Ghosh–Ramachandran's capacity-one area decay question

S. Ghosh and K. Ramachandran, A note on the Erdős minimal area problem, arXiv:2604.03036v3, beginning of Section 4:

This shows that while $\vartheta(K)=0$ may hold for all compacts $K$ with $\mathrm{cap}(K)=1$, the decay rate of $A_n(K)$ is not universal. It will be interesting to determine whether $A_n(K)$ can only decay like $\frac1n$ or $\frac1{\log n}$, or whether other rates are possible.

For compact $K\subseteq\mathbb C$, let $P_n(K)$ consist of the monic degree-$n$ polynomials whose zeros all lie in $K$. Write $\Lambda_p={z:|p(z)|<1}$ and $A_n(K)=\inf_{p\in P_n(K)}m(\Lambda_p)$, where $m$ is planar Lebesgue measure. The paper defines capacity as logarithmic capacity, equivalently the transfinite diameter.

The Lean claim says that for every nonempty compact $K$ of transfinite diameter one, the real-valued minimal-area sequence is either $\Theta(1/n)$ or $\Theta(1/\log n)$ as $n\to\infty$. Thus result : ¬ claim refutes the proposed two-rate restriction.

The closed unit disc and $I=[-2,2]$ both have capacity one. The text preceding the question cites a lower bound of order $1/\log n$ for the disc, whereas Theorem 4.1 proves $A_n(I)\leq c_1/n$. Lemma 4.6 gives period-$d$ sets with upper bounds of order $n^{-1/d}$ along multiples of $d$, without claiming these bounds are sharp. The declaration D5/S3/Analytic/Lemniscates/CapacityOneAreaDecayRate.result exhibits a capacity-one compact whose minimal area has neither of the two proposed asymptotic orders.

The literature readings identify arXiv:2609.06050 as treating $\vartheta(K)=0$ for capacity-one sets and explicitly giving no rate of decay. arXiv:2606.17097 concerns the disc and circle at order $1/\log n$. arXiv:2503.18270 does not treat isolated points. No answer to the Section 4 question was found in these readings. The source note is Library/Geometry/ghosh2026erdos.md.

Take $K=\overline{\mathbb D}\cup{4}$. For $s\geq2$, roots of unity in $K$ have Vandermonde product $s^{s/2}$. Every $s$-tuple in $K$ has Vandermonde product at most $s!,4^s\leq(4s)^s$: a repeated entry gives zero, and for an injective tuple at most one entry is $4$, so each term in the Vandermonde determinant has modulus at most $4^s$. Consequently its $s$-point diameter satisfies

$$ 1\leq d_s(K)\leq \exp!\left(\frac{2(\log4+\log s)}{s-1}\right)\longrightarrow1. $$

Use the monic competitor $p_{N,k}(z)=(z^N-1)(z-4)^k$, of degree $N+k$, with all roots in $K$. For $k\geq2$ and $N\geq k(k-2)+1$, its unit lemniscate is contained in

$$ {z:1-2^{-k}<|z|<1+2^{-k}} \ \cup\ {z:|z-4|<2^{-(k-2)}}. $$

The annulus has area $4\pi2^{-k}$ and the disc has area $16\pi2^{-2k}\leq16\pi2^{-k}$. Setting $N=k^2+1$ therefore gives the exact Lean bound

$$ A_{k^2+k+1}(K)\leq20\pi2^{-k}\qquad(k\geq2). $$

The compact $K=\overline{\mathbb D}\cup{4}$ is nonempty and has transfinite diameter one. Along $n=k^2+k+1$, the displayed area bound gives $nA_n(K)\to0$. It rules out an eventual positive lower bound of order $1/n$. Since $1/n=O(1/\log n)$, it also rules out an eventual positive lower bound of order $1/\log n$. Hence neither proposed $\Theta$ order holds. A subsequence bound suffices for this refutation; an all-degree sharp estimate is not needed.

The committed Lean module defines pointDiameter, TransfiniteDiameterOne, minimalArea and claim. Its only public theorem is result : ¬ claim. The private proofs pointDiameter_bounds and diameter_one establish the capacity condition; competitor_mem, competitor_cover, area_upper and subsequence_area_bound establish the admissible competitor and the bound $20\pi2^{-k}$. geometric_decay, no_reciprocal_bigO and reciprocal_nat_bigO_log give the asymptotic contradiction. The accepted axiom closure is ${\mathrm{propext},\mathrm{Classical.choice},\mathrm{Quot.sound}}$; there is no sorry or additional axiom.

What the settlement shows

1. Proved (Lean) — capacity and the two-rate refutation. $\overline{\mathbb D}\cup{4}$ has transfinite diameter one and $A_{k^2+k+1}(K)\leq20\pi2^{-k}$ for every $k\geq2$. Its area sequence is neither $\Theta(1/n)$ nor $\Theta(1/\log n)$.

2. Open in Lean — all-degree stretched-exponential order. An unformalized argument, which is not in the source paper and has not been independently checked, gives for all $n\geq4$

   $$ \frac{\pi}{4n^2},36^{-\sqrt n} \leq A_n(K)\leq48\pi,2^{-\sqrt n}, \qquad -\log A_n(K)=\Theta(\sqrt n). $$

   The lower bound uses the pairing product inequality of Geng–Qiu, arXiv:2609.06461, Proposition 2.3. Neither that inequality nor these all-degree bounds is proved in this Lean module.

3. Proved (Lean) — isolated-point mechanism at the certified scale. Adding the isolated point preserves capacity one, but the resulting compact has neither proposed decay type. For $z$ in the unit lemniscate of the competitor, the $k$ roots at $4$ force $|z^N-1|<2^{-k}$ whenever $|z-4|\geq2$, yielding the annular cover. Open in Lean — balanced-scale interpretation. The same unformalized argument (not in the source paper) gives a component near $4$ of scale $2^{-N/k}$ and balances it with the unit-circle level $2^{-k}$ at $k\asymp\sqrt n$. The Lean cover uses the coarser radius $2^{-(k-2)}$ under $N\geq k(k-2)+1$; it does not prove this sharper scale description.

4. Open — connected capacity-one sets. Which rates occur if $K$ is connected? The isolated-point example does not answer that restriction. Is every rate between stretched-exponential decay and $1/\log n$ realized by a capacity-one compact?

5. Open — sharpness of existing upper bounds. Are the interval's $1/n$ upper bound and the period-$d$ bounds $n^{-1/d}$ sharp? The present example determines neither of these orders.

The literature readings do not establish an exhaustive forward-citation list, so an answer outside those readings is not excluded. The Geng–Qiu pairing product inequality and the all-degree two-sided estimate are not formalized here. The balanced-scale interpretation and the neighbouring rate-realization and sharpness questions remain open in this Lean development.