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1 vote

Gronwall's inequality in discretized time

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 \...
Panda Jonas's user avatar
6 votes
Accepted

An inequality about factorial function

$\newcommand\p\partial$Let us show that the inequality in question holds for all integers $k,s,d$ such that \begin{equation} k\ge6s,\quad s\ge1,\quad 1\le d\le s+1. \tag{10}\label{10} \end{...
Iosif Pinelis's user avatar
5 votes
Accepted

Gronwall's inequality in discretized time

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 $...
Iosif Pinelis's user avatar
2 votes

A maximal inequality

$\newcommand\ol\overline$In accordance with the comment by the OP, consider \begin{equation*} p_{N,n}:=P\Big(\max_{k\in\ol{N,n}}\frac{|S_k|}{\sqrt{k/2}}\le1\Big), \end{equation*} where $\ol{A,B}:=\...
Iosif Pinelis's user avatar

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