Let $(X,d)$ be a complete metric space and let $f:X \to X$ be a
mapping such that, for some $p \geq 2$, the composite mapping
$f \circ f \circ \cdots \circ f$ with $p$ factors is a contraction.
Note that the mapping $f$ is not assumed to be continuous.
Show that $f$ has one and only one fixed point $x$.
Show that, given any $x_0 \in X$, the sequence $(x_n)_{n=0}^{\infty}$ defined by $x_{n+1}=f(x_n)$, $n \geq 0$,
converges to $x$.
I can prove the first part of this problem, but for the second part, I can only show that the subsequence $(x_{np})_{n=0}^{\infty}$ converges. Thanks for any help!
Supplement: Proof of part 1
Proof: (Existence) Since $f^p := \underbrace{f \circ f \circ \cdots \circ f}_{p}$ is a contraction, then by Banach fixed point theorem $f^p$ has a unique fixed point $x \in X$ such that $f^p(x)=x$. Thus
$$f(f^p(x))=f(x) \Rightarrow f^p(f(x))=f(x),$$
which implies that $f(x)$ is also a fixed point of $f^p$, by uniqueness we obtain $f(x)=x$.
(Uniqueness) Suppose that $f(x)=x$ and $f(x')=x'$, then
$$f^p(x)=f^{p-1}(f(x))=f^{p-1}(x)=\cdots=f(x)=x,$$
and similary $f^p(x')=x'$. We have $x=x'$ still by uniqueness of fixed point of $f^p$.
Best Answer
For each $k = 0, 1, \cdots, p-1$, the subsequence $\{ y_n\} = \{ x_{np+k}\}_{n=0}^\infty$ is given by
$$ y_{n+1} = f^p(y_n), \ \ y(0) = x_k.$$
Since $f^p$ is a contraction, $y_n$ converges to the unique fixed point $x$ of $f^p$, which is also the unique fixed point of $f$ as you shown. Hence $\{ x_{np+k}\}_{n=0}^\infty$ converges to $x$ for each $k = 0, 1, \cdots p-1$, and thus $\{x_n\}$ also converges to $x$.