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Dec 14, 2020 at 18:09 vote accept CommunityBot
Dec 8, 2020 at 12:52 comment added Matthew Daws there is some vector $S\alpha \in K$ and that $\alpha \mapsto S\alpha$ is linear and bounded, thus defining $S\in\mathcal B(H,K)$, with $\|S\| \leq \|T\| \|xi_0\| \|\xi_1\|$.
Dec 8, 2020 at 12:50 comment added Matthew Daws Just a slice map of normal functionals on a von Neumann algebra (covered in standard textbooks); here apply the "corner" trick and regard this as a subspace of $\mathcal B((H\oplus K)\otimes H)$. An ad hoc construction is the following: for $T\in\mathcal B(H\otimes H, K\otimes H)$ define $S=(\operatorname{id}\otimes\omega_{\xi_0,\xi_1})(T)$ by $\langle S\alpha, \beta \rangle = \langle T(\alpha\otimes\xi_0), \beta\otimes\xi_1 \rangle$. Notice that this satisfies $|\cdot| \leq \|T\| \|\xi_0\| \|\xi_1\| \|\alpha\| \|\beta\|$ and is sesquilinear, so the Riesz Thm shows that for fixed $\alpha$...
Dec 8, 2020 at 9:46 comment added user167952 Yes obviously I messed up the codomain. I meant what you wrote. How do you define that then? I can see how to define it on operators $S \otimes T$ with $S: H \to K$ and $T: H \to H$ bounded operators.
Dec 8, 2020 at 9:30 comment added Matthew Daws This is never done. It is applied to an element of $B(H\otimes H, K\otimes H)$ but that's easy to define.
Dec 7, 2020 at 20:31 comment added user167952 How to apply $\text{id}\otimes \omega_{\xi_0, \xi_0}$ to an element of $B(H \otimes H, K \otimes K)$?
Dec 7, 2020 at 14:13 comment added Matthew Daws Yes, correct. I've made the change
Dec 7, 2020 at 14:13 history edited Matthew Daws CC BY-SA 4.0
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Dec 4, 2020 at 21:56 history answered Matthew Daws CC BY-SA 4.0