# local bernstein type inequality for multivariate polynomials

Let's say $p(x_1,...,x_n)$ is an n-variate degree d homogenous polynomial. Assume $U \subset S^{n-1}$ and $vol(U) > 0$ is there any Bernstein type inequality saying

$$\max_{x \in U , y \in S^{n-1}} | \langle \nabla p(x) , y \rangle | \leq C(p) \max_{x \in U} | p(x) |$$

$C(p)$ could be any type of function depending on coefficients of $p$.

• No. Take $U=\{ x_0\}$ with $p(x_0)=0$. Apr 20 '15 at 2:01
• Thanks. I modified the question a little bit to avoid less dimensional $U$.
– alpx
Apr 20 '15 at 2:23
• @alpx: The modification doesn't do anything really. I can still just take a small $U$ containing a zero. Apr 20 '15 at 17:22

## 1 Answer

Take $p=2x_1x_2$, and $U=U_r$ a small disk of radius $r$ in $S^{n-1}$ in centered at $(1,0,\dotsc, 0)$. $\newcommand{\pa}{\partial}$ Then $\nabla p(x)=2x_2\pa_{x_1}+2x_1\pa_{x_2}$ so that

$$\sup_{x\in U_r}|\nabla p(x)|\geq 2.$$

On the other hand,

$$\sup_{x\in U_r}|p(x)|= O(r).$$ Thus

$$\lim_{r\to 0}\frac{\sup_{x\in U_r}|\nabla p(x)|}{\sup_{x\in U_r}|p(x)|}=\infty.$$

Thus the constant $C(p)$ has to depend on the region $U$ as well. On the other hand, you can find a constant that is independent of $p$, but dependent on $U$ and on the degree of $p$.

Denote by $H_d$ the space of degree $d$ homogeneous polynomials in $n$ variables. For $p\in H_d$ define

$$\Vert p\Vert_{U,0}:=\sup_{x\in U} |p(x)|,$$

$$\Vert p\Vert_{U,1}:=\sup_{x\in U}|\nabla p(x)|.$$

Because ${\rm vol}\,(U)>0$ we have

$$p\in H_d,\;\;\Vert p\Vert_{U,0}=0 \Longleftrightarrow p=0.$$

Indeed, the set $\{p=0\}\cap S^{n-1}$ is semialgebraic and, if it contains the set $U$ of nonzero volume (in $S^{n-1}$), it also contains an open subset of $S^{n-1}$. By unique continuation, $p$ must be identically zero on$S^{n-1}$.

Similarly, if $\Vert p\Vert_{U,1}=0$ we deduce that $\nabla p=0$ on $U$. Arguing as before we deduce that $\nabla p=0$ on $S^{n-1}$ and, in particular $p$ is constant on $S^{n-1}$. Since $p$ is a homogeneous polynomial of degree $d$, we deduce that it must have the form $p(x)= C|x|^{d/2}$. The gradient of this function is not zero if $d>0$. We thus have

$$d>0\;\; p\in H_d,\;\;\Vert p\Vert_{U,1}=0\Longleftrightarrow p=0.$$

Thus, when $d>0$, the functions $\Vert p\Vert_{U,0}$ and $\Vert p\Vert_{U,1}$ are norms on the finite dimensional vector space $H_d$ and, as such, they are equivalent.

• @ Liviu, Thanks! So for the example you gave, if we say $vol(U) > \epsilon$ this at least gives a constant $C(p)$ depending on $\epsilon$. I do agree depending on $U$ this constant may change, it may not be uniformly depending on $vol(U)$ only. For instance taking $U$ around $(\frac{1}{\sqrt{n}},\frac{1}{\sqrt{n}}, ... , \frac{1}{\sqrt{n}},)$ looks nicer.
– alpx
Apr 20 '15 at 14:38
• The argument I gave works for a given $p$ and $U$. For different $U$'s you will need more than the volume. In any case you need to specify precisely the depedencies of $C(p)$. A priori, it depends on $p$ and $U$. Apr 20 '15 at 16:50