# Gronwall’s inequality in discretized time

inequalitiesreal-analysis

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

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!

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 GrĂ¶nwall'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 $$C:=e^T$$. $$\quad\Box$$