Im stuck on the following part of this theorem:
The closed vector subspaces of $L^2(T)$ invariant for the bilateral shift $v=M_{z}$ (for $z: T \longrightarrow \mathbb{C}$ the inclusion map) are precisely the Weiner and Beurling spaces. If $K$ is a closed invariant subspace for $v$, then:
- $v(K)=K$ if and only if $K$ is Weiner.
- $v(K)=K$ if and only if $K$ is Beurling.
Proof: (proving 2)) Suppose that $v(K) \subset K$ properly. Then since $v(K)$ is closed, there exists a unit $\varphi \in K$ such that $\varphi \in v(K)^{\perp}$. Since $v^n(\varphi) \in v(K)$ for all $n>0$, it follows that: $$0 = \langle v^n(\varphi), \varphi\rangle = \int_{T} z^n|\varphi(z)|^2 \: d\lambda(z).$$ Since $T$ (the unit circle) is a finite measure space, $L^1(T) \subset L^2(T)$, it follows from above that $|\varphi|^2 \in H^1$, implying that $|\varphi|^2 = \alpha$ almost everywhere. However $||\varphi||_2=1$, it follows that $\alpha=1$. Hence, $\varphi \in L^{\infty}(T)$ is a unitary, and clearly $\varphi H^2 \subset K$...
Question: The text tells me this inclusion is clear, but I don't see it. Why is $\varphi H^2 \subset K$? I was trying to show that $\varphi f \in v(K)$ by showing that $$\langle \varphi f,g\rangle = 0 \: \: \forall g \in v(K)^{\perp}, \: f \in H^2$$ but I wasn't able to make any progress. Any suggestions wiuld be appreciated. Thank you.
If $e_n=z^n$, for $n\geq0$, is the standard basis for $H^2$, then $$ \varphi e_n = v^n(\varphi) \in K. $$