Bounding the value of a polynomial map by the distance from the variety

Question

Let $f_1,\ldots,f_r\in\Bbb C[x_1,\ldots,x_n]$ define a map $f\colon\Bbb C^n\to\Bbb C^r$. Let $Z:=Z(f_1,\ldots,f_r)\subseteq \Bbb C^n$ be the variety cut out by the $f_i$, and assume that $Z$ is nonempty. Define \begin{align*} d:\Bbb C^n &\longrightarrow \Bbb R_+ \\ x &\longmapsto \operatorname{dist}(x,Z)=\inf_{z\in Z}\|x-z\|, \end{align*} the distance from $Z$. I conjecture that $$\exists g\in\Bbb C[T]: \forall x\in\Bbb C^n: \|f(x)\|\ge |g(d(x))|.$$

My question is: Am I right? If so, can you name a reference or a result from which this follows?

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For $n=r=1$, the statement is certainly true. Let $f\in\Bbb C[x]$ and write $$f = u\cdot (x-\lambda_1)\cdots(x-\lambda_k)$$ for $u\in\Bbb C^\times$ and $\lambda_1,\ldots,\lambda_k\in\Bbb C$. Now, we have $Z=\{\lambda_1,\ldots,\lambda_k\}$ and $d(x)=\min_i |x-\lambda_i|.$ If we pick $g(T):=|u|\cdot T^k$, then $$|f(x)|\ge |u|\cdot|x-\lambda_1|\cdots|x-\lambda_k|\ge |u|\cdot d(x)^k = |g(d(x))|.$$

Answer 1

Here is a counter-example. Take $$f(x,y) = (x(1-x),xy-1)$$ So $Z = \{ (1,1) \}$. Consider now consider the family $(1/R,R)$ as $R \to \infty$. The distance from $(1/R,R)$ to $(1,1)$ goes to $\infty$, but $|f(1/R,R)| = |(1/R-1/R^2,0)|$ goes to $0$. So, if we were to have a bound of the form $|f(x,y)| > g(d(x,y))$, we'd have to have $\lim_{T \to \infty} g(T)=0$, so $g$ isn't polynomial.

The point, of course, is that $Z$ morally has a second point at $(0, \infty)$.

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The variant where all $f_i$ are homogeneous of the same degree is also false. Set $$f(w,x,y) = w^2 \qquad g(w,x,y) = x^2-wy.$$ As a set, $Z$ is the line $w=x=0$, so the distance from $(w,x,y)$ to $Z$ is $\sqrt{|w|^2+|x|^2}$. Now let $N$ be any positive integer and let $(w,x,y) = (r^{1+N},r,r^{1-N})$ as $r \to 0$. Then $d = \sqrt{r^2+r^{2N+2}} \sim r$, but $(f,g) = (r^{2N+2},0)$ so $|(f,g)| = r^{2N+2}$. We see that $|(f,g)|$ drops off faster than $d^{2N+2}$ and, since $N$ was arbitrary, there is no polynomial bound.