Timeline for Does mixing automatically imply this seemingly stronger "uniform modulo re-ordering" version of mixing?
Current License: CC BY-SA 4.0
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| Apr 5, 2022 at 19:03 | history | bounty ended | CommunityBot | ||
| Apr 4, 2022 at 15:39 | comment | added | Ronnie Pavlov | That seems right, good catch again! | |
| Apr 1, 2022 at 22:32 | comment | added | Julian Newman | In general, I expect that every $[0,1]$-valued function in $L^\infty([0,1])$ can be obtained as a weak limit of subsets of $[0,1]$. | |
| Apr 1, 2022 at 22:26 | comment | added | Julian Newman | I'm pretty sure that in general the weak limit of a sequence of 0-1 valued functions (i.e. sets) will not be a 0-1 valued function. For example, with $X=[0,1]$ and $\mu$ the Lebesgue measure, take $C_n$ to be the union of all intervals of the form $[\frac{2i}{2^n},\frac{2i+1}{2^n}]$ with $i \in \{0,1,\ldots,2^{n-2}-1,2^{n-2},2^{n-2}+2,\ldots,2^{n-1}-2\}$. I think the weak limit of $C_n$ is the function $\frac{1}{2}\mathbf{1}_{[0,\frac{1}{2})}+\frac{1}{4}\mathbf{1}_{[\frac{1}{2},1]}$. | |
| Apr 1, 2022 at 22:10 | comment | added | Ronnie Pavlov | That's definitely a good point, which I completely missed. Somehow I want to say that the weak limit needs to be a 0-1 valued function, but I guess I don't see why immediately. As you say, I guess it shouldn't affect the proof anyway. | |
| Mar 31, 2022 at 22:58 | comment | added | Julian Newman | Thanks again for providing this progress towards the problem. A perhaps minor point: your notation suggests that you believe the weak limit of $C_{m_n}$ to be a set $C$, rather than a more general $[0,1]$-valued density in $L^\infty(\mu)$. I wouldn't have thought that that's correct; but ultimately it makes no difference to the validity of your argument, as mixing w.r.t. sets implies mixing w.r.t. $L^\infty$ functions. | |
| Mar 30, 2022 at 19:23 | history | answered | Ronnie Pavlov | CC BY-SA 4.0 |