Timeline for Are all well behaved "mean" functions on $\mathbb{R}^+$ equivalent?
Current License: CC BY-SA 3.0
11 events
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| Dec 24, 2014 at 7:47 | comment | added | Yaakov Baruch | aaabbccc => aaabbcccaaabbccc => aaabaaabbcccbccc => aaaaababbcccbccc => aaaabcccababbccc => aaaabcccababcbcc => aaaabcccabcbabcc => aaaabcccabcbacbc => aaaabcccabcbaccb => aaaabcccabaccbcb => aaaabcccaabccbcb => aabcaaccaabccbcb => aabcccaaaabccbcb => aabcaabcccaacbcb => aabcaabcccaaccbb => aabcaabcccccaabb => aabcaabcccccabab => aabcaabcababcccc => aabcaabcabccabcc => aabcabccaabcabcc => aabcabcc | |
| Dec 24, 2014 at 0:17 | history | edited | Eric Wofsey | CC BY-SA 3.0 |
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| Dec 23, 2014 at 14:10 | comment | added | Eric Wofsey | Or actually, more simply: (7) automatically holds for the endpoints of the interval, and it holds for all other values of $a$ and $b$ by continuity and (5) (since $M(a,b)-a$ and $M(a,b)-b$ must have constant sign if we stay away from the diagonal). I guess this argument is slightly weaker than the one in my previous comment because it also requires (2). But assuming (2), it shows that (7) also follows from (5) and (8) in an open interval as long as you know that (7) holds for any single $a$ and $b$. | |
| Dec 23, 2014 at 13:49 | comment | added | Eric Wofsey | Oh, I see, that makes it a lot trickier than I thought. Incidentally, I think (7) follows from (5) and (8), at least on a closed interval. For instance, if $a<b<M(a,b)$, then by continuity and (5) $M(a,c)>c$ for all $c>a$. But now we get a contradiction by taking the limit of iterating $c\mapsto M(a,c)$. | |
| Dec 23, 2014 at 13:41 | comment | added | Yaakov Baruch | It's more complicated than that: you can go from aaabbccc to aaabbcccaaabbccc to (some work) to aaccacbbaaccacbb to aaccacbb and many other places in a similar way - just by using (1), (2) and (3). (1) is essential here. | |
| Dec 23, 2014 at 13:04 | comment | added | Eric Wofsey | I'm pretty sure (3) doesn't imply (4) in the free algebra on 3 generators: there seems to be no way to transform $M(M(M(a,a),M(a,b)),M(M(b,c),M(c,c)))$ that changes the composition of the two halves. | |
| Dec 23, 2014 at 12:52 | vote | accept | Yaakov Baruch | ||
| Dec 23, 2014 at 12:52 | comment | added | Yaakov Baruch | I was lost in a small computational nightmare trying to prove equivalence with 4 generators (which then would imply equivalence with any number, since strong associativity is defined on a set of 4 variables) and that's why I eventually started looking for the finite counterexample shown in my original post. - Thank you | |
| Dec 23, 2014 at 12:51 | comment | added | Yaakov Baruch | Awesome and eye-opening proof! In regards to weak associativity, I knew that (3) and (4) are equivalent in the free algebra on 2 generators (and I suspect they might be even in the free algebra on 3 generators!) What I hadn't realized until now was that (5), (7), (8) and the density of dyadics meant that one only needed equivalence in the free algebra on 2 generators! | |
| Dec 23, 2014 at 9:48 | history | edited | Eric Wofsey | CC BY-SA 3.0 |
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| Dec 23, 2014 at 9:05 | history | answered | Eric Wofsey | CC BY-SA 3.0 |