5
$\begingroup$

Let $H$ be a separable Hilbert space over $\mathbb{C}$, say $\ell_2$ for simplicity. Let $\mathcal{K}(H)$ denote the space of all compact operators on $H$ and $\mathcal{P}(H)$ the set of all finite rank orthogonal projections on $H$ (so $\mathcal{P}(H)\subset\mathcal{K}(H)$). Assume that $(x_n^*)_{n\in\mathbb{N}}$ is a sequence of bounded functionals on $\mathcal{K}(H)$ such that $\lim_{n\to\infty}x_n^*(P)=0$ for every $P\in\mathcal{P}(H)$.

Question: Is it true that $\lim_{n\to\infty}x_n^*(T)=0$ for every $T\in\mathcal{K}(H)$?

Equivalently, is $(x_n^*)_{n\in\mathbb{N}}$ norm bounded? Such a situation holds e.g. in von Neumann algebras (a result due to Darst '67) or C*-algebras of the form $C(K)$ where $K$ is the Stone space of a $\sigma$-complete Boolean algebra (Nikodym '33).

References:

R.B. Darst, On a theorem of Nikodym with applications to weak convergence and von Neumann algebras, Pacific J. Math. 23 (1967), no. 3, 473–477.

O. Nikodym, Sur les familles bornées de fonctions parfaitement additives d’ensemble abstrait, Monatsh. Math. Phys. 40 (1933), no. 1, 418–426.

$\endgroup$
15
  • 2
    $\begingroup$ Why compact projections? This is a confusing way of writing projections onto finite-dimensional subspaces. $\endgroup$ Jun 11 '16 at 21:19
  • $\begingroup$ [deleted previous comment; I missed the fact that we are not assuming $\sup_n \Vert x_n^*\Vert < \infty$ ] $\endgroup$
    – Yemon Choi
    Jun 11 '16 at 21:27
  • $\begingroup$ @DavidHandelman: I just mean only the projections which belong to $\mathcal{K}(H)$. $\endgroup$ Jun 11 '16 at 21:29
  • 1
    $\begingroup$ @NateEldredge: every element of a C*-algebra is a linear combination of two self-adjoint elements, so the set of self-adjoint operators is not a subspace (or, it is, but over $\mathbb{R}$, but I am concerned with the complex case). Hence, what you say is wrong. And in the case of finite dimensional Hilbert spaces, every operator is compact, so $\mathcal{K}(H)=\mathcal{B}(H)$ satisfies the demanded condition, since it is a von Neumann algebra. $\endgroup$ Jun 12 '16 at 17:09
  • 1
    $\begingroup$ As such, I will be very happy if the answer to your question is yes, but I am more inclined to suspect it will turn out to be no. $\endgroup$ Jun 12 '16 at 17:46
3
$\begingroup$

This is false. The dual $K(H)^*=B_1(H)$ is given by the trace class operators, and an $S\in B_1(H)$ acts on a $T\in K(H)$ by $(S,T)=\textrm{tr}\, ST$. Consider now $$ S_n = n2^{-n} \sum_{2^n\le j<2^{n+1}} \langle e_j , \cdot \rangle e_j . $$ The norm of $S_n$ as a functional is its trace norm, which equals $n$, so this sequence is unbounded. However, your condition holds: it suffices to check this for a rank one projection $T=\langle v, \cdot \rangle v$, and then $$ (S_n,T) = n2^{-n}\sum |\langle e_j, v\rangle |^2 \le n2^{-n}\to 0 , $$ as required.

$\endgroup$

Your Answer

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

Not the answer you're looking for? Browse other questions tagged or ask your own question.