Analytic contractions, nontangential limits, and the index of invariant subspaces

Abstract:

Let $\mathcal {H}$ be a Hilbert space of analytic functions on the open unit disc $\mathbb {D}$ such that the operator $M_{\zeta }$ of multiplication with the identity function $\zeta$ defines a contraction operator. In terms of the reproducing kernel for $\mathcal {H}$ we will characterize the largest set $\Delta (\mathcal {H}) \subseteq \partial \mathbb {D}$ such that for each $f, g \in \mathcal {H}$, $g \ne 0$ the meromorphic function $f/g$ has nontangential limits a.e. on $\Delta (\mathcal {H})$. We will see that the question of whether or not $\Delta (\mathcal {H})$ has linear Lebesgue measure 0 is related to questions concerning the invariant subspace structure of $M_{\zeta }$. We further associate with $\mathcal {H}$ a second set $\Sigma (\mathcal {H}) \subseteq \partial \mathbb {D}$, which is defined in terms of the norm on $\mathcal {H}$. For example, $\Sigma (\mathcal {H})$ has the property that $||\zeta ^{n}f|| \to 0$ for all $f \in \mathcal {H}$ if and only if $\Sigma (\mathcal {H})$ has linear Lebesgue measure 0. It turns out that $\Delta (\mathcal {H}) \subseteq \Sigma (\mathcal {H})$ a.e., by which we mean that $\Delta (\mathcal {H}) \setminus \Sigma (\mathcal {H})$ has linear Lebesgue measure 0. We will study conditions that imply that $\Delta (\mathcal {H}) = \Sigma (\mathcal {H})$ a.e.. As one corollary to our results we will show that if dim $\mathcal {H}/\zeta \mathcal {H} =1$ and if there is a $c>0$ such that for all $f \in \mathcal {H}$ and all $\lambda \in \mathbb {D}$ we have $||\frac {\zeta -\lambda }{1-\overline {\lambda }\zeta }f||\ge c||f||$, then $\Delta (\mathcal {H}) =\Sigma (\mathcal {H})$ a.e. and the following four conditions are equivalent: (1) $||\zeta ^{n} f||\nrightarrow 0$ for some $f \in \mathcal {H}$, (2) $||\zeta ^{n} f||\nrightarrow 0$ for all $f \in \mathcal {H}$, $f \ne 0$, (3) $\Delta (\mathcal {H})$ has nonzero Lebesgue measure, (4) every nonzero invariant subspace $\mathcal {M}$ of $M_{\zeta }$ has index 1, i.e., satisfies dim $\mathcal {M}/\zeta \mathcal {M} =1$.