2
$\begingroup$

Can someone please help me to answer the following problem?

Let $(e_k)$ be a total orthonormal sequence in a separable Hilbert space $H$ and let $T: H \to H$ be defined at $e_k$ by: $T(e_k) = e_{k+1} , k = 1, 2, \dots$ and then linearly and continuously extended to $H$. Find invariant subspaces, with justification.

This is my answer: $X_n=\operatorname{span}\{e_n,e_{n+1},\dots\}$ are invariant subspaces. But I don't know how to prove that they are the only invariant subspaces. Can someone help me?

$\endgroup$
7
  • $\begingroup$ Maybe it will help to observe that $X$ is invariant for $T$ iff $X^\perp$ is invariant for $T^*$, which is the right shift operator $T^*(e_k) = e_{k-1}$. $\endgroup$ Apr 26 '15 at 16:45
  • $\begingroup$ how it will help ? $\endgroup$
    – user172892
    Apr 26 '15 at 16:50
  • $\begingroup$ I don't know. That's why I said "maybe". $\endgroup$ Apr 26 '15 at 16:56
  • 1
    $\begingroup$ Do not try to blank out your post. $\endgroup$
    – user147263
    Apr 26 '15 at 17:57
  • 1
    $\begingroup$ Please do not change your question after it has received answers. $\endgroup$ Apr 26 '15 at 18:14
2
$\begingroup$

Your Operator is actually the unilateral forward shift Operator on the Hardy Space-$H^2(\mathbb{D})$. So By Beurling's Theorem all invariant subspace are given by $X = \phi H^2(\mathbb{D})$, where $\phi$ is an inner function. In detail See the follwing:

Your Operator is unitarily equivalent to the unilateral forward shift Operator on the Hardy Space-$H^2(\mathbb{D})$. If we define

$U: H^2(\mathbb{D})\rightarrow H$ by $U(z^n) = e_{n+1}$, for all $n \geq 0$

and extend it Linearly. Then $U$ is unitary operator intertwining $T$ and the forward shift operator $M_z$ on $H^2(\mathbb{D})$ i.e. $TU = UM_z$. Now By Beurling's Theorem all invariant subspace $W$ for $M_z$ are given by $W = \phi H^2(\mathbb{D})$, where $\phi$ is an inner function. And hence all invariant subspace for $T$ is given by $U(W)$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy