## prove part (b) of this integration inequality

(a) Suppose that $u(t)$ is a continuous solution of

$u(t)\leq f(t)+\int_{t_0}^t h(s)u(s)ds$, $t\geq t_0$,

where $f,h$ are continuous and $h(t)\geq 0$ for $t\geq t_0$. Show that

$u(t)\leq f(t)+\int_{t_0}^t f(s)h(s)\exp\left[\int_s^t h(\sigma)d\sigma\right]ds$, $t\geq t_0$.

This I have done! But then we have...

(b) Moreover, if $f(t)$ is increasing for $t\geq t_0$, show that

$u(t)\leq f(t)\exp\left[\int_{t_0}^t h(s)ds\right]$, $t\geq t_0$.

In order to do the exercise, we are given as a theorem the following formulation of the Gronwall inequality:

If $u(t)\leq M+\int_{t_0}^t h(s)u(s)ds$, $t\geq t_0$,

where $M\in\mathbb{R}$, $h(t)\geq 0$ for all $t\geq t_0$ and $h$ is continuous, and $u(t)$ is a continuous solution, then

$u(t)\leq M\exp\left[\int_{t_0}^t h(s)ds\right]$, $t\geq t_0$.

However, I believe this theorem may only be useful for part (a), which, as I said, I have already done. I provide it just in case.