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Let $f:{\mathbb R}\rightarrow{\mathbb R}$ be measurable and such that $$\int_{-\infty}^0(f(x)-a)^2dx+\int_0^{+\infty}(f(x)-b)^2dx<+\infty.$$ This, denoted as $E(a,b)$, is an affine space: $E(a,b):=f_0+L^2({\mathbb R})$ for any $f_0\in E(a,b)$. I am interested in the $L^2$-projection $\pi f$ of a given $f$ onto the close convex subset $K(a,b)$ which consists in non-decreasing function: $\pi f\in K(a,b)$ and $$\forall g\in K(a,b),\qquad \int_{\mathbb R}(\pi f-f)^2dx\le\int_{\mathbb R}(g-f)^2dx.$$

My guess (or a vague souvenir ?) is that $\pi f$ is the derivative of $\phi$, the convex enveloppe of a primitive $F$ of $f$.

This is probably well-known, and I should be delighted to receive a reference. Unless my guess is incorrect, in which case any help is welcome.

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    $\begingroup$ I can confirm your guess is true. Another formula for $\pi f$ is $$ \pi f(x) = \inf_{z > x} \sup_{y \leq x} \frac{1}{z-y} \int_y^z f(s) ~ds $$ I would also love a reference for it, especially if there is a short convex analysis proof. It took me a few days to work out all the details of a brute force proof and my proof is 12 pages long. $\endgroup$
    – Willie Wong
    Commented Mar 23, 2021 at 15:34
  • $\begingroup$ @WillieWong. Thank you so much for this close formula. I'll check soon that it coincides with $\phi'$. As to the fact that $\phi'$ is the projection of $f$, I have written up a one-page proof. $\endgroup$
    – Denis Serre
    Commented Mar 24, 2021 at 7:01
  • $\begingroup$ Yes, the hard thing for me was proving that the formula above has the right properties that you would expect of $\phi'$ (specifically that when $\pi f \neq f$ almost surely there exists an interval on which $\pi f$ is constant). After that the proof that it is the optimizer is short. Since the formula is the min-max of the difference quotient of the primitive, there's probably a shorter proof just by directly connecting to $\phi'$. $\endgroup$
    – Willie Wong
    Commented Mar 24, 2021 at 13:16
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    $\begingroup$ @WillieWong. Sketch of proof: on the one hand $\phi'\in K$ (obvious). Then because $K$ is convex and the distance is euclidian, it is enough to prove $\langle f-\phi',\phi'-g\rangle\ge0$ for every $g\in K$. This is the sum of integrals over disjoint integrals of term $\int\theta'hdx$, where $\theta=\psi-\phi\ge0$ vanishes on the boundaries, and $h=g-\phi'$ (here $\phi'$ is constant) is non-decreasing. Integration by parts give the correct sign. $\endgroup$
    – Denis Serre
    Commented Mar 24, 2021 at 14:19
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    $\begingroup$ In A Wasserstein approach to the one-dimensional sticky particle system, Natile and Savaré prove in Thm 3.1 that the projection of $f \in L_2(0, 1)$ to the set of nondecreasing functions (identified with their right-continuous representative) is given by $D^+ F^{**}$, where $F^{**}$ is the convex envelope of the primitive of $f$ and $D^+$ denotes the derivative from the right. $\endgroup$
    – ViktorStein
    Commented Sep 6 at 2:50

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