I want to prove that a function $f: S\to S$ where $S$ is a compact metric space and satisfying $d(f(x),f(y))<d(x,y)$ has a unique fixed point. My attempt :
We define $g(x)=d(x,f(x))$ which is continuous on a compact set and thus attains a minimum at, say, $\alpha$. There is two cases :
$g(\alpha) = 0$ and we are done
$g(\alpha)>0$. Then we have $g(f(\alpha)) = d(f(f(\alpha)),f(\alpha))<d(f(\alpha),\alpha)=g(\alpha)$ which is a contradiction
The uniqueness is clear from the property of $f$. I would like to know if it is correct and if there is a way to construct a sequence converging to a fixed point (as for the Banach fixed point theorem). Thank you
Your proof is correct. The same sequence $x_{n+1} := f(x_n)$ from Banach's theorem will converge to the unique fixed point $x^\ast$ for any choice $x_0 \in S$. To show this, apply the fact that compactness implies sequential compactness. Any converging subsequence will guarantee convergence of the whole sequence $(x_n)$ via the $L = 1$ Lipschitz condition on $f$.
Asymptotically, this sequence $(x_n)$ will converge more slowly than would its Banach analogues where $L < 1$. In practice, the convergence speed may, on occasion, even appear reversed for specific functions $f_1$, $f_{<1}$. For when the asymptotics finally kick in for $\tilde{x}_{n+1} = f_{<1}(\tilde{x}_n)$, the iterated value of $x_{n+1} = f_1(x_n)$ and its fixed point could be very close to equal.