Let $\Omega\subset \mathbb R^d$ be an open, convex domain, and consider the Hilbert space $L^2(\Omega)$. Each sum of convex functions is convex, hence the subset $Conv(\Omega)$ of all convex functions is convex. I also have the strong impression that $Conv(\Omega)$ is closed wrt the $L^2$-norm, in spite of the fact that its elements are continuous. So, there exists a projector of $L^2(\Omega)$ onto $Conv(\Omega)$. Can one say how it looks like?
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1$\begingroup$ Nice question. Even $\Omega = (0,1)$ seems interesting. $\endgroup$– Nik WeaverCommented Nov 30, 2015 at 12:50
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1$\begingroup$ What about the Fenchel bi-conjugate? That gives in a sense the "nearest" convex function....? $\endgroup$– SuvritCommented Nov 30, 2015 at 14:20
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1$\begingroup$ 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 $f$. $\endgroup$– Delio MugnoloCommented Dec 1, 2015 at 7:15
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1$\begingroup$ @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 $g\in Conv(\Omega)$. $\endgroup$– Delio MugnoloCommented Dec 2, 2015 at 14:08
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2$\begingroup$ 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 $f-f^{**}$). $\endgroup$– SuvritCommented Dec 3, 2015 at 0:11
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