Assume that convex $f: S \to \mathbb{R}$ with $L$-Lipschitz continuous gradient on some convex compact $S \subset \mathbb{R}^d$ is given. It would be very helpful if there existed function $F$ such that $F(x) = f(x)$ for all $x \in S$ and $F$ is convex and has $L$-Lipschitz continuous gradient on the whole $\mathbb{R}^d$. I suppose that $F(x) = \sup \{g(x)\;|\;g\text{ is convex, has }L\text{-Lipscitz continuous gradient and }g \leq f\}$ can be a candidate for such an "lower smooth convex envelope", but cannot see if it's so and how to prove it if it is. I would be grateful for any recommendations. Explicit construction is not necessary, at least existence would be enough.
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$\begingroup$ I would assume the "right" choice is something like $F(x) := \sup_{y \in S}\big( f(y) + \langle \nabla f(y),x-y \rangle\big)$. $\endgroup$– Dustin G. MixonCommented Oct 21, 2023 at 1:48
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$\begingroup$ @DustinG.Mixon, I was thinking about this function too, but I was not able to show that its gradient has the same Lipschitz constant. I stuck at the moment of proving non-expansiveness of $y^* =\arg\max_y\{\dotsi\}$ operator, which differs from projection in that it satisfies $\langle \nabla^2 f(y^*)[x-y^*],z-y^*\rangle \leq 0$, and non-expansiveness holds in some specific local norm associated with x,y, which has spectrum proportional to Hessian, so Lipschitz constant scales as well. I don't see how to avoid this :( $\endgroup$– Dmitry VilenskyCommented Oct 21, 2023 at 7:29
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$\begingroup$ As far as I am aware, the answer to the question is not known. $\endgroup$– Piotr HajlaszCommented Oct 21, 2023 at 18:38
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Such an extension is not always possible, a counterexample can be found in Section 2 of https://arxiv.org/abs/1812.02419v3.
