$\require{physics}\newcommand{\cbrt}[1]{\sqrt[3]{#1}}\newcommand{\sgn}{\text{sgn}}\newcommand{\ii}[1]{\textit{#1}}\newcommand{\eps}{\varepsilon}\newcommand{\EE}{\mathbb E}\newcommand{\PP}{\mathbb P}\newcommand{\Var}{\mathrm{Var}}\newcommand{\Cov}{\mathrm{Cov}}\newcommand{\pperp}{\perp\kern-6pt\perp}\newcommand{\LL}{\mathcal{L}}\newcommand{\pa}{\partial}\newcommand{\AAA}{\mathscr{A}}\newcommand{\BBB}{\mathscr{B}}\newcommand{\CCC}{\mathscr{C}}\newcommand{\DDD}{\mathscr{D}}\newcommand{\EEE}{\mathscr{E}}\newcommand{\FFF}{\mathscr{F}}\newcommand{\WFF}{\widetilde{\FFF}}\newcommand{\GGG}{\mathscr{G}}\newcommand{\HHH}{\mathscr{H}}\newcommand{\PPP}{\mathscr{P}}\newcommand{\Ff}{\mathcal{F}}\newcommand{\Gg}{\mathcal{G}}\newcommand{\Hh}{\mathbb{H}}\DeclareMathOperator{\ess}{ess}\newcommand{\CC}{\mathbb C}\newcommand{\FF}{\mathbb F}\newcommand{\NN}{\mathbb N}\newcommand{\QQ}{\mathbb Q}\newcommand{\RR}{\mathbb R}\newcommand{\ZZ}{\mathbb Z}\newcommand{\KK}{\mathbb K}\newcommand{\SSS}{\mathbb S}\newcommand{\II}{\mathbb I}\newcommand{\conj}[1]{\overline{#1}}\DeclareMathOperator{\cis}{cis}\newcommand{\abs}[1]{\left\lvert #1 \right\rvert}\newcommand{\norm}[1]{\left\lVert #1 \right\rVert}\newcommand{\floor}[1]{\left\lfloor #1 \right\rfloor}\newcommand{\ceil}[1]{\left\lceil #1 \right\rceil}\DeclareMathOperator*{\range}{range}\DeclareMathOperator*{\nul}{null}\DeclareMathOperator*{\Tr}{Tr}\DeclareMathOperator*{\tr}{Tr}\newcommand{\id}{1\!\!1}\newcommand{\Id}{1\!\!1}\newcommand{\der}{\ \mathrm {d}}\newcommand{\Zc}[1]{\ZZ / #1 \ZZ}\newcommand{\Zm}[1]{\left(\ZZ / #1 \ZZ\right)^\times}\DeclareMathOperator{\Hom}{Hom}\DeclareMathOperator{\End}{End}\newcommand{\GL}{\mathbb{GL}}\newcommand{\SL}{\mathbb{SL}}\newcommand{\SO}{\mathbb{SO}}\newcommand{\OO}{\mathbb{O}}\newcommand{\SU}{\mathbb{SU}}\newcommand{\U}{\mathbb{U}}\newcommand{\Spin}{\mathrm{Spin}}\newcommand{\Cl}{\mathrm{Cl}}\newcommand{\gr}{\mathrm{gr}}\newcommand{\gl}{\mathfrak{gl}}\newcommand{\sl}{\mathfrak{sl}}\newcommand{\so}{\mathfrak{so}}\newcommand{\su}{\mathfrak{su}}\newcommand{\sp}{\mathfrak{sp}}\newcommand{\uu}{\mathfrak{u}}\newcommand{\fg}{\mathfrak{g}}\newcommand{\hh}{\mathfrak{h}}\DeclareMathOperator{\Ad}{Ad}\DeclareMathOperator{\ad}{ad}\DeclareMathOperator{\Rad}{Rad}\DeclareMathOperator{\im}{im}\renewcommand{\BB}{\mathcal{B}}\newcommand{\HH}{\mathcal{H}}\DeclareMathOperator{\Lie}{Lie}\DeclareMathOperator{\Mat}{Mat}\DeclareMathOperator{\span}{span}\DeclareMathOperator{\proj}{proj}$ >[!claim] Gronwall >Let $g$ be a non-negative **bounded** measurable function on $[0,T]$. Suppose that there exist two constants $a,b\geq 0$ such that for every $t\in [0,T]$,$ g(t)\leq a + b\int_0^t g(s)ds. $Then, for every $t\in [0,T]$, $g(t) \leq a\exp(bt)$. >[!proof]- Iterate > This is clear, because one can simply iterate: > $ > g(t)\leq a + a(bt) + a\frac{(bt)^2}{2} + \dots + a\frac{(bt)^n}{n!} + b^{n+1}\underbrace{\int_0^t ds_1 \int_0^{s_1} ds_2 \dots \int_0^{s_n}ds_{n+1} ds_{n+1} g(s_{n+1})}_\star > $ > where if $0\leq g\leq K$ the error term is $\star \leq K \frac{(bt)^{n+1}}{(n+1)}!\to 0$. We win. There are other versions of this lemma where $a,b$ don't have to be constant. > [!claim] Gronwall Differential Form > Let $I = [0,\infty)$ or $[0,a)$ or $[0,a]$. Let $g, b$ be real-valued continuous functions on $I$. Suppose $g$ is differentiable in the interior of $I$, and$g'(t)\leq b(t)g(t)$on $I$. Then, for all $t\in I$,$g(t)\leq g(0)\exp\left(\int_0^t b(s)ds\right).$