Timeline for Is the module action $M\times M^*\to M^*$ jointly continuous?
Current License: CC BY-SA 3.0
11 events
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Feb 8, 2016 at 13:22 | vote | accept | ABB | ||
Feb 6, 2016 at 16:27 | comment | added | Yemon Choi | @NikWeaver Of course. Sorry, lack of sleep and coffee today. I think I was getting mixed up with some issue in my head to do with completed tensor products | |
Feb 6, 2016 at 16:17 | comment | added | Nik Weaver | Joint continuity means: continuous from the product topology. So take an arbitrary convergent net in $M\times M^*$. That is a net of the form $(a_i, f_i)$ which converges to some $(a,f)$, i.e., $a_i \to a$ and $f_i \to f$. And you want its image to converge, i.e., you want $a_if_i \to af$. | |
Feb 6, 2016 at 16:13 | comment | added | Yemon Choi | @NikWeaver oh, because you can enlarge the indexing set or refine to a subnet, or similar? | |
Feb 6, 2016 at 16:08 | comment | added | Nik Weaver | @YemonChoi: I think the statement of joint continuity is correct. | |
Feb 6, 2016 at 16:04 | answer | added | Nik Weaver | timeline score: 5 | |
Feb 6, 2016 at 15:54 | comment | added | Yemon Choi | A first start would be to see what happens if $M=\ell^\infty$ or $M=L^\infty[0,1]$... | |
Feb 6, 2016 at 15:54 | comment | added | Yemon Choi | Are you sure that for joint continuity it suffices to show that $a_if_i \to af$? I would have expected that one needs to consider something like the product net $(a_if_j)_{i,j\in (I\times J})$ with a suitable ordering. Could you please clarify whether you need joint continuity, or the seemingly weaker requirement that you have stated | |
S Feb 6, 2016 at 14:19 | history | suggested | Ali Taghavi |
I add two tags
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Feb 6, 2016 at 14:08 | review | Suggested edits | |||
S Feb 6, 2016 at 14:19 | |||||
Feb 6, 2016 at 14:04 | history | asked | ABB | CC BY-SA 3.0 |