Carathéodory's Theorem (Convex Analysis)

Theorem
Let $E \subset \R^\ell$.

Let $\mathbb x \in \operatorname{co} \left({E}\right)$.

Then $\mathbf x$ is a convex combination of affinely independent points of $E$.

In particular, $\mathbf x $ is a convex combination of at most $\ell+1$ points of $E$.

Proof
Since $\mathbb x \in \operatorname{co} \left({E}\right)$, $\mathbf x$ is a convex combination of points in $E$.

From the definition of convex combination:
 * $\displaystyle \mathbf x = \sum_{i \mathop = 1}^k \gamma_i \mathbf y_i$

with:
 * $\gamma_i \ge 0$


 * $\displaystyle \sum_{i \mathop = 1}^k \gamma_i = 1$

and:
 * $\mathbf y_i \in E$

Let $K \subset \N$ be the set of all possible $k$ such that $\mathbf x$ is a convex combination of $k$ elements of $E$.

By the Well-Ordering Principle, $K$ is well-ordered.

$K$ has a smallest element $k_s$ by the definition of well-ordered sets.

Suppose that the set $\left\{{\mathbf y_i: i \le k_s}\right\}$ is affinely dependent.

From the condition for affinely dependent set, there exists a set $\left\{{\alpha_i: i \le k_s}\right\}$ such that for some $\alpha_i > 0$:
 * $\displaystyle \sum_{i \mathop = 1}^{k_s} \alpha_i \mathbf y_i = 0$

and:
 * $\displaystyle \sum_{i \mathop = 1}^{k_s} \alpha_i = 0$

Pick the smallest $\dfrac {\gamma_j} {\alpha_j} > 0$.

Then:
 * $\displaystyle \mathbf x = \left({\sum_{i \mathop = 1}^{k_s} \gamma_i \mathbf y_i}\right) + 0 = \left({\sum_{i \mathop = 1}^{k_s} \gamma_i \mathbf y_i}\right) - \left({\frac {\gamma_j} {\alpha_j} \sum_{i \mathop = 1}^{k_s} \alpha_i \mathbf y_i}\right)$

Rearrange to get:


 * $\displaystyle \mathbf x = \left({\sum_{i \mathop = 1}^{k_s} \left({\gamma_i - \frac {\gamma_j} {\alpha_j} \alpha_i}\right) \mathbf y_i}\right)$

Since $\gamma_j > 0$ is made the smallest and $\alpha_j > 0$ is made the greatest:
 * $\dfrac {\alpha_i} {\alpha_j} \gamma_j \le \gamma_j \le \gamma_i$

for all $i \le k_s$.

Thus:
 * $\left({\gamma_i - \dfrac {\gamma_j} {\alpha_j} \alpha_i}\right) \ge 0$

For $i = j$:
 * $\left({\gamma_i - \dfrac {\gamma_j} {\alpha_j} \alpha_i}\right) = 0$

For $i \ne j$, let:
 * $\gamma'_{i'} \equiv \gamma_i - \dfrac {\gamma_j} {\alpha_j} \alpha_i$

We can express:
 * $\displaystyle x = \sum_{i' \mathop = 1}^{k_s-1} \gamma'_{i'} \mathbf y_i$

which contradicts the assumption that $k_s = \min K$.