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Let $X$ be a Hilbert space, and suppose that $f:X^2\rightarrow \mathbb{R}$ is a Lipschitz, supercoercive, convex function such that (for every $y \in X$) the set $$ \operatorname*{argmin}_{x\in X} f(x,y), $$ is single-valued and the map $$ y \mapsto \min_{x\in X} f(x,y), $$ is Lipschitz.

Is the map $$ y\mapsto \operatorname*{argmin}_{x\in X} f(x,y) , $$ Lipschitz? (If not what additional requirements are needed)?

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    $\begingroup$ Consider the case $X=\mathbb R$. If $g(x)$ is a conv x function with a very flat minimum at zero, then $f(x,y)=g(x)+xy$ has the property that the argmin is far from Lipschitz near $y=0$. $\endgroup$
    – Anthony Quas
    Commented Apr 19, 2019 at 21:41
  • $\begingroup$ If the map is strictly convex then this should rule this out no? $\endgroup$
    – ABIM
    Commented Apr 20, 2019 at 5:44
  • $\begingroup$ I have in mind $g(x)=x^{2n}$ for example. $\endgroup$
    – Anthony Quas
    Commented Apr 20, 2019 at 6:41
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    $\begingroup$ This reminds me of proximal mappings: there you have $f(x,y) = g(x) +\|x-y\|_2^2$ and the argmin over x is always 1-Lipschitz in y. $\endgroup$
    – Dirk
    Commented Apr 20, 2019 at 15:41
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    $\begingroup$ Maybe you should add your definition of Lipschitz (maybe you mean "locally Lipschitz"?) and of supercoercive (is it $\lim_{\|x\|\to\infty}f(x)/\|x\|=+\infty$ what you mean?) because they are incompatible with each other in the common usage $\endgroup$
    – Pietro Majer
    Commented Apr 20, 2019 at 21:07

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