Let $H$ be a Hilbert space and ${\mathcal A}$ a von Neumann algebra in $B(H)$. Let us endow ${\mathcal A}$ with the topology of pointwise convergence with the Hermitian adjoint operator, i.e. a net $A_i\in {\mathcal A}$ tends to 0 in ${\mathcal A}$ if
$$
\forall x\in H\qquad A_ix\overset{H}{\underset{i\to\infty}{\longrightarrow}}0 \quad\&\quad
A_i^*x\overset{H}{\underset{i\to\infty}{\longrightarrow}}0.
$$
The algebra ${\mathcal A}$ with this topology has totally bounded sets $S\subseteq {\mathcal A}$, which I believe, are described by the property
$$
\forall x\in H\qquad \text{the sets $\{Ax;\ A\in S\}$ and $\{A^*x;\ A\in S\}$ are totally bounded in $H$}
$$  
Let $u:{\mathcal A}\to{\mathbb C}$ be a linear functional, which is *continuous on every totally bounded set (equivalently, on each compact set) $S\subseteq {\mathcal A}$* in this sense (with respect to this topology induced from ${\mathcal A}$ to $S$).

Question:

> does $u$ belong to ${\mathcal A}_*$ (the predual of ${\mathcal A}$)?

**EDIT.** I must say, even the case of ${\mathcal A}=B(H)$ is not clear for me. And even if we replace the topology on ${\mathcal A}$ with the usual *strong operator topology* (i.e. if we remove the requirement $A^*_ix\to 0$). We also can change premises in other different ways, for example, we can think that ${\mathcal A}$ is commutative (for me that would be sufficient now). If anybody could cast some light on any weakened conditions, I would appreciate this very much. I need this for my current work, this is related to the study of [envelopes of topological algebras][1].


  [1]: https://arxiv.org/abs/1303.2424