# All continuous functions of finite order have a fixed point

I was attempting to do an exercise on my topology book. It states the following:

Let $$f:\mathbb{R} \mapsto \mathbb{R}$$ be a homeomorphism such that $$\exists p \in \mathbb{N}: f^p =$$ identity. Prove that $$f$$ has a fixed point.

I think I have a solution but I never really use the fact that $$f$$ is a homeomorphism. That is:

Let $$g(x)=f(x)-x$$. Then $$g$$ is a continuous function. Now, let's say that $$f(x)\geq x \ \forall x \in \mathbb{R}$$. Then $$f^p(x) \geq f(x) \geq x = f^p(x) \forall x$$, which implies $$f(x)=x \forall x$$.

In the same way, if $$f(x)\leq x \forall x$$ we have $$f^p(x) \leq f(x) \leq x = f^p(x) \implies f(x) = x$$.

So, if $$f(x)$$ is not the identity, then $$\exists x_1, x_2 \in \mathbb{R}: f(x_1) < x_1, f(x_2) > x_2$$, and because $$g$$ is continuous and $$g(x_1)<0, g(x_2)>0$$ then there's a point $$\bar{x}: g(\bar{x})=0 \implies f(\bar{x})=\bar{x}$$, so it is a fixed point.

I've not actually used the fact that $$f$$ is a homeomorphism, I only used the fact that it is continuous and has finite order, so is my proof wrong?

• I think your proof is fine. Note that being continuous and having finite order already implies that $f$ is a homeomorphism, since $f^{p-1}$ is a left- and right-inverse. – PhoemueX Jul 27 at 16:24
• That's a fair point indeed – DottorMaelstrom Jul 27 at 16:29

## 1 Answer

Your proof is correct, but in my opinion it could be formulated more transparently. In fact, you consider three cases.

Case 1: $$\forall x \in \mathbb{R} : f(x) \ge x$$.

Case 2. $$\forall x \in \mathbb{R} : f(x) \le x$$.

Case 3: Neither 1 nor 2 is satisfied. Then $$\exists x_1, x_2 \in \mathbb{R} : f(x_1) < x_1, f(x_2) > x_2$$. Now introduce $$g$$.