# Showing existence of a fixed point

So, I have a continuous function $$f$$ from $$\left[0,1\right]$$ to $$\left[0,1\right]$$ such that : $$\forall \, x_1,x_2\in\left[0,1\right]:\; \mid x_1-x_2\mid\;\leq \;\mid f(x_1)-f(x_2)\mid$$ It's easy enough to prove that $$f$$ is both surjective and injective so $$f$$ is a bijection, and also $$\{f(1),f(0)\}=\{0,1\}$$. Now I'm trying to prove that $$f$$ must have a fixed point. When drawing an example it seems obvious, but I'm not sure how to go about the proof, setting aside the trivial case when $$f(1)=1$$ and $$f(0)=0$$. Since $$f$$ is a bijection and if we assume we're in the case $$f(1)=0, f(0)=1$$ then $$f$$ is strictly decreasing but not sure how that would help. So could anyone give some kind of hint or an observation I might have missed ? Thanks.

Assume that we are in the case $$f(1)=0$$, $$f(0)=1$$. Consider the function $$g:[0,1]\to\mathbb{R},\; x\mapsto f(x)-x$$. Then, $$g$$ is also continuous. Note $$g(0)=f(0)=1$$ and $$g(1)=f(1)-1=-1$$. By the intermediate value theorem (Bolzano's theorem), there exists a $$x\in[0,1]$$ such that $$g(x)=f(x)-x=0$$, which completes your proof.
Think of of the link between a fix-point and the function $$g: [0,1] \rightarrow [0,1], x\mapsto x$$. If you were to find a fix-point what kind of equality would hold? Try to link that to $$g$$ and $$f$$.
You have already shown that $$f(x)$$ is bijective. Take a look at $$g(x)=f(x)-x$$ for the case $$f(1)=0$$, $$f(0)=1$$. If $$g(x)$$ has a zero, we are done by definition of a fixpoint. Now $$g(1)=-1$$, $$g(0)=1$$, so by continuity, there exists a $$t\in[0,1]$$, for which $$g(t)=0$$.
• Ok now I wish I had thought of that, it was really not that hard and I had done it before. But I don't see why the fact that $f$ is bijective matters ? Even if it's not bijective, a continuous mapping from $\left[0,1\right]$ to $\left[0,1\right]$ will still have a fixed point. Am I missing something again ? Mar 30 '20 at 18:00