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Consider the support function of a convex set $A\in \mathbb{R}^n$, defined as the map $$ x\in \mathbb{R}^n \mapsto \sup_{a\in A}x'a $$

Suppose now that $A$ is a subset of the $(n-1)$-dimensional simplex. In particular, $\forall a\equiv (a_1,...,a_n) \in A$, $a_i\in [0,1]$ and $\sum_{i=1}^n a_i=1$.

Question: Is it correct to say the the domain of the support function is the unit sphere in $\mathbb{R}^n$, i.e. $\{x\in \mathbb{R}^n: x'x=1\}$? Could you give an intuition of why correct or wrong?

A related question is here but I could not understand how to apply the answer to my question.

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  • $\begingroup$ Suppose $a\in\mathbb{R}^n$ is in the domain of the support function. Isn't the vector $2a$ in the domain too? $\endgroup$ – uniquesolution Aug 8 at 15:09
  • $\begingroup$ Thanks, sorry I don't get your comment, I'm a beginner of convex analysis. I've just read a book saying that this follows from the fact that the support function is positively homogeneous, but I don't get it. $\endgroup$ – STF Aug 8 at 15:17
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The domain of the support function is the set of points in $\mathbb{R}^n$ where it is finite. You raised the question whether the domain of the support function is confined to the unit sphere. My comment tells you that if any vector is in the domain, then you can multiply it by $2$ and it will still be in the domain, which goes to show that the domain is not confined to the unit sphere alone. What is true, however, is that if you know the value of the support function for every unit vector $u\in\mathbb{S}^{n-1}$, then you know it for all other vectors, because the support function is $1$-homogeneous, that is, $$h(t\vec{x})=th(\vec{x})\quad\forall t\geq 0$$

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