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Let's denote by $\cal{F}$ the family of all affine functions $f: \mathbb{R}^n \rightarrow\mathbb{R}$. Let $A\subset \mathbb{R}^n$.

What is a connection between the following definitions of convex hull of $A$:

$conv_1(A)=\{\sum_{i=1}^n \alpha_i a_i: \alpha_i \in [0,1], a_i \in A, \textrm{ for } i=1,...,n; \sum_{i=1}^n \alpha_i=1, n \in \mathbb{N} \},$

$conv_2(A)= \bigcap_{f \in \cal{F} } \{ x\in \mathbb{R}^n : |f(x)| \leq \sup_{y \in A} |f(y)| \}.$

Thanks.

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    Sorry for mistakes. In definition of $conv_1$ sum of coefficients should be 1. Moreover affine function have values in $R$.2012-03-13

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I guess the functions in $\mathcal{F}$ should take their values in $\mathbb{R}$. In that case, the two notions coincide when $A$ is compact.

The set $conv_2(A)$ is an intersection closed halfspaces and hence a closed and convex set. For every $a\in A$ and $f\in\mathcal{F}$, $|f(a)|\leq \sup_{y\in A}|f(y)|$, so $A\subseteq conv_2(A)$.

That $conv_2(A)$ contains no point not in the closure of $A$ is a bit more involved. It follows from the separating hyperplane theorem.

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    It is a standard fact the convex-hull (according to $conv_1$) of a compact set in $\mathbb{R}^n$ is compact. We want to show that a point $z$ not in the convex hull is not in $conv_2(A)$. For this we separate a closed ball around $z$ from the convex hull with a hyperplane. This gives us an affine function $f$ increasing in the direction of $z$, so f(z)>\sup_{y\in A}f(y), which shows that $z\notin conv_2(A)$.2012-03-13