Let $\phi:\mathbb{R}^2\rightarrow \mathbb{R}^2$ be a continuous map.

How do I prove that there exist $a>0$ and $x\in\mathbb{R}^2$ such that $\phi(x)=ax$?

What I know:
I thought maybe this can be proved by contradiction. So suppose that there is no $x$ such that $\phi(x)=ax$. How would I continue?
Also this question makes me think of Brouwer's fixed point theorem, can I use the strategy of his proof?

  • $\begingroup$ Can't you just apply Brouwer's fixed point theorem with the homeomorphism $re^{i\theta}\mapsto \frac{r}{1+r}e^{i\theta}$ of $\mathbb{C}\rightarrow D^2$? $\endgroup$
    – neth
    May 3 '16 at 20:51

Yes, you can reuse Brouwer's fixed point theorem. Let $\bar{\phi} : D^2 \to \mathbb{R}^2$ be the restriction of $\phi$ to $D^2 = \{ x \in \mathbb{R}^2 \mid \|x\| \le 1 \}$ (for whichever norm you want, usually the $L^2$ norm). Then since $D^2$ is compact, $\|\bar{\phi}\|$ reaches a maximum.

  • Either $\max \|\bar{\phi}\| = 0$, in which case $\bar{\phi}(0) = 0 = 1 \cdot 0$;
  • Or $a = \max \|\bar{\phi}\| > 0$, in which case the map $$x \mapsto \frac{\bar{\phi}(x)}{a}$$ is a map $D^2 \to D^2$ (because $\|\frac{\bar{\phi}(x)}{a}\| \le 1$). Hence by Brouwer's fixed point theorem, it has a fixed point $x \in D^2$, i.e. $\bar{\phi}(x)/a = x$. It follows that $\phi(x) = ax$.
  • $\begingroup$ $L^2$ norm is just the Euclidean right? $\endgroup$
    – user335050
    May 3 '16 at 13:53
  • $\begingroup$ I don't really understand what is happening in the case max$||\bar{\phi}||=0$, are all the points in $D^2$ fixed points then? $\endgroup$
    – user335050
    May 3 '16 at 13:56
  • 1
    $\begingroup$ @HenryJonesJr. Yes, $L^2$ is the Euclidean norm. When $\max \|\bar\phi\| = 0$, it means $\|\bar\phi(x)\| \le 0 \implies \bar\phi(x) = 0$ for all $x \in D^2$. Only $0$ is a fixed point in this case $\endgroup$ May 3 '16 at 14:34

Your Answer

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