Generalization of Banach's fixed point theorem

Solution 1:

I think your answer is fine, but personally, I would do this:

$f^n$ is a contraction map, with constant $K$, hence has a fixed point $p$

For a fixed $x$, let $x_1=f(x), x_2=f^2(x), ...$ and $x_n=f^n(x)$

$d(f^N(x), f^N(p)) \leq K^m d(p, f^l(x))=K^m d(p, x_l)\leq K^m\max\limits_{i=1}^n d(x_i,p) $. Here $l$ is the smallest positive integers such that $N=m n+l$ now, as $N\rightarrow \infty$, so does $m$, $K^m\max\limits_{i=1}^n d(x_i,l)\rightarrow 0$

EDIT: I realise, I didn't give a proof that $p$ is a fixed point proof of $f$, but your proof of this is how I would do it too.

Solution 2:

$f^k$ is a contraction, so by Banach fixed point theorem, $f^k$ has a UNIQUE fixed point $x^*$, ie: $f^k(x^*)=x^*$

Then: $f(f^k(x^*))=f(x^*)$, wich means that $f^k(f(x^*))=f(x^*)$ (here $f(x^*)$ is another fixed point of $f^k$).

By the uniqueness of the fixed point of $f^k$, we get: $f(x^*)=x^*$.

Finally, $x^*$ is the unique fixd point of $f$.