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.
2026-03-29 12:13:00.1774786380
Fixed point as limit
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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, $y=x_0$.