Research notes on every problem, and a library of the papers behind them. Built from the open erdos repository.
Updated
Claim. The answer is no. The question is Problem 7 of the 1958 paper of Erdős, Herzog and Piranian (Section 6): if all the zeros of lie in the open disc , , does the set have a component of diameter greater than ? The closed-disc hypothesis of the site's statement is Pommerenke's restatement (p. 98), which also takes closed and asks for a component of diameter at least . Pommerenke's unnumbered passage on p. 98, as printed: "The answer is negative for . To show this, let (). Then the set has components." Each component contains one zero , , and every point of the component of the zero satisfies , so the common diameter of the components tends to as ; for fixed and large no component has diameter or more. Every component of the open set lies in a component of , so the example refutes the question as posed (an observation recorded on the result page, not the paper's sentence). The complementary half is Theorem 3 (p. 99): when the zeros lie in , the component of containing has diameter for , for and for , each bound at least , so the paper answers Problem 7, in its restated closed form, affirmatively for . Two observations on the problem's open set, recorded on this page and not the paper's: for the bound carries over, because by the Gauss--Lucas theorem every critical point of lies in , where with equality only for , whose critical point is , so no critical point lies on the level set , each component of is the closure of one component of , and the open set, connected like , has diameter ; for the theorem bounds the closed component and does not decide the strict question for the open set. The statements are on the result pages example_p98 and theorem_3 of the source card pommerenke_1961_metric_properties_complex_polynomials.
Source. Ch. Pommerenke, On metric properties of complex polynomials, Michigan Math. J. 8 (1961), no. 2, 97--115, DOI 10.1307/mmj/1028998561; received November 26, 1960. The publisher's record dates the article to the year 1961 alone, and the page is named by the record's date.
Acceptance. Refereed: the paper appeared in the Michigan Mathematical Journal. Reviewed: the site's curator, T. F. Bloom, labels the problem disproved and credits the negative answer for to this example, with its components whose diameter tends to , and records the three bounds of Theorem 3 as the affirmative answer for . Nothing here is independently reviewed by this project.
Formalization. The Lean file Erdos1048 in Boris Alexeev's repository,
linked above at the commit that added it, declares itself a formalization
of a solution found by Pommerenke: the statement of Pommerenke's result was
given to Aristotle, the system of Harmonic, which formalized the proof. Its
main_result proves, for , that has exactly
connected components for every and that their diameters are
eventually below any , and not_erdos_1048 refutes a
statement erdos_1048 in which the diameter bound reads
. The author's thread post explains the change
from the strict inequality: with the strict form and allowed, the
polynomial , whose sublevel set is the unit disc of diameter exactly
, is already a counterexample. Refuting the non-strict form refutes the
strict one. The qualifier of the site's label DISPROVED (LEAN) and the
formal_proof attribute of the formal-conjectures statement file refer to
this development. This
corpus has not built or audited it, so no formalized evidence is listed.
Aristotle's separate disproof from the problem statement alone, with
, is recorded on
Alexeev's page.
Depends on. Nothing on the wiki. The example uses the binomial expansion of ; Theorem 3 uses Lemmas 1 and 2 of the paper and the fact that a continuum of capacity has diameter at least .