Ideal spaces of measurable operators affiliated to a semifinite von Neumann algebra. II

Let \(\tau \) $\tau$ be a faithful semifinite normal trace on a von Neumann algebra \(\mathcal {M}\) $\mathcal{M}$ , let \(S(\mathcal {M}, \tau )\) $S(\mathcal{M},\tau )$ be the \({}^*\) ${}^{\ast }$ -algebra of all \(\tau \) $\tau$ -measurable operators. Let \(\mu (t; X)\) $\mu (t;X)$ be the generalized singular value function of the operator \(X \in S(\mathcal {M}, \tau )\) $X\in S(\mathcal{M},\tau )$ . If \(\mathcal {E}\) $\mathcal{E}$ is a normed ideal space (NIS) on \((\mathcal {M}, \tau )\) $(\mathcal{M},\tau )$ , then * \(\begin{aligned} \Vert A\Vert _\mathcal {E}\le \Vert A+\textrm{i} B\Vert _\mathcal {E} \end{aligned}\) $$\begin{array}{c}\hfill {\Vert A\Vert }_{\mathcal{E}}\le {\Vert A+\text{i}B\Vert }_{\mathcal{E}}\end{array}$$ for all self-adjoint operators \(A, B \in \mathcal {E}\) $A,B\in \mathcal{E}$ . In particular, if \(A, B \in (L_1+L_{\infty })(\mathcal {M}, \tau )\) $A,B\in (L_1+L_{\infty })(\mathcal{M},\tau )$ are self-adjoint, then we have the (Hardy–Littlewood–Pólya) weak submajorization, \(A \preceq _w A+\textrm{i}B\) $A{⪯}_{w}A+\text{i}B$ . Inequality \((*)\) $(\ast )$ cannot be extended to the Shatten–von Neumann ideals \(\mathfrak {S}_p\) ${\mathfrak{S}}_p$ , \( 0< p <1\) $0<p<1$ . Hence, the well-known inequality \( \mu (t; A) \le \mu (t; A+\textrm{i} B)\) $\mu (t;A)\le \mu (t;A+\text{i}B)$ for all \(t>0\) $t>0$ , positive \(A \in S(\mathcal {M}, \tau )\) $A\in S(\mathcal{M},\tau )$ and self-adjoint \( B \in S(\mathcal {M}, \tau )\) $B\in S(\mathcal{M},\tau )$ cannot be extended to all self-adjoint operators \(A, B \in S(\mathcal {M}, \tau )\) $A,B\in S(\mathcal{M},\tau )$ . Consider self-adjoint operators \(X, Y\in S(\mathcal {M}, \tau )\) $X,Y\in S(\mathcal{M},\tau )$ , let K(X) be the Cayley transform of X. Then, \(\mu (t; K(X)-K(Y))\le 2 \mu (t; X-Y)\) $\mu (t;K(X)-K(Y))\le 2\mu (t;X-Y)$ for all \(t>0\) $t>0$ . If \(\mathcal {E}\) $\mathcal{E}$ is an F-NIS on \((\mathcal {M}, \tau )\) $(\mathcal{M},\tau )$ and \(X-Y\in \mathcal {E}\) $X-Y\in \mathcal{E}$ , then \(K(X)-K(Y)\in \mathcal {E}\) $K(X)-K(Y)\in \mathcal{E}$ and \(\Vert K(X)-K(Y)\Vert _\mathcal {E}\le 2 \Vert X-Y\Vert _\mathcal {E}\) ${\Vert K(X)-K(Y)\Vert }_{\mathcal{E}}\le 2{\Vert X-Y\Vert }_{\mathcal{E}}$ .