# Brouwer fixed point theorem : continuous case

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 ?

• 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. Commented Dec 22, 2021 at 15:59

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