Gronwall’s inequality in discretized time



We fix $T \in (0, \infty)$ and let $\TT := [0, T]$. For $n\in \NN^*$, let $\eps_n := T/n$ be the time step and $f_n : \TT \to \RR_+$ bounded measurable. We define $\tau^n_t := \lfloor t/ \eps_n \rfloor \eps_n$ for $t \in \TT$. We assume that
f_n(t) \le f_n(0) + \int_0^t f_n (\tau^n_s) \diff s,
\quad \forall t \in \TT.

Is there a constant $C>0$ independent of $n$ such that $\sup_{t \in \TT} f_n (t) \le C f_n (0)$?

Thank you so much for your elaboration!

Best Answer

Let $g_n(t):=f_n (\tau^n_t)$; here and in what follows, $t\in[0,T]$. Then $$g_n(t)=f_n (\tau^n_t)\le f_n(0)+\int_0^{\tau^n_t} g_n(s)\,ds \le f_n(0)+\int_0^t g_n(s)\,ds,$$ because $\tau^n_t\le t$ and $g_n\ge0$. So, by Grönwall's inequality, $$g_n(t)\le f_n(0)e^t$$ and hence $$f_n(t)\le f_n(0)+\int_0^t g_n(s)\,ds \le f_n(0)e^t\le Cf_n(0),$$ where $C:=e^T$. $\quad\Box$