Timeline for orthogonal projector onto the set of convex functions
Current License: CC BY-SA 3.0
13 events
| when toggle format | what | by | license | comment | |
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| Dec 3, 2015 at 0:42 | comment | added | Delio Mugnolo | Thanks! Although Willie Wong's comment makes the whole a bit hopeless... | |
| Dec 3, 2015 at 0:11 | comment | added | Suvrit | These decompositions are not unique, and finding the "best" such decomposition could give an idea in practice. I'd also prefer to move all of this to chat (except in chat, math does not work, so...); as for $\sin(x)$ have a look at this post and the comments to D. Speyer's answer: math.stackexchange.com/questions/13386 --- but on second thought, $f^{**}$ is perhaps the right object to consider....(or $f-f^{**}$). | |
| Dec 2, 2015 at 23:02 | comment | added | Delio Mugnolo | @Suvrit I am not sure I understand your decomposition, either. What about the sine function? How can it be possibly decomposed into a convex and a concave part? Btw, the above stated inequality shows that the key of the solution to the problem should be to understand the behaviour of $f-f^{**}$, or perhaps simply of $f-Pf$ ($P$ the projector). (Before we make a forum out of these comments, we may perhaps move to a chat) | |
| Dec 2, 2015 at 20:29 | comment | added | Suvrit | @DelioMugnolo ok that shows that maybe it may be better to not use binconjugates (in your example, $f^{**}\equiv -\infty$, hence useless), but perhaps to decompose $f(x) = f_{\text{cvx}} + f_{\text{cve}}$ into convex and concave parts (not always possible, but often it is). For instance, at least for twice differentiable functions, the Hessian will provide the way, but more generally not sure what recipe to follow... | |
| Dec 2, 2015 at 19:27 | comment | added | Delio Mugnolo | @Suvrit thanks, but I admit I don't quite get this. What about $f:x\mapsto -x^2$ on $\mathbb R$? How can a convex function be dominated pointwise by a concave function going to $-\infty$? Or does one have to give up the condition that $f^{**}$ is proper? | |
| Dec 2, 2015 at 14:41 | comment | added | Suvrit | Not sure about this; however, note that $f^{**} \le f$ (restricted to the domain of $f$), which is the sense of "nearest" that I meant. | |
| Dec 2, 2015 at 14:08 | comment | added | Delio Mugnolo | @Suvrit Is there actually a handier formula for $f^{**}$? In principle one should be able to check whether $f^{**}$ is the best approximation of $f$ in $Conv(\Omega)$ by checking whether $(f-f^{**},g-f^{**})\le 0$ for all $g\in Conv(\Omega)$. | |
| Dec 1, 2015 at 7:15 | comment | added | Delio Mugnolo | Thanks @Suvrit, I didn't know at all about the theory you mention, and it sounds promising. In a certain sense it sounds very "continuous", in fact it is not clear to me whether any connection with the $L^2$-norm may be expected - meaning that the $f^{**}$ should actually be the one convex function of minimal $L^2$-distance from $f$. | |
| Nov 30, 2015 at 14:20 | comment | added | Suvrit | What about the Fenchel bi-conjugate? That gives in a sense the "nearest" convex function....? | |
| Nov 30, 2015 at 13:08 | comment | added | Igor Rivin | What do you get if you just take convex quadratic functions? | |
| Nov 30, 2015 at 12:50 | comment | added | Nik Weaver | Nice question. Even $\Omega = (0,1)$ seems interesting. | |
| Nov 30, 2015 at 12:49 | history | edited | Delio Mugnolo | CC BY-SA 3.0 |
deleted 12 characters in body
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| Nov 30, 2015 at 12:29 | history | asked | Delio Mugnolo | CC BY-SA 3.0 |