January 2012. Let denote the unit circle and let be a probability measure on . Take the sequence and orthonormalize in the Hilbert space to produce . We have
by definition. The question is what is the size of if and we assume some additional information on the measure ? This is a classical question in approximation theory. The Steklov’s conjecture was: assume that is purely a.c. and has the weight uniformly bounded away from zero, i.e.
Is it true that uniformly in and ? The negative answer to this question was given by Rakhmanov. He proved that the possible growth is essentially up to an inverse logarithmic factor. Can one drop it? This is a nice problem. !!! SOLVED, see below. Another important class of measures extensively studied in the literature is the so-called Szego class. It is defined as follows: where the singular component is arbitrary and for one has
How large the polynomials can be in this case? Well, for a while the conjecture was that the polynomial entropy is bounded, i.e.
uniformly in . That, however, again tuned out to be wrong with the possible growth like . This result is sharp. Then, what restriction on the size of do we get from the orthogonality? Not clear to me. _______________________________________________________________________ August 2013. In the joint paper with A. Aptekarev and D. Tulyakov which you can find on my webpage, we proved the sharp estimates for the uniform norms of the orthogonal polynomials in the Steklov class. This class is defined by the condition that
for a.e. point on the circle where is some small positive constant. The world record so far was made by Rakhmanov about 30 years ago. The method we used turned out to be powerful enough to also give the sharp bounds for the polynomial entropy in this class. We hope to iterate the construction to prove the lower bounds over the subsequence. If we succeed, this will give the full solution to the famous Steklov’s conjecture. ______________________________________________________________________________ February 2014. Building on the previous paper (discussed above), we were able to settle a problem by Steklov. That involved a lot of technical work. Here is the main result: suppose is a probability measure on the circle which satisfies
and . That condition defines the Steklov’s class. If are the corresponding orthonormal polynomials, then the following upper bound is well-known and easy to prove:
In the paper, we prove Theorem. For every sequence , there is an absolutely continuous probability measure from the Steklov class such that
for some sequence . The proof is constructive and can be used when the measure satisfies different lower bounds. There are other interesting questions that naturally follow from the paper, some of them do not seem to be so hard anymore. The project on getting the uniform bound on the polynomials is mainly finished. What is left wide open is the problem on controlling the for a.e. provided the constructive information on measure is known. This is the so-called nonlinear Luzin’s conjecture and we know very little about it. ____________________________________________________________ November 6, 2014 Fix , large and such that . Consider the class of probability measures defined as where , are arbitrary points on , and . Let . In the recent paper, I proved the following estimate It implies, for example, that the maximizer in the Steklov problem has point masses. In the same paper, I also proved the following inequality improving the bound by Kos: if and , then . That was done by using the duality between two extremal problems for one of which the bound is easy to obtain if the generalization of the Halasz’ result is used. ___________________________________________________________________________________________________________________ May 19, 2015 The solution to a problem by Steklov discussed above allows to study other variational problems. For example, Then, the old idea by S.Bernstein and our method used for getting the lower bounds can be combined to get when is fixed and . In the case when the deviation of the weight is large, we have two results: 1. For every large , there is such that for every weight . 2. For every small , there is large and a weight such that . One can see that the growth can not be achieved by weights uniformly bounded in .
July 2015. We significantly simplified several proofs in the paper “On a problem by Steklov” and submitted a new version to arXiv. We also fixed some minor mistakes and typos.