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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?

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    $\begingroup$ We cannot say that $T$ itself is $q$-contractive. But what can you say about fixed points of $T$ and fixed points of $T^n$? $\endgroup$ – hardmath Jan 6 '14 at 14:35
  • $\begingroup$ Fixed points of T are fixed points of $T^n$. $\endgroup$ – math12 Jan 6 '14 at 14:37
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    $\begingroup$ That's a good start because now we know $T$ has at most one fixed point. Think about the existence of a fixed point of $T$; the list of candidates is short. $\endgroup$ – hardmath Jan 6 '14 at 14:42
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    $\begingroup$ Is it the same fixed point that $T^n$ does have? $\endgroup$ – math12 Jan 6 '14 at 14:49
  • $\begingroup$ If yes, why? I cannot see it clearly. $\endgroup$ – math12 Jan 6 '14 at 15:24
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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 $0 \le q \lt 1$.

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