[proofplan] We reformulate the IVP as a fixed-point problem for the Picard integral operator on a suitable complete metric space of continuous functions, then apply the Banach fixed-point theorem. The Lipschitz condition on $f$ makes the integral operator a contraction for sufficiently short time intervals. The unique fixed point is the unique solution. Continuous dependence on initial data follows from the contraction estimate. [/proofplan] [step:Reformulate the IVP as a fixed-point problem for the Picard operator] Define the time interval $I := [t_0 - T, t_0 + T]$ for a value of $T > 0$ to be chosen. Let $\overline{B}(x_0, r) := \{x \in \mathbb{R}^n : |x - x_0| \leq r\}$ for a radius $r > 0$ to be chosen. Define the complete metric space \begin{align*} \mathcal{X} := \{x \in C(I, \mathbb{R}^n) : x(t_0) = x_0, \; |x(t) - x_0| \leq r \text{ for all } t \in I\}, \end{align*} equipped with the supremum metric $d(x, y) := \sup_{t \in I} |x(t) - y(t)|$. Since $\mathcal{X}$ is a closed subset of the Banach space $(C(I, \mathbb{R}^n), \|\cdot\|_\infty)$, it is a complete metric space. Define the Picard integral operator $\mathcal{P}: \mathcal{X} \to C(I, \mathbb{R}^n)$ by \begin{align*} (\mathcal{P}x)(t) := x_0 + \int_{t_0}^{t} f(x(s)) \, d\mathcal{L}^1(s). \end{align*} A function $x \in \mathcal{X}$ is a solution of $\dot{x} = f(x)$, $x(t_0) = x_0$ if and only if $\mathcal{P}x = x$. [guided] The strategy is to apply the [Banach Fixed-Point Theorem](/theorems/???): find a complete metric space and a contraction mapping on it. The Picard operator $\mathcal{P}$ integrates the vector field along a candidate solution curve; a fixed point $\mathcal{P}x = x$ satisfies the integral formulation of the ODE, which is equivalent to the differential formulation. We define $\mathcal{X}$ as the space of continuous functions starting at $x_0$ and staying within a ball of radius $r$. This is a closed subset of $C(I, \mathbb{R}^n)$ (the limit of functions in $\overline{B}(x_0, r)$ lies in $\overline{B}(x_0, r)$ by continuity of the supremum), hence complete under $d(x, y) = \sup_{t \in I} |x(t) - y(t)|$. The Picard operator is \begin{align*} (\mathcal{P}x)(t) := x_0 + \int_{t_0}^{t} f(x(s)) \, d\mathcal{L}^1(s). \end{align*} Note that $(\mathcal{P}x)(t_0) = x_0$ for any $x$, so the initial condition is automatically satisfied. [/guided] [/step] [step:Choose $r$ and $T$ so that $\mathcal{P}$ maps $\mathcal{X}$ into itself] Since $f: E \to \mathbb{R}^n$ is continuous and $x_0 \in E$ with $E$ open, there exists $r > 0$ such that $\overline{B}(x_0, r) \subset E$. Since $f$ is continuous on the compact set $\overline{B}(x_0, r)$, it attains its maximum: define $M := \max_{x \in \overline{B}(x_0, r)} |f(x)|$. Choose $T > 0$ satisfying both $T \leq r / M$ and $T < 1/L$, where $L$ is the Lipschitz constant of $f$ at $x_0$. For any $x \in \mathcal{X}$ and $t \in I$, \begin{align*} |(\mathcal{P}x)(t) - x_0| = \left|\int_{t_0}^{t} f(x(s)) \, d\mathcal{L}^1(s)\right| \leq M |t - t_0| \leq M T \leq r. \end{align*} Since $\mathcal{P}x$ is continuous (as $f \circ x$ is continuous and integration preserves continuity) and $(\mathcal{P}x)(t_0) = x_0$, we have $\mathcal{P}x \in \mathcal{X}$. Hence $\mathcal{P}: \mathcal{X} \to \mathcal{X}$. [/step] [step:Show $\mathcal{P}$ is a contraction using the Lipschitz condition] Let $x, y \in \mathcal{X}$. Since $f$ satisfies a Lipschitz condition with constant $L$ on $\overline{B}(x_0, r)$ — that is, $|f(x) - f(y)| \leq L|x - y|$ for all $x, y \in \overline{B}(x_0, r)$ — we estimate for each $t \in I$: \begin{align*} |(\mathcal{P}x)(t) - (\mathcal{P}y)(t)| &= \left|\int_{t_0}^{t} \bigl(f(x(s)) - f(y(s))\bigr) \, d\mathcal{L}^1(s)\right| \\ &\leq \int_{t_0}^{t} |f(x(s)) - f(y(s))| \, d\mathcal{L}^1(s) \\ &\leq \int_{t_0}^{t} L |x(s) - y(s)| \, d\mathcal{L}^1(s) \\ &\leq L \cdot d(x, y) \cdot |t - t_0| \\ &\leq LT \cdot d(x, y). \end{align*} Taking the supremum over $t \in I$ gives $d(\mathcal{P}x, \mathcal{P}y) \leq LT \cdot d(x, y)$. Since $T < 1/L$, the contraction constant $q := LT$ satisfies $q < 1$, so $\mathcal{P}$ is a contraction on $(\mathcal{X}, d)$. [guided] The Lipschitz condition enters here and nowhere else. It converts the integral operator's estimate from a bound on the sup-norm of $f$ (which gave self-mapping) to a bound on the difference of $f$-values (which gives contraction). Without Lipschitz continuity, we cannot bound $|f(x(s)) - f(y(s))|$ by $L|x(s) - y(s)|$, and $\mathcal{P}$ need not be a contraction. For $x, y \in \mathcal{X}$ and $t \in I$: \begin{align*} |(\mathcal{P}x)(t) - (\mathcal{P}y)(t)| &\leq \int_{t_0}^{t} |f(x(s)) - f(y(s))| \, d\mathcal{L}^1(s) \leq L \int_{t_0}^{t} |x(s) - y(s)| \, d\mathcal{L}^1(s). \end{align*} Since $|x(s) - y(s)| \leq d(x, y)$ for all $s$, the integral is at most $L \cdot d(x, y) \cdot |t - t_0| \leq LT \cdot d(x, y)$. Taking the supremum gives $d(\mathcal{P}x, \mathcal{P}y) \leq LT \cdot d(x, y)$. The condition $LT < 1$ ensures the contraction constant $q = LT$ is strictly less than $1$. Why choose $T < 1/L$? The contraction factor $LT$ depends on both the Lipschitz constant and the length of the time interval. By making $T$ small enough, we force $LT < 1$. This is the mechanism that makes the theorem local: the solution is guaranteed only on a short interval, and extending it requires reapplying the theorem with the endpoint as a new initial condition. [/guided] [/step] [step:Apply the Banach fixed-point theorem to obtain existence and uniqueness] The [Banach Fixed-Point Theorem](/theorems/???) states that a contraction mapping on a complete metric space has a unique fixed point. Since $\mathcal{P}: \mathcal{X} \to \mathcal{X}$ is a contraction on the complete metric space $(\mathcal{X}, d)$, there exists a unique $x \in \mathcal{X}$ with $\mathcal{P}x = x$. This $x$ is the unique solution to $\dot{x} = f(x)$, $x(t_0) = x_0$ on the interval $I = [t_0 - T, t_0 + T]$. Equivalently, the Picard iterates $x_0^{(0)}(t) := x_0$ and $x^{(k+1)} := \mathcal{P}x^{(k)}$ converge uniformly to $x$ on $I$, with the error estimate $d(x^{(k)}, x) \leq \frac{q^k}{1-q} d(x^{(1)}, x^{(0)})$. [/step] [step:Establish continuous dependence on initial data] Let $x_0, \tilde{x}_0 \in E$ with $|\tilde{x}_0 - x_0|$ sufficiently small, and let $x(t) = \varphi(t - t_0, x_0)$ and $\tilde{x}(t) = \varphi(t - t_0, \tilde{x}_0)$ be the corresponding solutions on a common interval $I' \subset I$. For $t \in I'$, \begin{align*} |x(t) - \tilde{x}(t)| &= \left|x_0 - \tilde{x}_0 + \int_{t_0}^{t} \bigl(f(x(s)) - f(\tilde{x}(s))\bigr) \, d\mathcal{L}^1(s)\right| \\ &\leq |x_0 - \tilde{x}_0| + L \int_{t_0}^{t} |x(s) - \tilde{x}(s)| \, d\mathcal{L}^1(s). \end{align*} Define $\psi: I' \to [0, \infty)$ by $\psi(t) := |x(t) - \tilde{x}(t)|$. The inequality above reads $\psi(t) \leq |x_0 - \tilde{x}_0| + L \int_{t_0}^{t} \psi(s) \, d\mathcal{L}^1(s)$. By the [Grönwall Inequality](/theorems/???), \begin{align*} |x(t) - \tilde{x}(t)| \leq |x_0 - \tilde{x}_0| \, e^{L|t - t_0|} \end{align*} for all $t \in I'$. In particular, $\varphi(t - t_0, x_0)$ depends continuously on $(x_0, t)$: as $\tilde{x}_0 \to x_0$, $\sup_{t \in I'} |\varphi(t - t_0, \tilde{x}_0) - \varphi(t - t_0, x_0)| \to 0$. [guided] The Grönwall inequality is the standard tool for converting an integral inequality of the form $\psi(t) \leq C + L\int_{t_0}^t \psi(s) \, d\mathcal{L}^1(s)$ into an exponential bound $\psi(t) \leq C e^{L|t - t_0|}$. We verify that its hypotheses are satisfied: $\psi$ is continuous and non-negative, and $C = |x_0 - \tilde{x}_0| \geq 0$ and $L > 0$ are constants. The [Grönwall Inequality](/theorems/???) then gives \begin{align*} |x(t) - \tilde{x}(t)| \leq |x_0 - \tilde{x}_0| \, e^{L|t-t_0|}. \end{align*} This estimate says that nearby initial conditions produce nearby solutions, with the distance growing at most exponentially. The exponential rate $L$ is the Lipschitz constant of the vector field — the larger the Lipschitz constant, the faster nearby solutions can diverge. On the interval $|t - t_0| \leq T < 1/L$, the exponential factor is bounded by $e^{LT} < e$, so the dependence on initial data is uniformly controlled. [/guided] [/step]