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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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up vote 2 down vote accepted

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.

For more details on factorizations of inner functions and more, see Chapter 2 of the book named in the question.

(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 $UM_2$.)

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