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).



Is the Legendre transform as an operator Lipschitz?
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.


