GPT 6 Astra Ultra attempt on Math-6269
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Contents
Erdős Problem 1215: unit-circle zeros can force arbitrarily long paths
Statement and endpoint convention
For polynomials $P$ with $P(0)=1$ and all zeros on $|z|=1$, can a universal constant bound the length of a path from $0$ to the unit circle along which $|P|<1$? As usual, the initial endpoint is excepted from the strict inequality, since $|P(0)|=1$. We disprove the intended assertion, even with the weaker requirement $|P|\le1$. The known negative answer is due to Mac Lane. We reconstruct the approximation and labyrinth argument explicitly.Approximating exponentials by polynomials with unit-circle zeros
We first prove a useful approximation fact. For every polynomial $h(z)=\sum_{m=1}^M c_mz^m$ with $h(0)=0$, there are polynomials $P_d$, of arbitrarily large integer degrees $d$, with $P_d(0)=1$, all zeros on the unit circle, and \[ P_d(z)\longrightarrow e^{h(z)} \quad\text{uniformly on every closed disk }|z|\le r<1. \tag{1} \] Define a positive density on $[0,2\pi]$ by \[ w_d(t)=\frac1{2\pi}\left(d-2\operatorname{Re} \sum_{m=1}^M mc_me^{imt}\right), \] taking $d>2\sum m|c_m|+1$. Its total mass is $d$. Put $W_d(t)=\int_0^t w_d(v)\,dv$, and choose \[ t_j=W_d^{-1}(j-\tfrac12)\quad(1\le j\le d),\qquad P_d(z)=\prod_{j=1}^d(1-ze^{-it_j}). \] All the required root and normalization conditions hold.For $|z|\le r<1$, use the analytic logarithm \[ G_z(t)=\log(1-ze^{-it})=-\sum_{m\ge1}\frac{z^me^{-imt}}m. \] Its first two derivatives in $t$ are uniformly bounded in $z$. Orthogonality of the exponentials gives exactly \[ \int_0^{2\pi}G_z(t)w_d(t)\,dt=h(z). \tag{2} \] Indeed, the constant part of $w_d$ integrates to zero, its $m$th Fourier moment is $-mc_m$ for $m\le M$, and all higher moments vanish.
Let $\tau_d=W_d^{-1}$ on $[0,d]$. Since $w_d\asymp d$ and $w_d'=O_h(1)$, \[ \tau_d'=O_h(d^{-1}),\qquad \tau_d''=O_h(d^{-3}). \] Thus $(G_z\circ\tau_d)''=O_{h,r}(d^{-2})$. The midpoint-rule error on each unit interval is $O_{h,r}(d^{-2})$; summing $d$ intervals and using $2$ yields \[ \sum_{j=1}^dG_z(t_j)=h(z)+O_{h,r}(d^{-1}). \] Exponentiation proves $1$. This supplies the required approximation with roots constrained to the circle, rather than invoking unconstrained polynomial approximation for $P$.
A forbidden spiral
Fix $0<a<b<1$ and a positive integer $k$. The simple compact arc \[ \Gamma_k=\left\{t\exp\left(2\pi i k\frac{t-a}{b-a}\right): a\le t\le b\right\} \] winds $k$ times between the two circles. Its complement, and the complement of $\Gamma_k\cup\{0\}$, are connected. The Runge polynomial approximation theorem applies to the function equal to $0$ near the isolated point $0$ and equal to $3$ near $\Gamma_k$. Approximate it closely by a polynomial $g$, and put $h=g-g(0)$. Then $h(0)=0$ and, for example, $\operatorname{Re}h>2$ on $\Gamma_k$. By $1$, for sufficiently large $d$ the associated polynomial satisfies \[ |P_d(z)|>2\qquad(z\in\Gamma_k). \tag{3} \] The use of Runge here concerns the unrestricted auxiliary polynomial $h$; the preceding lemma supplies the separate circle-root restriction on $P_d$.Every admissible path has large length
Let a continuous path start at $0$ and reach $|z|=1$, with $|P_d|\le1$ throughout. By $3$ it avoids $\Gamma_k$. Take its first time reaching radius $b$, and its last preceding time at radius $a$. The intervening segment lies in the closed annulus $a\le|z|\le b$. Write it in polar coordinates with a continuous lifted argument $\theta$.On that segment the continuous real function \[ \theta-2\pi k\frac{|z|-a}{b-a} \] never belongs to $2\pi\mathbb Z$, because membership would put the path on $\Gamma_k$. It consequently stays within one interval between consecutive multiples of $2\pi$. Its change has magnitude less than $2\pi$, whereas the second term changes by $2\pi k$. Therefore the net change in $\theta$ is greater than $2\pi(k-1)$. If the path is rectifiable, its length is at least \[ a\,\operatorname{Var}(\theta)\ge2\pi a(k-1). \] This follows either from arc-length parametrization and $|d\theta|\le|dz|/a$, or by polygonal approximation in the annulus. A nonrectifiable path already has infinite length.
Since $k$ is arbitrary, no universal length bound exists. One may also choose $b$ arbitrarily small before choosing $k$: arbitrarily large length can be forced inside any prescribed neighborhood of the origin.