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The von Neumann algebra generated by a non-closable operator

The answer to Question 1. is positive. Namely, consider the polar decomposition of your operator $M=U|M|$ and define $X:= U f(|M|)$, where $f:[0,\infty) \to [0,\infty)$ is a bounded increasing functio …
Mateusz Wasilewski's user avatar