[proofplan] We prove local existence of solutions to $\dot{X} = f(X)$ by reducing to a globally bounded case and applying the Arzela--Ascoli theorem. In Case 1, we assume $f$ is bounded on all of $\mathbb{R}^n$: we construct Euler polygon approximations, verify uniform boundedness and equicontinuity (Lipschitz with constant $M$), extract a uniformly convergent subsequence via [Arzela--Ascoli](/theorems/66), and pass the limit through the integral formulation using uniform continuity of $f$. In Case 2 (general $U$ and $f$), we project $f$ onto a compact ball around $X_0$ to create a bounded extension, apply Case 1, and restrict the solution to a short time interval where it remains inside $U$. [/proofplan] [step:Reduce to the globally bounded case (Case 1 setup)] Assume first that $U = \mathbb{R}^n$ and that $f: \mathbb{R}^n \to \mathbb{R}^n$ is [continuous](/page/Continuity) and bounded: \begin{align*} M := \sup_{x \in \mathbb{R}^n} |f(x)| < \infty. \end{align*} The general case will be recovered in the final step by a truncation argument. [/step] [step:Construct Euler polygon approximations] For each $k \in \mathbb{N}$, define the piecewise-linear Euler approximation $X_k: [0, \infty) \to \mathbb{R}^n$ inductively on intervals of length $1/k$. On $[0, 1/k]$: \begin{align*} X_k(t) = X_0 + t\,f(X_0). \end{align*} On $[j/k,\, (j+1)/k]$ for $j \ge 1$: \begin{align*} X_k(t) = X_k\!\left(\frac{j}{k}\right) + \left(t - \frac{j}{k}\right) f\!\left(X_k\!\left(\frac{j}{k}\right)\right). \end{align*} Define the piecewise-constant companion: \begin{align*} Y_k: [0, \infty) &\to \mathbb{R}^n \\ t &\mapsto X_k\!\left(\frac{j}{k}\right) \quad \text{for } \frac{j}{k} \le t < \frac{j+1}{k}. \end{align*} By construction, $X_k'(t) = f(Y_k(t))$ on each open subinterval, so integrating yields: \begin{align*} X_k(t) = X_0 + \int_0^t f(Y_k(s)) \, d\mathcal{L}^1(s). \end{align*} [guided] The Euler polygon method replaces the ODE $\dot{X} = f(X)$ with a sequence of piecewise-linear approximations. On each subinterval $[j/k, (j+1)/k]$, the slope is frozen at the value $f(X_k(j/k))$, producing a polygonal path. As $k \to \infty$, the mesh size $1/k \to 0$ and the approximations should converge to a true solution. For each $k \in \mathbb{N}$, define $X_k: [0, \infty) \to \mathbb{R}^n$ inductively. Set $X_k(0) = X_0$. On the first subinterval $[0, 1/k]$: \begin{align*} X_k(t) = X_0 + t\,f(X_0). \end{align*} On the general subinterval $[j/k,\, (j+1)/k]$ for $j \ge 1$, having already defined $X_k(j/k)$: \begin{align*} X_k(t) = X_k\!\left(\frac{j}{k}\right) + \left(t - \frac{j}{k}\right) f\!\left(X_k\!\left(\frac{j}{k}\right)\right). \end{align*} This constructs a continuous piecewise-linear path whose slope on each subinterval equals $f$ evaluated at the left endpoint. The companion piecewise-constant [function](/page/Function) $Y_k: [0, \infty) \to \mathbb{R}^n$ is defined by $Y_k(t) = X_k(j/k)$ for $t \in [j/k, (j+1)/k)$. By construction, $X_k'(t) = f(Y_k(t))$ on each open subinterval $(j/k, (j+1)/k)$. Integrating the derivative recovers the integral representation: \begin{align*} X_k(t) = X_0 + \int_0^t f(Y_k(s)) \, d\mathcal{L}^1(s). \end{align*} This integral form is the key to passing limits: it avoids differentiating the $X_k$ (which converge only uniformly, not in $C^1$) and instead reduces convergence to interchanging a limit with integration, which is justified by uniform convergence of the integrands. [/guided] [/step] [step:Verify uniform boundedness and equicontinuity of $\{X_k\}$] **Lipschitz continuity.** For any $t, \tau \ge 0$: \begin{align*} |X_k(t) - X_k(\tau)| = \left|\int_\tau^t f(Y_k(s)) \, d\mathcal{L}^1(s)\right| \le M\,|t - \tau|. \end{align*} In particular, the family $\{X_k\}$ is [equicontinuous](/page/Equicontinuity) on any bounded interval. **Uniform boundedness on $[0, 1]$.** For all $t \in [0, 1]$: \begin{align*} |X_k(t)| \le |X_0| + |X_k(t) - X_0| \le |X_0| + M\,t \le |X_0| + M. \end{align*} The [sequence](/page/Sequence) $\{X_k\}$ satisfies both hypotheses of the [Arzela--Ascoli Theorem](/theorems/66) on $[0, 1]$. [guided] The [Arzela--Ascoli Theorem](/theorems/66) requires two conditions: uniform boundedness and equicontinuity. We verify both for the family $\{X_k\}_{k \in \mathbb{N}}$ on the compact interval $[0, 1]$. **Equicontinuity.** Using the integral representation $X_k(t) = X_0 + \int_0^t f(Y_k(s)) \, d\mathcal{L}^1(s)$ and the bound $|f| \le M$: \begin{align*} |X_k(t) - X_k(\tau)| = \left|\int_\tau^t f(Y_k(s)) \, d\mathcal{L}^1(s)\right| \le \int_{\min(t,\tau)}^{\max(t,\tau)} |f(Y_k(s))| \, d\mathcal{L}^1(s) \le M\,|t - \tau|. \end{align*} This is a uniform Lipschitz bound (constant $M$, independent of $k$), which gives equicontinuity: for any $\delta > 0$, choosing $|t - \tau| < \delta/M$ ensures $|X_k(t) - X_k(\tau)| < \delta$ for all $k$. **Uniform boundedness.** For all $t \in [0, 1]$ and all $k$: \begin{align*} |X_k(t)| \le |X_0| + |X_k(t) - X_0| \le |X_0| + M\,t \le |X_0| + M. \end{align*} So the family is uniformly bounded by $|X_0| + M$. Note: the [Arzela--Ascoli Theorem](/theorems/66) as stated on Androma applies to real-valued functions on compact metric spaces. For $\mathbb{R}^n$-valued functions, we apply it to each component $X_k^{(i)}: [0, 1] \to \mathbb{R}$ ($i = 1, \ldots, n$), extracting a subsequence for the first component, then a further subsequence for the second, and so on. After $n$ extractions, we obtain a single subsequence along which all components converge uniformly, hence the $\mathbb{R}^n$-valued sequence converges uniformly. [/guided] [/step] [step:Extract a uniformly convergent subsequence and show $Y_{k(i)} \to X$ uniformly] By the [Arzela--Ascoli Theorem](/theorems/66), there exists a subsequence $(X_{k(i)})_{i \in \mathbb{N}}$ converging [uniformly](/page/Uniform%20Convergence) on $[0, 1]$ to a continuous function $X: [0, 1] \to \mathbb{R}^n$: \begin{align*} \sup_{t \in [0,1]} |X_{k(i)}(t) - X(t)| \to 0 \quad \text{as } i \to \infty. \end{align*} The companion sequence $Y_{k(i)}$ also converges uniformly to $X$. For $t \in [j/k, (j+1)/k)$: \begin{align*} |X_k(t) - Y_k(t)| = \left|X_k(t) - X_k\!\left(\frac{j}{k}\right)\right| \le M \cdot \frac{1}{k}. \end{align*} By the triangle inequality: \begin{align*} \sup_{t \in [0,1]} |Y_{k(i)}(t) - X(t)| \le \frac{M}{k(i)} + \sup_{t \in [0,1]} |X_{k(i)}(t) - X(t)| \to 0. \end{align*} [guided] The [Arzela--Ascoli Theorem](/theorems/66) extracts from the equicontinuous, uniformly bounded family $\{X_k\}$ a subsequence $(X_{k(i)})_{i \in \mathbb{N}}$ converging uniformly on $[0, 1]$ to a continuous limit $X: [0, 1] \to \mathbb{R}^n$. We also need the companion sequence $Y_{k(i)}$ to converge to $X$. Recall that $Y_k(t) = X_k(j/k)$ for $t \in [j/k, (j+1)/k)$ — it is the piecewise-constant function that evaluates $X_k$ at the left endpoint of each subinterval. The gap between $X_k$ and $Y_k$ is controlled by the Lipschitz bound: \begin{align*} |X_k(t) - Y_k(t)| = |X_k(t) - X_k(j/k)| \le M|t - j/k| \le M/k. \end{align*} This bound is uniform in $t$ and tends to $0$ as $k \to \infty$. By the triangle inequality: \begin{align*} \sup_{t \in [0,1]} |Y_{k(i)}(t) - X(t)| \le \underbrace{\sup_{t \in [0,1]} |Y_{k(i)}(t) - X_{k(i)}(t)|}_{\le M/k(i) \to 0} + \underbrace{\sup_{t \in [0,1]} |X_{k(i)}(t) - X(t)|}_{\to 0 \text{ by Arzela--Ascoli}} \to 0. \end{align*} So $Y_{k(i)} \to X$ uniformly on $[0, 1]$. This is needed in the next step to pass the limit through $f$. [/guided] [/step] [step:Pass to the limit in the integral equation to obtain a solution] Since the trajectories $X_{k(i)}$ and $Y_{k(i)}$ remain in the closed ball $\overline{B}(X_0, |X_0| + M)$, which is [compact](/page/Compact%20Space), and $f$ is continuous on this compact set, $f$ is [uniformly continuous](/page/Uniform%20Continuity) there (continuous functions on compact sets are uniformly continuous). Combined with $Y_{k(i)} \to X$ uniformly, we obtain $f(Y_{k(i)}(\cdot)) \to f(X(\cdot))$ uniformly on $[0, 1]$. Passing to the [limit](/page/Limit) in the integral representation: since $f(Y_{k(i)}(\cdot)) \to f(X(\cdot))$ uniformly on $[0, t]$, the uniform convergence theorem for integrals gives \begin{align*} X(t) = \lim_{i \to \infty} \left(X_0 + \int_0^t f(Y_{k(i)}(s)) \, d\mathcal{L}^1(s)\right) = X_0 + \int_0^t f(X(s)) \, d\mathcal{L}^1(s). \end{align*} By the [Fundamental Theorem of Calculus](/theorems/632), $X$ is [differentiable](/page/Derivative) and satisfies $\dot{X}(t) = f(X(t))$ with $X(0) = X_0$. This completes Case 1. [guided] The passage from discrete approximation to continuous solution requires interchanging a limit with an integral. We justify this rigorously. The trajectories $\{X_{k(i)}\}$ all lie in the compact set $\overline{B}(X_0, |X_0| + M)$ (by the uniform bound), so $\{Y_{k(i)}\}$ does too. Since $f$ is continuous on a compact set, it is uniformly continuous: for every $\epsilon > 0$ there exists $\eta > 0$ such that $|x - y| < \eta$ implies $|f(x) - f(y)| < \epsilon$. Since $Y_{k(i)} \to X$ uniformly, for large $i$ we have $|Y_{k(i)}(s) - X(s)| < \eta$ for all $s \in [0, 1]$, whence $|f(Y_{k(i)}(s)) - f(X(s))| < \epsilon$ for all $s$. This gives: \begin{align*} \left|\int_0^t f(Y_{k(i)}(s)) \, d\mathcal{L}^1(s) - \int_0^t f(X(s)) \, d\mathcal{L}^1(s)\right| \le \int_0^t |f(Y_{k(i)}(s)) - f(X(s))| \, d\mathcal{L}^1(s) \le \epsilon \cdot t. \end{align*} Therefore $X(t) = X_0 + \int_0^t f(X(s)) \, d\mathcal{L}^1(s)$. Since $f \circ X$ is continuous (composition of continuous functions), the [Fundamental Theorem of Calculus](/theorems/632) implies $X$ is $C^1$ and $\dot{X}(t) = f(X(t))$ with $X(0) = X_0$. [/guided] [/step] [step:Recover the local result for general $U$ and $f$ via projection] Now let $U \subseteq \mathbb{R}^n$ be an arbitrary open set with $X_0 \in U$, and let $f: U \to \mathbb{R}^n$ be continuous (not necessarily bounded). Choose $R > 0$ such that $\overline{B}(X_0, R) \subseteq U$. Define the nearest-point projection: \begin{align*} \Pi: \mathbb{R}^n &\to \overline{B}(X_0, R) \\ x &\mapsto \begin{cases} x & \text{if } |x - X_0| \le R, \\ X_0 + R\,\dfrac{x - X_0}{|x - X_0|} & \text{if } |x - X_0| > R. \end{cases} \end{align*} Define the modified vector field $\tilde{f} := f \circ \Pi: \mathbb{R}^n \to \mathbb{R}^n$. Since $\Pi$ is continuous and $f$ is continuous on $\overline{B}(X_0, R)$, the composition $\tilde{f}$ is continuous on $\mathbb{R}^n$. Moreover, $\tilde{f}$ is bounded because $\overline{B}(X_0, R)$ is compact: \begin{align*} \sup_{x \in \mathbb{R}^n} |\tilde{f}(x)| = \sup_{y \in \overline{B}(X_0, R)} |f(y)| < \infty. \end{align*} By Case 1, there exists a $C^1$ solution $X: [0, 1] \to \mathbb{R}^n$ to $\dot{X} = \tilde{f}(X)$ with $X(0) = X_0$. Since $X$ is continuous and $X(0) = X_0 \in B(X_0, R)$, by continuity there exists $\delta > 0$ such that $X(t) \in B(X_0, R) \subseteq U$ for all $t \in [0, \delta]$. On this interval, $\Pi(X(t)) = X(t)$, so $\tilde{f}(X(t)) = f(X(t))$ and $X$ solves the original ODE on $[0, \delta]$. [guided] The general case reduces to Case 1 by a truncation trick. The obstacle is that $f: U \to \mathbb{R}^n$ may be unbounded and $U$ may not be all of $\mathbb{R}^n$. We fix both problems at once by projecting onto a compact ball. Since $U$ is open and $X_0 \in U$, there exists $R > 0$ with $\overline{B}(X_0, R) \subseteq U$. The nearest-point projection $\Pi: \mathbb{R}^n \to \overline{B}(X_0, R)$ is defined by: \begin{align*} \Pi(x) = \begin{cases} x & \text{if } |x - X_0| \le R, \\ X_0 + R\,\dfrac{x - X_0}{|x - X_0|} & \text{if } |x - X_0| > R. \end{cases} \end{align*} This map is continuous (it is the identity inside the ball and the radial projection outside). The modified vector field $\tilde{f} := f \circ \Pi: \mathbb{R}^n \to \mathbb{R}^n$ is continuous (composition of continuous maps) and bounded: since $\Pi$ maps all of $\mathbb{R}^n$ into the compact set $\overline{B}(X_0, R)$, and $f$ is continuous on this compact set, we have $\sup_{x \in \mathbb{R}^n} |\tilde{f}(x)| = \sup_{y \in \overline{B}(X_0, R)} |f(y)| < \infty$. Case 1 now applies to $\tilde{f}$: there exists a $C^1$ solution $X: [0, 1] \to \mathbb{R}^n$ to $\dot{X} = \tilde{f}(X)$ with $X(0) = X_0$. Since $X(0) = X_0$ and $X$ is continuous, there exists $\delta > 0$ such that $|X(t) - X_0| < R$ for all $t \in [0, \delta]$. On this interval, $X(t) \in B(X_0, R)$, so $\Pi(X(t)) = X(t)$ and therefore $\tilde{f}(X(t)) = f(\Pi(X(t))) = f(X(t))$. Hence $X$ solves the original ODE $\dot{X} = f(X)$ on $[0, \delta]$ with $X(0) = X_0$. [/guided] [/step]