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  "body": "\n## Proof\n\nWe argue by induction on $m$, with a pointwise bound. Write $d = d(\\gamma, \\delta)$.\n\n1. Lemma. Let $c \\le e$, let $h : [c, e] \\to \\R^n$ be integrable, and let $\\phi : [c, e] \\to \\R$ be integrable with $\\norm{h(s)} \\le \\phi(s)$. Then $\\norm{\\int_c^e h(s)\\, ds} \\le \\int_c^e \\phi(s)\\, ds$.\n2. Proof of the lemma. Let $v = \\int_c^e h(s)\\, ds$ and assume $v \\ne 0$. By [[linearity of the integral|080RWVVSB6]], $\\norm{v}^2 = \\sum_i v_i \\int_c^e h_i = \\int_c^e v \\cdot h(s)\\, ds$.\n3. By the [[Cauchy–Schwarz inequality|0RFWW56Y22]], $v \\cdot h(s) \\le \\norm{v} \\norm{h(s)} \\le \\norm{v} \\phi(s)$. By [[monotonicity of the integral|0MBWQKM8QR]], $\\norm{v}^2 \\le \\norm{v} \\int_c^e \\phi$. Dividing by $\\norm{v}$ proves the lemma.\n4. Claim: for each $m \\ge 0$ and each $t \\in J$, $\\norm{(P^m \\gamma)(t) - (P^m \\delta)(t)} \\le \\frac{(K \\abs{t - t_0})^m}{m!}\\, d$.\n5. For $m = 0$, $P^0$ is the identity, and $\\norm{\\gamma(t) - \\delta(t)} \\le d$ by the definition of $d$.\n6. Assume the claim for $m$. Let $\\gamma_m = P^m \\gamma$ and $\\delta_m = P^m \\delta$. They lie in $X$, so their values lie in the ball $B$ of radius $a$ around $x_0$.\n7. For $s \\in J$, condition (1) of the [[Picard–Lindelöf hypotheses|0J7HF2PF1H]] says $f(s, \\cdot)$ is $K$-[[Lipschitz|055VVSN00V]] on $B$. With the claim for $m$, $\\norm{f(s, \\gamma_m(s)) - f(s, \\delta_m(s))} \\le K \\frac{(K \\abs{s - t_0})^m}{m!}\\, d$.\n8. By the definition of the [[Picard operator|00WAVB68F9]], $(P^{m+1} \\gamma)(t) - (P^{m+1} \\delta)(t) = \\int_{t_0}^{t} \\left(f(s, \\gamma_m(s)) - f(s, \\delta_m(s))\\right) ds$.\n9. Let $t \\ge t_0$. The lemma with the bound of step 7 gives the estimate $\\int_{t_0}^{t} \\frac{K^{m+1} (s - t_0)^m}{m!}\\, d\\, ds = \\frac{(K (t - t_0))^{m+1}}{(m + 1)!}\\, d$.\n10. For $t \u003c t_0$, the integral is $-\\int_t^{t_0}$, and the same computation with $t_0 - s$ gives $\\frac{(K (t_0 - t))^{m+1}}{(m+1)!}\\, d$. This proves the claim for $m + 1$.\n11. Each $t \\in J$ has $\\abs{t - t_0} \\le T$. Taking the supremum over $t$ in the claim gives $d(P^m \\gamma, P^m \\delta) \\le \\frac{(K T)^m}{m!}\\, d(\\gamma, \\delta)$.\n12. Let $q_m = (K T)^m / m!$. If $K T = 0$, then $q_1 = 0 \u003c 1$.\n13. Otherwise choose an integer $N \\ge 2 K T$. For $m \\ge N$, $q_{m+1} = q_m \\frac{K T}{m + 1} \\le q_m / 2$. So $q_{N + j} \\le q_N / 2^j$, which is less than $1$ for large $j$.\n14. In both cases there is $m \\ge 1$ with $q_m \u003c 1$. By step 11, $P^m$ is a [[contraction|0MT2S0ZX4Y]] of $X$ with constant $q_m$.\n\n## Idea\n\nEach application of $P$ integrates the previous error bound once more. Integrating $\\abs{s - t_0}^m / m!$ produces $\\abs{t - t_0}^{m+1}/(m+1)!$. The factorial eventually beats the power $(K T)^m$, so a high enough iterate shrinks distances.\n",
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