# Proof of a corollary of the Banach Fixed Point Theorem

If $(X,d)$ is a complete metric space, and $f: X \rightarrow X$ is a continuous function, show that if $f^{N}$ is a contraction (for some $N > 0$),then $\exists! x \in X$ such that $f(x) = x$.

I have showed the uniqueness part of the proof. I'm currently having issues with the proof for existance of such point $x \in X$. I don't know whether the proof I have is correct or not. My sketch goes as follows:

Since $f^{N}$ is a contraction, by Banach Fixed Point Theorem, $\exists! x \in X$ such that $f^{N}(x) = x$. Let this point be $x'$. Now, assume that the statement we want to prove is not true. Hence, $\forall x \in X$, $f(x) \neq x$. Thus, this implies $d(f(x), x) > 0$, $\forall x \in X$. Using the fact that $f^{N}$ is a contraction, we obtain (for $\alpha \in (0,1)$):

$d(f^{N+1}(x'), f^{N}(x')) = d(f^{N}(f(x')), f^{N}(x')) \leq \alpha d(f(x'), x')$

Since we had $f^{N}(x') = x'$, it follows $d(f^{N+1}(x'), f^{N}(x')) = d(f(x'), x') \leq \alpha d(f(x'), x')$, so $(1-\alpha)d(f(x'), x') \leq 0$. Then, if $d(f(x'), x') = 0$, we obtain a contradiction as $f(x') \neq x'$ (which was our assumption). Conversely, if $1-\alpha \leq 0$, we have $1 \leq \alpha$, which will be another contradiction as $\alpha \in (0,1)$ since this is the contraction constant for $f^{N}$. Thus, by contradiction, then $\exists! x \in X$ such that $f(x) =x$.

Is this proof suitable for this theorem? I'm not sure if it's correct or not. If it's not correct, which would be the mistakes done in this proof and how to avoid them?

Thank you for all the help.

Here's a smaller one that we did in class: By the Banach fixed point theorem: Choose $x_0 \in X$, then $\exists x: x = \lim_{n\to\infty}f^{kn}(x_0)$. Then as the fixed point is unique, $$x = \lim_{n\to\infty}f^{nk}(f(x_0))=\lim_{n\to\infty}f(f^{nk}(x_0))=f(\lim_{n\to\infty}f^{nk}(x_0))=f(x),$$(the last equality follows from f being continuous )