Results 111 to 120 of about 254 (149)

5-ALA induced PpIX fluorescence spectroscopy in neurosurgery: a review. [PDF]

open access: yesFront Neurosci
Gautheron A   +5 more
europepmc   +1 more source

Domination of Berezin Transform

Vietnam Journal of Mathematics, 2014
zbMATH Open Web Interface contents unavailable due to conflicting licenses.
Das, Namita, Sahoo, Madhusmita
openaire   +1 more source

The Berezin Transform and Its Applications

Acta Mathematica Scientia, 2021
zbMATH Open Web Interface contents unavailable due to conflicting licenses.
openaire   +1 more source

Berezin Transform for Solvable Groups

Acta Applicandae Mathematica, 2004
The authors study the Berezin transform for the multi-weighted Bergman spaces \(H^2_\nu(D)=\{h\in L^2(D,d\mu_\nu): h\) holomorphic\} of Gindikin type on homogeneous cones and Siegel domains \(D,\) depending on a vector parameter \(\nu=(\nu_1,\ldots,\nu_r).\) Using an explicit kernel representation for the associated eigen-operators they provide its ...
Arazy, Jonathan, Upmeier, Harald
openaire   +2 more sources

Berezin Transform, Mellin Transform and Toeplitz Operators

Complex Analysis and Operator Theory, 2010
Let \(B\) be the Berezin transform associated with the Bergman space. The authors of the article under review improve a theorem of \textit{P. Ahern} [J. Funct. Anal. 215, No. 1, 206--216 (2004; Zbl 1088.47014)]. Namely, they show that, if \(u\in L^1\) and \(Bu\) is a harmonic function, then \(u\) itself is harmonic.
Čučković, Željko, Li, Bo
openaire   +1 more source

Fixed Points of Poly-Berezin Transforms

Complex Analysis and Operator Theory, 2019
zbMATH Open Web Interface contents unavailable due to conflicting licenses.
Lijia Ding, Wei He
openaire   +2 more sources

Singular Berezin Transforms

Complex Analysis and Operator Theory, 2007
We give examples of pseudoconvex Reinhardt domains where the Berezin transform has integral kernel with singularities and, hence, fails to be a smoothing map. On the other hand, we show that this can never happen for a plane domain – in fact, then the Bergman kernel is always either identically zero or strictly positive everywhere on the diagonal – and
openaire   +2 more sources

Asymptotic expansions of Berezin transforms

Indiana University Mathematics Journal, 2000
Let \(\Omega\) be an irreducible bounded symmetric (i.e. Cartan) domain in its Harish-Chandra realization, so that its Bergman kernel with respect to the normalized Lebesgue measure \(dm\) is equal to \(h(x,y)^ {-p}\), where \(h\) is the Jordan determinant and \(p\) the genus of \(\Omega\). For \(\nu>p-1\), the weighted Bergman spaces \(L^2_{\text{hol}}
Arazy, J., Ørsted, B.
openaire   +3 more sources

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