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  "body": "\n## Proof\n\nWe build Euler polygons, extract a uniformly convergent subsequence with the Arzelà–Ascoli theorem, and show that the limit solves the integral equation.\n\n1. On $\\mathbb{R}^n$ we use the norm $\\abs{x} = \\max_i \\abs{x_i}$, and on $\\mathbb{R} \\times \\mathbb{R}^n$ the norm $\\max(\\abs{t}, \\abs{x})$. Integrals of vector functions are taken coordinatewise. For continuous $v$ on $[c, d]$, the [[triangle inequality for integrals|0NR82KMYKD]] and [[monotonicity of the integral|0MBWQKM8QR]] in each coordinate give $\\abs{\\int_c^d v} \\le \\int_c^d \\abs{v}$.\n2. Since $D$ is open, there are $a, b \u003e 0$ such that $R = [t_0 - a, t_0 + a] \\times \\set{x : \\abs{x - x_0} \\le b}$ lies in $D$.\n3. The set $R$ is closed and bounded, so it is compact by the [[Heine–Borel theorem|0PHT6NBYPB]]. Its image under the continuous map $\\abs{f}$ is [[compact|0K9AAJ8G51]], hence bounded. Choose $M \u003e 0$ with $\\abs{f(p)} \\le M$ for all $p \\in R$.\n4. By the [[Heine–Cantor theorem|06B8BNJCD6]], $f$ is uniformly continuous on $R$.\n5. Let $h = \\min(a, b/M)$ and $X = [t_0 - h, t_0 + h]$. For an integer $N \\ge 1$, let $t_k = t_0 + k h / N$ for $k = -N, \\dots, N$.\n6. Define $x_N : X \\to \\mathbb{R}^n$ by $x_N(t_0) = x_0$ and, step by step outward from $t_0$, $x_N(t) = x_N(t_k) + (t - t_k)\\, f(t_k, x_N(t_k))$. For $k \\ge 0$ this is used for $t \\in [t_k, t_{k+1}]$, and for $k \\le 0$ for $t \\in [t_{k-1}, t_k]$.\n7. We show by induction on $\\abs{k}$ that this is well defined and that $\\abs{x_N(t) - x_0} \\le M \\abs{t - t_0}$. Assume the bound holds at $t_k$. Then $\\abs{x_N(t_k) - x_0} \\le M h \\le b$ and $\\abs{t_k - t_0} \\le a$, so $(t_k, x_N(t_k)) \\in R$.\n8. So $f(t_k, x_N(t_k))$ is defined and has norm at most $M$. On the next piece, $\\abs{x_N(t) - x_0} \\le M \\abs{t_k - t_0} + M \\abs{t - t_k} = M \\abs{t - t_0}$, since $t_k$ lies between $t_0$ and $t$. This completes the induction.\n9. Each $x_N$ is continuous and piecewise linear, with slopes of norm at most $M$. Splitting $[t', t]$ at the nodes and using the triangle inequality gives $\\abs{x_N(t) - x_N(t')} \\le M \\abs{t - t'}$ for all $t, t' \\in X$.\n10. By steps 8 and 9, each coordinate family $\\set{x_{N, i} : N \\ge 1}$ is bounded by $\\abs{x_0} + b$. It is [[equicontinuous|0MW2WXTHMC]], with $\\delta = \\varepsilon / M$ for each $\\varepsilon \u003e 0$.\n11. By the [[Arzelà–Ascoli theorem|0381S8Z5P0]], the closure of the family of first coordinates is compact in $C(X)$. By [[sequential compactness|0K8SGQSV7G]], some subsequence of $(x_{N,1})$ converges in $C(X)$, that is [[uniformly|08FPABDQ4C]].\n12. Repeating step 11 for coordinates $2, \\dots, n$, each time inside the previous subsequence, gives $N_1 \u003c N_2 \u003c \\cdots$ and a continuous $z : X \\to \\mathbb{R}^n$ with $\\sup_{t \\in X} \\abs{x_{N_j}(t) - z(t)} \\to 0$.\n13. Letting $j \\to \\infty$ in step 8 gives $\\abs{z(t) - x_0} \\le M \\abs{t - t_0} \\le b$. So $(t, z(t)) \\in R \\subseteq D$ for $t \\in X$, and $z(t_0) = x_0$.\n14. Let $\\varepsilon \u003e 0$. By step 4, choose $\\eta \u003e 0$ with $\\abs{f(p) - f(q)} \u003c \\varepsilon$ for $p, q \\in R$ with distance less than $\\eta$.\n15. Let $e_N(t) = x_N(t) - x_0 - \\int_{t_0}^t f(s, x_N(s))\\, ds$ for $t \\in X$. The integrand is continuous, since $x_N$ is continuous.\n16. Let $t \\ge t_0$ lie in $[t_k, t_{k+1}]$. By step 6, $x_N(t) - x_0$ is the sum of the integrals of the constants $f(t_j, x_N(t_j))$ over the pieces $[t_j, t_{j+1}]$ for $j \u003c k$ and over $[t_k, t]$.\n17. So, by [[linearity of the integral|080RWVVSB6]], $e_N(t)$ is the sum over these pieces of the integrals of $f(t_j, x_N(t_j)) - f(s, x_N(s))$, with $s$ in the piece of $t_j$.\n18. For $s$ in that piece, $\\abs{s - t_j} \\le h / N$ and, by step 9, $\\abs{x_N(s) - x_N(t_j)} \\le M h / N$. If $N$ is so large that $h/N \u003c \\eta$ and $Mh/N \u003c \\eta$, each integrand has norm less than $\\varepsilon$.\n19. By step 1 the pieces of total length $t - t_0$ then give $\\abs{e_N(t)} \\le \\varepsilon h$. The case $t \\le t_0$ is the same. So $\\sup_X \\abs{e_N} \\to 0$ as $N \\to \\infty$.\n20. For large $j$, $\\abs{x_{N_j}(s) - z(s)} \u003c \\eta$ for all $s \\in X$, so $\\abs{f(s, x_{N_j}(s)) - f(s, z(s))} \u003c \\varepsilon$. By step 1, $\\abs{\\int_{t_0}^t \\left( f(s, x_{N_j}(s)) - f(s, z(s)) \\right) ds} \\le \\varepsilon h$ for all $t \\in X$.\n21. For $t \\in X$, $z(t) - x_0 - \\int_{t_0}^t f(s, z(s))\\, ds$ equals $\\left( z(t) - x_{N_j}(t) \\right) + e_{N_j}(t) + \\int_{t_0}^t \\left( f(s, x_{N_j}(s)) - f(s, z(s)) \\right) ds$.\n22. By steps 12, 19 and 20, all three terms tend to $0$ as $j \\to \\infty$. Hence $z(t) = x_0 + \\int_{t_0}^t f(s, z(s))\\, ds$ for each $t \\in X$.\n23. The map $s \\mapsto f(s, z(s))$ is continuous on $X$. Write $\\int_{t_0}^t = \\int_{t_0 - h}^t - \\int_{t_0 - h}^{t_0}$. By the [[first fundamental theorem of calculus|063H5KJC69]] in each coordinate, $z$ is differentiable at each $t \\in (t_0 - h, t_0 + h)$ with $z'(t) = f(t, z(t))$.\n24. Let $I = (t_0 - h, t_0 + h)$, an open interval that contains $t_0$. By steps 13 and 23, $z$ restricted to $I$ solves the [[initial value problem|075VBY4K4X]]: $(t, z(t)) \\in D$ and $z'(t) = f(t, z(t))$ for $t \\in I$, and $z(t_0) = x_0$.\n\n## Idea\n\nEuler polygons follow the field $f$ in small straight steps, so they almost solve the equation. Their slopes are bounded by $M$, which makes the family equicontinuous, and the Arzelà–Ascoli theorem gives a uniformly convergent subsequence. Uniform continuity of $f$ on a compact box then shows that the limit solves the integral equation exactly.\n",
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