Beurling's theorem

Remarks

Proof idea: $S$ is an isometry and $\bigcap_n S^n M = \set{0}$, so $M \ominus S M$ is not zero. Take $\theta$ of norm $1$ in $M \ominus S M$. Then $\theta \perp S^n \theta$ for $n \ge 1$, which says that all Fourier coefficients of $\abs{\theta^*}^2$ vanish except the one of index $0$; hence $\abs{\theta^*} = 1$ almost everywhere and $\theta$ is inner. A decomposition of $M$ into the orthogonal pieces $S^n (M \ominus S M)$ then shows $M = \theta H^2$.

Example: the functions in $H^2$ that vanish at a point $a \in \mathbb{D}$ form a closed invariant subspace; it equals $\theta H^2$ with $\theta(z) = \frac{z - a}{1 - \bar a z}$.

A consequence: $f \in H^2$ is cyclic for $S$, meaning that the polynomial multiples of $f$ are dense, if and only if $f$ is an outer function.

Going deeper

For a student

Beurling's theorem was among the first complete descriptions of the invariant subspaces of a natural operator on an infinite-dimensional space. Together with the inner-outer factorization, it shows that the invariant subspace generated by $f$ is $I H^2$, where $I$ is the inner factor of $f$. Lax extended it to vector-valued Hardy spaces, which is the Beurling–Lax theorem; the model theory of contractions builds on it. Its dual form describes the invariant subspaces of the backward shift as the model spaces $H^2 \ominus \theta H^2$.