Consider a complete metric space $(X,d)$ and $T\colon X\to X$. Suppose there exists $n\in\mathbb{N}$ such that the n-th power of $T$ is $q$-contractive. Show that then $T$ has exactly one fixed point $\overline{x}\in X$.
The n-th power is defined inductively: $$ T^{n+1}(x):=T(T^{n}(x)), n\in\mathbb{N}. $$ And for a function $T\colon X\to X$, $q$-contractive means $$ \exists 0\leq q<1~\forall~x,y\in X: d(T(x),T(y))\leq q d(x,y). $$
Now to the proof.
I think I have to show, that $T$ is q-contractive, because then it follows with Banach, that $T$ does have exactly one fixed point. So I have to show, that there exists a $0\leq q<1$ so that for all $x,y\in X$ it is $$ d(T(x),T(y))\leq q\cdot d(x,y). $$ And most likely I have to use the q-contractivity of $T^n$, i.e. that there exists a $0\leq q <1$ so that for all $x,y\in X$ it is $$ d(T^n(x),T^n(y))\leq q\cdot d(x,y). $$
Can you help me?
The attempt to show $T$ is $q$-contractive is doomed, as we will show by an example below.
However any fixed point of $T$ is also a fixed point of $T^n$, and there is only one of these. So if we can show $T$ has a fixed point, we are done.
Let $x \in X$ be the unique fixed point of $T^n$, and consider $T^n(T(x))=T(T^n(x))=T(x)$. But now $T(x)$ is a fixed point of $T^n$, so $T(x) = x$ and $x$ is also a fixed point of $T$.
To show $T$ itself need not be $q$-contractive, consider $T:X\rightarrow X$ on $X=[-1,1]$ defined by $X(x) = |x|$ if $x \lt 0$ and $X(x) = x/2$ if $x \ge 0$. Then $T^2$ is $\frac{1}{2}$-contractive, but $T$ is not $q$-contractive for any $0 \le q \lt 1$.