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Products of Volterra type operator and composition operator from \(H^\infty \)and Bloch spaces to Zygmund spaces. (English) Zbl 1145.47022
The Zygmund space \(\mathcal{Z}\) is the set of all analytic functions \(f\) on the unit disc \(\mathbb D\) such that \(\| f\| _{\mathcal{Z}}=| f(0)| +| f'(0)| +\sup_z(1-| z| ^2)| f''(z)| <+\infty,\) endowed with such a norm. If \(\lim_{| z| \to 1}(1-| z| ^2)| f''(z)| =0,\) then \(f\) is said to belong to the little Zygmund space \(\mathcal{Z}_0.\) Given an analytic function \(g\in H(\mathbb D),\) two types of Volterra integral operator are defined according to \(J_g(f)(z)=\int_0^z fg'\) and \(I_g(f)(z)=\int_0^z f'g\) for \(f\in H(\mathbb D).\) The authors consider \(\mathcal{Z}\) and \( \mathcal{Z}_0\) valued compositions of these operators and composition operators \(C_\varphi\) whose symbol \(\varphi\) is an analytic self-map of \(\mathbb D.\) They compare their boundedness or compactness regarding the domain space, specifically, \(H^\infty,\) the Bloch space \(\mathcal{B}, \) or the little Bloch space \(\mathcal{B}_0 .\) As a sample of the paper results, let us quote the following (Theorem 1): Set \(T=C_\varphi \circ I_g.\) Then \(T:H^\infty \to \mathcal{Z}\) is bounded if and only if \(T:\mathcal{B} \to \mathcal{Z}\) is bounded if and only if \(T:\mathcal{B}_0\to \mathcal{Z}\) is bounded. Similarly, for the compact case.

MSC:
47B33 Linear composition operators
47B38 Linear operators on function spaces (general)
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