# Invariant Subspaces (Hardy Space)

Suppose $M_1$ and $M_2$ are invariant subspaces of the unilateral shift U such that $M_1$ subset $M_2$ and $M_1$ is of codimension strictly larger than $1$ in $M_2$. Show that there exists $M$ invariant under $U$ satisfying $M_1 \subset M \subset M_2$ where the inclusions are strict. All subspaces are closed.

This problem is from the Springer GTM: "An introduction to operators on the Hardy-Hilbert space".

Edit: Perhaps I can take $M := U M_2$? Maybe I should give that some more thought.

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To solve this you can use the function theoretic description of the invariant subspaces of the shift known as Beurling's theorem. Identify your Hilbert space with the Hardy space $H^2$ on the disk such that $U$ is identified with "multiplication by $z$". Beurling's theorem says that each invariant subspace of $U$ has the form $\phi H^2$ for a so-called inner function $\phi$, i.e., a bounded analytic function on the disk whose radial (or non-tangential) limit function has modulus 1 a.e. on the circle.
In your problem, there are inner functions $\phi_1$ and $\phi_2$ such that $M_1=\phi_1 H^2$ and $M_2=\phi_2 H^2$. You have $\phi_1\in M_1\subset M_2$, so $\phi_1=\phi_2 f$ for some $f\in H^2$. The modulus of $f$ on the circle is 1 a.e. because $f=\phi_1/\phi_2$ a.e., and thus $f$ is an inner function. Suppose, for the sake of argument, that you can write $f=gh$ for some nonconstant inner functions $g$ and $h$, and let $M=\phi_2 g H^2$. Then $M_1\subset M\subset M_2$. I claim that the inclusions are strict, and more specifically that $\phi_2 g$ is in $M\setminus M_1$ and $\phi_2$ is in $M_2\setminus M$. This amounts to the same thing as saying that $1/h$ and $1/g$ (respectively) are not in $H^2$. Note that $1/h$ and $1/g$ have modulus 1 a.e. on the circle, so if they were in $H^2$ they would in fact be in $H^\infty$ and bounded by 1 on the disk. But $h$ and $g$ are bounded by 1 on the disk and nonconstant, so this is impossible, showing that in fact $1/h$ and $1/g$ are not in $H^2$ as claimed.
It remains to be seen why $f$ has such a factorization, and this is where the hypothesis about codimension is used. Every inner function has a factorization into a Blaschke product times a singular inner function. If the singular part of $f$ is nontrivial, it can be factored nontrivially by scaling the corresponding singular measure by numbers between 0 and 1. If the Blaschke part of $f$ has more than one factor, then it factors. Given that $M_1\neq M_2$, $f$ is not constant, so the only other possibility is that $f$ is a Blaschke product with a single factor, a.k.a. a holomorphic automorphism of the disk. I claim that this would imply that $M_1$ has codimension 1 in $M_2$, and more specifically $M_2=M_1 + \mathbb{C}\phi_2$. To see this, let $\alpha$ be the zero of $f$, and let $G=\phi_2 H$ be an element of $M_2$. Then $G=\phi_2 f\frac{H-H(\alpha)}{f}+\phi_2 H(\alpha)\in M_1+\mathbb{C}\phi_2$. Q.E.D.
(As for the edit, you can't always take $M=UM_2$. Let $0<|\alpha|<1$, and let $\phi_\alpha$ be the holomorphic automorphism of the disk that swaps 0 and $\alpha$, $\phi_\alpha(z)=\frac{\alpha-z}{1-\overline{\alpha}z}$. Suppose that $M_2=H^2$ and $M_1=\phi_\alpha^2 H^2$. Then your hypotheses are met, but $M_1$ is not contained in $UM_2$ because $\phi_\alpha^2$ is not in $UM_2$.)