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

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