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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 !

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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, $$\| P_n q P_n \| \leq \sum_{|k|\leq n} \| P_n q(k) P_n\| + \|\sum_{|k|>n} P_n q(k) P_n\| = (2n+1) 2^{-n} + 2^{-n}.$$

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