5
$\begingroup$

What is the state of the Krzyż Conjecture? It states for that for all $f:\mathbb{D}\to \bar{\mathbb{D}}$ holomorphic and non-vanishing, the coefficients $a_n$ in the power series of $f$ are at most $2/e$ in absolute value seen from the maximal function $$f(z)=e^{\frac{z^n+1}{z^n-1}}$$

I have read The Krzyż Conjecture Revisited where the authors (in 2014) say the problem is open for coefficients $n>5$. A paper by S. Krushkal claims to have proved it, but in the former paper Krushkal's work is mentioned:

Krushkal's unpublished preprint, in spite of some gaps, contains a wealth of geometric and analytic ideas which may be useful for a further study of problems of this type

I could not spot any obvious holes but it has yet to be published in a reputable journal as far as I know, so I wonder if someone with a more penetrating and experienced eye could verify or point the "gaps".

$\endgroup$
6
  • 1
    $\begingroup$ On my opinion the paper "Krzyz conjecture revisited" correctly reflects the state of the problem. I did not learn of any new advances since it was published. $\endgroup$ Apr 5, 2018 at 14:54
  • $\begingroup$ S. Krushkal has another paper on arxiv related to Zalcman's conjecture (link: arxiv.org/abs/1109.4646). But it is unknown whether this has survived the scrutiny of peer reviews. Besides, I was not able to find his email address. $\endgroup$
    – Chee
    Aug 30, 2018 at 20:43
  • $\begingroup$ Here is his profile on ResearchGate: researchgate.net/profile/Samuel_Krushkal $\endgroup$
    – Chee
    Aug 30, 2018 at 20:46
  • 1
    $\begingroup$ I recall that the Riemman mapping theorem was proved using an extremal sequence, and that the proof of the Bieberbach conjecture used Loewner equation. Could someone please explain in terms of strategy, why attacking Krzyz conjecture needs a different one? $\endgroup$
    – Chee
    Aug 30, 2018 at 20:51
  • $\begingroup$ adding an arxiv paper, claiming a proof of a variant of the conjecture: arxiv.org/abs/1803.09718 $\endgroup$
    – Chee
    May 30, 2019 at 15:08

0

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.