Let $f: \mathbb{R} \to \mathbb{R}$ be a function such that $|f(x) - f(y)| \leq c|x - y|$, with $0 < c < 1$. Prove that for all $x_0, (f^n(x_0))_n$ converges to a unique fixed point.
Note: This is the Contraction Mapping Theorem, of which proofs are readily available. My question is to verify or critique my proof.
Lemma 1: By induction, for all $n \geq 1, j \geq 0, |f^{n+j-1}(x) - f^{n+j}(x)| \leq c^j|f^{n-1}(x) - f^{n}(x)|.$
Lemma 2: For any $\varepsilon > 0$ there exists $n$ such that $|f^{n-1}(x) - f^{n}(x)| < \varepsilon $. This follows directly from Lemma 1, since $c^n \to 0$ for large $n$.
We now show that for any $\varepsilon > 0$, there exists $n$ such that for all $m,|f^n(x) - f^{n+m}(x)| < \varepsilon$. Choose $n$ such that $|f^{n-1}(x) - f^{n}(x)| < (1-c)\varepsilon$. Then for all $m$:
$$ \begin{align*} |f^n(x) - f^{n+m}(x)| & \leq \sum_{j = 1}^{m} |f^{n+j-1}(x) - f^{n+j}(x)| \quad \text{(by triangle inequality)} \\ & \leq \sum_{j = 1}^{m} c^j|f^{n-1}(x) - f^{n}(x)| \quad \text{(by Lemma 1)} \\ & < \sum_{j = 1}^{m}c^j(1-c)\varepsilon \\ & < \frac 1 {1-c} {(1-c)\varepsilon} \\ & < \varepsilon. \end{align*} $$
Thus, $(f^n(x))_n$ is a Cauchy sequence and converges to a limit $\ell$. Since $f$ is clearly continuous, $\ell := \lim f^n(x) = \lim f(f^n(x)) = f(\lim f^n(x)) = f(\ell)$ is a fixed point.
This fixed point $\ell$ is unique. Assume $f(a) = a, f(b) = b, a \neq b$. Then $|a - b| = |f(a) - f(b)| \leq c|a - b|$, a contradiction. Thus, $f^n(x) \to \ell$ for all $x$.
Questions:
1. Is the proof correct? Please provide any critique or suggestions, on the proof or its writing.
2. I know that the my initial attempt to prove uniqueness of fixed point ($|f^n(x_0) - f^n(x_1)| \leq c^n|f(x_0) - f(x_1)| \to 0$) can't be used to build a proof, but request some clarification as to why not. If any two points in $\mathbb{R}$ end up going to the same point $y$, isn't that $y$ a fixed point for any $x_0$? The counter example is $x \mapsto k + \frac 1 x$, suggesting that the points can become arbitrarily close but not fixed; but I'd appreciate some clarification here.
Perhaps the answer to #2 is: The last line shows that any two points in $\mathbb{R}$ end up arbitrarily close. That does not mean they end up fixed. Only once we've already shown that in the limit each one individually is fixed, and yet are arbitrarily close, can we conclude that they are equal. Is this correct? How should I clarify the last line to capture this?