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I need to find some conditions on a function $h:\mathbb{R}^+\times \mathbb{R}^n\rightarrow\mathbb{R}^n$ to ensure that it goes, in norm, to zero faster than $\|z\|$ uniformly in time, i.e. $$ \lim_{\|z\|\to 0}\sup_{t\geq 0}\frac{\|h(t,z)\|}{\|z\|}=0.$$

I know that for any fixed time $t\in\mathbb{R}^+$ the limit goes to zero, but in general this convergence is not uniform in time.

My function $h$ has even the property that $h(t,0)=0$ for any $t\in\mathbb{R}^+$. Therefore the first property I thought for $h$ in order to satisfy this condition is that it is so that $\|h(t,x)\|\leq \|x\|^p$ where $p>1$, which however does not give me relevant information about it.

Do you suggest other possible conditions to ensure this kind of behaviour?

If it might help, I need this condition in order to define the linearization of a non-autonomous dynamical system, where $h(t,z)$ is the nonlinear part of my ODE.

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    $\begingroup$ Do you know about moduli of continuity? Maybe they're what you're looking for. en.wikipedia.org/wiki/Modulus_of_continuity $\endgroup$
    – giobrach
    Mar 27, 2021 at 9:16
  • $\begingroup$ Thank you, I knew about uniform continuity and the modulus, but I did not think about it to solve my problem $\endgroup$
    – Dadeslam
    Mar 30, 2021 at 8:33

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An equivalent formulation of the property is:

There is a function $F\colon\mathbb R^n\to\mathbb R_+$ satisfying $F(0)=0$ and $F'(0)=0$ such that $\lVert h(t,z)\rVert\le F(z)$ for all $t\ge0$ and all $z$ in a neighborhood $U\subseteq\mathbb R^n$ of $0$.

(Indeed, if your property is satisfied, then $$F(z)=\min\{1,\sup_{t\ge0}\lVert h(t,z)\rVert\}$$ is such a function with the set $U=\{z\in\mathbb R^n:F(z)<1\}$ (which is a neighborhood of $0$ by the mere definition of the limit). Conversely, given such a pair $(F,U)$, there holds $$0\le\sup_{t\ge0}\frac{\lVert h(t,z)\rVert}{\lVert z\rVert}\le\underbrace{\frac{\lVert F(z)\rVert}{\lVert z\rVert}}_{\to0\quad\text{as $z\to0$}}\quad\text{for all $t\ge0$, $z\in U$,}$$ and so the middle term tends to $0$ as $z\to0$.)

The choices $F(z)=C\lVert z\rVert^p$ are of course the simplest examples of such functions as you observed. You can do finer choices like $F(z)=f(\lVert z\rVert)$ with $f(0)=f'(0)=0$. However, as proved above, you cannot do much better, in principle.

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  • $\begingroup$ Thank you, you confirmed my ideas and explained them clearly $\endgroup$
    – Dadeslam
    Apr 2, 2021 at 16:59

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