2009
DOI: 10.1002/mana.200810169
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Generalized Hankel operators on the Fock space

Abstract: Key words Hankel operator, Fock space MSC (2000) 47B35, 32A36In this paper we study generalized Hankel operators of the form H l z k :The investigations in this article extend the ones in [11] and [6], where the special cases l = 0 and l = 1 are considered, respectively. The main result is that the operators H l z k are not bounded for l < k − 1. The proof relies on a combinatoric argument and a generalization to general conjugate holomorphic L 2 symbols, generalizing arguments from [6], seems possible and is … Show more

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Cited by 7 publications
(14 citation statements)
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“… Abstract In this paper we study boundedness of generalized Hankel operators of the form \documentclass{article}\usepackage{amssymb,mathrsfs}\begin{document}\pagestyle{empty}${\rm H}_{{\overline{z}}^k}^l: {\mathscr F}^2\big (|z|^2\big )\rightarrow L^2\big (|z|^2\big )$\end{document} and thereby extend our results from 10. Here, \documentclass{article}\usepackage{amssymb}\begin{document}\pagestyle{empty}$\rm H_{{\overline{z}}^k}^l(f):=(\rm Id-\rm P_l)\big (\overline{z}^k f\big )$\end{document} and P l is the projection onto \documentclass{article}\usepackage{amssymb}\begin{document}\pagestyle{empty}$A^2_l\big (\mathbb C,| z|^2\big ):=\rm cl\big (\rm span\lbrace \overline{z} ^mz^n\,|\,m,n\in \mathbb N,\,m\le l\rbrace \big )$\end{document}.…”
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confidence: 54%
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“… Abstract In this paper we study boundedness of generalized Hankel operators of the form \documentclass{article}\usepackage{amssymb,mathrsfs}\begin{document}\pagestyle{empty}${\rm H}_{{\overline{z}}^k}^l: {\mathscr F}^2\big (|z|^2\big )\rightarrow L^2\big (|z|^2\big )$\end{document} and thereby extend our results from 10. Here, \documentclass{article}\usepackage{amssymb}\begin{document}\pagestyle{empty}$\rm H_{{\overline{z}}^k}^l(f):=(\rm Id-\rm P_l)\big (\overline{z}^k f\big )$\end{document} and P l is the projection onto \documentclass{article}\usepackage{amssymb}\begin{document}\pagestyle{empty}$A^2_l\big (\mathbb C,| z|^2\big ):=\rm cl\big (\rm span\lbrace \overline{z} ^mz^n\,|\,m,n\in \mathbb N,\,m\le l\rbrace \big )$\end{document}.…”
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confidence: 54%
“…However, it should be mentioned that the remaining case is much more difficult to prove, as already mentioned in 10. One of the two main purposes of the present paper is to close this gap ( k = l + 1).…”
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confidence: 70%
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