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Given a polynomial $p(x_1,\ldots, x_k)$ in $k$ variables with maximum degree $n$, and $x_1,\ldots, x_k \in [0,1]$. Suppose $\max_{x \in [0,1]^k} p(x) = 1$, can we get an upper bound on the probability $\mathbb{P}(|p(x)| < \epsilon)$ where we treat the product space $[0,1]^k$ as a natural probability space? Using techniques from this paper: one can show that the above probability is small provided $\epsilon$ is of tower exponential order $e^{-e^n}$. But I need $\epsilon$ to be of poly-exponential order, i.e., $e^{-(nk)^c}$. Looking at the polynomial $x_1^n x_2^n \ldots x_k^n$, one deduce by central limit theorem that $\mathbb{P}(|p(x)| < \epsilon)$ is small provided $\epsilon = \mathcal{O}(e^{-kn})$. But I can't prove that this polynomial has the the highest probability of staying near zero.

If needed, one can also bound $\epsilon$ in terms of the total degree, i.e., the sum of degrees of all the monomial terms in $p$.

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up vote 2 down vote accepted

I believe that the result you want follows from the results in this very cool paper. (see, in particular page 9). There are probably improvements since then...

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That's exactly the literature I am looking for. Thank you so much! – John Jiang Dec 28 '11 at 21:30

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