# Convex combination

Assume that $I$ is a countable set, and we have $u_i\in \mathbb{R}^n$ for $i\in I$. Suppose that $v=\sum_{i\in I} a_i u_i$ and $\sum_{i\in I}a_i=1$ and $a_i\geq 0$.

Can one show that there exists a finite $J\subseteq I$ and $b_j$ for $j\in J$ such that $\sum_{j\in J}b_j=1$, $b_j\geq 0$, and $v=\sum_{j\in J} b_j u_j$?

Is this some well-known result?

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Carathéodory's theorem almost says that you can get away with just $n+1$ elements in $J$. Except here you start with an infinite convex combination, and then the theorem does not apply, at least not without further work. – Harald Hanche-Olsen Oct 4 '12 at 10:41
I second joriki's comment. Do you not know about accepting answers to questions you post on this site? – Gerry Myerson Oct 4 '12 at 12:34

Without loss of generality, assume $a_i\gt0$ and let $I=\mathbb N$. For any subset $A\subseteq\mathbb N$, call
$$W_A=\frac{\sum_{i\in A}a_iu_i}{\sum_{i\in A}a_i}$$
the weighted average for $A$. If the $u_i$ all lie in some proper affine subspace of $\mathbb R^n$, we can restrict to that subspace, so without loss of generality we can assume that their affine span is $\mathbb R^n$. Then at some finite index $k$ there are positive contributions from $n+1$ affinely independent vectors $A=\{u_{i_1},\dotsc,u_{i_{n+1}}\}$, and $W_A$ is therefore in the interior of their convex hull $H$. Thus, since both $\sum_ia_i$ and $\sum_ia_iu_i$ converge (so their tails converge to zero) there is some $l\ge k$ such that the weighted average remains inside $H$ if the tail $\{i\in\mathbb N\mid i\ge l\}$ is added to $A$. Then the desired finite sum can be obtained by replacing the contributions from the tail by corresponding linear combinations of $u_{i_1},\dotsc,u_{i_{n+1}}$.
I don't understand the argument. Let $A^\prime=\{i_1,\dots, i_{n+1}\} \cup\{i\in\mathbb{N}: i\geq l\}$. How do we know that $v=W_{A^\prime}$? – Byron Schmuland Jul 24 '14 at 0:21