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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.

Damian Sobota
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    Why compact projections? This is a confusing way of writing projections onto finite-dimensional subspaces. – David Handelman Jun 11 '16 at 21:19
  • [deleted previous comment; I missed the fact that we are not assuming $\sup_n \Vert x_n^*\Vert < \infty$ ] – Yemon Choi Jun 11 '16 at 21:27
  • @DavidHandelman: I just mean only the projections which belong to $\mathcal{K}(H)$. – Damian Sobota Jun 11 '16 at 21:29
  • But aren't those precisely the projections whose images are finite dimensional? – Nate Eldredge Jun 12 '16 at 01:40
  • @NateEldredge: why do you think this is equivalent? Finite rank operators are just dense in $\mathcal{K}(H)$, so you need to take the closure to obtain entire $\mathcal{K}(H)$. Hence, actually, this is the same problem as going from finite rank projections to all compact operators. Or I miss something? – Damian Sobota Jun 12 '16 at 10:26
  • And yes, the compact projections are precisely those which are finite rank. – Damian Sobota Jun 12 '16 at 10:44
  • Does "projection" here mean orthogonal projection? – Nate Eldredge Jun 12 '16 at 15:41
  • @NateEldredge: Yes, of course! I meant orthogonal projections. I have edited the post. – Damian Sobota Jun 12 '16 at 16:10
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    @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. – Damian Sobota Jun 12 '16 at 17:09
  • Oh, we are over $\mathbb{C}$. Ok, back to the drawing board. – Nate Eldredge Jun 12 '16 at 17:13
  • Ok, let me try again with my comment from yesterday. If the answer to your question is yes, then the answer to my (currently unresolved) Q2 here is no. Let $X = \mathcal{K}(H)$ and let $E \subset X$ be the space of finite rank operators. Let ${x_n^} \subset X^$ be a sequence of bounded functionals, and suppose $x_n^(T) \to 0$ for all $T \in E$. In particular, $x_n^(P) \to 0$ for all compact projections $P$ (since they have finite rank). (cont'd) – Nate Eldredge Jun 12 '16 at 17:42
  • If the answer to your question is yes, then we may conclude that $x_n^(T) \to 0$ for every $T \in X$. Since ${x_n^}$ was arbitrary, in the terminology of my question, we have that $E$ determines weak-* convergence. Yet $E$ is meager in $X$, since the set $E_n$ of operators having rank at most $n$ is closed and nowhere dense, and $E = \bigcup_n E_n$. My question Q2 was whether a subspace determines weak-* convergence iff it is nonmeager, so the answer to Q2 would be no. – Nate Eldredge Jun 12 '16 at 17:45
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    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. – Nate Eldredge Jun 12 '16 at 17:46
  • Would you be able to give the full citations to the papers of Darst and Nikodym? I am wondering if they might be relevant to my other question. – Nate Eldredge Jun 12 '16 at 18:29
  • @NateEldredge: Done. In fact, concerning your problem Q2 the literature is quite vast, I think. I'll contact you later via mail (at unco.edu). – Damian Sobota Jun 12 '16 at 18:44

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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.

Christian Remling
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