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I wanted to show that if $f:X\to X$ is a function from a complete metric space to itself and if $f^k$ is a contraction, then $f$ has a unique fixed point (say $p$) and for any $x$ in $X$ $f^n(x)\longrightarrow p$.

What I did is use what I know from Banach as follows: Take $x$ arbitrary in $X$. Then consider the sequence $f^n(x)$, which I denote by $x_n$. Let $x^{(i)}$ be the sequence $x_i, x_{i+k}, x_{i+2k}...$ for $i$ between 0 and $k - 1$. As $f^k$ is a contraction, we can apply Banach and so it has a fixed point (say $l$) and all the $x^{(i)}$ are convergent to $l$.

Then $x_n$ is convergent to $l$ (for any $\epsilon$ exists $N$ such that $d_X(x_n, l) < \epsilon$, this $N$ being the maximum of the corresponding N's obtained from each of the $x^{(i)}$).

Since $x$ was arbitrary and $l$ is unique, this means that $p=l$ and the proof is completed.

Am I doing something wrong or missing anything?

I just want to ensure I understand this properly, this is not a homework or something similar.

Thank you!

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I think your answer is fine, but personally, I would do this:

$f^n$ is a contraction map, with constant $K$, hence has a fixed point $p$

For a fixed $x$, let $x_1=f(x), x_2=f^2(x), ...$ and $x_n=f^n(x)$

$d(f^N(x), f^N(p)) \leq K^m d(p, f^l(x))=K^m d(p, x_l)\leq K^m\max\limits_{i=1}^n d(x_i,p) $. Here $l$ is the smallest positive integers such that $N=m n+l$ now, as $N\rightarrow \infty$, so does $m$, $K^m\max\limits_{i=1}^n d(x_i,l)\rightarrow 0$

EDIT: I realise, I didn't give a proof that $p$ is a fixed point proof of $f$, but your proof of this is how I would do it too.

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$f^k$ is a contraction, so by Banach fixed point theorem, $f^k$ has a UNIQUE fixed point $x^*$, ie: $f^k(x^*)=x^*$

Then: $f(f^k(x^*))=f(x^*)$, wich means that $f^k(f(x^*))=f(x^*)$ (here $f(x^*)$ is another fixed point of $f^k$).

By the uniqueness of the fixed point of $f^k$, we get: $f(x^*)=x^*$.

Finally, $x^*$ is the unoque fixd point of $f$.

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