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Goldberg 1979 sets which modulus entire function has
section_1: Gol'dberg's proof of Hayman's conjecture: if the set where an entire function has modulus greater than some c > 0 has finite planar measure, then the integral of r dr / ln ln M(r,f) to infinity converges.
section_2: For every continuous positive nondecreasing Phi on [0, infinity) with the integral of r dr / Phi(r) from 1 to infinity convergent, some entire f has ln ln M(r,f) = O(Phi(r)) and E(c) of finite measure for every c > 0.
section_3: For every m > 0 there are entire functions for which the set of c > 0 with E(c) of finite planar measure is exactly [m, infinity), and others for which it is exactly (m, infinity), which answers Erdos's question no.
Golʹdberg, A. A., Sets on which the modulus of an entire function has a lower bound. Sibirsk. Mat. Zh. 20 (1979), no. 3, 512--518, 691.
In Russian. The paper solves Problem 2.40 of Hayman's "New problems" (Canterbury symposium, 1973; published 1974): if f is a non-constant entire function and the planar measure of E(c) = {z : |f(z)| > c} is finite for some c, what is the minimal growth of f? Hayman conjectured that the integral of r dr / ln ln M(r) over [r_0, infinity) converges and that this is best possible. Erdos's variant asked whether finiteness of |E(c)| for one c implies the same for E(c') with c' < c; Gol'dberg inserts the word some into the question (finiteness for some c' < c), since the answer to that form is already negative, and the answer to the for-all form (finiteness for every c' < c) is then a fortiori negative too. Part 1 proves Hayman's convergence claim: using the inequality ln^+ ln^+ M(er,f) >= pi times the integral of dt/l(t) over A(r) minus K (inequality (1)), which Pfluger and Arima proved independently by Carleman's method, where l(r) is the largest arc length of the part of the circle |z| = r lying in E(c), together with the Cauchy-Bunyakovsky inequality on dyadic blocks (steps (3)-(5)), he derives that |E(c)| < infinity implies convergence of the integral of r dr / ln ln M(r,f) (formula (6)). Part 2 shows (6) cannot be sharpened: for any continuous positive nondecreasing Phi with the integral of r dr / Phi(r) convergent (7), there exists an entire f with ln ln M(r,f) = O(Phi(r)) and |E(c)| < infinity for all c > 0, a lemma for which is credited to V. S. Boichuk. Part 3 shows, by Keldysh's approximation theorem, that the set of c with |E(c)| finite can be [m, infinity) or (m, infinity) for any m > 0. A closing note (p. 517), written while the paper was with the editors, records that G. Camera's 1977 London doctoral thesis also proved Hayman's conjecture as established in Parts 1 and 2. The manuscript was received on 6 July 1977.
Source: https://www.mathnet.ru/eng/smj3876. No notice is printed on any of the seven pages, and the hosting site's terms of use state that its materials "are fully copyrighted by Steklov Mathematical Institute, Russian Academy of Sciences, and/or by other copyright holder" and that reproduction or republication "requires written permission of the copyright holder" (https://www.mathnet.ru/php/agreement.phtml?option_lang=eng, read 2026-10-02), every other right reserved.
Read status: claims checked. The statements of Parts 1--3 and the remark on Erdős's question were read clause by clause on the page images of pp. 512--517; the proofs were read in outline only, the results they import were not checked against their sources, and nothing here is independently reviewed.
Contents
The paper numbers its parts 1°, 2°, 3° and labels no theorem; the result pages take those section numbers. Throughout, , is its planar measure and .
- Section 1° (pp. 512--513, formula (6) on p. 513): if is entire and for some , then .
- Section 2° (statement p. 513, lemma p. 514, construction pp. 515--517): for every continuous positive nondecreasing on with there is an entire with as and for every .
- Section 3° (p. 517): with , the cases and occur, and for every there are entire functions with and with . The result page also records the remark of p. 512 on Erdős's question.
Bears on. #1118, whose two questions are the two the paper quotes from Hayman's Problem 2.40, the second attributed there to Erdős (p. 512). Section 1° proves the growth bound Hayman conjectured for the first question, and section 2° shows it cannot be sharpened in the sense stated there. For the second question the paper states (p. 512) that the answer is negative even when only some is asked for; the functions of section 3° with have and for every .
No file of this source is held: no license on record permits its redistribution, and the card cites the edition it names above.