Convergence of functionals on compact projections on a separable Hilbert space

Question

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.

Answer 1

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.