# invariant subspace of a Hardy space

Let $T$ be the unit circle and $H^1=\{f\in L^1(T): \int_0^{2\pi} f(e^{it})\chi_n(e^{it})dt=0 \text{ for } n>0\}$ where $\chi_n(e^{it})=e^{int}$. Let $M$ be a closed subspace of $H^1$. Then $\chi_1 M\subset M$ if and only if $M=\phi H^1$ for some inner function $\phi$.

We say $\psi$ is an inner function if $\psi\in H^\infty$ and $|\psi|=1$ a.e.

This is a problem from Banach algebra Techniques in Operator theory by Ronald Douglas.

I was able to show that if $M=\phi H^1$ for some inner function $\phi$ then $\chi_1M\subset M$. For the other direction, I tried going through Beurling's theorem but I get stuck.

I also tried writing $M$ as $M_1M_2$ where $M_1$ and $M_2$ are both subsets of $H^2$ but that got me nowhere.

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i posted what an inner function is –  john Dec 31 '12 at 21:49
Here's an idea; not guaranteed to work. As you know, the space $H^1$ can be identified with a certain space of holomorphic functions on unit disk $\mathbb D$. For $0<r<1$ consider the set $M_r=\{f(rz): f\in M\}$. This is a subspace of $H^2$, which is shift-invariant but not closed. Take its closure; still shift-invariant. Now Beurling's theorem applies and gives you a certain inner function $\phi_r$. Try to figure out what happens as $r\to 1$. –  user53153 Jan 1 '13 at 6:19
In Arveson's book A short course on spectral theory, there is a an exercise where we have to prove the following: Let $M\subset L^2$ a closed nonzero vector subspace, $U\colon L^2\to L^2$ an unitary operator such that $\bigcap_{n>0} U^nM=\{0\}$. There exists an inner function $v$ such that $M=vH^2$. Maybe I can write the step of the proof of the theorem. –  Davide Giraudo Jan 1 '13 at 11:33

An idea: maybe you can show that if $x_1M \subset M$ then $(x_1M \cap H^2) \subset (M\cap H^2)$ (+ remaining a CLOSED subspace, which I have no idea how to show). Once there, you could apply Beurling's Thm and you know that $M \cap H^2$ is of the form $\phi H^2$. By density of $H^2$ as a subspace of $H^1$, you can probably show that the space $M$ is also of the form $\phi H^1$.