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Let $p:[0,\infty) \times \mathbb{R}^n \to \mathbb{R}$ be a smooth convex function for all $x \in \mathbb{R}^n$.

We define its Legendre transform (or convex conjugate) as

$$ p^*(t,y)=\sup_{x \in \mathbb{R^n}}\left\{ x \cdot y - p(t,x)\right\} $$

I know differentiability and convexity are dual concepts, so convexity of $p$ implies differentiability of $p^*$. However, I should be able to get smoothness of $p^*$ and Hessian $D^2_y p^*$ bounded away from zero and $+\infty$. How do I do this?

Thanks in advance for any helpful comments, ideas, etc.

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up vote 1 down vote accepted

Correction: strict convexity of $p$ implies differentiability of $p^*$.

Think in terms of the gradient map $\nabla p\colon\mathbb R^n\to\mathbb R^n$. The Hessian of $p$ is the Jacobian determinant of $\nabla p$. Now, $\nabla p^*$ is $(\nabla p)^{-1}$, the inverse map. The Jacobian determinant of the inverse map is, of course, the reciprocal of the Jacobian determinant of $\nabla p$. Thus, the inequality $c\le \det D^2 p^*\le C$ is equivalent to $C^{-1}\le \det D^2 p\le c^{-1}$.

There are some sufficient conditions for $C^{-1}\le \det D^2 p\le c^{-1}$... For example: if $p(x)-\epsilon |x|^2$ is convex for some $\epsilon>0$, you have a lower bound. And if the second derivatives are bounded, you have an upper bound. Of course, these are only sufficient conditions, they are not necessary.

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Leonid, any idea where I might find a proof of the fact that strict convexity of $p$ implies differentiability of $p^*$? Thx! – chango Jun 10 '12 at 11:09
@mathscat A popular reference book on the subject is Convex Analysis by Rockafellar. Theorem 26.6 describes the relation between strict convexity and differentiability. – user31373 Jun 10 '12 at 15:47

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