1
$\begingroup$

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 $|x|\leq 1$?

$\endgroup$
1
  • 1
    $\begingroup$ 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. $\endgroup$
    – fedja
    Jun 7, 2017 at 2:13

1 Answer 1

2
$\begingroup$

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.

$\endgroup$
1
  • $\begingroup$ Thanks for the answer ! Does this theorem extend also to rational functions (quotient of polynomials of degree at most n each) or is it restricted only to polynomials ? $\endgroup$
    – Lior Eldar
    Jun 7, 2017 at 12:15

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.