# Bounding function by random sampling

Let $f(x): \mathbf{C} \to \mathbf{C}$ be a complex valued polynomial of degree $n$, and suppose that for a random point $z$ sampled uniformly from the unit disk $|z|\leq 1$ we have $$\mathbf{P}_{z}( |f(z)| \leq \epsilon ) \geq 1 - n^{-k},$$ for some large positive $k>0$. Under what conditions can we conclude that $|f(x)|\leq \epsilon$ for all $|x|\leq 1$?

• A pointless nitpick: if you want the same $\varepsilon$ in both conditions, then the answer is obviously "Never".(take $f(z)=qz^n$ with $q$ slightly bigger than $\varepsilon$) If you are willing to relax the conclusion a bit, the problem gets meaningful. However, in that case you'd better decide what exactly you want before engaging other people's brainpower. Jun 7, 2017 at 2:13

As pointed out by fedja, one can not expect the inequality yo hold with the same epsilon. However, the following holds : if $\mathbf{P}_{z}( |f(z)| \leq \epsilon ) \geq 1 - \delta$, with $\delta \ll n^{-2} (\log n)^{-2}$, then $$\forall z \in \mathbb D, \ |f(z)| \leq \epsilon \left(1 + O\left(n \sqrt{\delta} \log \frac{1}{\delta}\right) \right)$$ holds. In the other direction, Fedja's comment yields for each $\delta < 1$ an example where the sup norm is $\geq \epsilon( 1 + \frac{1}{2} n \delta)$.
Proof: assume we have proved an equality of the form $|| f||_{L^{\infty}} \leq C_{n,p} || f||_{L^{p}}$ for some $p$. Then, writing $$C_{n,p}^{-p}|| f||_{L^{\infty}}^p \leq || f||_{L^{p}}^p \leq \epsilon^p + \delta || f||_{L^{\infty}}^p,$$ and assuming $\delta C_{n,p}^p < 1$, one gets $$|| f||_{L^{\infty}} \leq \epsilon \frac{C_{n,p}}{\left( 1 - \delta C_{n,p}^p \right)^{\frac{1}{p}}}.$$ Thus, one just needs good estimates on $C_{n,p}$. Let us start with $p=2$. If $f(z) = \sum_{i=0}^n a_i z^i$, then $$|| f||_{L^{\infty}} \leq \sum_{i=0}^n |a_i| \ \ \text{and} \ \ || f||_{L^{p}}^2 = \sum_{i=0}^n \frac{|a_i|^2}{i+1},$$ so that $$C_{n,2}^2 = \sum_{i=0}^n (i+1) = \frac{1}{2} (n+1) (n+2) \asymp n^2.$$ Now, if $p = 2k$ for some integer $k$, applying the previous inequality to the polynomial $f^k$ shows that one can choose $$C_{n,2k}^{2k} = C_{nk,2}^2 \asymp n^2 k^2.$$ One then takes $k \asymp n^{-1} \delta^{-\frac{1}{2}}$ so that $\delta C_{n,p}^p \leq \frac{1}{2}$ for $p=2k$. We then have $$C_{n,p} = 1 + O\left(n \sqrt{\delta} \log \frac{1}{\delta}\right),\\ \left( 1 - \delta C_{n,p}^p \right)^{\frac{1}{p}} = 1 + O\left(n \sqrt{\delta} \right),$$ hence the result.