## prove part (b) of this integration inequality

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

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

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

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

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

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

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

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

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

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

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

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.