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