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Timeline for Minimiser of a certain functional

Current License: CC BY-SA 4.0

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Jan 18 at 14:59 comment added Nate River @fedja Good advice tbh :P
Jan 18 at 13:20 comment added fedja @NateRiver Don't divide the hide of a bear not yet killed, as they say in Russia ;-) First see if it works, and then we'll discuss such details. :-)
Jan 18 at 3:14 comment added Nate River @fedja I was revisiting this question, and was hoping to write an alternative proof of the Komlos theorem using the ideas in your answer, as suggested by Pietro Mayer. If it goes well, I plan to try submitting it to a journal. Would you like to coauthor the publication if it is successful? If that would be too much trouble for you, I could also just give a (very heavy) acknowledgement. Seems like the least I could do…
Aug 3, 2021 at 4:27 comment added Nate River Komlos’s theorem seems rather difficult to prove - I think that makes your answer even more impressive!
Aug 2, 2021 at 13:51 comment added Pietro Majer Of course, and I wouldn't be surprised if you answer (which I'm still reading) contains as a byproduct an alternative proof of that Komlós theorem (which is a 10 page paper: link.springer.com/article/10.1007%2FBF02020976 ).
Aug 2, 2021 at 13:48 comment added fedja @PietroMajer Indeed. I didn't know about the Komlos theorem when solving this problem. It looks like a very useful tool. However it doesn't look like its full proof is much shorter than my contraption :-)
Aug 2, 2021 at 13:38 comment added Pietro Majer An alternative proof for Part 2 (existence of a minimizing sequence converging a.e.) Let $h_n$ be any minimizing sequence for $F$. As observed, it is bounded in $L_1$, so by the Komlós Theorem, up to extracting a subsequence, it is a.e. converging in Cesaro sense to some $h\in L^1$. Since $F$ is a convex functional, the sequence of the Cesaro means is still a minimizing sequence.
Aug 2, 2021 at 13:33 comment added fedja @NateRiver I just tried to solve the problem in $L^2$ first. There weak convergence is guaranteed, but doesn't seem to drop the value of the functional immediately, so I needed the norm convergence of a minimizing sequence. Fortunately, if one has a nested sequence of bounded closed convex sets in a Hilbert space, the smallest norm elements of the sets converge in norm (the proof is the same as above just using the parallelogram identity). I tried to mimic that idea in $L^1$ (which required strict convexity of something) and got the a.e. convergence this way, which turned out to be sufficient.
Aug 2, 2021 at 8:08 history bounty ended Nate River
Aug 2, 2021 at 8:08 vote accept Nate River
Aug 2, 2021 at 8:08 comment added Nate River Very nicely done… can I ask how you got the idea to use $\Phi$ in that way? I have not done much convex analysis so am not familiar with many of the techniques.
Aug 2, 2021 at 7:45 comment added fedja @NateRiver Exactly as you said: $U_k=u_{n_k}+v_{n_k}\chi_{\cup_{q>k}E_{n_q}}$, $V_k=v_{n_k}\chi_{E_{n_k}\setminus \cup_{q>k}E_{n_q}}=v_{n_k}\chi_{G_k}$.
Aug 2, 2021 at 4:04 comment added Nate River Wow, this looks impressive. This part was a bit confusing to me - “ Then $\|v_{n_k}\chi_{\cup_{q>k}E_{n_q}}\|_1\to 0$ and, adding these parts to $u_{n_k}$, we see that we can (passing to a subsequence) assume that our minimizing sequence can be represented as $h_n=U_n+V_n$ where $U_n\to h$ in $L^1$ and $V_n$ have disjoint supports $G_n$.” Do you mean to add $v_{n_k}\chi_{\cup_{q>k}E_{n_q}}$ to $u_{n_k}$? I am not sure how to get the desired representation from this.
Aug 2, 2021 at 2:38 history edited fedja CC BY-SA 4.0
Corrected a stupid mistake
Aug 2, 2021 at 2:31 history undeleted fedja
Aug 2, 2021 at 2:31 history edited fedja CC BY-SA 4.0
Corrected a stupid mistake
Aug 2, 2021 at 2:08 history deleted fedja via Vote
Aug 2, 2021 at 2:02 history answered fedja CC BY-SA 4.0