Gronwall Inequality

This is used to establish a explicit exponential upper bound on a Riemann-integrable function that satisfies an implicit integral inequality

$f(t) \le a + b\int_0^t f(x)dx$

Let $u(t) = a + b\int_0^t f(x) dx$ . Then $u'(t) = bf(t) \le bu(t)$ . Multiplying by $e^{-bt}$ does not change ordering, and so $e^{-bt} u'(t) - be^{-bt} u(t) = (e^{-bt} u)' \le 0$ . Hence $e^{-bt} u(t)$ is a non-increasing function of $t$ and so $e^{-bt} u(t) \le u(0) = a$ . Hence $f(t) \le u(t) \le ae^{bt}$ .

There are other variants, usually fit to purpose for which they are used. For example, consider

$f(t) \le a + b\int_0^t f(x) g(x) dx$

Repeating the above programme, let $u(t) = a + b\int_0^t f(x) g(x) dx$ where $b,g > 0$ . Then $u'(t) = b f(t) g(t) \le b u(t) g(t)$ . Then

$\left(u(t) e^{-b\int_0^t g(x) dx} \right)' = (u'(t) - bu(t) g(t))\left( e^{-b \int_0^t g(x) dx} \right)\le 0$ and so $u(t) e^{-b \int_0^t g(x) dx}$ is non-increasing and so $u(t) e^{-b \int_0^t g(x) dx} \le a$ . Hence $f(t) \le ae^{b \int_0^t g(x) dx}$

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