0
$\begingroup$

Let (X,d) be a complete metric space and $f^n$ be a contraction mapping with $x_0$ as it's fixed point. I have already proved that $x_0$ is also a fixed point of f, I need to prove that: $$\forall x \in X \lim_{k \rightarrow \infty} f^k (x) = x_0$$ If i take only $\lim_{nk \rightarrow \infty} f^{nk}(x)$ then it's easy as the powers of $f^n$ are also contraction mappings, but I'm not sure it's correct.

$\endgroup$
1
$\begingroup$

Hint: Let $x\in X$. Prove that the sequence $(f^n(x))_{n\in\mathbb N}$ is a Cauchy sequence. Since your space is complete, it converges to some $y$. Then prove that $f(y)=y$. Since $f(x_0)=x_0$ and $f$ is a contraction, $y=x_0$.

$\endgroup$

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.