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Fix $a \in \mathbb R^n$ and let $\|\cdot\|$ be any norm on $\mathbb R$ (e.g $\ell_1$ norm). For any $a \in \mathbb R^n$, it is clear that the function $f_a(x) := \|x-a\|_2 + \|x\|$ is strictly convex and has a unique minimizer $x(a)$.

Question. Given $a,b \in \mathbb R^n$, can $\|x(a)-x(b)\|_2$ be bounded in terms of some norm of $a-b$ ?

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First of all, the function $f_a$ is not strictly convex, and hence, you should expect multiple minimizers. As such, non-expansiveness (even in some generalized sense) does not seem likely. Consider the one-dimensional case where $$f_a(x) = |x-a| + |x|$$ for which $\operatorname{argmin} f_a = [0,a]$. So you can select minimizers which depend continuously on $a$, but this may not be what you want…

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  • $\begingroup$ Indeed, Indeed I should have checked before... $\endgroup$
    – dohmatob
    Commented Oct 13, 2022 at 11:14
  • $\begingroup$ Do you know the answer to my question mathoverflow.net/q/434486/121665 ? You have many posts in convex analysis so I hope you might have some answers. $\endgroup$
    – Piotr Hajlasz
    Commented Mar 9, 2023 at 0:57
  • $\begingroup$ No, I don't. I've seen the question but don't know anything in this direction… I even did not know this characterization via second differences. $\endgroup$
    – Dirk
    Commented Mar 9, 2023 at 9:33

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