# Elementary proof of growth estimate for a polynomial via size from its zero set

The paper Asymptotic properties of polynomials and algebraic functions of several variables by Gorin contains the following.

Lemma 3.1. Let $$f\in \mathbb R[x_1,\dots,x_n]$$. Suppose $$f$$ has a root in the interior of the unit circle. Then there exists a positive constant $$c$$ such that $$\|f(x)\|\geq c \cdot d(x,f^{-1}(0))^{\deg f},$$ where $$d$$ denotes the standard metric on $$\mathbb R^n$$.

The proof involves a previous theorem which is itself rather involved.

Question. Is there a quick direct proof of the above lemma?

• The statement is weird. The stated bound on $\|f(x)\|$ is invariant under shifts, hence the condition on a root in the unit circle should be redundant – you can always shift the polynomial so that e.g. the origin is a root. If $f$ has any root at all, that is. – Emil Jeřábek Jun 11 '19 at 8:22
• @EmilJeřábek I agree the condition is redundant, but it's part of the statement in the reference, so I figured I should write it down. – Arrow Jun 11 '19 at 8:25
• Sure. I just mentioned it because it raises the possibility that there is a typo in the original paper. – Emil Jeřábek Jun 11 '19 at 8:32
• @EmilJeřábek I guess the evaluation point is supposed to be in the disk (or some other compact) too. The statement is clearly false in the entire space. – fedja Jun 11 '19 at 19:57

The Russian original text clearly states that the exponent is not $$\deg f$$ but just some $$\alpha>0$$. For $$\alpha=\deg f$$ is it not true: a polynomial $$f(x,y)=x^{2n}+(x-y^n)^2$$ has a unique zero $$(0,0)$$, but $$f(\delta^n,\delta)=\delta^{2n^2}$$.
I remember obtaining a cubic upper bound for $$\alpha$$ in the case of a polynomial in two variables with a single zero (but the estimate holds only in a disk centred at the root, not on the whole plane). I may try to find it if you wish.