Given a function $g(x)$ defined over $\Omega = [a,b]$ with the following properties:
- $\Omega$ stable by $g$ : $g(\Omega) \subset \Omega \iff \forall x \in \Omega, g(\Omega) \in \Omega$
- $\exists K,K<1, \forall x \forall y, |g(x)-g(y)|<K|x-y|$
How do I show that the function $g(x)-x$ changes sign over $\Omega$ and deduct that exists a fixed point over $\Omega$? I am inclined to use the fact that $(g(a)-a).(g(b)-b)<0$, but I am not quite sure how to get there.
$g(a) \in [a,b] \implies g(a) \ge a$. If $g(a)=a$, then $a$ is a fixed point.
$g(b) \in [a,b] \implies g(b) \le b$. If $g(b)=b$, then $b$ is a fixed point.
Otherwise, $g(a)-a>0$ and $g(b)-b<0$.
Finally, $|g(x)-g(y)|<K|x-y|$ implies that $g$ is continuous and you can apply the intermediate value theorem to $f(x)=g(x)-x$, as planned.