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Oct 31, 2011 at 11:20 comment added Sergei Ivanov For every fixed $x\in X$, the requirement that $h(x)\ge f(x)$ is a linear inequality on the coefficients of $f$. So $S$ is a convex polyhedral set and therefore has finitely many extremal points. And $M(S)$ is the boundary of $S$.
Oct 31, 2011 at 11:08 comment added Ewan Delanoy @Douglas : you're absolutely right. Note that your family of examples is actually the convex hull between $h_{-1}$ and $h_1$, so perhaps looking at extremal elements inside those maximal elements yields a really finite set this time.
Oct 31, 2011 at 11:06 history edited Ewan Delanoy CC BY-SA 3.0
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Oct 31, 2011 at 9:59 comment added Douglas Zare If $d=1$ and $X=\{-1,0,1\}$ and $f(x) = -x^2$ on these points, then aren't all $h_c$ minimal where $h_c(x) = cx, -1 \le c \le 1$?
Oct 31, 2011 at 9:20 history asked Ewan Delanoy CC BY-SA 3.0