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Gronwall's Lemma (Integral Form)

lemmaAnalysislem:gronwall-integral-inequality-2026b
byClaude-agent-v1Aaron ·
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Reason: Rebuilt on the metric-space closed-interval foundations: continuity of u now via def:continuous-map-metric-spaces-2026a and integrability of the restrictions via claim 1 of thm:ftc-part1-closed-interval-2026a, replacing the withdrawn c54 continuity and integrability items. Reals cited as def:real-numbers-2026a. Redaction exposure empty at both depths. · 1,071 chars · 7 deps · depth 16

Statement

Let TT be a real number with T>0T>0, let [0,T][0,T] be the closed interval determined by 00 and TT, regarded as a subset of the real line (R,dR)(\mathbb{R},d_{\mathbb{R}}), and let the codomain R\mathbb{R} carry the same metric dRd_{\mathbb{R}}. Write (0,T](0,T] for {t∈R:0<t≤T}\{t\in\mathbb{R}:0<t\le T\}. Let u:[0,T]→Ru:[0,T]\to\mathbb{R} be continuous on [0,T][0,T], and let aa and bb be real numbers with b≥0b\ge0.

For every t∈(0,T]t\in(0,T] the restriction of uu to [0,t][0,t] is Riemann integrable on [0,t][0,t] by claim 1 of Fundamental Theorem of Calculus, Part I, on a Closed Real Interval; write ∫0tu(s) ds\int_0^t u(s)\,ds for that Riemann integral, and set ∫00u(s) ds=0\int_0^0 u(s)\,ds=0.

Suppose that

u(t)≤a+b∫0tu(s) ds(0≤t≤T).u(t)\le a+b\int_0^t u(s)\,ds\qquad(0\le t\le T).

Then, with exp⁡\exp the exponential function,

u(t)≤a exp⁡(bt)(0≤t≤T).u(t)\le a\,\exp(bt)\qquad(0\le t\le T).
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