Proof for fixed point under specific circumstances

Question

Prove that every $f \in C(I, \mathbb{R})$ with $I := [-1, 1] \subset \mathbb{R}$ and $f(I) \subseteq I$ has a fixed point.


This would be true if $f$ is a contraction on $I$, since then Banach's fixed point theorem would apply. For $f$ to be a contraction on $I$, the following has to be true: $$\forall a, b \in I: \exists L<1: \vert f(b)-f(a) \vert \leq L \cdot \vert b - a \vert$$ Moving the $\vert b - a \vert$ to the left side: $$\frac{\vert f(b) - f(a) \vert}{\vert b - a \vert} \leq L < 1$$ Estimating the left side using the supremum gives: $$\sup_{a, b \in I} \frac{\vert f(b) - f(a) \vert}{\vert b - a \vert} < 1$$ Intuitively this seems to be true, but I do not know how to prove it formally.

Answer 1

There 's no need for $f$ to be a contraction; set $g(x):=f(x)-x, x\in [a,b]$. Then $g(a)=f(a)-a\geq0$ and $g(b)=f(b)-b\leq0$, for $f([a,b])\subset[a,b]$, thus $g(a)g(b)\leq0$. By Bolzano Theorem $\exists ξ\in (a,b):g(ξ)=0\iff f(ξ)=ξ$. If $f$ is additionally a contraction, then the fixed is unique; let $x,x'$ two distinct fixed points of $f$. Then $|x-x'|=|f(x)-f(x')|\leq L|x-x'|\implies1\leq L$, which is false, because $L\in (0,1)$.