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!


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.


$f^k$ 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 $f$.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.