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  "body": "\n## Proof\n\nHere $\\norm{\\cdot}$ is the Euclidean norm, and $Df(z)$ is the Jacobian matrix of $f$ at $z$. Its entries $\\partial f_i / \\partial x_j$ are continuous on $U$, since $f$ is continuously differentiable.\n\n1. Since $U$ is open, there is $r \u003e 0$ such that the [[open ball|0DC3MPR7C9]] of radius $r$ around $x_0$ lies in $U$. Let $a = r/2 \u003e 0$.\n2. The closed ball $B = \\set{x : \\norm{x - x_0} \\le a}$ lies in that open ball, so $B \\subseteq U$. It is closed, bounded and nonempty, so it is compact by the [[Heine–Borel theorem|07PFYV65MB]].\n3. The function $z \\mapsto \\norm{f(z)}$ is continuous on $U$. By [[the extreme value theorem on compact sets|0V4DET59NW]], it has a maximum $L \\ge 0$ on $B$. So $\\norm{f(x)} \\le L$ for $x \\in B$.\n4. The function $z \\mapsto \\left(\\sum_{i, j} \\left(\\partial f_i / \\partial x_j (z)\\right)^2\\right)^{1/2}$ is continuous on $U$. Let $K \\ge 0$ be its maximum on $B$.\n5. Let $z \\in B$ and $h \\in \\R^n$. Entry $i$ of $Df(z) h$ is the dot product of row $i$ of $Df(z)$ with $h$. The [[Cauchy–Schwarz inequality|0RFWW56Y22]] bounds it, and summing the squares gives $\\norm{Df(z) h} \\le K \\norm{h}$.\n6. So the operator norm of $Df(z)$ is at most $K$ for each $z \\in B$.\n7. Let $x, y \\in B$ and $0 \\le s \\le 1$. Then $\\norm{x + s(y - x) - x_0} \\le (1 - s) \\norm{x - x_0} + s \\norm{y - x_0} \\le a$. So the segment $[x, y]$ lies in $B \\subseteq U$.\n8. By the [[mean value inequality|02CYB184RC]] and step 6, $\\norm{f(y) - f(x)} \\le K \\norm{y - x}$ for all $x, y \\in B$.\n9. Let $\\varepsilon = a/(L + 1) \u003e 0$ and $J = [t_0 - \\varepsilon, t_0 + \\varepsilon]$. Let $g : J \\times \\R^n \\to \\R^n$ be any function with $g(t, x) = f(x)$ for $t \\in J$ and $x \\in B$.\n10. Condition (1) of the [[Picard–Lindelöf hypotheses|0J7HF2PF1H]]: for each $t \\in J$, $x \\mapsto g(t, x) = f(x)$ is $K$-[[Lipschitz|055VVSN00V]] on $B$, by step 8.\n11. Condition (2): for each $x \\in B$, $t \\mapsto g(t, x) = f(x)$ is constant, hence continuous on $J$.\n12. Condition (3): $\\norm{g(t, x)} = \\norm{f(x)} \\le L$ for $t \\in J$ and $x \\in B$, by step 3.\n13. Condition (4): here $\\max(t_{\\max} - t_0, t_0 - t_{\\min}) = \\varepsilon$, and $L \\varepsilon = L a/(L + 1) \\le a$.\n14. So $g$ satisfies the hypotheses at $(t_0, x_0)$ on $J$ with the constants $a, L, K$.\n\n## Idea\n\nA small closed ball around $x_0$ is compact. On it, the continuous functions $\\norm{f}$ and $\\norm{Df}$ are bounded, which gives the bound $L$ and, through the mean value inequality, the Lipschitz constant $K$. A short time interval then keeps $L \\varepsilon \\le a$.\n",
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