[proofplan] Starting from an arbitrary $x_0 \in M$, define iterates $x_{n+1} := f(x_n)$. The contraction condition forces consecutive iterates to approach each other at a geometric rate, which makes the [sequence](/page/Sequence) Cauchy via a geometric series estimate. Completeness provides a [limit](/page/Limit) $z$, and Lipschitz [continuity](/page/Continuity) of $f$ shows $z$ is a fixed point. The contraction condition gives uniqueness, and passing the Cauchy estimate to the limit yields the convergence rate. [/proofplan] [step:Establish geometric decay of consecutive distances by induction] Fix an arbitrary $x_0 \in M$ and define the [sequence](/page/Sequence) of iterates by $x_{n+1} := f(x_n)$ for $n \geq 0$. [claim:Geometric Decay] For all $n \geq 0$, $d(x_{n+1}, x_n) \leq \lambda^n \, d(x_1, x_0)$. [/claim] [proof] We proceed by induction on $n$. The base case $n = 0$ reads $d(x_1, x_0) \leq d(x_1, x_0)$, which holds. For the inductive step, suppose $d(x_n, x_{n-1}) \leq \lambda^{n-1} d(x_1, x_0)$. Then \begin{align*} d(x_{n+1}, x_n) &= d(f(x_n), f(x_{n-1})) \leq \lambda \, d(x_n, x_{n-1}) \leq \lambda \cdot \lambda^{n-1} \, d(x_1, x_0) = \lambda^n \, d(x_1, x_0), \end{align*} where the first inequality uses the contraction condition and the second uses the inductive hypothesis. [/proof] [/step] [step:Show the iterates form a Cauchy sequence via a geometric series bound] Let $m > n \geq 0$. The triangle inequality gives \begin{align*} d(x_m, x_n) &\leq \sum_{k=n}^{m-1} d(x_{k+1}, x_k) \leq \sum_{k=n}^{m-1} \lambda^k \, d(x_1, x_0) \leq \left(\sum_{k=n}^{\infty} \lambda^k\right) d(x_1, x_0) = \frac{\lambda^n}{1 - \lambda} \, d(x_1, x_0). \end{align*} The final equality uses the geometric [series](/page/Series) formula $\sum_{k=n}^{\infty} \lambda^k = \lambda^n / (1 - \lambda)$, valid since $\lambda \in [0, 1)$. As $n \to \infty$, $\lambda^n \to 0$, so $d(x_m, x_n) \to 0$ uniformly in $m > n$, which is the [Cauchy condition](/page/Cauchy%20Sequence). [guided] We bound the distance between the $n$-th and $m$-th iterates by summing the consecutive distances. Applying the triangle inequality: \begin{align*} d(x_m, x_n) &\leq \sum_{k=n}^{m-1} d(x_{k+1}, x_k). \end{align*} By the geometric decay from the previous step, $d(x_{k+1}, x_k) \leq \lambda^k d(x_1, x_0)$, so \begin{align*} d(x_m, x_n) &\leq d(x_1, x_0) \sum_{k=n}^{m-1} \lambda^k. \end{align*} We bound the partial sum by the tail of the full geometric series: \begin{align*} \sum_{k=n}^{m-1} \lambda^k &\leq \sum_{k=n}^{\infty} \lambda^k = \frac{\lambda^n}{1 - \lambda}, \end{align*} which converges because $\lambda \in [0, 1)$. Therefore $d(x_m, x_n) \leq \frac{\lambda^n}{1 - \lambda} d(x_1, x_0) \to 0$ as $n \to \infty$, confirming the Cauchy condition. [/guided] [/step] [step:Pass to the limit and verify the fixed-point equation $f(z) = z$] Since $(x_n)$ is Cauchy in the complete [metric space](/page/Metric%20Space) $(M, d)$, there exists $z \in M$ with $\lim_{n \to \infty} x_n = z$. The contraction condition implies $f$ is Lipschitz [continuous](/page/Continuity) with constant $\lambda < 1$, hence continuous. Therefore \begin{align*} f(z) &= f\!\left(\lim_{n \to \infty} x_n\right) = \lim_{n \to \infty} f(x_n) = \lim_{n \to \infty} x_{n+1} = z, \end{align*} where the second equality uses continuity of $f$ and the third uses $x_{n+1} = f(x_n)$. [/step] [step:Prove uniqueness of the fixed point] Suppose $z' \in M$ satisfies $f(z') = z'$. Then \begin{align*} d(z, z') &= d(f(z), f(z')) \leq \lambda \, d(z, z'). \end{align*} Rearranging gives $(1 - \lambda) \, d(z, z') \leq 0$. Since $1 - \lambda > 0$ and $d(z, z') \geq 0$, this forces $d(z, z') = 0$, so $z = z'$. [/step] [step:Derive the convergence rate by passing the Cauchy estimate to the limit] Taking $m \to \infty$ in the estimate $d(x_m, x_n) \leq \frac{\lambda^n}{1 - \lambda} d(x_1, x_0)$ and using [continuity](/page/Continuity) of the metric ($x_m \to z$ implies $d(x_m, x_n) \to d(z, x_n)$): \begin{align*} d(x_n, z) &= \lim_{m \to \infty} d(x_m, x_n) \leq \frac{\lambda^n}{1 - \lambda} \, d(x_1, x_0) \end{align*} for all $n \geq 0$. The error after $n$ iterations decays geometrically with ratio $\lambda$. [/step]