How to show something is a contraction? 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?
 A: 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 $S x_0 = x_0$.
A: Here's another way around, which is from an exercise in  Fixed point theory and applications Agarwal et al (2001).
In general, it's not necessarily true $\operatorname{S}$ is a contraction with respect to $d$. But it can be, provided you find a new metric $\sigma$, which is defined as: $$\sigma(x,y): = d(x,y)+\frac{1}{\beta}d(\operatorname{S}x, \operatorname{S}y)+ \ldots+\frac{1}{\beta^{m-1}}d(\operatorname{S}^{m-1}x, \operatorname{S}^{m-1}y)$$ 
It's not difficult to show it's indeed qualified as a metric.And we have:
$$\sigma(\operatorname{S}x, \operatorname{S}y) = \beta \sigma(x,y) -\beta d(x,y) + \frac{1}{\beta^{m-1}}d(\operatorname{S}^{m}x, \operatorname{S}^{m}y)$$ Hence $\sigma(\operatorname{S}x, \operatorname{S}y) \leq \beta \sigma(x,y)$.
