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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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