Timeline for Fixed points and their continuity
Current License: CC BY-SA 3.0
11 events
| when toggle format | what | by | license | comment | |
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| Jan 25, 2014 at 10:54 | comment | added | Michał Kukieła | You may find arxiv.org/abs/1210.6496 interesting. | |
| Jan 24, 2014 at 5:56 | history | edited | Ricardo Andrade | CC BY-SA 3.0 |
added top level tag; minor correction and rewording of suggested edit; readded some of the information in the original question
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| S Jan 24, 2014 at 5:50 | history | edited | Ricardo Andrade | CC BY-SA 3.0 |
added top level tag; minor correction and rewording of suggested edit
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| S Jan 24, 2014 at 5:50 | history | suggested | Włodzimierz Holsztyński | CC BY-SA 3.0 |
an equivalent but more formal reformulation
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| Jan 24, 2014 at 5:36 | review | Suggested edits | |||
| Jan 24, 2014 at 5:50 | |||||
| Jan 24, 2014 at 5:34 | vote | accept | LSZ | ||
| Jan 24, 2014 at 5:17 | answer | added | Noam D. Elkies | timeline score: 9 | |
| Jan 24, 2014 at 5:11 | comment | added | LSZ | Yes. Exactly. I forgot mentioning the range of $f(x,y)$. | |
| Jan 24, 2014 at 4:05 | comment | added | Daniele Zuddas | Do you mean that $f : [0,1] \times [0,1] \to [0,1]$ and searching for a continuous $g : [0,1] \to [0,1]$ such that $f(g(y), y) = g(y)$ for all $y \in [0,1]$, is it? | |
| Jan 24, 2014 at 3:57 | review | First posts | |||
| Jan 24, 2014 at 4:40 | |||||
| Jan 24, 2014 at 3:37 | history | asked | LSZ | CC BY-SA 3.0 |