The logarithmic single layer on a circle changes by a scale term of rank at most one

The circle $S_1$ is compact. Since $\mathrm{\mid }x-y\mathrm{\mid }\le 2$ on it, every probability energy is at least $-\log 2$. Normalized arclength has finite energy: writing $x=e^{i\theta }$ and $y=e^{i\phi }$ gives $\mathrm{\mid }x-y\mathrm{\mid }=2\mathrm{\mid }\sin ((\theta -\phi )\mathrm{/}2)\mathrm{\mid }$, whose logarithmic singularity is integrable because ${\int }_0^{ϵ}\mathrm{\mid }\log t\mathrm{\mid }dt<\mathrm{\infty }$. Thus $I(S_1)$ is finite and $0<\mathrm{cap}(S_1)<\mathrm{\infty }$. Apply [F1] with $K=S_1$ and $\mathrm{\Gamma }=S_1$ to obtain both scaling identities in the Example. The functional $\lambda \mapsto {\int }_{S_1}\lambda ds$ has one-dimensional range, and its output is a constant function, so the trace correction is rank at most one. Its scalar factor vanishes precisely at $s=1$; otherwise the constant density gives a nonzero correction, proving rank one.

For $\lambda =1\mathrm{/}(2\pi )$ on the unit circle, ${\int }_{S_1}\lambda ds=(1\mathrm{/}(2\pi ))(2\pi )=1$. Substituting $s=e$ into the correction gives $-(\log e)\mathrm{/}(2\pi )=-1\mathrm{/}(2\pi )$. Any zero-total-charge density has no scale correction by the same formula.