Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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.

share|improve this question
1  
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

1 Answer 1

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 $\phi H^1$.

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.