Timeline for Two commuting mappings in the disk
Current License: CC BY-SA 2.5
8 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Nov 7, 2009 at 9:34 | comment | added | Jose Capco | I wasn't able to delete after several posts.. and things get being added to the features I am able to do here. I think this is a *overlflow.net system, it's a bit confusing and annoying. I wasn't even able to make comments before. But now I can at least delete :) | |
| Nov 7, 2009 at 7:47 | comment | added | Ori Gurel-Gurevich | I think that you should avoid editing your answers the way you did (erasing everything and writing a new answer). It's better to delete the answer and write a new one. | |
| Nov 6, 2009 at 18:10 | comment | added | Andy Putman | The trouble with proving this via an approximation scheme is that you would have to prove that if $f$ and $g$ commute, then you can approximate $f$ and $g$ by polynomials $f'$ and $g'$ that commute. This sounds unlikely. | |
| Nov 6, 2009 at 18:03 | history | edited | Jose Capco | CC BY-SA 2.5 |
deleted 590 characters in body
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| Nov 6, 2009 at 14:07 | history | edited | Jose Capco | CC BY-SA 2.5 |
added 451 characters in body
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| Nov 6, 2009 at 13:48 | comment | added | Jose Capco | Ok sorry.. Kent is correct.. Brouwers fixed point theorem, my argument does not work because the function that I "drew" below the diagonal must have values in -0.5 an 0.5 respectively so it will necessarily cross the diagonal.. | |
| Nov 6, 2009 at 12:58 | comment | added | Autumn Kent | By the Brouwer fixed point theorem, for every f there is an x such that f(x) = x, so the answer is yes when g is the identity. | |
| Nov 6, 2009 at 12:55 | history | answered | Jose Capco | CC BY-SA 2.5 |