Corollary of Banach's fixed point theorem

Question

I tried to prove this corollary of BFPT. Here's the statement :

"Let $\left( E, d \right)$ be a complete metric space and $f : E \longrightarrow E$ a mapping such that $f^p$ is a contraction for some $p \in \mathbb{N} ~ \backslash \left\lbrace 0 \right\rbrace$. Show that $f$ has a unique fixed point $x_0$ and that for any $x \in E$, the sequence $\left( f^n (x) \right)_{n \in \mathbb{N}}$ converges to $x_0$."

I'm not sure if what I did is correct or not. Proving that $x_0$ is the limit of $\left( f^n (x) \right)_{n \in \mathbb{N}}$ for any given $x \in E$ is the part that makes be doubt...

Here's my whole work :

By Banach's fixed point theorem, it exists a unique $x_0 \in E$ such that $f^p(x_0)=x_0$. We also have $f(f^p(x_0))=f(x_0)=f^p(f(x_0))$ hence $f(x_0)$ is a fixed point of $f^p$, so by unicity $f(x_0)=x_0$. Let $y_0 \in E$ be an another fixed point of $f$. By induction we have $f^p(y_0)=f^{p-1}(f(y_0))=f^{p-1}(y_0)= \ldots =f(f(y_0))=f(y_0)=y_0$ thus $x_0=y_0$. This shows that $f$ has a unique fixed point.

If $x=x_0, ~ \left( f^n (x) \right)_{n \in \mathbb{N}}$ trivially converges to $x_0$. For the next step, suppose $x \neq x_0$.

Let $k_p \in \left[ 0,1 \right[$ be the contraction constant of $f^p$. We define the sequence $\left( u_n \right)_{n \in \mathbb{N}}$ as $u_n := f^{np}(x)$. We clearly have $u_{n+1}=f^p(u_n)$. Moreover,

\begin{align} d(u_i,u_{i+1})=d \left( f^{pi}(x), f^{p(i+1)}(x) \right) = d \left( f^p \left(f^{p(i-1)}(x) \right),f^p \left(f^{pi}(x) \right) \right) & \underset{\vdots}{\leqslant} k_p \cdot d \left( f^{p(i-1)}(x), f^{pi}(x) \right) \\ & \leqslant k_p^i \cdot d ( u_0 , u_1 ) \end{align}

For any $n,m \in \mathbb{N},~m>n$,

\begin{align} d(u_n,u_m) \leqslant \sum_{i=n}^{m-1} d(u_i,u_{i+1}) = \sum_{i=n}^{m-1} k_p^i \cdot d ( u_0 , u_1 ) &= k_p^n \cdot d(u_0,u_1) \sum_{i=0}^{m-n-1} k_p^i \\ &\leqslant k_p^n \cdot d(u_0,u_1) \sum_{i \geqslant 0} k_p^i \\ &= \frac{k_p^n}{1-k_p} d(u_0 , u_1) \end{align}

Let $\epsilon > 0$. We choose $N \in \mathbb{N}$ such that

$$ k_p^{N} < \frac{\epsilon (1-k_p)}{d (u_0,u_1 )}$$

For any $m>n \geqslant N, ~ d(u_n,u_m) < \epsilon$ hence the sequence $\left( u_n \right)_{n \in \mathbb{N}}$ is Cauchy in $\left( E,d \right)$. It converges to a limit in $E$ that we denote as $a$. But,

$$ a = \underset{n \longrightarrow +\infty}{\text{lim}} u_n = \underset{n \longrightarrow +\infty}{\text{lim}} f^p (u_n)=f^p \left(\underset{n \longrightarrow +\infty}{\text{lim}} u_n \right)=f^p(a)$$

Thus $a=x_0$. Let $r=np+q, ~ 0 \leqslant q \leqslant p-1$. If $m \longrightarrow +\infty$,

\begin{align} d(u_n,x_0) \leqslant \frac{k_p^n}{1-k_p} d(u_0 , u_1) &\Longleftrightarrow d\left(f^{np}(x),x_0\right) \leqslant \frac{k_p^n}{1-k_p} d \left(x , f^p(x) \right) \\ & \Longrightarrow d\left(f^{np+q}(x),x_0\right) \leqslant \frac{k_p^n}{1-k_p} d \left(f^q(x) , f^{p+q}(x) \right) \\ & \Longrightarrow d\left(f^r(x),x_0\right) \leqslant \frac{k_p^{r/q}}{1-k_p} d \left(f^q(x) , f^{p+q}(x) \right) \end{align}

When $r \longrightarrow +\infty,~ d\left(f^r(x),x_0\right) \longrightarrow 0$ hence $\left( f^r (x) \right)_{r \in \mathbb{N}}$ converges to $x_0$.

Answer 1

Your proof seems ok, although perhaps one can write a slightly shorter argument:

For any $n\in\mathbb{Z}_{>0}$, there are unique $q_n=q_{p,n}\in \mathbb{Z}_{\geq0}$ and $r_n=r_{p,n}\in \{0,1,...,p-1\}$ such that

$$n = q_n p + r_n.$$

Note that $\lim_{n\to\infty} q_n=\infty$.

Then one has the following estimate by using the triangular inequality with $f^{r_n}(x)$ as the middle term:

\begin{align*} d(f^n(x),x_0) \leq \operatorname{Lip}(f^p)^{q_n}\max_{1\leq i<p} d(f^i(x),x) + d((f^p)^{q_n}(x),x_0). \end{align*}

Here $\operatorname{Lip}(g)$ is the Lipschitz constant of the function $g$ (see e.g. Question on the distortion of a metric embedding and Lipschitz maps), hence is a number less than one. Thus one has that $d(f^n(x),x_0) $ is bounded by a term that decays exponentially fast as $n$ grows.