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 :
- $\Vert f_n \Vert_2 =1$
- $\displaystyle \forall \varepsilon >0, \lim_{n \rightarrow \infty} \int_0^\varepsilon |f_n|^2 = 1$.
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 counterexample 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 $\mathrm{ev}_0$ (the sequence $f_n$ is exactly a sequence of vector state converging weakly to the character $\mathrm{ev}_0$, and $\Vert f_n q \Vert_2$ is the evaluation of $q$ on the state corresponding to $f_n$. )
Thanks !