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In section 2.6.3 of "Convex Optimization" by Boyd and Vandenberghe (p. 55) it is stated that

If $\lambda\succ_{K^*}0$ and $x$ minimizes $\lambda^T z$ over $z\in S$, then $x$ is minimal.

This is complemented a bit later with the assertion

Provided the set $S$ is convex, we can say that for any minimal element $x$ there exists a non-zero $\lambda\succeq_{K^*}0$ such that $x$ minimizes $\lambda^T z$ over $z\in S$.

In my application, $K$ is the non negative orthant, $S$ is indeed convex and $\lambda^T z$ can be shown to have a unique minimum over $z\in S$ for any $\lambda\succeq 0$.

Can I say that $x$ is minimal if and only if $x$ minimizes $\lambda^T z$ over $z\in S$ for some non zero $\lambda\succeq 0$?

Patricio
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  • Please add a little bit more context. Is there any relation between $K$ and $S$? What does it mean "$x$ is minimal"? – gerw Dec 13 '23 at 10:13
  • @gerw Context is that of Boyd and Vabderberghe: $K\subseteq\mathbb R^n$ is a proper cone and $K^$ its dual. $\prec_{k^}$ is the generalized inequality induced by $K^*$. $S\subseteq\mathbb R^n$ is just any set. $x\in S$ is a minimal element of $S$ if $\nexists y\in S\mid y\preceq_K x$ – Patricio Dec 18 '23 at 09:21
  • Your current assumptions imply that $S$ is a singleton, since the function $z \mapsto 0^\top z$ has a unique minimizer over $S$... – gerw Dec 18 '23 at 10:18
  • @gerw. You are right, I should write $\lambda\succeq 0$ and $\lambda\ne 0$. I'll edit – Patricio Dec 18 '23 at 14:19

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