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Consider the following fragment from Takesaki's book "Theory of operator algebras", chapter IX Non-commutative integration, Section 3 on p187-188:

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I have trouble understanding the equality $$\mathfrak{D}(\mathfrak{H}, \psi) = \mathscr{L}(L^2(\mathcal{N})_{\mathcal{N}}, \mathfrak{H}_{\mathcal{N}})\mathfrak{B}_\psi.$$

For instance, let $x \in \mathscr{N}_\psi$ (i.e. $\psi(x^*x)< \infty$) and $a\in \mathscr{L}(L^2(\mathcal{N})_{\mathcal{N}}, \mathfrak{H}_{\mathcal{N}})$. Then we should be able to prove that $$a \Lambda_\psi(x)\in \mathfrak{D}(\mathfrak{H}, \psi)$$ but this is not clear to me. Could someone explain? Thanks in advance!

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  • $\begingroup$ I don’t know enough modular theory to say for sure, but I believe it should be the case that $\eta_\psi(x)y = x\eta’_\psi(y)$, for any $x \in \mathfrak{n}_\psi$ and $y \in \mathfrak{n}_\psi^\ast$, from which the result quickly follows. $\endgroup$
    – David Gao
    Commented Jun 15 at 17:22
  • $\begingroup$ @DavidGao Thanks for your comment. Could you explain why this equality holds? I don't quite understand it. $\endgroup$
    – Andromeda
    Commented Jun 16 at 21:18

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Again, I’m not really familiar with modular theory, so I can’t really do this fully rigorously. However, since this question still has no answer at this point and this is a bit too long for a comment, I’m writing an answer here to present my understanding of how this probably should work. I advise the OP to take my answer with a grain of salt, and mostly treat this as a proof idea to be filled in with details by themself.

For $x, y$ satisfying suitable assumptions - presumably $x \in \mathfrak{n}_\psi$, $y \in \mathfrak{n}_\psi^\ast$, and both elements are entire (i.e., $\sigma_\alpha(x)$ and $\sigma_\alpha(y)$ are well-defined for all $\alpha \in \mathbb{C}$), we have,

$$\begin{split} \eta_\psi(x)y &= Jy^\ast J\eta_\psi(x)\\ &= Jy^\ast S\Delta^{-\frac{1}{2}}\eta_\psi(x)\\ &= Jy^\ast S\eta_\psi(\sigma_{\frac{1}{2}i}(x))\\ &= Jy^\ast \eta_\psi(\sigma_{\frac{1}{2}i}(x)^\ast)\\ &= S\Delta^{-\frac{1}{2}}\eta_\psi(y^\ast \sigma_{-\frac{1}{2}i}(x^\ast))\\ &= S\eta_\psi(\sigma_{\frac{1}{2}i}(y^\ast \sigma_{-\frac{1}{2}i}(x^\ast)))\\ &= S\eta_\psi(\sigma_{\frac{1}{2}i}(y^\ast)x^\ast)\\ &= \eta_\psi(x\sigma_{\frac{1}{2}i}(y^\ast)^\ast)\\ &= xS\eta_\psi(\sigma_{\frac{1}{2}i}(y^\ast))\\ &= xJ\Delta^{\frac{1}{2}}\eta_\psi(\sigma_{\frac{1}{2}i}(y^\ast))\\ &= xJ\eta_\psi(y^\ast)\\ &= x\eta_\psi’(y) \end{split}$$

Using some approximation argument (presumably weak$^\ast$ density of entire elements and the Kaplansky density theorem), one should be able to deduce then that $\eta_\psi(x)y = x\eta_\psi’(y)$ for all $x \in \mathfrak{n}_\psi$, $y \in \mathfrak{n}_\psi^\ast$. The desired result immediately follows.

(Heuristically, $\eta_\psi(x)y = x\eta_\psi’(y)$ is just saying that $(x\psi^{\frac{1}{2}})y = x(\psi^{\frac{1}{2}}y)$.)

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    $\begingroup$ In this context it is convenient to work with the approximants $x_n=\sqrt{n/\pi}\int_{-\infty}^\infty e^{-nt^2}\sigma^\phi_t(x)\,dt$. These elements are entire analytic and converge to $x$ in all the relevant topologies (i.e. in the strong$^\ast$ topology, but also $\Lambda_\psi(x_n)\to \Lambda_\psi(x)$ if $x\in \mathfrak n_\psi$ etc.) $\endgroup$
    – MaoWao
    Commented Jun 18 at 12:07

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