[proofplan] The proof is constructive. Starting from any $x_0 \in X$, we iterate the contraction to form $x_n = f^n(x_0)$. The contraction inequality gives $d(x_{n+1}, x_n) \leq \lambda^n d(x_1, x_0)$, making the sequence Cauchy with geometric decay. Completeness provides a limit $z$, and continuity of $f$ (inherited from the Lipschitz bound) shows $f(z) = z$. Uniqueness follows by applying the contraction inequality to two hypothetical fixed points. [/proofplan] [step:Iterate the contraction and establish geometric decay of consecutive distances] Choose any $x_0 \in X$ and define $x_n = f(x_{n-1})$ for $n \geq 1$, so that $x_n = f^n(x_0)$. The contraction inequality gives \begin{align*} d(x_{n+1}, x_n) = d(f(x_n), f(x_{n-1})) \leq \lambda\, d(x_n, x_{n-1}). \end{align*} By induction, $d(x_{n+1}, x_n) \leq \lambda^n d(x_1, x_0)$ for all $n \geq 0$. [/step] [step:Prove the sequence $(x_n)$ is Cauchy using the geometric series bound] For $m > n \geq 0$, 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) = \lambda^n\, d(x_1, x_0) \sum_{j=0}^{m-n-1} \lambda^j \leq \frac{\lambda^n}{1 - \lambda}\, d(x_1, x_0). \end{align*} The geometric series $\sum \lambda^j$ converges because $\lambda < 1$. Since $\lambda^n \to 0$ as $n \to \infty$, for any $\varepsilon > 0$ there exists $N$ with $\lambda^N d(x_1, x_0)/(1 - \lambda) < \varepsilon$, and then $d(x_m, x_n) < \varepsilon$ for all $m > n \geq N$. Therefore $(x_n)$ is Cauchy. [guided] The key estimate bounds the distance between any two terms $x_m$ and $x_n$ (with $m > n$) by a geometric tail. We telescope the distance using the triangle inequality: \begin{align*} d(x_m, x_n) \leq \sum_{k=n}^{m-1} d(x_{k+1}, x_k). \end{align*} Each consecutive distance satisfies $d(x_{k+1}, x_k) \leq \lambda^k d(x_1, x_0)$ by the inductive bound from the previous step. Substituting: \begin{align*} d(x_m, x_n) \leq d(x_1, x_0) \sum_{k=n}^{m-1} \lambda^k = \lambda^n d(x_1, x_0) \sum_{j=0}^{m-n-1} \lambda^j. \end{align*} Since $0 \leq \lambda < 1$, the partial geometric sum is bounded by the full series: $\sum_{j=0}^{m-n-1} \lambda^j \leq \sum_{j=0}^\infty \lambda^j = 1/(1-\lambda)$. This gives the uniform bound $d(x_m, x_n) \leq \lambda^n d(x_1, x_0)/(1-\lambda)$, which depends only on $n$ (not on $m$) and tends to zero as $n \to \infty$. This is the Cauchy property. [/guided] [/step] [step:Use completeness to obtain a limit and verify it is a fixed point] Since $X$ is complete and $(x_n)$ is Cauchy, there exists $z \in X$ with $x_n \to z$. The contraction $f$ is Lipschitz continuous with constant $\lambda < 1$, hence continuous. Applying $f$ to both sides of $x_n \to z$: \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*} Therefore $z$ is a fixed point of $f$. [/step] [step:Prove uniqueness of the fixed point] Suppose $z$ and $w$ are both fixed points of $f$. Then \begin{align*} d(z, w) = d(f(z), f(w)) \leq \lambda\, d(z, w). \end{align*} Rearranging: $(1 - \lambda)\, d(z, w) \leq 0$. Since $1 - \lambda > 0$ and $d(z, w) \geq 0$, this forces $d(z, w) = 0$, so $z = w$. [/step]