slug: minadiaz-rubin-2025-chebyshev-faber-equipotential
bibkey: minadiaz2025chebyshev
doi: null
url: https://arxiv.org/abs/2505.03967v1
triage: theorem
motivation_gids:
  - D5/S3/Analytic/Chebyshev/FaberEquipotentialRefutation.result

Miña-Díaz–Rubin’s Chebyshev–Faber equipotential question

Problem

E. Miña-Díaz and O. Rubin, Chebyshev polynomials on equipotential curves, arXiv:2505.03967v1, Section 2, Problem 2.2:

Suppose the compact set $K$ is such that for some $n\in \mathbb{N}$, $T_{n,K}={\hat{F}}_n$, where ${\hat{F}}_n$ is the $n$th monic Faber polynomial associated to the map $\varphi$ that satisfies (1.2). Is it true then that $T_{n,L_r}={\hat{F}}_n$ for all $r>1$?

Here $\Omega$ is the unbounded component of $\mathbb{C}\setminus K$, $g_{\Omega }(z,\infty )=\log |\varphi (z)|$, and $\varphi (z)=cz+c_0+c_{-1}/z+\cdots$, with $c=\mathrm{cap}(K{)}^{-1}$. When $\Omega$ is simply connected, $\psi ={\varphi }^{-1}$. The equipotential curve is $L_r=\{z\in \Omega :|\varphi (z)|=r\}=\psi (\{|w|=r\})$. The polynomial $F_n$ is the polynomial part of ${\varphi }^n$ at infinity, ${\hat{F}}_n=c^{-n}{F}_n$, and $T_{n,E}$ is the unique monic degree-$n$ sup-norm minimizer on $E$.

The Lean claim quantifies over exterior maps $\psi$ differentiable and injective on $|w|>1$, a positive capacity $\kappa =1/c$, compact complement $K$, and a bounded correction $\psi (w)-\kappa w$ for $|w|>2$. Its IsMonicFaber requires a monic degree-$n$ polynomial $F$ with $F(\psi (w))-(\kappa w{)}^n\to 0$ at infinity. Its IsChebyshev compares every uniform real bound for every monic degree-$n$ competitor. For compact nonempty sets this expresses sup-norm minimization. The implication asks whether such an $F$ Chebyshev on $K$ remains Chebyshev on every $L_r$, $r>1$.

Motivation

Theorem 2.1 gives $T_{n,L_r}\to {\hat{F}}_n$ as $r\to \infty$. Equality at every finite radius is stronger than this limit. The declaration D5/S3/Analytic/Chebyshev/FaberEquipotentialRefutation.result proves ¬ claim with $n=1$, $\kappa =1$ and $r=2$.

Gap

Preregistration #15412 identifies the tier-1 question and its refutation. Rubin’s Chebyshev polynomials in the complex plane and on the real line, arXiv:2411.14175v2, revised 2026-02-18, still lists it as Problem 5.6. Hübner–Rubin, arXiv:2405.05067, gives rotation-symmetric examples with $T=\hat{F}=z^{\ell }$ preserved on equipotentials. Miña-Díaz–Rubin–Wennman, arXiv:2509.22588, concerns $n\to \infty$ asymptotics and does not supply the fixed-degree propagation assertion.

Xiao–Qiu, Acta Mathematica Sinica, Chinese Series 53(2) (2010), 323–328, gives an example where Chebyshev polynomials on a Julia set and its equipotentials differ. Whether that example has $T=\hat{F}$ on the Julia set is ASSUMED-UNVERIFIED because its full text was not available. The centre criterion is discussed in Triage. These references do not establish an exhaustive priority claim. The source note is Library/Geometry/minadiaz2025chebyshev.md.

Route

Take

$$\psi (w)=w+\frac{1}{4w}+\frac{1}{100}(w^{-2}-w^{-4}),K=\mathbb{C}\setminus \psi (\{|w|>1\}).$$

For $|u|,|v|\ge 1$, the proof establishes

$$|\psi (u)-\psi (v)-(u-v)|\le \frac{31}{100}|u-v|.$$

This proves injectivity on the closed exterior and two-sided bounds $|\psi (u)-\psi (v)|\le (131/100)|u-v|$ and $|u-v|\le 2|\psi (u)-\psi (v)|$. Differentiability and the complex open mapping theorem make the open exterior image open; the two-sided bounds make the closed exterior image closed. Connectedness of the exterior of a disc then gives $K\subseteq \overline{B}(0,5/4)$ and compactness. Injectivity on the closed exterior puts $\psi (\pm 1)=\pm 5/4$ in $K$. The two opposite endpoints force every degree-one monic competitor to have norm at least $5/4$, so $z$ is Chebyshev on $K$. Also $\psi (w)-w\to 0$, giving ${\hat{F}}_1=z$.

The private circle_bound proves, for every $r\ge 1$ and $|w|=r$,

$$\left|\psi (w)-\frac{1}{100}(r^{-2}-r^{-4})\right|\le r+\frac{1}{4r}.$$

At $r=2$ the competitor $q(z)=z-3/1600$ therefore has norm at most $17/8=3400/1600$. But $\psi (2)=3403/1600$ lies on $L_2$, so $z$ has strictly larger norm there. The exact gap is $3/1600$.

Falsifier

The single admissible exterior map, compact complement, degree-one Faber polynomial and radius-two competitor falsify the universal propagation assertion. A norm comparison without injectivity, compactness, the Faber normalization or the initial Chebyshev property would not address the source question; the proof supplies each. No classification of the minimizer on every outer curve is required for this radius-two contradiction.

Evidence

The module’s public definitions are IsChebyshev, IsMonicFaber and claim; its only public theorem is result : ¬ claim. Supporting declarations are private, including perturbation_lipschitz, injective_closed, exterior_bilipschitz, compact_complement, circle_algebra_bound, circle_complex_bound, circle_bound, chebyshev_of_disc and faber. The norm-square estimate in circle_algebra_bound controls the cross term and residual term on a normalized circle. The proof uses exact rational arithmetic, the frozen D5.S3.AnalyticClosure.ComplexPowerDifference.norm_pow_sub_pow_le at $H=1$, and the pinned Mathlib topological and complex-analytic results. The consumed helpers powers_lipschitz, perturbation_lipschitz, injective_closed and exterior_bilipschitz are bind-only; the per-declaration inventory is in the Lean header. Its accepted axiom closure is $\{\mathrm{propext},Classical.choice,Quot.sound\}$; there is no sorry or additional axiom.

The odd-cubic computation in Triage covers $a\in \{-1,1,2\}$ and $r\in \{1.5,2,3\}$, with agreement to $10^{-7}$. No experiment program or certified numerical error bound is available; this computed observation is separate from the Lean counterexample.

Triage

What the settlement shows

1. Proved (Lean) — degree-one failure. The map above has $T_{1,K}={\hat{F}}_1=z$, but $z$ is not Chebyshev on $L_2$. The proof establishes Lipschitz injectivity, compactness and the enclosing-circle norm bound, including its general $r\ge 1$ form.

2. Open in Lean (analytic argument) — every radius. Write $c_r=(1/100)(r^{-2}-r^{-4})$ and $R_r=r+1/(4r)$. The enclosing-disc bound and $\psi (\pm r)=c_r\pm R_r$ identify the minimal enclosing circle as the circle of centre $c_r$ and radius $R_r$. This gives $T_{1,L_r}=z-c_r\ne z$ for every $r>1$. The full minimizer identification and universal failure are analytic consequences not stated as Lean theorems in this module.

3. Open here (not formalized; analytic argument) — every degree. For each positive integer $d$, set $K_d=\{z:z^d\in K\}$. Rotation averaging, or the Kamo–Borodin / Novello–Schiefermayr–Zinchenko composition rule, gives the proposed extension

   $$T_{d,K_d}={\hat{F}}_d=z^d,T_{d,L_r(K_d)}=z^d-c_{r^d}.$$

   The Green-function normalization is $g_{K_d}(z)=d^{-1}{g}_K(z^d)$; hence radius $r$ for $K_d$ maps to radius $r^d$ for $K$. The composition and Faber identifications are not formalized here.

4. Open in Lean (analytic argument) — minimal-circle interpretation and centre drift for every $r>1$. Degree-one minimization is the minimal enclosing circle problem. The perturbation $b(w^{-2}-w^{-4})$, $b=1/100$, contributes zero to the centre at $r=1$ and a positive $c_r$ for $r>1$; it need not vanish pointwise on the unit circle. Thus the initial equality does not fix the outer centres. Literature-attested (paper, Theorem 2.1; not formalized here) — the limiting statement survives. Since $c_r\to 0$, the analytic family is consistent with $T_{n,L_r}\to {\hat{F}}_n$.

5. Open here (composition argument not formalized) — Julia-set criterion. For a monic polynomial $P(z)=z^m+a_{m-1}{z}^{m-1}+\cdots$, $m\ge 2$, and centre $h$ of the minimal enclosing circle of $J_P$, the composition rule gives $T_{m^k,J_P}=P^{\circ k}-h$, while ${\hat{F}}_{m^k}=P^{\circ k}+a_{m-1}/m$, for $k\ge 1$. A counterexample through these degrees therefore needs $h=-a_{m-1}/m$ and an outer centre that drifts. A rotational symmetry of order at least two about that centre preserves it on equipotentials and prevents this mechanism. Computed — numerical Julia-set probe. For the odd cubics $z^3+az$ with $a\in \{-1,1,2\}$ on levels $r\in \{1.5,2,3\}$, the numerical probe found agreement to $10^{-7}$. There is no certified error bound; this is not a uniform result for all cubics.

6. Open — additional hypotheses. Does equality propagate when $K$ has rotational symmetry of order greater than $n$? Does it propagate for an analytic boundary if $T_{n,K}={\hat{F}}_n$ for every $n$? These sufficient-condition questions are not settled by the Lean counterexample. The rotation-symmetric literature examples supply surviving special cases.

ASSUMED-UNVERIFIED

The full text of Xiao–Qiu (2010) was not available. Whether its example satisfies $T=\hat{F}$ on the Julia set, equivalently the relevant enclosing-circle centre condition $h=-a_{m-1}/m$, was not determined. Its reported difference between Julia-set and equipotential Chebyshev polynomials alone does not settle Problem 2.2.

No DOI is recorded, and exhaustive priority is not established. Failure at every $r>1$, the every-degree pullback extension, the Julia-set composition analysis and the additional- hypothesis questions have no Lean theorem here. The odd-cubic numerics have no experiment program or certified error bound.