Does limit of "approximate zero set" converge to the zero set?

[Vulture1991]

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

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

I really appreciate any help or hint.
Thank you.

 

[caffeinemachine]

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

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

I really appreciate any help or hint.
Thank you.

If one assumes that $f$ is a proper map then the conclusion holds. By proper I mean $f^{-1}(K)$ is compact whenever $K\subseteq \mathbf R^n$ is compact. Let's prove this.

Assume by way of contradiction that there is a proper map $f:\mathbf R^n\to \mathbf R^m$ such that
$$\lim_{\epsilon\to 0} \max_{x\in \mathcal Z_\epsilon} \text{dist}(x, \mathcal Z)$$
is greater than $0$. Call this limit $\alpha$.

So for each positive natural number $n$ we can find $x_n\in \mathcal Z_{1/n}$ such that $\text{dist}(x_n, \mathcal Z)\geq \alpha$. But each $x_n$ lies is $\mathcal Z_1$, which by assumption of properness is a compact set. Thus the sequence $(x_n)$ has a convergent subsequence $(x_{n_k})_{k=1}^\infty$ which, say, converges to $x_0$. It follows that $\text{dist}(x_0, \mathcal Z)\geq \alpha$.

But for each $N>0$, the points in the sequence $(x_{n_k})$ eventually lie in $\mathcal Z_{1/N}$ which is a closed set, since it is compact. Thus $x_0\in Z_{1/N}$ for each $N$. Therefore $f(x_0)\leq 1/N$ or all $N$ which shows that $f(x_0)=0$. But then $\text{dist}(x_0, \mathcal Z)=0$ and we have a contradiction.

 

[math771]

It is difficult to give a definitive answer without knowing more about the function f, but here are some thoughts:

- It is not clear what you mean by "assume any condition on the function f." Do you mean any specific condition, or any condition at all? In general, without any assumptions on the function f, it is not possible to guarantee that the limit as \epsilon\rightarrow 0 is equal to 0. This is because for any given \epsilon, there may exist points in \mathcal{Z}_\epsilon that are arbitrarily far from \mathcal{Z}, making the maximum distance non-zero.

- Your example function f(x) does not satisfy the given conditions, as it is not defined for all x\in\mathbb{R}^m. However, if we modify it to be defined for all x\in\mathbb{R}^m by setting f(x)=0 for x>0, then it does satisfy the conditions. In this case, the limit as \epsilon\rightarrow 0 is equal to 0, since for any \epsilon>0, the maximum distance between a point in \mathcal{Z}_\epsilon and \mathcal{Z} is 0.

- If we impose some regularity conditions on the function f, such as continuity or differentiability, then it may be possible to show that the limit as \epsilon\rightarrow 0 is equal to 0. For example, if f is continuously differentiable and \mathcal{Z} is a compact set, then by the mean value theorem, we can show that the distance between a point in \mathcal{Z}_\epsilon and \mathcal{Z} is bounded by \epsilon times the norm of the gradient of f at that point. This suggests that if the norm of the gradient of f is small near \mathcal{Z}, then the limit as \epsilon\rightarrow 0 is equal to 0.

- Another approach could be to use convexity or Lipschitz continuity of f to show that the set \mathcal{Z}_\epsilon becomes closer and closer to \mathcal{Z} as \epsilon\rightarrow 0. For example, if f is convex and \mathcal{Z} is a singleton, then it can be shown that the limit as \epsilon\rightarrow 0 is equal to 0. This is because for any point x\in\math