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 canonic 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
It is probably easier to answer the second question before the first.
Answer to the second question.
The argument in the OP already shows that ${\textsf{conv}}(e_i;i\in A) \subseteq F(A)$. Conversely, suppose that $p\in F(A)$. Since $p\in \Delta_n$ we can write $(*) \ : \ p=(p_1,p_2,\ldots,p_n)$ with $\sum_{i=1}^{n}p_i=1$. By definition of $F(A)$, we have $p_i=0$ for any $i\not\in A$, so that $(*)$ reduces to $p=\sum_{i\in A}p_ie_i$ with $\sum_{i\in A}p_i=1$ and this is clearly in ${\textsf{conv}}(e_i;i\in A)$.
Answer to the first question.
Let $\cal F$ be a face of $\Delta_n$. By induction on $r$, we have that if $x_1,x_2,\ldots,x_r \in \Delta_n$ and $\lambda_1,\lambda_2,\ldots, \lambda_r \in{\mathbb R}_{+}$ with $\sum_{k=1}^r \lambda_k=1$ and $\sum_{k=1}^r \lambda_k x_k \in {\cal F}$, that all the $x_k$ are in $\cal F$.
It follows that for any $p\in {\cal F}$ and $i\in{\textsf{supp}}(p)$, we have $e_i\in {\cal F}$. Now, let us put $A=\bigcup_{p\in {\cal F}} {\textsf{supp}}(p)$. By what we have just shown, we have $e_i\in {\cal F}$ for every $i\in A$. If $p\in F(A)$, then $p=\sum_{i\in A}p_ie_i$ with $\sum_{i\in A}p_i=1$ by the second question above, we see that $p\in {\cal F}$ because $\cal F$ is convex. This shows the inclusion $F(A) \subseteq {\cal F}$. The reverse inclusion is clear by definition of $A$, so in the end we have ${\cal F}=F(A)$ as wished.