\(L^1\) factorizations for some perturbations of the unilateral shift (Q2708183)
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scientific article
| Language | Label | Description | Also known as |
|---|---|---|---|
| English | \(L^1\) factorizations for some perturbations of the unilateral shift |
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Statements
8 October 2002
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\(L^1\)-factorization
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absolutely continuous contraction
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Bourgain's solution of the Douglas and Rudin problem
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Fourier coefficients
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unilateral shift
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Hardy space
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\(L^1\) factorizations for some perturbations of the unilateral shift (English)
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It is said that an operator \(T\) on a Hilbert space \({\mathcal H}\) factorizes an integrable function \(f\in L^1(\mathbb{T})\), if there exist \(x,y\in{\mathcal H}\) such that the Fourier coefficients of \(f\) are equal to \(\langle T^n x,y\rangle\) where it is assumed that \(T^n= (T^*)^{|n|}\) whenever \(n<0\). Let \(S\) denote the unilateral shift on the Hardy space \(H^2\), \(S(f)(z)= zf(z)\).NEWLINENEWLINENEWLINEThe following theorem is proved: If \(T= S\oplus A\), where \(A\in{\mathcal L}({\mathcal H})\) factorizes some function \(f_0\in L^\infty(\mathbb{T})\) such that \(1/f_0\) is also in \(L^\infty(\mathbb{T})\), then \(T\) factorizes all functions \(f\in L^1(\mathbb{T})\). The proof essentially depends on \textit{J. Bourgain's} solution of the Douglas and Rudin problem on factorization of a bounded function into the classes \(H^\infty\), \(\overline H^\infty\) [Pac. J. Math. 121, 47-50 (1986; Zbl 0609.30035)]. As a corollary it is obtained that if \(A\) has an eigenvalue inside the unit disk, then \(T\) factorizes all functions from \(L^1(\mathbb{T})\).
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