Commun. Korean Math. Soc. 2017; 32(3): 779-787
Online first article May 2, 2017 Printed July 31, 2017
https://doi.org/10.4134/CKMS.c160158
Copyright © The Korean Mathematical Society.
Jaesung Lee
Sogang University
Let $m$ be the Lebesgue measure on $\mathbb C$ normalized to $m(D)=1$, $\mu$ be an invariant measure on $D$ defined by $d \mu(z) = (1-|z|^{2})^{-2} \ dm(z)$. For $f \in L^{1}(D^{n} , m \times \cdot \cdot \cdot \times m) , \ Bf$ the Berezin transform of $f$ is defined by, $$(Bf)(z_{1} , \ldots, z_{n})\! =\! \int_{D} \cdot \cdot \cdot \int_{D} f \big( \varphi_{z_{1}}(x_{1}), \ldots , \varphi_{z_{n}}(x_{n}) \big) dm(x_{1}) \cdots dm(x_{n}).$$ We prove that if $f \in L^{1}(D^{2},\mu\times\mu)$ is radial and satisfies $ \int \int_{D^{2}} f d\mu \times d\mu = 0, $ then for every bounded radial function $\ell$ on $D^2$ we have \[\lim_{n \rightarrow \infty} \int\!\!\int_{D^{2}} (B^{n} f) (z,w) \ell(z,w) d\mu(z) d\mu(w) = 0. \] Then, using the above property we prove $n$-harmonicity of bounded function which is invariant under the Berezin transform. And we show the same results for the weighted the Berezin transform in the polydisc.
Keywords: Berezin transform, bidisc, harmonic function
MSC numbers: Primary 32A70; Secondary 47G10
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