Timeline for $(S\otimes T)^{it}= S^{it}\otimes T^{it}$ for unbounded operators
Current License: CC BY-SA 4.0
3 events
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| Sep 26, 2023 at 20:59 | comment | added | Christian Remling | @MatthiasLudewig: Yes, of course. The OP suggested to assume that $S,T>0$ and mentions functional calculus, so this would seem to be the intended interpretation. (I also have no idea how we would define $S^{it}$ for general $S$.) | |
| Sep 26, 2023 at 20:51 | comment | added | Matthias Ludewig | This only works if $S$ and $T$ are self-adjoint. | |
| Sep 26, 2023 at 18:04 | history | answered | Christian Remling | CC BY-SA 4.0 |