Timeline for Additional conditions under which separately continuous implies jointly continuous
Current License: CC BY-SA 4.0
9 events
| when toggle format | what | by | license | comment | |
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| Aug 13 at 18:14 | history | edited | YCor | CC BY-SA 4.0 |
fixed English
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| Aug 13 at 18:06 | answer | added | realgreathero | timeline score: 4 | |
| Nov 22, 2020 at 18:07 | answer | added | Liaqat Ali Khan | timeline score: 0 | |
| Mar 15, 2018 at 12:52 | answer | added | MasleniZZa | timeline score: 5 | |
| Feb 2, 2018 at 11:19 | comment | added | user493456 | Unfortunately, in my setting there are no additional algebraic conditions | |
| Feb 2, 2018 at 7:15 | comment | added | Taras Banakh | Maybe you have some algebraic conditions on $f$? Like $X,Y$ are compact topological groups and $f(\cdot,y)$ is a homomorphism for every $y$. Then you can try to derive the joint continuity from the existence of many continuity points (given by the Namioka Theorem)? | |
| Feb 1, 2018 at 18:36 | comment | added | Pietro Majer | Since $X$ and $Y$ are compact, a necessary ans sufficient (and a bit trivial) condition is that all $f(\cdot,y)$ are continuous, and all $f(x,\cdot)$ are equicontinuous (or viceversa). | |
| Feb 1, 2018 at 11:26 | review | First posts | |||
| Feb 1, 2018 at 11:44 | |||||
| Feb 1, 2018 at 11:23 | history | asked | user493456 | CC BY-SA 3.0 |