[proofplan]
We reduce the problem to the [Banach Contraction Mapping Principle](/theorems/71). Since $g = f^m$ is a contraction on the complete metric space $(X, d)$, it has a unique fixed point $\bar{x}$. We show $\bar{x}$ is also a fixed point of $f$ by observing that $f(\bar{x})$ is itself a fixed point of $g$ (via commutativity of $f$ with $g$), so uniqueness forces $f(\bar{x}) = \bar{x}$. Uniqueness of the fixed point of $f$ follows because any fixed point of $f$ is automatically a fixed point of $g$. Convergence of iterates $f^n(x_0) \to \bar{x}$ is established by writing $n = km + r$ and bounding $d(f^n(x_0), \bar{x})$ by $\lambda^k \max_{0 \le r < m} d(f^r(x_0), \bar{x})$, where $\lambda \in [0,1)$ is the contraction constant of $g$.
[/proofplan]
[step:Apply the Banach Contraction Mapping Principle to $g = f^m$ to obtain a fixed point of $g$]
Define $g := f^m: X \to X$. By hypothesis, $g$ is a contraction on the complete metric space $(X, d)$: there exists $\lambda \in [0, 1)$ such that $d(g(x), g(y)) \le \lambda\, d(x, y)$ for all $x, y \in X$. By the [Banach Contraction Mapping Principle](/theorems/71), $g$ has a unique fixed point $\bar{x} \in X$, i.e., $f^m(\bar{x}) = \bar{x}$.
[/step]
[step:Show $\bar{x}$ is a fixed point of $f$ via uniqueness of the fixed point of $g$]
Consider $f(\bar{x})$. Since $f$ commutes with $g = f^m$ (both are iterates of $f$):
\begin{align*}
g(f(\bar{x})) = f^m(f(\bar{x})) = f(f^m(\bar{x})) = f(\bar{x}).
\end{align*}
Thus $f(\bar{x})$ is a fixed point of $g$. Since $g$ has a unique fixed point $\bar{x}$, we conclude $f(\bar{x}) = \bar{x}$.
[guided]
The key identity is $f^m(f(\bar{x})) = f(f^m(\bar{x}))$. This holds because composition of a map with itself is associative: $f^m \circ f = f^{m+1} = f \circ f^m$. Spelling this out explicitly:
\begin{align*}
g(f(\bar{x})) = f^m(f(\bar{x})) = f^{m+1}(\bar{x}) = f(f^m(\bar{x})) = f(g(\bar{x})) = f(\bar{x}),
\end{align*}
where the last equality uses $g(\bar{x}) = f^m(\bar{x}) = \bar{x}$. So $f(\bar{x})$ satisfies $g(f(\bar{x})) = f(\bar{x})$, meaning $f(\bar{x})$ is a fixed point of $g$.
Now the [Banach Contraction Mapping Principle](/theorems/71) guarantees that $g$ has a **unique** fixed point, namely $\bar{x}$. Since $f(\bar{x})$ is also a fixed point of $g$, uniqueness forces:
\begin{align*}
f(\bar{x}) = \bar{x}.
\end{align*}
Note the structure of the argument: we do not compute $f(\bar{x})$ directly. Instead, we show that $f(\bar{x})$ satisfies the same fixed-point equation $g(y) = y$ as $\bar{x}$, and then uniqueness does the work. This is a standard technique -- when a contraction has a unique fixed point, any element shown to satisfy the fixed-point equation must equal that fixed point.
[/guided]
[/step]
[step:Prove uniqueness: any fixed point of $f$ is a fixed point of $g$]
Suppose $y \in X$ satisfies $f(y) = y$. Iterating $m$ times:
\begin{align*}
f^m(y) = f^{m-1}(f(y)) = f^{m-1}(y) = \cdots = y.
\end{align*}
So $y$ is a fixed point of $g = f^m$. By the uniqueness of $g$'s fixed point, $y = \bar{x}$.
[/step]
[step:Establish convergence $f^n(x_0) \to \bar{x}$ via the contraction estimate on $g$]
Let $x_0 \in X$. Write $n = km + r$ with $k = \lfloor n/m \rfloor \ge 0$ and $0 \le r < m$. Since $f^m(\bar{x}) = \bar{x}$, iterating gives $f^{km}(\bar{x}) = \bar{x}$. Define $y_r := f^r(x_0)$. Then:
\begin{align*}
d(f^n(x_0), \bar{x}) = d(f^{km}(f^r(x_0)), f^{km}(\bar{x})) = d(g^k(y_r), g^k(\bar{x})).
\end{align*}
Since $g$ is a contraction with constant $\lambda \in [0,1)$, iterating the contraction estimate $k$ times gives $d(g^k(y_r), g^k(\bar{x})) \le \lambda^k\, d(y_r, \bar{x})$. Therefore:
\begin{align*}
d(f^n(x_0), \bar{x}) \le \lambda^k \max_{0 \le r < m} d(f^r(x_0), \bar{x}).
\end{align*}
The maximum $M := \max_{0 \le r < m} d(f^r(x_0), \bar{x})$ is a finite constant depending only on $x_0$, $f$, $m$, and $\bar{x}$. As $n \to \infty$, $k = \lfloor n/m \rfloor \to \infty$, so $\lambda^k \to 0$ (since $\lambda < 1$), giving $d(f^n(x_0), \bar{x}) \to 0$.
[guided]
The idea is to group the iterates of $f$ into blocks of size $m$. Writing $n = km + r$ with $0 \le r < m$, we have $f^n = f^{km} \circ f^r = g^k \circ f^r$. The remainder $r$ takes only finitely many values ($0, 1, \ldots, m-1$), so the distances $d(f^r(x_0), \bar{x})$ are bounded by a finite maximum $M$.
For the contraction estimate: since $g$ has Lipschitz constant $\lambda < 1$, the $k$-th iterate $g^k$ has Lipschitz constant $\lambda^k$. This gives:
\begin{align*}
d(g^k(y_r), g^k(\bar{x})) \le \lambda^k\, d(y_r, \bar{x}) \le \lambda^k M.
\end{align*}
As $n \to \infty$, $k \to \infty$ and $\lambda^k \to 0$, so $d(f^n(x_0), \bar{x}) \le \lambda^k M \to 0$. This shows convergence for every initial point $x_0 \in X$.
[/guided]
[/step]
