Math-6172 · mathematics · Mathematics

Math-6172

Erdős Problem #1118

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Statement

Suppose $f$ is a nonconstant entire function and $E(c)=\{z:|f(z)|>c\}$ has finite planar area for some $c$. Necessarily $c>0$. The known sharp growth condition is convergence, for sufficiently large $R_0$, of \[ \int_{R_0}^{\infty}\frac{r}{\log\log M(r,f)}\,dr. \tag{1} \] Writing the lower endpoint as zero is shorthand: the denominator need not even be positive at small radii. Camera and Gol'dberg proved the sharp theorem. We prove necessity of $1$ from a precisely stated Carleman--Tsuji estimate, and show the stronger pointwise consequence $\log\log M(r,f)/r^2\to\infty$. The sharp converse and threshold examples remain unresolved in this attempt.

Restatement from Math-6172-A1 r1, with the source manuscript’s qualifications.

Original problem on erdosproblems.com

Statement licence: CC-BY-4.0 · Source file

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Math-6172-A1 · No claim recorded · LLM-generated · gpt 6 astra ultra ·

GPT 6 Astra Ultra attempt on Math-6172

Erdős Problem 1118: finite-area tracts force rapid growth

Statement and scope

Suppose $f$ is a nonconstant entire function and $E(c)=\{z:|f(z)|>c\}$ has finite planar area for some $c$. Necessarily $c>0$. The known sharp growth condition is convergence, for sufficiently large $R_0$, of \[ \int_{R_0}^{\infty}\frac{r}{\log\log M(r,f)}\,dr. \tag{1} \] Writing the lower endpoint as zero is shorthand: the denominator need not even be positive at small radii. Camera and Gol'dberg proved the sharp theorem. We prove necessity of $1$ from a precisely stated Carleman--Tsuji estimate, and show the stronger pointwise consequence $\log\log M(r,f)/r^2\to\infty$. The sharp converse and threshold examples remain unresolved in this attempt.

The external tract inequality

Choose $c_1>c$ which is not the modulus of a critical value of $f$, and let $U$ be a component of $\{|f|>c_1\}$. Such a component exists by Liouville's theorem and is unbounded: a bounded one contradicts the maximum modulus principle with boundary modulus $c_1$. Its complement is unbounded because it has finite area. The regular choice of level gives the usual smooth local boundary condition for a direct tract.

Let $\theta(t)$ be the angular measure of $U\cap\{|z|=t\}$. Since $U$ is open, connected and unbounded, $\theta(t)>0$ for all sufficiently large $t$. Put $\theta^*(t)=\theta(t)$ unless the whole circle lies in $U$, in which case put $\theta^*(t)=\infty$. We use the standard estimate \[ \log\log M(r,f)\ge \pi\int_{r_0}^{r/2}\frac{dt}{t\theta^*(t)}-C \qquad(r\text{ sufficiently large}). \tag{2} \] This is the Carleman--Tsuji direct-tract inequality, in the form $1.7$ of the revised Aspenberg--Bergweiler paper (equation $1.6$ in the original arXiv version), Entire functions with Julia sets of positive measure, https://analysis.math.uni-kiel.de/bergweiler/papers/89.pdf. Applying their formula to $f/c_1$ changes only a bounded term at large $r$. Inequality $2$ is external; it is not the sharp finite-area theorem being investigated.

Controlling the exceptional full circles

Polar coordinates give \[ A:=\int_{r_0}^{\infty}t\theta(t)\,dt<\infty. \] Let $B$ be the set of radii $t\ge r_0$ whose entire circles belong to $U$. Then $\int_B t\,dt\le A/(2\pi)$. For large $r$, the set \[ J_r=[r/4,r/2]\setminus B \] has length at least $r/8$, since the removed length is at most $4A/(2\pi r)$. Define \[ a(r)=\int_{r/4}^{r/2}t\theta(t)\,dt. \] Cauchy--Schwarz, applied only on $J_r$, yields \[ \int_{r/4}^{r/2}\frac{dt}{t\theta^*(t)} =\int_{J_r}\frac{dt}{t\theta(t)} \ge\frac{|J_r|^2}{\int_{J_r}t\theta(t)\,dt} \ge\frac{r^2}{64a(r)}. \] Therefore $2$ implies, after increasing the lower threshold, \[ \log\log M(r,f)\ge c_0\frac{r^2}{a(r)} \tag{3} \] for some positive constant $c_0$. The fixed subtractive constant in $2$ is absorbed because $a(r)\le A$ and $r^2/A\to\infty$. Since $a(r)\to0$, $3$ proves \[ \frac{\log\log M(r,f)}{r^2}\longrightarrow\infty. \] In particular such a function has infinite order. The treatment of $B$ matters: replacing $\theta^*$ by $\theta$ without checking full circles would not be justified.

The integral necessity

From $3$, nonnegativity and Tonelli's theorem, \[ \begin{split} \int_{R_0}^{\infty}\frac{r}{\log\log M(r,f)}\,dr &\le c_0^{-1}\int_{R_0}^{\infty}\frac{a(r)}r\,dr\\ &\le c_0^{-1}\int_{R_0/4}^{\infty} t\theta(t)\left(\int_{2t}^{4t}\frac{dr}r\right)dt\\ &\le c_0^{-1}(\log2)A<\infty. \end{split} \] Changing $R_0$ only changes a finite initial integral. This establishes the full necessary growth condition, relative to $2$.

What the threshold question requires

The set $T=\{c>0:|E(c)|<\infty\}$ is upward closed, since $E(c')\subseteq E(c)$ when $c'>c$. Thus a nonempty $T$ is $(m,\infty)$ or $[m,\infty)$ for some $m\ge0$, intersected with $(0,\infty)$. This elementary classification does not determine whether the endpoint can belong to $T$. Gol'dberg's construction realizes both alternatives for every $m>0$, so a smaller level need not exist. That construction, and examples demonstrating sharpness of $1$, are not supplied here. In particular the necessity proof cannot be presented as a proof of either existence assertion.

Assessment and sources

The finite-area growth obstruction, its pointwise strengthening and the integral necessity are proved from the stated analytic input. The two existence portions remain missing. Sources: https://www.erdosproblems.com/1118; G. Camera, On the minimum rate of growth of certain classes of integral and subharmonic functions, PhD thesis $1977$; A. A. Gol'dberg, Sets on which the modulus of an entire function has a lower bound, Sibirsk. Mat. Zh. 20 $1979$, 512--518; and the primary Carleman--Tsuji formulation linked above. No novelty is claimed.

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