Gronwall's inequality in discretized time

Question

$ \newcommand{\RR}{\mathbb{R}} \newcommand{\TT}{\mathbb{T}} \newcommand{\NN}{\mathbb{N}} \newcommand{\PP}{\mathbb{P}} \newcommand{\EE}{\mathbb{E}} \newcommand{\FF}{\mathbb{F}} \newcommand{\PPP}{\mathscr{P}} \newcommand{\KKK}{\mathscr{K}} \newcommand{\eps}{\varepsilon} \newcommand{\diff}{\mathop{}\!\mathrm{d}} $

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!

Answer 1

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 $C:=e^T$. $\quad\Box$

Answer 2

Another (more direct) proof without using Gronwall's inequality (or any continuity assumptions.)

For all $k\in \{ 0,\cdots,n \}$, denote $t_k=k\varepsilon_n$. The assumption says that, if $t_k \leq t \leq t_{k+1}$, then $$ f(t)\leq f(0) + \sum_{p=0}^{k-1}f(t_p)\varepsilon_n + f(t_p)(t-t_k). $$ In particular, for $k=0$, one has $$ f(t)\leq f(0) + f(0)t \leq f(0)(1+\varepsilon_n) := f(0)A_0. $$ More generally, one has (by direct induction), for $t_k \leq t \leq t_{k+1}$ $$ f(t)\leq f(0)A_k, $$ where $A_k$ satisfies $A_{k}=(1+\varepsilon_n)A_{k-1}$. Hence (and since the $A_k$'s are increasing) $$ \sup_{[0,T]} f\leq f(0)A_{n-1} = f(0)\left(1+\frac{T}{n}\right)^n \leq f(0)e^T, $$ as $1+x\leq e^x$ for every real number $x$. Therefore, one could take $C=e^T$.