Timeline for Strongly continuous semigroup: continuous or continuous componentwise?
Current License: CC BY-SA 3.0
5 events
| when toggle format | what | by | license | comment | |
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| Apr 20, 2018 at 19:21 | comment | added | Jochen Glueck | Maybe it is worthwhile to point out the following two small remarks: (i) To see that @Nate Eldredge's argument implies joint strong continuity of operator multiplication on bounded sets, one actually has to take $(A_n)$ and $(B_n)$.to be nets (not only sequences). (ii) On the other hand, operator multiplication is jointly sequentially strongly continuous on the entire operator space (not only on sets which are a priori bounded) since a convergent operator sequence is automatically bounded due to the uniform boundedness theorem. | |
| Apr 20, 2018 at 15:28 | comment | added | Nate Eldredge | @Sascha: Simpler. If $A_n \to A$, $B_n \to B$ strongly, write $A_n B_n -A B = (A_n B_n - A_n B) + (A_n B - AB)$, and for the first term, use the fact that the sequence $A_n$ is bounded. | |
| Apr 20, 2018 at 14:53 | comment | added | Sascha | @ I assume the continuity of multiplication on bounded sets follows somehow from the open mapping theorem or is it simpler? | |
| Apr 20, 2018 at 14:49 | vote | accept | Sascha | ||
| Apr 20, 2018 at 14:36 | history | answered | Mateusz Wasilewski | CC BY-SA 3.0 |