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Bergman completeness is not a quasi-conformal invariant. (English) Zbl 1264.32006

In 2011 B.-Y. Chen [“An essay on Bergman completeness”, Preprint (2011)] asked the question whether the Bergman completeness is a quasi-conformal invariant of Riemann surfaces. The paper under review gives an example showing that the answer to the question is negative.
Recall that a bounded planar domain \(D\) is called Bergman complete if any Cauchy sequence with respect to the Bergman distance is convergent to some point in \(D\) with respect to the standard topology in \(D\). A homeomorphism \(f\) defined on a planar domain \(D\) is called \(L\) (\(L>1\)) quasi-conformal if it is differentiable almost everywhere and \[ \left|\frac{\partial f}{\partial\bar z}\right|\leq\frac{L-1}{L+1}\left|\frac{\partial f}{\partial z}\right|. \]
The paper uses the methods from the paper by P. Pflug and W. Zwonek [Arch. Math. 80, No. 5, 536–552 (2003; Zbl 1037.30009)], whereas the original idea of the example is based on W. Zwonek’s paper [Bull. Pol. Acad. Sci., Math. 50, No. 3, 297–311 (2002; Zbl 1016.32004)].
For \(r\in(0,1/4)\), \(t\in(0,1/2)\) let \[ A^{r,t}:=\bigcup_{k=1}^{\infty}\{r^ke^{i\theta}:|\theta|<2\alpha_k\}\cup\{0\}, \] where \(\sin\alpha_k=e^{-t^{-k}}\). Put \[ D^{r,t}:=\mathbb D\setminus A^{r,t}, \] where \(\mathbb D\) is the unit disc in \(\mathbb C\). Finally, for \(a>1/2\) let \[ f(z):=z^a\bar z^{a-1}. \] The author shows that for some parameters \(r\), \(t\), and \(a\) (e.g. \(r=1/8\), \(t=1/32\), \(a=2/3\)) the function \(f:D^{r,t}\to D^{r^{2a-1},t}\) is quasi-conformal, \(D^{r,t}\) is Bergman complete, but \(D^{r^{2a-1},t}\) is not Bergman complete.

MSC:

32F45 Invariant metrics and pseudodistances in several complex variables
32A25 Integral representations; canonical kernels (Szegő, Bergman, etc.)
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References:

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