Timeline for Does the norm on a sequence space have to be monotone?
Current License: CC BY-SA 4.0
6 events
| when toggle format | what | by | license | comment | |
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| Feb 4, 2019 at 10:43 | comment | added | Matthew Daws | Sorry, I kept editing my comment. I think I understand finally: the construction given in the answer would work over $\mathbb C$. But to address the original question, we need that there is a $\rho$ occurring. This is equivalent to $\|(x_n)\| = \| \ ( |x_n| ) \ \|$ for any $(x_n)$ in the space (and that if $( |x_n| )$ is in the space, then so is $(x_n)$). This is why we need to be able to multiply by arbitrary unit modulus sequences! | |
| Feb 3, 2019 at 14:11 | comment | added | Dap | @MatthewDaws: no, you need to be able to multiply by unit complex numbers without changing the norm - this is hidden in the $\pm$ detail erz mentioned above. For example $n\mapsto \exp(in)$ would need to have the same norm as $x\mapsto 1.$ | |
| Feb 3, 2019 at 10:18 | comment | added | Matthew Daws | It seems to me that this would also work with complex scalars. Is that right? | |
| Feb 3, 2019 at 9:34 | comment | added | erz | Thank you! Perhaps it is worth adding that in order to complete the solution one also has to show that $\|\cdot\|$ is indeed a norm, and is symmetric with respect to taking $\pm$ of the entries of the sequences. However, this is easy to see. | |
| Feb 3, 2019 at 9:32 | vote | accept | erz | ||
| Feb 3, 2019 at 6:12 | history | answered | Dap | CC BY-SA 4.0 |