TITLE: A question on regularity of the Legendre transform
QUESTION [7 upvotes]: Let $f(x)$ be a strictly convex real-valued $C^{\infty}$ function on an open neighborhood of the origin in $\mathbb R^n$ with
$f (0, \ldots , 0)= \partial_j f(0, \ldots , 0)=0$ for all $j$. If the second derivative $f''(0, \ldots , 0)$ is non-degenerate as a quadratic form, then it is easy to see that the Legendre transform $\tilde f(\xi)$ of $f$ is $C^{\infty}$ in some neighborhood of the origin. On the other hand, if $f''(0, \ldots , 0)$ is degenerate, then  $\tilde f(\xi)$ is never in $C^{1,1}$ in a neighborhood of the origin. For instance, if $f(x_1, x_2 ) = \frac 12 x_1^2 + \frac 14 x_2^4$, then
$\tilde f(\xi_1, \xi_2 ) = \frac 12 \xi_1^2 + \frac 34 |\xi_2|^{4/3}$.
Does anyone know a reference for this fact?

REPLY [2 votes]: Jean-Baptiste Hiriart-Urruty has pointed out that Theorem 4.2.2 in his book (together with C. Lemaréchal), Convex analysis and minimization algorithms, Volume 2, implies the assertion in the question.   That theorem  says (in particular) the following.  If $f(x)$ is convex and the gradient $f'(x)$ is Lipschitz continuous, then $\tilde f(\xi)$ is strongly convex.
To see that this implies the assertion just note that $\tilde{\tilde f} = f$ if $f$ is convex and that
if $f$ is convex and $f''(x)$ is degenerate, then $f$ is not strongly convex.