Proof of Gronwall's Lemma (Integral Form)

Define $v:[0,T]\to\mathbb{R}$ by $v(t)=\int_0^t u(s)\,ds$, with $v(0)=0$ by the stated convention. The hypothesis reads $u(t)\le a+b\,v(t)$ for $0\le t\le T$.

Case $b=0$. By item 1 of Basic Properties of the Exponential Function, $\exp(0)=1$, so for every $t\in[0,T]$, $u(t)\le a=a\,\exp(bt)$. Assume from now on $b>0$.

Step 1: regularity of $v$. The function $u$ is continuous at every point of $[0,T]$, by hypothesis and Continuity on a Closed Interval. By Fundamental Theorem of Calculus, Part I in One Dimension applied with $f=u$ on $[0,T]$, the integral defining $v(t)$ is well defined for every $t\in[0,T]$ and $v$ is differentiable at every $t\in(0,T)$ with $v'(t)=u(t)$ (with the same degenerate-interval convention $v(0)=0$, so $v$ is the function $F$ of that theorem).

By the extreme value theorem there are $x_{\min},x_{\max}\in[0,T]$ with $u(x_{\min})\le u(s)\le u(x_{\max})$ for all $s\in[0,T]$; put $K=\max\bigl(|u(x_{\min})|,|u(x_{\max})|\bigr)$, so $|u(s)|\le K$ on $[0,T]$. Let $0\le t_0<t_1\le T$. By additivity of the Riemann integral on adjacent intervals (Additivity of the Riemann Integral on Adjacent Intervals), together with the convention $v(0)=0$ in the case $t_0=0$,

and by the mean value theorem for integrals there is $c^*\in[t_0,t_1]$ with $\int_{t_0}^{t_1}u(s)\,ds=u(c^*)\,(t_1-t_0)$. Hence $|v(t_1)-v(t_0)|\le K\,(t_1-t_0)$ for all $0\le t_0<t_1\le T$. Given any $t^*\in[0,T]$ and $\varepsilon>0$, taking $\delta=\varepsilon/(K+1)$ in Continuity at a Point shows that $v$ is continuous at $t^*$; thus $v$ is continuous at every point of $[0,T]$.

Step 2: the auxiliary function decreases. Define $\chi:[0,T]\to\mathbb{R}$ by

By the scaled-exponential lemma with $c=-b$, the function $t\mapsto\exp(-bt)$ on $\mathbb{R}$ is differentiable at every real point with derivative $-b\exp(-bt)$ and continuous at every real point. The defining conditions of Derivative at an Interior Point and Continuity at a Point quantify only over points of the domain, so the restriction of this function to $[0,T]$ is continuous at every point of $[0,T]$ and differentiable, with the same derivative, at every interior point of the interval $[0,T]$.

By claim 4 of the one-dimensional rules lemma applied with $E=[0,T]$ (constants are continuous, and sums and products of continuous functions are continuous), together with Step 1, $\chi$ is continuous at every point of $[0,T]$, i.e., continuous on $[0,T]$.

Let $t\in(0,T)$; then $t$ is an interior point of $[0,T]$, since $0<t<T$ with $0,T\in[0,T]$. By claim 2 of the rules lemma, $t\mapsto a/b+v(t)$ is differentiable at $t$ with derivative $v'(t)=u(t)$ (Step 1), and by claim 3 (the product rule),

By item 2 of Basic Properties of the Exponential Function, $\exp(-bt)>0$, and $u(t)-a-b\,v(t)\le0$ by hypothesis; hence $\chi'(t)\le0$ for every $t\in(0,T)$.

Step 3: conclusion. Fix $t\in(0,T]$. The function $\chi$ is continuous at every point of $[0,t]$ and differentiable at every point of $(0,t)\subseteq(0,T)$. By the mean value theorem applied to $\chi$ on $[0,t]$ (with the interval $I=[0,T]$), there is $\xi\in(0,t)$ with

Since $\chi(0)=\exp(0)\bigl(a/b+v(0)\bigr)=a/b$, using $\exp(0)=1$ and $v(0)=0$, we obtain, for every $t\in[0,T]$ (the case $t=0$ holding with equality),

By items 1 and 2 of Basic Properties of the Exponential Function, $\exp(bt)>0$ and $\exp(-bt)\exp(bt)=\exp(0)=1$; multiplying the last display by $\exp(bt)$ therefore gives

hence $b\,v(t)\le a\exp(bt)-a$, and by the hypothesis,

$\blacksquare$