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 claim 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 $b>0$.

We use repeatedly that if $x\le y$ are real numbers and $\lambda>0$, then $\lambda x\le\lambda y$: this is claim 10 of Elementary Order Arithmetic in an Ordered Field when $x<y$, and immediate when $x=y$.

Step 1: regularity of $v$. Since $0<T$ and $u$ is continuous on $[0,T]$, Fundamental Theorem of Calculus, Part I, on a Closed Real Interval applies with $0$ and $T$ as the endpoints of the closed interval there and with $u$ as the continuous function there. The function $F$ produced there is exactly $v$, since both are given by $t\mapsto\int_0^t u(s)\,ds$ with the same convention at $t=0$. Claim 2 of that theorem gives that $v$ is continuous on $[0,T]$, and claim 3 gives that $v$ is differentiable at every $t\in(0,T)$ with $v'(t)=u(t)$.

Step 2: the auxiliary function decreases. Define $E:\mathbb{R}\to\mathbb{R}$ by $E(t)=\exp(-bt)$, and define $\chi:[0,T]\to\mathbb{R}$ by

$$\chi(t)=E(t)\Bigl(\frac{a}{b}+v(t)\Bigr).$$

Apply Derivative and Continuity of the Scaled Exponential Function with $c=-b$: claim 1 gives that $E$ is differentiable at every real point with $E'(t)=-b\exp(-bt)$, and claim 2 gives that $E$ is continuous on $\mathbb{R}$. Both $\mathbb{R}$ and $[0,T]$ are intervals and $[0,T]\subseteq\mathbb{R}$, so claim 1 of Restriction Stability of Continuity and of the Derivative shows that the restriction $E|_{[0,T]}$ is continuous on $[0,T]$, and claim 2 of that lemma shows that $E|_{[0,T]}$ is differentiable, with the same derivative as $E$, at every interior point of $[0,T]$. Every $t\in(0,T)$ is such an interior point, since $0,T\in[0,T]$ and $0<t<T$.

The map on $[0,T]$ with constant value $a/b$ is continuous on $[0,T]$ by claim 1 of Continuity of Sums and Products of Real-Valued Functions on a Metric Space, applied at each point of $[0,T]$. Hence, together with Step 1, claim 5 of that theorem shows first that $a/b+v$ is continuous on $[0,T]$ and then that the product $\chi=E|_{[0,T]}\cdot(a/b+v)$ is continuous on $[0,T]$.

Let $t\in(0,T)$, an interior point of $[0,T]$. By claim 1 of Sum, Constant Multiple, and Product Rules for One-Dimensional Derivatives the function on $[0,T]$ with constant value $a/b$ is differentiable at $t$ with derivative $0$, so by claim 2 the function $a/b+v$ is differentiable at $t$ with derivative $v'(t)=u(t)$ (Step 1); by claim 3 (the product rule) applied to $E|_{[0,T]}$ and $a/b+v$,

$$\chi'(t)=-b\exp(-bt)\Bigl(\frac{a}{b}+v(t)\Bigr)+\exp(-bt)\,u(t)=\exp(-bt)\bigl(u(t)-a-b\,v(t)\bigr).$$

By claim 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 $t\in(0,T]$. Since $[0,t]\subseteq[0,T]$, claim 1 of Restriction Stability of Continuity and of the Derivative shows that $\chi|_{[0,t]}$ is continuous on $[0,t]$, and claim 2 of that lemma shows that $\chi|_{[0,t]}$ is differentiable, with the same derivative as $\chi$, at every interior point of $[0,t]$; every $s\in(0,t)$ is such a point and lies in $(0,T)$, so $(\chi|_{[0,t]})'(s)=\chi'(s)\le0$ by Step 2. As $0<t$, Mean Value Theorem on a Closed Real Interval applied to $\chi|_{[0,t]}$ on $[0,t]$ yields $\xi\in(0,t)$ with

$$\chi(t)-\chi(0)=\chi'(\xi)\,(t-0).$$

Since $\chi'(\xi)\le0$ and $t-0>0$, the right-hand side is at most $0$, so $\chi(t)\le\chi(0)$. 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),

$$\exp(-bt)\Bigl(\frac{a}{b}+v(t)\Bigr)\le\frac{a}{b}.$$

By claims 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

$$\frac{a}{b}+v(t)\le\frac{a}{b}\exp(bt),$$

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

$$u(t)\le a+b\,v(t)\le a\exp(bt)\qquad(0\le t\le T).$$

$\blacksquare$