# Suppose $S^m$ is a contraction on a complete metric space $(X,d)$. I want to show that this implies $S$ has a unique fix-point. [duplicate]

This question already has an answer here:

Let $(X,d)$ be a complete metric space and let $S: X \rightarrow X$ be a mapping. Suppose there exist $m \ge 1$ such that $\underbrace {S^m = S \circ S \circ \dots \circ S}_{\text {m times}}$ is a contraction, i.e. there exist $0\le\beta<1$ such that $d(S^m(x),S^m(y)) \le \beta d(x,y)$ for all $x,y \in X$.

I want to show that this implies that $S$ has a unique fix-point, i.e. there exist a unique $z \in X$ such that $S(z) = z$.

I've tried to apply Banach fix-point theorem which imply that $S^m$ has a unique fix-point. Also, I've tried consider if $S$ has an inverse. Last, I've tried to come up with a proof by contradiction.

Can anyone guide or show how the above could be proven ?

## marked as duplicate by Martin Sleziak, Ali Caglayan, Claude Leibovici, N. F. Taussig, ShaunFeb 13 '15 at 13:30

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

• Investigate the relations between the fixed points of $S$ and of $S^m$. – Daniel Fischer Feb 13 '15 at 10:16
• But how do I know in the first place that $S$ has a fix-point ? – Shuzheng Feb 13 '15 at 10:21
• You deduce that. First you go hypothetical, "if $x$ is a fixed point of $S$, then ...". Then from what you get, you'll be able to deduce that $S$ has a unique fixed point. – Daniel Fischer Feb 13 '15 at 10:23
• Show that if $z$ is a fixed point of $S^m$ then $S(z)$ is also a fixed point of $S^m$. – user 59363 Feb 13 '15 at 10:28
• This is what I've trouble with. I've proved that $S$ has $k\le 1$ fixed points, but having trouble with the existence part. – Shuzheng Feb 13 '15 at 10:57