# Thread: Does limit of "approximate zero set" converge to the zero set?

1. ## Does limit of "approximate zero set" converge to the zero set?

Let $\displaystyle f:\mathbb{R}^m\rightarrow\mathbb{R}^m$.
Define the zero set by $\displaystyle \mathcal{Z}\triangleq\{x\in\mathbb{R}^m | f(x)=\mathbf{0}\}$ and an $\displaystyle \epsilon$-approximation of this set by $\displaystyle \mathcal{Z}_\epsilon\triangleq\{x\in\mathbb{R}^m|~ ||f(x)||\leq\epsilon\}$ for some $\displaystyle \epsilon>0$. Clearly $\displaystyle \mathcal{Z}\subseteq \mathcal{Z}_\epsilon$. Can one assume any condition on the function $\displaystyle f$ so that
$\displaystyle \lim_{\epsilon\rightarrow 0}~\max_{x\in \mathcal{Z}_\epsilon}~\text{dist}(x, \mathcal{Z})=0,$
holds?

I know in general this doesn't hold by this example (function of a scalar variable):
\displaystyle f(x)=\left\{\begin{align} 0,\quad{x\leq 0}; \\ 1/x,\quad x>0. \end{align} \right.

I really appreciate any help or hint.
Thank you.

2. ## Re: Does limit of "approximate zero set" converge to the zero set?

If $f$ is a continuous bijection, this follows trivially. You can probably relax the condition to monotone and continuous and still arrive at the conclusion.

3. ## Re: Does limit of "approximate zero set" converge to the zero set?

Actually $\displaystyle f$ here is the gradient of a non-convex function $\displaystyle g$, i.e. $\displaystyle f=\nabla g$ which is not monotone, and the zero set is the set of critical points. However, I assume $\displaystyle g$ is $\displaystyle \mathcal{C}^\infty$.
If $f$ is a continuous bijection, this follows trivially. You can probably relax the condition to monotone and continuous and still arrive at the conclusion.
Actually $\displaystyle f$ here is the gradient of a non-convex function $\displaystyle g$, i.e. $\displaystyle f=\nabla g$ which is not monotone, and the zero set is the set of critical points. However, I assume $\displaystyle g$ is $\displaystyle \mathcal{C}^\infty$.