Let $g$ be a nonnegative Borel-measurable function, that is locally integrable on $[0, \infty)$. Assume that $g$ satisfies for all $t \geq 0$ the inequality $g(t) \leq a + b \int^t_0 g(s) ds$, where $a, b \geq 0$. I want to show that $g(t) \leq ae^{bt}$.
A hint to me was given that I should solve the inhomogeneous integral equation $$g(t) = a + b\int^t_0 g(s) ds − p(t)$$ for a nonnegative function $p$. But I have no clue how I should prove this and how this helps me proving the first statement. Thanks for any hints and help.
A little context; this is a question asked in a stochastic integration course but I don't see a real connection here.
This inequality is called the Grownwall inequality in the ode theory. Following the hint, you may compute the first order derivative of the both sides of the inhomogeneous integral equation, and convert the integral equation to an ordinary diffrential equation. Then you can solve the ODE and obtain the analytic solution. The following is another way to show the inequality. First we let $$ G\left(t\right)=\int_{0}^{t}g\left(s\right)ds $$ Substitute this equation into the inequality $$ G'\left(t\right)\le a+bG\left(t\right) $$ Multiply by $ \exp\left(-bt\right) $
$$ \frac{d \left(\exp\left(-bt\right)G\left(t\right)\right)}{dt} \le a\exp\left(-bt\right) $$ \begin{equation} \begin{aligned} \exp\left(-bs\right)G\left(s\right)\Big|_{s=0}^{s=t} & \le -\frac{a}{b}exp\left(-bs\right)\Big|_{s=0}^{s=t} \\ &= \frac{a}{b}\left(1-\exp\left(-bt\right) \right) \end{aligned} \end{equation} Integrate the inequality $$ G\left(t\right) \le \frac{a}{b}\left(exp\left(bt\right)-1\right) $$ \begin{equation} \begin{aligned} g\left(t\right) & \le a+bG\left(t\right) \\ & \le a\exp\left(bt\right) \end{aligned} \end{equation}