Let $\mathfrak{M}$ be a von Neumann algebra acting on a Hilbert space $\mathcal{H}$, and let $\psi$ be a positive normal linear functional on $\mathfrak{M}$. According to [1, p. 69] the so-called lineal of $\psi$ (or set of $\psi$-bounded vectors) is defined to be $$D(\mathcal{H}, \psi) := \{x \in \mathcal{H} \ : \ \Vert ax \Vert_\mathcal{H} \le C_x \psi(a a^\ast) \ \text{for all $a \in \mathfrak{M}$}\} \ .$$ The book [1] continues with the observation that if there exists a vector $\varPsi \in \mathcal{H}$ such that $\psi(a) = \langle \varPsi, a \varPsi \rangle$ for all $a \in \mathfrak{M}$, that is, if $\psi$ is a vector state, then the lineal of $\psi$ takes the form $$D(\mathcal{H}, \psi) = \mathfrak{M}^\prime \varPsi = \{a^\prime \varPsi \ : \ a^\prime \in \mathfrak{M}^\prime\} \ ,$$ where $\mathfrak{M}^\prime$ denotes the commutant of $\mathfrak{M}$.
I have two questions about this:
Q1: Is the definition of $D(\mathcal{H}, \psi)$ given above indeed the most useful one? To me, it would seem more natural to set $D(\mathcal{H}, \psi) := \{x \in \mathcal{H} \ : \ \Vert ax \Vert_\mathcal{H}^2 \le C_x \psi(a^\ast a) \ \text{for all $a \in \mathfrak{M}$}\}$.
Q2: How can one prove the assertion that $D(\mathcal{H}, \psi) = \mathfrak{M}^\prime \varPsi$ for a vector state $\psi$? It is clear to me that $\mathfrak{M}^\prime \varPsi \subset D(\mathcal{H}, \psi)$, but I have trouble showing the reverse inclusion $D(\mathcal{H}, \psi) \subset \mathfrak{M}^\prime \varPsi$.
(Regarding the second set inclusion in Q2, my initial (maybe naive) idea was trying to show that $x \in D(\mathcal{H}, \psi)$ can be written as $\sqrt{C_x} \varPsi$, where $C_x > 0$ is as in the definition of the lineal of $\psi$.)
I would greatly appreciate any insights!
[1] Ohya, M. and Petz, D. Quantum Entropy and Its Use. Corrected Second Printing. Springer, Berlin, Heidelberg, 2004.
Here's an outline how to do this. I'll let you check some of the detail.
Let $x\in D(\mathcal{H}, \psi)$ with $\psi(a) = \langle \varPsi, a \varPsi \rangle$. So $\forall a\in\mathfrak{M}: \Vert ax\Vert \le \sqrt{C_x}\Vert a\varPsi\Vert$.
Let $$H_\varPsi = \overline{\{a\varPsi: a\in \mathfrak{M}\}}$$ $$H_x = \overline{\{ax: a\in \mathfrak{M}\}}$$ Then $H_\varPsi, H_x$ are closed subspaces of $\mathcal{H}$. Since $\mathfrak{M}$ contains bounded operators, it is a routine matter to check that $H_\varPsi, H_x$ are $\mathfrak{M}$-invariant subspaces. Moreover, since $\mathfrak{M}$ contains the adjoint of any member of it - the orthogonal complements $H_\varPsi^\perp, H_x^\perp$ are also $\mathfrak{M}$-invariant. ($\varPsi, x$ are called cyclic vectors for $H_\varPsi, H_x$ respectively.)
Now define a linear operator $T$ on $\{a\varPsi: a\in \mathfrak{M}\}$ by $$ T(a\varPsi) = ax\in H_x$$
For start we need to check that $T$ is well-defined on this subspace. If $a\varPsi=0$ then $ax=0$ because of the inequality $\Vert ax\Vert \le \sqrt{C_x}\Vert a\varPsi\Vert$. I'll leave it to you to check that $T$ is a linear map and so it is well defined. Moreover we see that $\Vert T\Vert \le \sqrt{C_x}$ on the subspace $\{a\varPsi: a\in \mathfrak{M}\}$.
Since $T$ is a bounded linear operator, it may be extended uniquely (via Cauchy sequences) to $H_\varPsi$, and it still maps into $H_x$ on this larger domain because $H_x$ is closed.
Finally we extend $T$ to $\mathcal{H}$ by defining $T(v)=0$ for $v\in H_\varPsi^\perp$.
For $\tilde{a}\in \mathfrak{M}$ and $v\in H_\varPsi^\perp$, $\tilde{a}v\in H_\varPsi^\perp$, so $Tv = 0, \tilde{a}Tv = 0, T\tilde{a}v = 0$. In other words $\tilde{a}$ commutes with $T$ on $H_\varPsi^\perp$.
I'll leave it to you to show that $\tilde{a}$ commutes with $T$ on $H_\varPsi$ (starting from the subspace $\{a\varPsi: a\in \mathfrak{M}\}$ of course.)
So... $T\in \mathfrak{M}^\prime$ and ... $T(\varPsi) = x$. $\square$
PS: The spaces $H_\varPsi, H_x$ are isomorphic to the spaces constructed by the GNS construction from the vector states associated with $\varPsi, x$.