Research Problems in Function Theory — Problem 3.16
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$\mathbb N=\{1,2,\ldots\}$, and $\mathbb Z,\mathbb Q,\mathbb R,\mathbb C$ denote the integers, rationals, reals and complex numbers. Any other conventions are specified locally.For compact $K\subset\mathbb R^n$, $n\geq3$, define Newtonian capacity by \[\operatorname{cap}(K)=\left(\inf_{\mu\in\mathcal P(K)}\iint |x-y|^{2-n}\,d\mu(x)\,d\mu(y)\right)^{-1},\] where $\mathcal P(K)$ denotes probability measures supported on $K$, $1/\infty=0$, and $\operatorname{cap}(\varnothing)=0$. A set is polar if each of its compact subsets has capacity zero. For a compact $E$ set $c(P,r)=\operatorname{cap}(E\cap\overline B(P,r))$ and \[T(E)=\left\{P\in E:\int_0^1\frac{c(P,r)}{r^{n-1}}\,dr<\infty\right\}.\]
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Can one prove directly, without invoking Kellogg's theorem, that $T(E)$ is polar for every compact $E\subset\mathbb R^n$, and thereby clarify the sharpness of the displayed integrated Wiener criterion? [S2]Short English statement
Give a direct proof that the points where a compact set satisfies the thinness integral form a polar set, preferably explaining why the integral condition is sharp.Sources
- S1 ULAM.AI and the UnsolvedMath Contributors, UnsolvedMath: A Curated Collection of Open Mathematics Problems, record 2303016 (AMR-022-3016). CC BY 4.0. https://huggingface.co/datasets/ulamai/UnsolvedMath.
- S2 W. K. Hayman and E. F. Lingham, Research Problems in Function Theory, arXiv:1809.07200v2 (2018), Problem 3.16. https://arxiv.org/abs/1809.07200.
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