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Let $X$ be a separable Fréchet space and $\varphi,\psi:X\to \mathbb{R}$ be a lower semi-continuous and convex function with $\psi$ bounded below and coercive. Consider the infimal convolution $$ \varphi\mathbin\square \psi(x):= \operatorname*{arginf}_{x\in X} \varphi(x-y) + \psi(y) $$ e.g. if $(X,\|\cdot\|_X)$ is a Banach space then one may take $\psi=\|\cdot\|_X^2$ and obtain the proximal operator as a special case.

Under what conditions is $\varphi\mathbin\square \psi$ a construction, i.e. $L<1$-Lipschitz?

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  • $\begingroup$ What does "arginf" mean? Maybe you mean instead $$ \varphi\square \psi(x):= \inf_{y\in X} \varphi(x-y) + \psi(y) $$ ? $\endgroup$
    – Pietro Majer
    Commented Jun 21 at 22:39
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    $\begingroup$ Ah so, I mean a selection into the set of accumulation points of the $\epsilon$-minimizers; and I assume its non-empty. Tbh, one can possibly consider the Banachian case with $\psi$ being its norm squared. $\endgroup$
    – Math_Newbie
    Commented Jun 22 at 17:39
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    $\begingroup$ I changed $\displaystyle \operatorname{arginf}_{x\in X}$ to $\displaystyle \operatorname*{arginf}_{x\in X}$ by changing \operatorname{arginf}_{x\in X} to \operatorname*{arginf}_{x\in X}. Also, I changed $\varphi\square\psi$ to $\varphi\mathbin\square\psi$ by putting \mathbin\square where \square had been. $\endgroup$
    – Michael Hardy
    Commented Jun 22 at 20:30
  • $\begingroup$ Thanks @MichaelHardy a lot :) $\endgroup$
    – Math_Newbie
    Commented Jul 5 at 2:25

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Here are just some thoughts. I think it is a matter of curvature, so let us assume that $\varphi$ and $\psi$ are smooth. Then, $y(x)$ solves the optimality condition $$ \psi'(y(x)) = \varphi'(x - y(x)). $$ Differentiating again, we get $$ \nabla^2\psi(y(x)) y'(x) = \nabla^2\varphi(x - y(x)) (I - y'(x)) $$ thus, $$ y'(x) = \big( \nabla^2\psi(y(x)) + \nabla^2\varphi(x-y(x)))^{-1} \nabla^2\varphi(x-y(x)) $$ and you want to bound the norm of this guy by something smaller than $1$. This should work if $\psi$ has positive curvature and if the curvature of $\varphi$ is bounded.

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