GPT 6 Astra Ultra attempt on Math-6098

Erdős Problem 1044: a checked attempt

GPT_6_Astra_Ultra

24 September 2026

Which infimum is determined

For the infimum over all positive degrees, the answer is \[ \boxed{\inf_f\ \max_{U\text{ a component of }\{|f|<1\}} \operatorname{length}(\partial U)=2.} \tag{1} \] Here $f$ is monic and all its roots lie in the closed unit disk. Degree is allowed to vary. Determining the minimum separately at every fixed degree is a different, stronger question, and is not asserted below.

We reconstruct the proof in Quanyu Tang, On Erdős Problem #1044, Sections 2--4, https://raw.githubusercontent.com/QuanyuTang/erdos-problem-1044/main/On_Erd%C5%91s_Problem_1044.pdf. A diameter theorem of Pommerenke is explicitly retained as an external input. We give the boundary-length step in a form that counts geometric boundary length without counting overlapping parametrized arcs twice.

A boundary-length lemma

Let $U$ be a bounded connected open set whose boundary is a finite union of rectifiable analytic arcs. Then \[ \operatorname{length}(\partial U)\ge2\operatorname{diam}(\overline U). \tag{2} \] To see this, project onto a direction parallel to a pair of diameter-realizing points of $\overline U$. The projection of $U$ is an interval whose length is its diameter. For almost every coordinate in this interval, the corresponding perpendicular line meets $U$ in a nonempty bounded open set. Its first and last endpoints are two distinct points of $\partial U$. The one-dimensional projection formula on the boundary arcs gives \[ \operatorname{length}(\partial U) \ge\int\#\{\partial U\text{ on the perpendicular line at }t\}\,dt \ge2\operatorname{diam}(\overline U). \] One can obtain the first inequality directly by dividing the arcs at projection-critical points and integrating $|dx|\le ds$; endpoints and intersections form a set of zero arclength and can be counted once. For polynomial sublevel components the boundary is a compact real algebraic curve, hence has the required finite rectifiable-arc decomposition. The diameter of the closure also equals the diameter of the boundary: a diameter-realizing point in the interior could be moved slightly away from its partner.

Every component of $\{|f|<1\}$ contains a root. Otherwise $\log|f|$ would be harmonic in that bounded component with boundary value zero, contrary to its strictly negative interior values. In particular there are at most $\deg f$ components.

The one external diameter input

We use the following consequence of Pommerenke, On metric properties of complex polynomials, Michigan Math. J. 8 $1961$, Theorem 3, https://doi.org/10.1307/mmj/1028998561, in the form invoked in Tang's proof of Theorem 4.3:

If a monic polynomial has all roots in the closed unit disk and $|f(0)|<1$, then the component of $\{|f|<1\}$ containing zero has diameter strictly greater than one.

Only this weak consequence is imported, not the desired arclength conclusion. For maximum root modulus $r>0$, the more precise piecewise diameter bounds imply it; when $r=0$, the polynomial is $z^n$ and the diameter is two directly. Tang's cited use of Pommerenke is the source checked for this input; an independent proof of the diameter theorem is not provided here.

If $|f(0)|<1$, apply this input and $2$ to get a component with boundary length greater than two. The only remaining possibility is $|f(0)|=1$, since the product of the root moduli is at most one. Equality forces every root to have modulus one. The nonconstant harmonic function $\log|f|$ near zero has value zero there, so every neighbourhood of zero contains a point of $\{|f|<1\}$. There are only finitely many components; hence zero belongs to the boundary of one component $U$.

That component contains a root $a$ with $|a|=1$. Since $a$ is an interior point of $U$, the set $U$ contains a point $b$ with $|b|>1$. Thus the maximum of $|z|$ on $\overline U$, attained on $\partial U$, is strictly greater than one. Together with $0\in\partial U$ this gives $\operatorname{diam}(\overline U)>1$. Equation $2$ again gives boundary length greater than two. We have proved a strict lower bound for each individual polynomial, and hence a lower bound of two for the infimum.

An explicit sequence approaching two

Take $f_n(z)=z^n-1$. Its roots have modulus one and its strict unit sublevel set has $n$ congruent petals. The closure of the petal containing the positive real root has the polar boundary \[ z=\rho(\theta)e^{i\theta},\qquad \rho(\theta)=(2\cos(n\theta))^{1/n},\qquad -\frac\pi{2n}\le\theta\le\frac\pi{2n}. \] The endpoints are both zero. Differentiating gives \[ \sqrt{\rho^2+(\rho')^2} =2(2\cos(n\theta))^{1/n-1}. \] Consequently every petal boundary has length \[ L_n=\frac{2^{1/n}}n B\left(\frac1{2n},\frac12\right),\qquad B(x,y)=\int_0^1 t^{x-1}(1-t)^{y-1}\,dt. \tag{3} \] For $0<a\le1/2$, \[ B(a,1/2)=\frac1a+ \int_0^1t^{a-1}\left((1-t)^{-1/2}-1\right)\,dt =\frac1a+O(1). \] The remainder is uniformly bounded: near zero the parenthesized factor is $O(t)$, and near one it is integrable, while $t^{a-1}$ is bounded there. Substituting $a=1/(2n)$ in $3$ yields $L_n\to2$. This proves the matching upper bound in $1$. The global infimum is approached but not attained, by the strict lower bound already proved.

Assessment

The global degree-varying infimum and its nonattainment are proved relative to the stated Pommerenke diameter input. The length computation and passage to the infimum are supplied. No optimality assertion is made for $z^n-1$ at every fixed $n$.

Completion rate estimate: 100%.