Vector function going to zero faster than the norm of its variable uniformly in time 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.
 A: 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.
