[proofplan] We prove existence and uniqueness of a fixed point for a contraction on a complete metric space. For uniqueness, the contraction inequality forces any two fixed points to coincide. For existence, we pick an arbitrary initial point $x_0 \in X$, form the iterates $x_{n+1} = f(x_n)$, show the contraction constant $\lambda$ makes consecutive distances decay geometrically, use the geometric series to establish the Cauchy property, and then invoke completeness to extract a limit whose fixed-point property follows from continuity of $f$. [/proofplan] [step:Show that any two fixed points must coincide] Suppose $x, y \in X$ both satisfy $f(x) = x$ and $f(y) = y$. By the contraction property: \begin{align*} d(x, y) = d(f(x), f(y)) \le \lambda \, d(x, y). \end{align*} Rearranging gives $(1 - \lambda) \, d(x, y) \le 0$. Since $\lambda \in [0,1)$ we have $1 - \lambda > 0$, and since $d(x,y) \ge 0$ by the metric axioms, it follows that $d(x,y) = 0$, hence $x = y$. [guided] We want to show that if a fixed point exists, it is unique. Suppose $x$ and $y$ are both fixed points: $f(x) = x$ and $f(y) = y$. Substituting into the contraction inequality: \begin{align*} d(x, y) = d(f(x), f(y)) \le \lambda \, d(x, y). \end{align*} Rearranging: $(1 - \lambda) \, d(x, y) \le 0$. Since $\lambda < 1$, the factor $1 - \lambda$ is strictly positive, and $d(x,y) \ge 0$ by definition of a metric. The only way a non-negative number multiplied by a positive number can be $\le 0$ is if $d(x,y) = 0$. By the separation axiom of a metric, $d(x,y) = 0$ implies $x = y$. Note that this argument uses only the contraction property -- completeness is not needed for uniqueness. [/guided] [/step] [step:Construct the Picard iteration sequence and bound consecutive distances] Choose an arbitrary $x_0 \in X$ and define the sequence $(x_n)_{n=0}^\infty$ by $x_{n+1} := f(x_n)$. We prove by induction that for all $n \ge 0$: \begin{align*} d(x_{n+1}, x_n) \le \lambda^n \, d(x_1, x_0). \end{align*} The base case $n = 0$ is the identity $d(x_1, x_0) \le \lambda^0 \, d(x_1, x_0)$. For the inductive step, assume $d(x_n, x_{n-1}) \le \lambda^{n-1} \, d(x_1, x_0)$. Then: \begin{align*} d(x_{n+1}, x_n) = d(f(x_n), f(x_{n-1})) \le \lambda \, d(x_n, x_{n-1}) \le \lambda \cdot \lambda^{n-1} \, d(x_1, x_0) = \lambda^n \, d(x_1, x_0). \end{align*} [guided] The strategy for existence is to build an approximating sequence and show it converges. We define the Picard iterates: pick any starting point $x_0 \in X$ and set $x_{n+1} := f(x_n)$. The key observation is that the contraction property makes each successive pair of iterates closer together by a factor of $\lambda$. We formalise this with induction. We claim $d(x_{n+1}, x_n) \le \lambda^n \, d(x_1, x_0)$ for all $n \ge 0$. **Base case** ($n = 0$): $d(x_1, x_0) \le \lambda^0 \, d(x_1, x_0) = d(x_1, x_0)$. This holds as an equality. **Inductive step**: Assume $d(x_n, x_{n-1}) \le \lambda^{n-1} \, d(x_1, x_0)$. Since $x_{n+1} = f(x_n)$ and $x_n = f(x_{n-1})$, the contraction inequality gives: \begin{align*} d(x_{n+1}, x_n) = d(f(x_n), f(x_{n-1})) \le \lambda \, d(x_n, x_{n-1}) \le \lambda \cdot \lambda^{n-1} \, d(x_1, x_0) = \lambda^n \, d(x_1, x_0). \end{align*} This geometric decay is the engine of the proof: it ensures the iterates bunch together and will let us sum a geometric series in the next step. [/guided] [/step] [step:Establish the Cauchy estimate via the geometric series] For $n \ge 0$ and $m \ge 1$, apply the triangle inequality $m - 1$ times and use the bound from the previous step: \begin{align*} d(x_{n+m}, x_n) &\le \sum_{j=0}^{m-1} d(x_{n+j+1}, x_{n+j}) \\ &\le \sum_{j=0}^{m-1} \lambda^{n+j} \, d(x_1, x_0) \\ &= \lambda^n \, d(x_1, x_0) \sum_{j=0}^{m-1} \lambda^j \\ &\le \lambda^n \, d(x_1, x_0) \cdot \frac{1}{1 - \lambda}. \end{align*} The last inequality uses the convergence of the geometric series $\sum_{j=0}^{\infty} \lambda^j = \frac{1}{1-\lambda}$, which is valid since $0 \le \lambda < 1$. As $n \to \infty$, $\lambda^n \to 0$, so for any $\varepsilon > 0$ we can choose $N$ large enough that $\frac{\lambda^N}{1-\lambda} d(x_1, x_0) < \varepsilon$. Then for all $n \ge N$ and all $m \ge 1$, $d(x_{n+m}, x_n) < \varepsilon$. Therefore $(x_n)$ is a [Cauchy sequence](/page/Cauchy%20Sequence). [guided] We need to show the iterates form a [Cauchy sequence](/page/Cauchy%20Sequence). This means controlling $d(x_{n+m}, x_n)$ for arbitrary $m \ge 1$. The idea is to telescope the distance along the chain $x_n, x_{n+1}, \ldots, x_{n+m}$ and then sum the geometric bounds. By the triangle inequality applied repeatedly: \begin{align*} d(x_{n+m}, x_n) \le \sum_{j=0}^{m-1} d(x_{n+j+1}, x_{n+j}). \end{align*} Substituting the bound $d(x_{n+j+1}, x_{n+j}) \le \lambda^{n+j} \, d(x_1, x_0)$: \begin{align*} d(x_{n+m}, x_n) \le d(x_1, x_0) \sum_{j=0}^{m-1} \lambda^{n+j} = \lambda^n \, d(x_1, x_0) \sum_{j=0}^{m-1} \lambda^j. \end{align*} Since $0 \le \lambda < 1$, the partial geometric sum is bounded by the full series: $\sum_{j=0}^{m-1} \lambda^j \le \sum_{j=0}^{\infty} \lambda^j = \frac{1}{1-\lambda}$. Therefore: \begin{align*} d(x_{n+m}, x_n) \le \frac{\lambda^n}{1 - \lambda} \, d(x_1, x_0). \end{align*} The right-hand side depends only on $n$ (not on $m$) and tends to $0$ as $n \to \infty$, since $\lambda^n \to 0$. Given any $\varepsilon > 0$, choose $N$ so that $\frac{\lambda^N}{1-\lambda} d(x_1, x_0) < \varepsilon$; then $d(x_{n+m}, x_n) < \varepsilon$ for all $n \ge N$ and all $m \ge 1$. This is the definition of a Cauchy sequence. [/guided] [/step] [step:Invoke completeness and verify the limit is a fixed point] Since $X$ is complete, the [Cauchy sequence](/page/Cauchy%20Sequence) $(x_n)$ converges to some $\bar{x} \in X$. We verify that $\bar{x}$ is a fixed point. The contraction $f$ is Lipschitz continuous with Lipschitz constant $\lambda$, hence in particular continuous. Therefore: \begin{align*} f(\bar{x}) = f\!\left(\lim_{n \to \infty} x_n\right) = \lim_{n \to \infty} f(x_n) = \lim_{n \to \infty} x_{n+1} = \bar{x}. \end{align*} Combined with the uniqueness established in the first step, $\bar{x}$ is the unique fixed point of $f$. [guided] Completeness of $X$ guarantees that every Cauchy sequence converges. Since $(x_n)$ is Cauchy, there exists $\bar{x} \in X$ with $x_n \to \bar{x}$. To show $f(\bar{x}) = \bar{x}$, we need to pass the map $f$ through the limit. A contraction mapping is Lipschitz continuous: for any $a, b \in X$, $d(f(a), f(b)) \le \lambda \, d(a, b)$. In particular, if $x_n \to \bar{x}$ then $d(f(x_n), f(\bar{x})) \le \lambda \, d(x_n, \bar{x}) \to 0$, so $f(x_n) \to f(\bar{x})$. But $f(x_n) = x_{n+1}$, and $x_{n+1} \to \bar{x}$ (it is the same convergent sequence shifted by one index). Therefore: \begin{align*} f(\bar{x}) = \lim_{n \to \infty} f(x_n) = \lim_{n \to \infty} x_{n+1} = \bar{x}. \end{align*} So $\bar{x}$ is a fixed point. Uniqueness was proved in the first step: any two fixed points must coincide. Hence $\bar{x}$ is the unique fixed point of $f$ in $X$. [/guided] [/step]