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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:

is $u$ a restriction on ${\mathcal A}$ of some functional $v\in B(H)_*$ (from the predual of $B(H)$)?

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.

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.

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:

is $u$ a restriction on ${\mathcal A}$ of some functional $v\in B(H)_*$ (from the predual of $B(H)$)?

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:

is $u$ a restriction on ${\mathcal A}$ of some functional $v\in B(H)_*$ (from the predual of $B(H)$)?

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.

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 this mean that $u$ is a restriction on ${\mathcal A}$ of some functional $v\in B(H)_*$ (from the predual of $B(H)$)?

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:

is $u$ a restriction on ${\mathcal A}$ of some functional $v\in B(H)_*$ (from the predual of $B(H)$)?