May 06, 2018 / by Matthew Kvalheim / In news

### A nice proof of Grönwall's inequality

Let $J$ denote an interval of the real line of the form $[a,\infty)$ or $[a,b]$ or $[a,b)$, with $a<b$. Let $u$ and $\beta$ be continuous real-valued functions on $J$, and let $\alpha$ be a Riemann integrable real-valued function on $J$. Assume also that $\beta$ is nonnegative. A fairly general version of Grönwall’s inequality states that if

[\forall t \in J: u(t) \leq \alpha(t) + \int_a^t \beta(s)u(s)\, ds,]

then

[\forall t \in J: u(t) \leq \alpha(t) + \int_a^t \alpha(s)\beta(s) e^{\int_{s}^{t}\beta(r)\,dr}\,ds.]

Remark: no assumptions are needed on the signs of the functions $\alpha$ or $u$.

Here is a nice proof of Grönwall’s inequality, which I learned from Hans Lundmark here.

Define $I(t):= \int_a^t \beta(s)u(s)\, ds$. Then $\dot{I}:=\frac{d}{dt}I = \beta u \leq \beta(\alpha + I),$ with the last inequality following since $\beta \geq 0$. (Here we used the continuity of $u,\beta$ to invoke part 1 of the fundamental theorem of calculus, FTC-1 hereafter.) Hence

[\dot{I}-\beta I \leq \alpha \beta.]

Define the antiderivative $B(t):= \int_{a}^t\beta(s)\,ds.$ (Here continuity of $\beta$ is again required, in anticipation of another invocation of FTC-1.) Multiplying both sides of the above inequality by the integrating factor $e^{-B(t)}$, we obtain

[\frac{d}{dt}\left(e^{-B(t)}I(t)\right) \leq e^{-B(t)}\alpha(t)\beta(t).]

Integrating both sides from $a$ to $t$ (this is where we require integrability of $\alpha$) and using the fact that $B(a) = 0$, we obtain

[I(t) \leq \int_a^t e^{B(t)-B(s)}\alpha(s)\beta(s)\, ds.]

Since $u(t) \leq \alpha(t) + I(t)$, we obtain

[u(t) \leq \alpha(t) + \int_a^t e^{B(t)-B(s)}\alpha(s)\beta(s)\, ds = \alpha(t) + \int_a^t \alpha(s)\beta(s) e^{\int_{s}^{t}\beta(r)\,dr}\, ds.]

$\square$

Note that if we additionally assume that $\alpha$ is nondecreasing, then we obtain

[u(t) \leq \alpha(t)\left(1 + \int_a^t \beta(s) e^{\int_{s}^{t}\beta(r)\,dr}\, ds\right) = \alpha(t)\left(1 - \int_a^t \frac{d}{ds}e^{\int_{s}^{t}\beta(r)\,dr}\, ds\right),]

and therefore we obtain the special case

[u(t) \leq \alpha(t)e^{\int_{a}^{t}\beta(r)\,dr}.]