# Gronwall Inequality

There are a number of different statements of Gronwall’s inequality. In this post, we will consider only one of them, perhaps the weakest of all.

Proposition 1 Let ${f(t)}$ be a non-negative continuous function on ${\left[ a,b \right]}$ such that there are positive constants ${C}$ and ${K}$ satisfying

$\displaystyle \begin{array}{rcl} f(t)\le C + K\int_{a}^{t}f(s)ds \end{array}$

for all ${t\in\left[ a,b \right]}$. Then,

$\displaystyle \begin{array}{rcl} f(t)\le Ce^{K(t-a)} \end{array}$

for all ${t \in \left[ a,b \right]}$.

Proof: Define ${U(t) = C + K\int_{a}^{t}f(s)ds}$. Note that, by definition, ${f(t)\le U(t)}$ and ${U}$ is a strictly positive differentiable function. Also, we have ${U'(t) = Kf(t)\le KU(t)}$. In other words, ${\frac{U'(t)}{U(t)}\le K}$ which means the relative rate of change of ${U}$ is less than ${K}$. Hence, the growth of ${U}$ is slower than an exponential function with relative rate of change ${K}$. That is ${U(t) \le U(a) e^{K(t-a)}}$ (if you did not like this reasoning, you may integrate both sides of the previous inequality from ${a}$ to ${t}$). So, we have the desired result ${f(t) \le U(t) \le U(a)e^{K(t-a)}= Ce^{K(t-a)}}$. $\Box$