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Do Legendre transforms have any applications that do not appeal to convexity?

What is the intuitive interpretation of the Legendre transform of a non-convex function?

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I can answer your second question.

Recall that by convex duality, applying the Legendre transform twice on a convex (and coercive) function gives you back the original function. Now even if you plug in a non-convex function $W$ instead, the Legendre transform $W^*$ will be convex and coercive (one can see that by the definition) and so is $V:=(W^*)^*$. So clearly the convex duality will not carry over to the non-convex case. However one can show (under appropriate assumptions of course) that $V$ is the convexification or convex hull of $W$, that is roughly (here $W:\mathbb{R}\rightarrow\mathbb{R}$) the function whose graph is the boundary curve of the convex hull of the set $\{(x,y)\in\mathbb{R}^2\,|\,y\ge f(x)\}$. By convex duality we conclude that $V^*=((W^*)^*)^*=W^*$, so the Legendre transform of a non-convex function is the Legendre transform of it's convexification.

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Thanks! Is it possible to give me some references that would allow me to study these points in greater detail? – Per Oct 14 '12 at 10:02
E.g. Evans' book "Partial Differential Equations" contains a proof of convex duality (in a chapter on Hamilton-Jacobi equations I think). For the non-convex case, I think looking at $(W^*)^*$ for the example $W(p)=(1-p^2)^2$ is instructive (Hint: don't try calculating $W^*$ first, it's a mess and doesn't help). – IHaveAStupidQuestion Oct 14 '12 at 11:47

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