Erdős 问题 #1040 的一个解法
A solution to the Erdős Problem #1040
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中文总结 AI 辅助
本文证明对容量为1的任意紧集，零点在该集合中的首一多项式的单位叶形线面积下确界为零，从而结合已有结果完全解决了Erdős问题1040中的消失性问题。
中文摘要 AI 辅助
对于紧集 $K\subset\mathbb{C}$，设 $\vartheta(K)$ 为所有零点在 $K$ 中的首一多项式（允许任意次数和重零点）的单位叶形线（lemniscate）的平面面积的下确界。我们证明当 $\operatorname{cap}(K)=1$ 时，$\vartheta(K)=0$，且对 $K$ 无需任何正则性假设。证明使用了一个中心调和多项式，该多项式在 $K$ 的多项式包络中除一个面积任意小的集合外均为正。外部调和测度的傅里叶平均将该多项式实现为关于平衡测度具有有界密度的带符号测度的对数势。正扰动和用经验测度进行的 $L^1$ 逼近随后产生所需的多项式。这将 Krishnapur、Lundberg 和 Ramachandran 的光滑边界结果推广到任意容量为 1 的紧集。结合 Ghosh 和 Ramachandran 的容量大于 1 的定理以及对无界集的一个初等论证，可得对每个闭无限集 $F\subset\mathbb{C}$，若其超越直径至少为 1，则 $\vartheta(F)=0$，从而回答了 Erdős 问题 1040 中的消失性问题。
英文摘要
For a compact set $K\subset\mathbb{C}$, let $\vartheta(K)$ be the infimum of the planar areas of the unit lemniscates of all monic polynomials with zeros in $K$, allowing arbitrary degree and repeated zeros. We prove that $\vartheta(K)=0$ whenever $\operatorname{cap}(K)=1$, with no regularity assumption on $K$. The proof uses a centered harmonic polynomial that is positive on all but a set of arbitrarily small area in the polynomial hull of $K$. A Fourier average of exterior harmonic measures realizes this polynomial as the logarithmic potential of a signed measure having bounded density with respect to the equilibrium measure. A positive perturbation and an $L^1$ approximation by empirical measures then produce the required polynomials. This extends the smooth-boundary result of Krishnapur, Lundberg, and Ramachandran to arbitrary compact sets of capacity one. Together with the capacity-greater-than-one theorem of Ghosh and Ramachandran and an elementary argument for unbounded sets, it follows that $\vartheta(F)=0$ for every closed infinite set $F\subset\mathbb{C}$ of transfinite diameter at least one, answering the vanishing question in Erdős Problem 1040.
