(x) - G(x) \right\Vert &= \left\Vert \frac{P(x)}{\delta/2+1} - G(x) \right\Vert\\ &=\frac{1}{\delta/2+1}\left\Vert (P(x) - G(x)) - (\delta/2) G(x) \right\Vert\\ &\le \frac{1}{\delta/2+1}\left(\left\Vert (P(x) - G(x))\right\Vert + (\delta/2) \left\Vert G(x) \right\Vert\right)\\ &\lt \frac{1}{\delta/2+1}\left(\delta/2+ \delta/2 \right)\lt \delta \end{aligned}$$

And if $\overline{P}$ was having a fixed point $x_0$, you'll get the contradiction

$$ \left\Vert x_0 - G(x_0) \right\Vert=\left\Vert \overline{P}(x_0) - G(x_0) \right\Vert \lt \delta$$


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