Faces of a simplex, convex hull of the canonical basis of \$\mathbb{R}^n\$

[Samwise1]

Let $I_n := \{1,2,...,n \}, \ p \in \Delta_n = \{(p_1, ..., p_n) \ | \ p_i \ge 0, \sum_{i=1}^n =1\}$

$ \text{supp (p)}= \{ i \in I_n \ | \ p_i \neq 0\}$

For a convex set $C$ we define $F$ to be its face if $F$ is convex and $\forall x,y \in C, \lambda \in (0,1) : \lambda x + (1- \lambda ) y \in F \Rightarrow x,y \in F$

For $A \subset I_n$ we define $F(A) : = \{ p \in \Delta_n \ | \ \text{supp(p)} \subset A \}$

I have two things to prove:

$1) F \text{ is a face } \iff \exists A \subset I_n : F = F(A)$

Here, $\Leftarrow$ is no problem. When it comes to $\Rightarrow$, I've been thinking about indirect proof. That is, let's assume that $F$ is a face but for every $A \subset I_n, \ F(A) \neq F$. But then, if we take $A$ to be a subset of canonical basis of $\mathbb{R}^n$ we get a contradiction. Is that correct?

$2) \text{conv} (\{ e_i, i \in A \} ) = F(A)$

here $e_i$ are elements of the canonical basis of $\mathbb{R}^n$

Now, when it comes to $\subset, \ F(A)$ is a convex set and evidently $\{ e_i \ , \ i \in A \} \subset F(A)$ so $F(A)$ is one of the convex sets containing $\{ e_i \ , \ i \in A \} $, so the interesection of those sets must be a subset of $F(A)$.

I have problems proving the other inclusion.

Could you help me with that?

Thank you

 

[jvicens]

!For the first part, your idea of using an indirect proof is good. However, instead of assuming that $F(A) \neq F$ for all $A \subset I_n$, you can assume that $F(A) \neq F$ for some specific $A$, say $A=\{1,2\}$. Then, you can use the definition of a face to show that $F$ must contain all convex combinations of elements in $F(A)$, which would lead to a contradiction.

For the second part, one direction is as you mentioned. For the other direction, we need to show that any $p \in F(A)$ can be written as a convex combination of elements in $\{e_i \ | \ i \in A\}$. Since $p \in F(A)$, $\text{supp}(p) \subset A$. This means that $p$ can be written as a convex combination of elements in $A$, i.e. $p = \sum_{i \in A} \lambda_i e_i$ for some $\lambda_i \geq 0$ with $\sum_{i \in A} \lambda_i = 1$. Since $A \subset I_n$, we can extend this to a convex combination of all the canonical basis elements, i.e. $p = \sum_{i \in I_n} \gamma_i e_i$ where $\gamma_i = \lambda_i$ for $i \in A$ and $\gamma_i = 0$ for $i \in I_n \setminus A$. This shows that $p \in \text{conv}(\{e_i \ | \ i \in A\})$, completing the proof.