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  "body": "\n## Proof\n\nWe show that the Picard–Lindelöf theorem gives local solutions on time steps of a length that does not depend on the starting point. Then we glue finitely many steps on each compact subinterval.\n\n1. A solution on a subinterval $[c, d] \\subseteq J$ is a curve $y : [c, d] \\to \\mathbb{R}^n$ with $y'(t) = f(t, y(t))$ at each $t \\in [c, d]$, using one-sided derivatives at $c$ and $d$. A solution on $J$ is defined in the same way. If $J$ is a single point, the constant $x_0$ is the only solution. So let $J$ contain more than one point.\n2. Uniqueness on a compact piece. Let $[c, d] \\subseteq J$ with $c \\le t_0 \\le d$, and let $y, z$ be solutions on $[c, d]$ with $y(t_0) = z(t_0) = x_0$.\n3. If $t_0 \u003c d$, [[continuous dependence on initial data|0G3DBXRRQJ]] on $[t_0, d]$, with $U = \\mathbb{R}^n$, gives $\\|y(t) - z(t)\\| \\le 0 \\cdot e^{K(t - t_0)} = 0$. So $y = z$ on $[t_0, d]$.\n4. If $c \u003c t_0$, let $g(t, x) = -f(-t, x)$ on $(-J) \\times \\mathbb{R}^n$, where $-J = \\set{-t : t \\in J}$. It is continuous and $K$-Lipschitz in $x$. The curves $\\tilde y(t) = y(-t)$ and $\\tilde z(t) = z(-t)$ satisfy $\\tilde y'(t) = -y'(-t) = g(t, \\tilde y(t))$, and the same for $\\tilde z$.\n5. Applying step 3 to $g$ on $[-t_0, -c]$ gives $\\tilde y = \\tilde z$ there. So $y = z$ on $[c, t_0]$, and hence on all of $[c, d]$.\n6. A growth bound. By the [[Lipschitz condition|055VVSN00V]] and the [[triangle inequality|0FJZMBX0ZZ]], $\\|f(t, x)\\| \\le \\|f(t, 0)\\| + K \\|x\\|$ for $t \\in J$ and $x \\in \\mathbb{R}^n$.\n7. One step. Fix $[c, d] \\subseteq J$ with $c \u003c d$. The map $t \\mapsto \\|f(t, 0)\\|$ is continuous, so by the [[extreme value theorem|0WS1A8E2P9]] it has a maximum $m$ on $[c, d]$. Let $h = 1 / (1 + K)$.\n8. Let $t_1 \\in [c, d)$, $y_1 \\in \\mathbb{R}^n$ and $t_2 = \\min(t_1 + h, d)$. Let $A = m + K \\|y_1\\| + 1 \u003e 0$, let $B$ be the closed ball of radius $A$ about $y_1$, and let $L = m + K \\|y_1\\| + K A$.\n9. For $t \\in [t_1, t_2]$ and $x \\in B$, step 6 gives $\\|f(t, x)\\| \\le m + K (\\|y_1\\| + A) = L$. Also $L = A - 1 + KA \\le (1 + K) A$, so $L (t_2 - t_1) \\le L h \\le A$.\n10. So the restriction of $f$ to $[t_1, t_2] \\times \\mathbb{R}^n$ satisfies the [[Picard–Lindelöf hypotheses|0J7HF2PF1H]] at $(t_1, y_1)$ with constants $A, L, K$. Condition (1) is the global Lipschitz bound, (2) holds as $f$ is continuous, and (3) and (4) are step 9.\n11. By the [[Picard–Lindelöf theorem|0P6KRD5B7A]], there is a solution $\\alpha$ on $[t_1, t_2]$ with $\\alpha(t_1) = y_1$. Now let $t_1 \\in (c, d]$. Apply steps 7 to 10 to $g$ of step 4, and reverse time as in step 4. This gives a solution on $[\\max(t_1 - h, c), t_1]$ with value $y_1$ at $t_1$.\n12. Existence on a compact piece. Let $[c, d] \\subseteq J$ with $c \\le t_0 \\le d$ and $c \u003c d$. Let $s_k = \\min(t_0 + k h, d)$ for $k \\ge 0$.\n13. Suppose $t_0 \u003c d$. By step 11 at $(t_0, x_0)$ there is a solution $\\beta_1$ on $[t_0, s_1]$ with $\\beta_1(t_0) = x_0$.\n14. Given a solution $\\beta_k$ on $[t_0, s_k]$ with $s_k \u003c d$, step 11 at $(s_k, \\beta_k(s_k))$ gives a solution $\\alpha$ on $[s_k, s_{k+1}]$ with $\\alpha(s_k) = \\beta_k(s_k)$. Let $\\beta_{k+1}$ be $\\beta_k$ on $[t_0, s_k]$ and $\\alpha$ on $[s_k, s_{k+1}]$.\n15. At $s_k$ the left derivative of $\\beta_{k+1}$ is $f(s_k, \\beta_k(s_k))$ and the right derivative is $f(s_k, \\alpha(s_k))$. These are equal, so $\\beta_{k+1}$ is differentiable at $s_k$, and $\\beta_{k+1}$ is a solution on $[t_0, s_{k+1}]$.\n16. Since $h \u003e 0$ is fixed, $s_k = d$ once $k \\ge (d - t_0)/h$. So after finitely many steps we get a solution on $[t_0, d]$.\n17. In the same way, with the backward steps of step 11, we get a solution on $[c, t_0]$ with value $x_0$ at $t_0$. Both one-sided derivatives at $t_0$ equal $f(t_0, x_0)$. So the glued curve $x_{[c, d]}$ is a solution on $[c, d]$ with $x_{[c, d]}(t_0) = x_0$.\n18. Exhausting $J$. Choose $c_m, d_m \\in J$ for $m \\ge 1$ with $c_m \\le t_0 \\le d_m$, $c_m \u003c d_m$, $c_m$ non-increasing and $d_m$ non-decreasing. If $J$ has a least element, let each $c_m$ be it. Otherwise let $c_m \\to \\inf J$. Treat $d_m$ in the same way with the upper end of $J$.\n19. Then every $t \\in J$ lies in some $[c_m, d_m]$. If $t$ is an interior point of $J$, then $c_m \u003c t \u003c d_m$ for all large $m$.\n20. Let $x_m = x_{[c_m, d_m]}$ from step 17. For $m' \\ge m$, the restriction of $x_{m'}$ to $[c_m, d_m]$ is a solution with value $x_0$ at $t_0$. By steps 2 to 5 it equals $x_m$.\n21. So $x(t) = x_m(t)$, for any $m$ with $t \\in [c_m, d_m]$, defines a curve $x : J \\to \\mathbb{R}^n$ with $x(t_0) = x_0$.\n22. If $t$ is interior to $J$, then $x = x_m$ on the open interval $(c_m, d_m)$ around $t$ for a large $m$, so $x'(t) = f(t, x(t))$. If $t$ is an end point of $J$ in $J$, then $t = c_m$ or $t = d_m$ for all $m$, and the one-sided derivative is that of $x_m$.\n23. So $x$ is a solution of the [[initial value problem|075VBY4K4X]] $x' = f(t, x)$, $x(t_0) = x_0$ on all of $J$.\n24. If $y$ is another such solution on $J$, its restriction to each $[c_m, d_m]$ equals $x_m$ by steps 2 to 5. By step 19 these intervals cover $J$, so $y = x$. Hence the solution on $J$ exists and is unique.\n\n## Idea\n\nFor a general field, Picard–Lindelöf gives a time step that may shrink as the solution grows, and the steps can add up to a finite time. With a global Lipschitz bound, the field grows at most linearly, and a ball of radius about $\\|y_1\\|$ gives a step of length $1/(1+K)$ from every starting point. Finitely many such steps cover any compact time interval.\n",
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