Proposition 15.31.16.DD2C5D7B64EC Suppose, for some $C>0$ and $\delta >0$,
\[ \mu (B_{\mathrm{ch}}(x,r))\leq Cr^\delta \qquad (x\in Y,\ 0<r\leq 1). \]
Then $Q_\mu $ is finite and continuous, so theorem 15.31.15 applies. In particular it applies to
\[ \mathrm d\mu (\vartheta )=(1+\epsilon \cos \vartheta ) \frac{\mathrm d\vartheta }{2\pi },\qquad 0<\epsilon <1, \]
on the circle $[1:\mathrm e^{\mathrm i\vartheta }]$. This measure and its product $\nu $ are singular with respect to smooth volume, and $Q_\mu $ is nonconstant on its support.
Proof.
Since $K=-\log d_{\mathrm{ch}}^2$, layer-cake integration gives
\[ \sup _ x\int _{\{ K(x,y)>L\} }K(x,y)\, \mathrm d\mu (y) \leq C\left(L+\frac2\delta \right)\mathrm e^{-\delta L/2}. \]
Consequently the continuous truncated-kernel potentials converge uniformly to $Q_\mu $. Weighted arclength on the circle has the ball bound with $\delta =1$. On that circle the kernel is $1+\log \sin ^2((\vartheta -\varphi )/2)$, whose first cosine coefficient is $-2$. Integrating its $L^1$ Fourier series against $1+\epsilon \cos \varphi $ gives
\[ Q_\mu ([1:\mathrm e^{\mathrm i\vartheta }]) =1-2\log 2-\epsilon \cos \vartheta . \]
This proves nonconstancy, and the support has smooth curve volume zero. Its product with $X_0$ also has smooth volume zero.
$\square$
Pome
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