Carathéodory's Theorem (Convex Analysis)

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

Let $\mathbb x \in \map {\operatorname {conv} } E$, where $\map {\operatorname {conv} } E$ denotes the convex hull of $E$.

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

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

Proof
Since $\mathbb x \in \map {\operatorname {conv} } E$, $\mathbf x$ is a convex combination of points in $E$.

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

with:
 * $\gamma_i \ge 0$


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

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 $\set {\mathbf y_i: i \le k_s}$ is affinely dependent.

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

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

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

Then:
 * $\ds \mathbf x = \paren {\sum_{i \mathop = 1}^{k_s} \gamma_i \mathbf y_i} + 0 = \paren {\sum_{i \mathop = 1}^{k_s} \gamma_i \mathbf y_i} - \paren {\frac {\gamma_j} {\alpha_j} \sum_{i \mathop = 1}^{k_s} \alpha_i \mathbf y_i}$

Rearrange to get:


 * $\ds \mathbf x = \paren {\sum_{i \mathop = 1}^{k_s} \paren {\gamma_i - \frac {\gamma_j} {\alpha_j} \alpha_i} \mathbf y_i}$

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:
 * $\paren {\gamma_i - \dfrac {\gamma_j} {\alpha_j} \alpha_i} \ge 0$

For $i = j$:
 * $\paren {\gamma_i - \dfrac {\gamma_j} {\alpha_j} \alpha_i} = 0$

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

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

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