# Convex hull of unions

I would like to find a representation for convex hulls co$(\cdot)$ (see wikipedia for the definition of the convex hull) in normed spaces. Let $A,B\subset X$ be bounded and convex subsets of a normed space $X$, then $$co(A\cup B)=\bigcup_{t\in[0,1]}tA+(1-t)B$$ where $tA=\{ta: a\in A\}$ for $t\in[0,1]$ and where $co(A)=\{\sum_{i=1}^nt_ix_i:t_i\ge0, \sum_{i=1}^nt_i=1\text{ and }x_i\in X\}$. I have convinced myself (intuitively) that this equality holds, but I do not know how to write it formally down.

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What is your definition of $co(S)$ –  no identity Apr 14 '12 at 20:35
@Norbert : You should know from the question that it is the convex hull of $S$. –  Patrick Da Silva Apr 14 '12 at 20:38
Pick an $x$ in the rhs, show it is in the lhs, then repeat the other way around. Alternatively, pick a point in the rhs and show that it must be in any convex set containing $A \cup B$ (hence it would be the convex hull, by definition). –  copper.hat Apr 14 '12 at 20:40
@PatrickDaSilva You should know that there are different definitions of convex hull –  no identity Apr 14 '12 at 20:40
@Norbert : aren't they all equivalent? I mean, isn't the intersection of all convex sets containing some subset $S$ the same as the set of all convex combinations of elements of $S$? –  Patrick Da Silva Apr 14 '12 at 20:41

Let $C_1$ be the left hand side above, and let $C_2$ be the right hand side. It should be clear that $C_2 = \{ t a+(1-t)b | a \in A, b \in B, t \in [0,1]\}$.

First choose $x \in C_2$. Since $x = t a+(1-t)b$, by definition of $co$, $x \in C_1$.

Now choose $x \in C_1$. If $x \in A \cup B$, then $x \in C_2$ trivially, so suppose $x \notin A \cup B.$ By definition, $x= \sum_{i=0}^n t_i x_i$, where $x_i \in A \cup B$, $t_i \in [0,1]$, and $\sum_{i=0}^n t_i = 1$. Let $I_A = \{i | x_i \in A\}$, and similarly for $I_B$. Let $t =\sum_{i \in I_A} t_i$, then it should be clear that $1-t = \sum_{i \in I_B} t_i$. Both $t$ and $1-t$ are non-zero since $x \notin A \cup B.$

Since $A$ is convex, then it contains the point $a = \frac{1}{t}\sum_{i \in I_A} t_i x_i$, and similarly, B contains the point $b = \frac{1}{1-t}\sum_{i \in I_B} t_i x_i$. Finally, since $x=t a +(1-t)b$, it is clear that $x \in C_2$.

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How can one delete one's own comments? –  copper.hat Apr 14 '12 at 21:43
I didn't read the part of the question which said that $A$ and $B$ were convex sets. I am sorry for my comment. Good answer. +1 –  Patrick Da Silva Apr 14 '12 at 21:43
No issue, this is how it works. Proofs are social processes :-). –  copper.hat Apr 14 '12 at 21:46
just after the "n mins ago" there's supposed to be an $\times$ that you can click on. –  Patrick Da Silva Apr 14 '12 at 22:10