Function eventually contractive implies it has unique fixed point

Question

Let $(X,d)$ be a complete metric space, and let $f:X\to X$ be eventually contractive (there exists $p\in\mathbb{N}$ such that $g:=f^{\circ p}$ is contractive). Then $f$ has a unique fixed point.

Is there a way to prove this without using the Banach contraction theorem? Would appreciate a hint.

Answer 1

I suppose $f$ is continuous...

If $f^p$ is a contraction, then $(f^{np}(x))_n$ converges to the fixed point $x_0$ of $f^p$ for every $x\in X$. But then $f^{np+1}(x)$ converges to $f(x_0)$. But it also has to converge to $x_0$ (take $x'= f(x)$ as initial point). We conclude that $f(x_0) = x_0$.

Note that the fixed point will be the limit of $f^n(x)$ for every initial point $x$. The convergence rate is of the order $q^{\frac{n}{p}}$.