1
$\begingroup$

(X,d) is a complete metric space. If $f:X\rightarrow X$ is such that there exists $n\in \mathbb{N}$ for which $f^{n}$ is contracting, show that $f$ admits a unique fixed point.

I think the method is to prove that $f^{n}$ contracting $\Rightarrow $ $f$ is contracting, hence we can apply the Banach fixed point theorem. But I have no idea how to prove this. Maybe by a recursion?

$\endgroup$
1
  • 1
    $\begingroup$ Welcome to MSE. A question should be written in such a way that it can be understood even by someone who did not read the title. $\endgroup$
    – José Carlos Santos
    Oct 13, 2021 at 5:58

1 Answer 1

Reset to default
6
$\begingroup$

By the Banach fixed point theorem, $f^n$ has a unique fixed point $x$. Then $f(x)$ is also a fixed point of $f^n$, and so by uniqueness $f(x)=x$.

$\endgroup$
1
  • 1
    $\begingroup$ I dont get how you come to the conclusion that $f(x)$ is also a fixed point of $f^{ n}$. EDIT: GOT IT IM SO DUMB $\endgroup$
    – SerIddi
    Oct 13, 2021 at 14:50

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.