The question is below:
$U$ is the unilateral shift (i.e. right shift) on Hardy-Hilbert space $\mathbb{H}^2$, and $A$ is a bounded operator on $\mathbb{H}^2$ such that
$Lat\, U\subset Lat\, A$
where the $Lat\, X$ denote the set of all invariant subspace of $X$. Show that AU=UA
(This is an exercise in GTM 237 Chapter 2 (exercise 2.2).)
Here is my thoughts: suppose $f\in \mathbb{H}^2$, then $ AUf(z)=A(zf(z))=zg(z) $ for some $g(z)\in \mathbb{H}^2$.
Since $U$ acts as multiply by $z$ and $z\mathbb{H}^2$ is an invariant space of $U$, we can define an operator B:f to g Hence $AU=UB$, it is sufficient to prove that $A=B$. But this doesn't use the condition $Lat\, U\subset Lat\, A$ fully, just $z\mathbb{H}^2\in Lat\, A$. So I feel it is far away from the answer... Besides, I tried some other methods such as Bureling’s theorem and the inner-outer function decomposition but got nothing.
Any help or hint? Thank you!
PS: there is a hint on this book: Consider Adjoint.