Abstract
It is known that Beurling’s theorem concerning invariant subspaces is not true in the Bergman space (in contrast to the Hardy space case).However, Aleman, Richter, and Sundberge proved that every cyclic invariantasubspace in the Bergman space Lp(D), 0 < p < +1, is generated by its extremal function. This implies, in particular, that for every zero-based invariantsubspace in the Bergman space the Beurling’s theorem stands true. Here, wecalculate the reproducing kernel of the zero-based invariant subspace Mninathe Bergman space L2(D) where the associated wandering subspace Mnis one-dimensional, and spanned by the unit vector Gn(z) =zMn p n + 1zn
Highlights
IntroductionIt is known that every zero-based invariant subspace is cyclic
Let D denote the open unit disk in the complex plane
We calculate the reproducing kernel of the zero-based invariant subspace Mn in the Bergman space is one-dimensional, L2a(D) where the and spanned by associated wandering subpspace the unit vector Gn(z) = n +
Summary
It is known that every zero-based invariant subspace is cyclic. In the Hardy spaces, by Beurling’s Theorem, every invariant subspace other than the trivial one f0g is generated by an inner function (which is an extremal function in that context). Every invariant subspace of the Hardy space is cyclic. For the Bergman space L2a(D), the Beurling-type Theorem holds true and every invariant subspace M is generated by M zM , that is, M = [M zM ] = [M \ (zM )?]: In [1], the author proved that every zero-based invariant subspace of Lpa(D) is generated by its extremal function. We calculate the reproducing kernel of the wandering subspace Mn zMn of the zero-based invariant subspace Mn in the Bergman space L2a(D)
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