Is the Legendre transform as an operator Lipschitz?

Question

Let $C_{lsc}(\mathbb{R}^n)$ be the space of lower semicontinuous convex functions $\mathbb{R}^n \to \mathbb{R}$. The Legendre-Fenchel (LF) transform of $f \in C_{lsc}(\mathbb{R}^n)$ is: $$ f^*(y) := \sup_{x \in \mathbb{R}^n} (\langle x, y \rangle - f(x)) $$

It is known that the LF transform is continuous and an involution on $C_{lsc}(\mathbb{R}^n)$ (Wijsman 1963). I want to know if the LF transform is Lipschitz. That is, given $f$ and $g$ lsc, is there a way to bound $\|f^* - g^*\|$ by $\|f-g\|$ (under a standard norm)?

I haven't found any sources that do this, so my suspicion is that the LF transform is not Lipschitz. Does anyone know if this is true, or if not a simple counterexample?

It is worth noting that Attouch and Wets ("Isometries of the Legendre-Fenchel transform") constructed norms under which the LF transform is an isometry -- however these norms are not particularly useful for me. I am looking for any results using standard norms (any of the p-norms).

Answer 1

This is basically true for sup norm by Fenchel's inequality. Indeed, for all $y$,

$$ f^*(y) = \sup_x\left( \langle x,y\rangle - f(x) \right) \leq \sup_x\left( g(x)+g^*(y) - f(x) \right) \leq \|f-g\|_{\infty} + g^*(y). $$

The same is true when the roles of $f,g$ are reversed, giving essentially what you want. One should take care, though, in writing the final conclusion because it is possible for $f^*(y) = g^*(y) = +\infty$, despite $\|f-g\|_{\infty}=0$ (e.g., take $f=g=0$). That said, if $\|f-g\|_{\infty}<\infty$, then it is clear from above that $$ \{y : f^*(y) = +\infty\} = \{y : g^*(y) = +\infty\}, $$ so it is reasonable to write $$ \|f^*-g^*\|_{\infty} \leq \|f-g\|_{\infty}, $$ provided you adopt the convention $|(+\infty)-(+\infty)| = 0$ for handling the indeterminate form that can appear in defining the sup norm on the LHS.