Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

Hello !

I Would like to know if the following is true :

Let $\mathcal{H}$ be the complex hilbert space $L^2([0,1])$ for the Lebesgue measure. Let $q$ be the orthogonal projection on the subspace of $\mathcal{H}$ spammed by the $(exp(\pm 2 i \pi 2^k x ))_{k \in \mathbb{N}}$.

Let $f_n$ be a sequence of elements of $\mathcal{H}$ such that :

  • $\Vert f_n \Vert_2 =1$
  • $\displaystyle \forall \epsilon >0, \lim_{n \rightarrow \infty} \int_0^\epsilon |f_n|^2 = 1$.

Then :

$\lim_{n \rightarrow \infty} \Vert q f_n \Vert_2 = 0$

In the first place I thought it will be false, but I haven't been able to found a counterexemple and in second thought It is reasonable because functions in the image of $q$ have quasi-periodicity property, and hence it shouldn't be possible to produce of sequence with the desired properties inside the image of $q$.

The motivation for the question lie in the paper of F.W. Shultz "Pure states as a dual object for C-algebras". In the end of this paper, the author give an exemple of a non perfect type one $C^ * $ algebra : the $C^* $ algebra generated in $B(\mathcal{H})$ by the compact operator and $\mathcal{C}([0,1])$ acting on $\mathcal{H}$ by multiplication. The pure state of this algebra are :

  • The vectors state from the action on $\mathcal{H}$
  • the character of $\mathcal{C}([0,1])$ extended to $C$ by sending all the compact operator to $0$.

Hence the atomic part of the enveloping algebra is $B(\mathcal{H}) \oplus l^{\infty}([0,1])$ (where $l^2$ is for the counting measure on the discret set.) and (if I'm understanding well) the author affirm without proof that $(q,0)$, is continuous on the stat space. I'm not convinced by this fact, and I would like to understand this point better, and my question is (if I'm not mistaken) equivalent to this continuity at the character $ev_0$ (the sequence $f_n$ is exactly a sequence of vector state converging weakly to the character $ev_0$, and $\Vert f_n q \Vert_2$ is the evaluation of $q$ on the state corresponding to $f_n$. )

Thanks !

share|improve this question

1 Answer 1

up vote 4 down vote accepted

It is true. More precisely, if $P_n$ is the orthogonal projection on the space of functions supported in $[0,1/2^n]$, I claim that $\| P_n q P_n \| \leq (2n+2) 2^{-n}$. This implies what you are asking for.

Here is a proof. For $k \in \mathbb Z$, let $e_k$ be the function $e^{sgn(k) 2i\pi 2^{|k|}}$ and $q(k)$ the orthogonal projection on $\mathbb C e_k$, so that $q=\sum_k q(k)$. $P_n e_k$ has norm $2^{n/2}$, and the family $(P_n e_k)_{|k|>n}$ is orthogonal. Hence, $$\| P_n q P_n \| \leq \sum_{|k|\leq n} \| P_n q(k) P_n\| + \|\sum_{|k|>n} P_n q(k) P_n\| = (2n+1) 2^{-n} + 2^{-n}.$$

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.