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

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 $\vert\vert\zeta ^{n}f\vert\vert \to 0$ 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 $\vert\vert\frac{\zeta -\lambda }{1-\overline{\lambda }\zeta }f\vert\vert\ge c\vert\vert f\vert\vert$, then $\Delta (\mathcal{H}) =\Sigma (\mathcal{H})$ a.e. and the following four conditions are equivalent:

(1) $\vert\vert\zeta ^{n} f\vert\vert\nrightarrow 0$ for some $f \in \mathcal{H}$,

(2) $\vert\vert\zeta ^{n} f\vert\vert\nrightarrow 0$ 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 $\mathcal{M}/\zeta \mathcal{M} =1$.