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Source. Section 3° (p. 517, statement and proof), with the remark on Erdős's question in the opening paragraph (p. 512), of A. A. Gol'dberg, Sets on which the modulus of an entire function has a lower bound (Russian), Sibirsk. Mat. Zh. 20 (1979), no. 3, 512--518, 691, the edition named on the source card.

Statement

Setting (pp. 512, 517). For an entire function ff and c>0c>0, E(c)={z:∣f(z)∣>c}E(c)=\{z:|f(z)|>c\} and ∣E(c)∣|E(c)| is its planar measure. Section 3° considers T={c>0:∣E(c)∣<∞}T=\{c>0:|E(c)|<\infty\}, which is an up-set, since E(c)E(c) shrinks as cc grows.

Result of 3° (p. 517). The cases T=∅T=\emptyset and T=(0,∞)T=(0,\infty) both occur; the paper calls this evident and points, for the second, to the function of section 2°. For every m>0m>0 there is an entire function with T=[m,∞)T=[m,\infty), and an entire function with T=(m,∞)T=(m,\infty).

Erdős's question (p. 512). The paper states Erdős's question in the form: if E(c)E(c) has finite measure, does E(c′)E(c') have finite measure for some c′<cc'<c? It says that the author inserted the word some, because in that form the answer is already negative, so that it is negative a fortiori in the form asking about all c′<cc'<c. The case T=[m,∞)T=[m,\infty) above is such an example: ∣E(m)∣<∞|E(m)|<\infty, while ∣E(c′)∣=∞|E(c')|=\infty for every c′<mc'<m.

Read depth. Claims checked: the statement, the remark of p. 512 and the construction were read on the page images of pp. 512 and 517. The approximation theorem the construction cites was not checked against its source, and nothing here is independently reviewed.

Proof pointer

Page 517, outlined here. Let D\mathcal D be the union of the unit disc and the thin region {x<0, ∣y∣<(1+x2)−1}\{x<0,\ |y|<(1+x^2)^{-1}\}, which has finite area, and E=C∖DE=\mathbb C\setminus\mathcal D. On EE take ψ(z)=m(1−14z−1/4)2\psi(z)=m(1-\tfrac14z^{-1/4})^2, with the branch of z1/4z^{1/4} positive on (1,∞)(1,\infty). By Keldysh's approximation theorem (cited from Mergelyan's 1952 survey, p. 59, Theorem 1.3) there is an entire ff with ∣f(z)−ψ(z)∣<m16exp⁡(−∣z∣1/4)|f(z)-\psi(z)|<\tfrac{m}{16}\exp(-|z|^{1/4}) on EE. A direct estimate gives ∣f∣<m|f|<m on EE, so E(m)⊂DE(m)\subset\mathcal D and ∣E(m)∣<∞|E(m)|<\infty; and ∣f(z)∣→m|f(z)|\to m uniformly as z→∞z\to\infty in EE, so for c<mc<m the set E(c)E(c) contains all of EE outside a large disc and ∣E(c)∣=∞|E(c)|=\infty. Thus T=[m,∞)T=[m,\infty). For T=(m,∞)T=(m,\infty) the same argument is run with ψ(z)=m(1+14z−1/4)2\psi(z)=m(1+\tfrac14z^{-1/4})^2.

Dependencies

M. V. Keldysh's approximation theorem, cited from S. N. Mergelyan, Uniform approximations of functions of a complex variable (Russian), Uspekhi Mat. Nauk 7 (1952), no. 2, 31--122.

Bears on

  • Problem 1118: the problem's second question asks whether finite measure of E(c)E(c) forces finite measure of E(c′)E(c') for some c′<cc'<c. The entire functions with T=[m,∞)T=[m,\infty) have ∣E(m)∣<∞|E(m)|<\infty and ∣E(c′)∣=∞|E(c')|=\infty for every c′<mc'<m, so the answer is no; the paper states the negative answer on p. 512.