If we let $X$ be a complete metric space, and let $S:X\to X$ be a map, such that $S^m$ is a contraction. We now want to show, that $S$ has a unique fixed point
This is what I've thought so far:
Due to Banachs fixed point theorem is it enough to show, that $S$ is a contraction. Due to $S^m$ being a contraction, we know this about $S^m$(the definition of being a contraction): $$\exists\beta, 0\le\beta\lt1:d(S^mx,S^my)\le\beta d(x,y),\forall x,y\in X$$
I'm not really sure how to show that S is a contraction.. Any ideas ad to how to approach this?
If $x,y$ are fixed points of $S$, then $d(S^n x, S^n y) = d(x,y)\leq \beta d(x,y)$, from which we conclude that $d(x,y) = 0$. Hence there is at most one fixed point.
Since $S^n$ is a contraction, it has a unique fixed point $x_0$. Then we have $x_k = S^k x_0$. Since $x_{k+n} = x_k$, we see that each $x_0,...,x_{n-1}$ is a fixed point of $S^n$, from which it follows that $x_k = x_0$ for all $k$, and hence $S x_0 = x_0$.