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  "body": "\n## Proof\n\nWe find the curve as the unique fixed point of the Picard operator, with the Banach fixed point theorem.\n\n1. If $t_{\\min} = t_{\\max}$, then $J = \\set{t_0}$. The equation at $t_0$ says $\\alpha(t_0) = x_0$, so exactly one curve works. From now on let $t_{\\min} \u003c t_{\\max}$, and let $T = \\max(t_{\\max} - t_0, t_0 - t_{\\min})$.\n2. $f$ is continuous on $J \\times B$. Indeed, $\\norm{f(s, x) - f(s', x')} \\le K \\norm{x - x'} + \\norm{f(s, x') - f(s', x')}$, by condition (1) of the [[Picard–Lindelöf hypotheses|0J7HF2PF1H]]. The last term tends to $0$ as $s \\to s'$, by condition (2).\n3. Let $\\pi(x) = x$ if $\\norm{x - x_0} \\le a$, and $\\pi(x) = x_0 + a (x - x_0)/\\norm{x - x_0}$ otherwise. Then $\\pi : \\R^n \\to B$ is continuous, since both formulas are continuous and agree where $\\norm{x - x_0} = a$. If $a = 0$, then $\\pi$ is the constant $x_0$.\n4. Let $\\tilde f(t, x) = f(t, \\pi(x))$. By steps 2 and 3, $\\tilde f$ is continuous on $J \\times \\R^n$. It equals $f$ on $J \\times B$, so it satisfies the hypotheses with the same constants $a, L, K$.\n5. For each continuous $\\gamma : J \\to \\R^n$, the map $s \\mapsto \\tilde f(s, \\gamma(s))$ is [[continuous|0NRQPQX3VA]], so it is [[integrable|06SESV7GQY]]. So the [[Picard operator|00WAVB68F9]] $P$ of $\\tilde f$ at $(t_0, x_0)$ is defined.\n6. Let $X$ be the set of continuous $\\gamma : J \\to B$, with $d(\\gamma, \\delta) = \\sup_{t \\in J} \\norm{\\gamma(t) - \\delta(t)}$. It is nonempty, since it contains the constant curve $x_0$.\n7. $X$ is complete. The curves in $X$ are bounded, and the space of bounded continuous maps $J \\to \\R^n$ is [[complete|09GRCZVR5E]], since [[$\\R^n$ is complete|0ABP2F2ZP7]]. A Cauchy sequence $(\\gamma_k)$ in $X$ converges uniformly to a continuous $\\gamma$.\n8. For $t \\in J$, $\\norm{\\gamma(t) - x_0} \\le \\norm{\\gamma(t) - \\gamma_k(t)} + a$, and the first term tends to $0$. So $\\gamma \\in X$, and $\\gamma_k \\to \\gamma$ in $X$.\n9. Lemma. If $h : [c, e] \\to \\R^n$ and $\\phi : [c, e] \\to \\R$ are integrable with $\\norm{h} \\le \\phi$, then $\\norm{\\int_c^e h} \\le \\int_c^e \\phi$.\n10. Proof of the lemma: for $v = \\int_c^e h \\ne 0$, [[linearity|080RWVVSB6]] gives $\\norm{v}^2 = \\int_c^e v \\cdot h$. By [[Cauchy–Schwarz|0RFWW56Y22]] and [[monotonicity|0MBWQKM8QR]], this is at most $\\norm{v} \\int_c^e \\phi$.\n11. $P$ maps $X$ into $X$. Let $\\gamma \\in X$. By condition (3), $\\norm{\\tilde f(s, \\gamma(s))} \\le L$. The lemma gives $\\norm{(P\\gamma)(t) - (P\\gamma)(t')} \\le L \\abs{t - t'}$, so $P\\gamma$ is continuous.\n12. The lemma also gives $\\norm{(P\\gamma)(t) - x_0} \\le L \\abs{t - t_0} \\le L T \\le a$, by condition (4). So $P\\gamma \\in X$.\n13. By [[the contraction lemma for the Picard operator|08XDXP86XE]], applied to $\\tilde f$, there is $m \\ge 1$ such that $P^m$ is a contraction of $X$.\n14. By the [[Banach fixed point theorem|0VK6VEH345]], $P^m$ has exactly one fixed point $\\alpha \\in X$.\n15. Then $P^m(P\\alpha) = P(P^m \\alpha) = P\\alpha$, and $P\\alpha \\in X$. So $P\\alpha$ is a fixed point of $P^m$, and $P\\alpha = \\alpha$ by uniqueness.\n16. If $\\beta \\in X$ and $P\\beta = \\beta$, then $P^m \\beta = \\beta$, so $\\beta = \\alpha$. Thus $\\alpha$ is the only fixed point of $P$ in $X$.\n17. For $\\gamma \\in X$, the values $\\gamma(s)$ lie in $B$, where $\\tilde f = f$. So $(P\\gamma)(t) = x_0 + \\int_{t_0}^{t} f(s, \\gamma(s))\\, ds$.\n18. Hence the continuous curves $J \\to B$ that satisfy the integral equation are exactly the fixed points of $P$ in $X$. By step 16, there is exactly one, namely $\\alpha$.\n\n## Idea\n\nThe integral equation says that $\\alpha$ is a fixed point of the Picard operator. On continuous curves with values in $B$, the operator itself may not contract, but a high iterate does. The Banach fixed point theorem then gives one fixed point of the iterate, and it is also the only fixed point of $P$.\n",
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