# Is the Legendre transform as an operator Lipschitz?

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).

• What is a standard norm? Dec 8, 2022 at 22:57
• Any of the L^p norms Dec 8, 2022 at 23:05

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.
• Lipschitz constant of 1 would then imply that LF is an isometry (on $C_{lsc}(\mathbb{R}^n)$), by the involution property $f^{**} = f$? This makes sense geometrically actually now that I think about it. Thank you! Dec 10, 2022 at 1:42