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Let $\Gamma_0$ denote the set of lower-semi-continuous convex functionals on a Hilbert space $H$. What exactly is the image of $\Gamma_0$ under the proximal operator $$ \begin{aligned} &\Gamma_0\rightarrow \{f:H \rightarrow H\}\\ & f \mapsto \left[\operatorname{argmin}_{h \in H} \|x-h\|_H^2 +\frac1{2}f(h) \right]. \end{aligned} $$ (Not its co-domain, but what characterizes it's image?)

Necessary: Here are a few necessary conditions I've noted so far.

  • The functions in its image must have a convex domain, since the Moreau envelope is convex, and $Prox_f$ associates the outputs of the Moreau envelope with elements of its domain.
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    $\begingroup$ There should be a square norm in the definition of the proximal map. $\endgroup$
    – cheyp
    Commented Feb 4, 2019 at 8:51
  • $\begingroup$ @cheyp Thanks, for pointing that out :) $\endgroup$
    – ABIM
    Commented Feb 4, 2019 at 8:53
  • $\begingroup$ I do not believe that your first condition is necessary: If $f$ is the indicator function of $\{0\}$, then the proximal operator is $x \mapsto 0$ and this is not invertible. Second, the domain of a proximal operator is always $H$, i.e., it is trivially convex. $\endgroup$
    – gerw
    Commented Feb 4, 2019 at 9:57
  • $\begingroup$ True, that's a good example. I'll edit the question. $\endgroup$
    – ABIM
    Commented Feb 4, 2019 at 10:45
  • $\begingroup$ The prox equals identity minus the gradient of the Moreau envelope of $f$ (which is a strongly convex function). Does that help? $\endgroup$
    – Dirk
    Commented Feb 4, 2019 at 14:04

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I am not sure that it is going to work but it seems to me that this is the Legendre transform: $$\|x-h\|^2+\frac{1}{2}f(h)=\|x\|^2-2\Re\langle x,h\rangle+\|h\|^2+\frac{1}{2}f(h) $$ (Where we see $H$ as a $\mathbb{R}$ linear space and $\Re\langle .,.\rangle$ is the real scalar product. So if we call $$h_\min =\text{argmin} (-\Re\langle 2x,h\rangle+\|h\|^2+\frac{1}{2}f(h)) $$ and $$-g(2x)=\min_h (-\Re\langle 2x,h\rangle+\|h\|^2+\frac{1}{2}f(h)) \\= -\Re\langle 2x,h_\min\rangle+\|h_\min\|^2+\frac{1}{2}f(h_\min)$$Then $g$ is the Legendre transform of $h\rightarrow \|h\|^2+\frac{1}{2}f(h)$. And there is the formula $$h_\min =\partial_x g $$ So I thing one should first caracterize the Legendre transform of the set $[\|h\|^2+\frac{1}{2}f(h), f\in \Gamma_0]$ and take its differential.

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