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This question already has an answer here:

Let $(X,d)$ be a complete metric space and $$f \ :\ X\rightarrow X$$ be a map such that , for some positive integer $k$ , $$f\circ f\circ .....\circ f(\ k\ \ fold\ \ composition\ \ with\ \ itself\ )$$ is a contraction.

Then $f$ has a unique fixed point .

How should I approach this problem $?$

Sorry for the lack of efforts here .

Please give me some hints as to how to begin the thinking $?$

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marked as duplicate by user10354138, Xander Henderson, Shailesh, Tianlalu, Ben Dec 17 '18 at 6:17

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Use Banach fixed-point theorem then $f^k$ has unique fix point $\alpha$ i.e $f^k (\alpha)=\alpha$ then $f(f^k (\alpha))=\color{red}{f^k(f(\alpha))}=f(\alpha)$thus $f(\alpha)$ is also fixed point, uniqueness implies $f(\alpha)=\alpha$

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(I will give a full answer, I am not feeling subtle today)

Let $T = f\circ \ldots \circ f =: f^{(k)} $ (k fold composition of f).

Then surely $T$ is a contraction map?

We will show that one of $f$'s fixed points is in fact $T$'s fixed point:

Let $x$ be the unique fixed point of $T$. Then $f(x) = f(T^m(x))$ for any $m$.

But, $T$ is a stack of $f$'s. So they must commute.

$ f(x) = T^m(f(x))$ for any $m$.

Let $m \to \infty$. So, in particular, the whole space gets crushed to a single point $x$ by $T^m$.

So $f(x) = x$.

Next, showing that this point is unique:

Let $z$ be any fixed point of $f$:

$f(z) = z$

For any m: $T^m(z) = z$ (remember $T$ is just a stack of $f$'s)

Let $m \to \infty .$

Then $x =z$ QED.

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