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This stack exchange question got me thinking about quasiconvex analysis.

Given a compact,convex subset $X\subset \mathbb{R}^n$ and a quasiconvex function $f:X\rightarrow \mathbb{R}$

Define the double Legendre-Fenchel transform of $f$, written $f^{**}$, given by: $$\operatorname{epi} (f^{**})=\operatorname{co}(\operatorname{epi}(f))$$ where $\operatorname{epi}(f)\in X\times\mathbb{R}$ is the epigraph of f, and $\operatorname{co}$ is the convex hull of a set.

Assume $f$ has a unique minimum

I know that bad things can happen here if $X$ is not compact, but for the closed and bounded case, it seems like $f^{**}$ can also only have a single minimum on $X$, and this will also minimize $f$. Does someone have a counter-example or a proof?

My intuition has failed me often enough in the past, so I thought I would ask before making an ass of myself.

EDIT: I decided to provide a (simple) example which is indicative of my intuition on this matter. Consider $f(x)=\sqrt{\left| x\right| }$. If we take the domain to be the entire real axis, then $f^{**}\equiv 0$. But for any closed interval $[a,b]$ , $(f\left| _{[a,b]} \right.)^{**}$ fulfills the qualities that I ask for in my question.

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  • $\begingroup$ I think you need take closure to $f^{**}$ be well-defined i.e $\operatorname{epi} (f^{**})= \operatorname{cl} \operatorname{co}(\operatorname{epi}(f))$ $\endgroup$
    – Red shoes
    Jun 5, 2017 at 5:04
  • $\begingroup$ I proved your claim, it is true for any lsc function defined on compact interval, and quasiconvexity does not play any role! $\endgroup$
    – Red shoes
    Jun 5, 2017 at 5:37
  • $\begingroup$ but I'm lazy to type here.. $\endgroup$
    – Red shoes
    Jun 5, 2017 at 5:49
  • $\begingroup$ @Ashkan I was specifically interested in the quasiconvex case. Unless you also proved that lsc is a necessary condition, your result doesn't really interest me. It was a long time ago that I asked this question, but I think your result was known to me at the time. $\endgroup$
    – Tim Seguine
    Jun 6, 2017 at 9:47
  • $\begingroup$ once you define $f^{**}$ is this form $\operatorname{epi} (f^{**})=\operatorname{co}(\operatorname{epi}(f))$ you need assume $f$ is lsc, other wise $\operatorname{co}(\operatorname{epi}(f))$ it is not closed set... and so there is no function like $f^{**}$ such that $\operatorname{epi} (f^{**})=\operatorname{co}(\operatorname{epi}(f))$... so lsc is zero condition here. BTW I am interested to see your proof for lsc, or any reference about it if you have seen somewhere , since mine is rather complicated ... want to see simple proof, thanks $\endgroup$
    – Red shoes
    Jun 6, 2017 at 19:24

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