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I have always defined the convex hull of a set $X$ to be the smallest convex set that contains $X$.

I am currently reading about holomorphic convexity, and the author has introduced a new definition of the convex hull. That is, the convex hull of a set $X$ is the intersection of all half spaces containing $X$, or, in other words, it is the set of points at which any real linear function takes values not exceeding its maximum on $X$.

Can someone help characterise the equivalence of these three definitions?

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    So then it would be hard to define the convex hull of $\mathbb{R}^n$ (since no half-spaces contain it). – Michael Jun 13 '17 at 23:25
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    @Michael No, the intersection of an empty family of subsets is the whole space. – Robert Israel Jun 13 '17 at 23:26
  • I suppose one could use that convention, it seems strange to just assume that without mentioning it. – Michael Jun 13 '17 at 23:29
  • @Michael Why thinking at once to an extreme case and presenting it as a major flaw to the definition given by theOP, knowing that this definition is a good one ? – Jean Marie Jun 14 '17 at 00:41
  • @JeanMarie : I'm not following your comment. Usually it is good to think of extreme cases to see if things make sense. The definition did not make sense to me and I felt clarification was needed. The Robert Israel answer shows further closedness assumptions are also needed. Are you upset that I did not understand how the case $\mathbb{R}^n$ could be treated? – Michael Jun 14 '17 at 01:58
  • @Michael Don't consider that I have something against you at all. My point was only educational, and I should have said it at first. Your answer would be perfect if you say at first to this student(supposed so...) : yes, in general it is a valid definition, but you should also consider particular cases: what happens if.. – Jean Marie Jun 14 '17 at 09:39
  • @JeanMarie : Thanks for the kind clarification. However, I feel it would only be a valid definition if we remove the extreme case of $\mathbb{R}^n$ with a convention about intersecting no sets, and we remove the extreme cases as in Robert's answer by allowing a "half space" to be a set of the form ${a^T x < b}$, ${a^T x \leq b}$, or ${a^T x < b} \cup A$, where $A$ is a convex subset of the boundary set ${a^Tx = b}$. – Michael Jun 14 '17 at 13:33

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This is in a locally convex topological vector space over $\mathbb R$. The equivalence is for the closed convex hull, using closed half-spaces, closed convex sets, and continuous linear functionals. That is, the closed convex hull of $X$ is the intersection of all closed convex sets containing $X$, and this is also the intersection of all closed half-spaces containing $X$.

Counterexample to your statement: In $\mathbb R^2$, let $X$ be the union of the open half-space $\{(x,y): x > 0\}$ and the point $(0,0)$. This is a convex set; its convex hull is itself. But any half-space containing $X$ contains the whole $y$ axis, so the intersection of these half-spaces is $\{(x,y): x \ge 0\}$.

A closed half-space is a closed convex set, and the intersection of closed convex sets is a closed convex set, so the intersection of all half-spaces containing $X$ is a closed convex set containing $X$.

Conversely, if $C$ is a closed convex set and $p$ a point not in $C$, the Hahn-Banach separation theorem says there is a closed half-space containing $C$ but not $p$. So any point in the intersection of all closed half-spaces containing $X$ is in the closed convex hull of $X$.

Robert Israel
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