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  "body": "\n## Statement\n\nLet $f$ satisfy the [[Picard–Lindelöf hypotheses|0J7HF2PF1H]] at $(t_0, x_0)$ on $J = [t_{\\min}, t_{\\max}]$ with constants $a, L, K$.\nLet $X$ be the set of continuous curves $\\gamma : J \\to \\mathbb{R}^n$ with $\\|\\gamma(t) - x_0\\| \\le a$ for each $t \\in J$, with the distance $d(\\gamma, \\delta) = \\sup_{t \\in J} \\|\\gamma(t) - \\delta(t)\\|$.\nLet $P : X \\to X$ be the [[Picard operator|00WAVB68F9]] and let $T = \\max(t_{\\max} - t_0,\\ t_0 - t_{\\min})$.\nThen for each integer $m \\ge 0$ and all $\\gamma, \\delta \\in X$,\n$$d\\left(P^m \\gamma, P^m \\delta\\right) \\le \\frac{(K T)^m}{m!}\\, d(\\gamma, \\delta).$$\nIn particular, there is an $m \\ge 1$ such that $P^m$ is a contraction of $X$: $(KT)^m / m! \u003c 1$.\n\n## Remarks\n\nProof idea: by induction on $m$, $\\|(P^m \\gamma)(t) - (P^m \\delta)(t)\\| \\le \\frac{(K |t - t_0|)^m}{m!}\\, d(\\gamma, \\delta)$ for each $t \\in J$. The induction step integrates the bound of the previous step.\nThe space $X$ is a complete metric space. So the Banach fixed point theorem applies to $P^m$, and it gives a unique fixed point of $P$.\nIf $K T \u003c 1$, then $P$ itself is a contraction. The power $P^m$ avoids the need for a short time interval.\n\n## Explanations\n\n### L1 human\n\nA contraction brings any two points closer by a fixed factor smaller than one.\nThe Picard operator may not do this in one step, but after enough steps it does, because $m!$ grows faster than $(KT)^m$.\nThat is enough to get a unique fixed point.\n",
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