Maximal Faces Of The Nonnegative Cone

Describe the maximal faces of $P_{n,2d}$. Show that the maximal faces of $P_{n,2d}$ have codimension exactly $n$ in $\mathbb{R}_{n,2d}$. Hint: Use Exercise Dual Faces.

Solution:
For $x \in \mathbb{R}^n$ we define $F_x:= \{ f \in P_{n,2d} \mid f(x)=0 \}$.
Claim: The set $\{F_x \mid x \in \mathbb{R}^n\}$ is the set of all maximal faces of $P_{n,2d}$.
Let $F$ be a maximal face. Exercise 1.2 (b) implies that $F$ ist exposed. Thus there exists an $l \in P_{n,2d}^*$ such that
$F=\{f \in P_{2,d} \mid l(f)=0 \}$. Write $l=\lambda_1 l_{v_1} + \dots \lambda_r l_{v_r}$ for points $v_i$ and positive $\lambda_i$. Since $F \subseteq F_{v_1}$ and $F$ maximal, we have $F=F_{v_1}$.
Conversely, each $F_x$ is contained in a maximal face $F$. It is $F=F_v$ for a $v \in \mathbb{R}^n$ and thus $F_x=F_v$ maximal.

Claim: The maximal faces have codimension $n$.
Let $F=F_x$ be a maximal face. Each $f \in F_x$ attains a minimum in $x$. Hence $F$ is contained in the linear space $H:=\{ f \in \mathbb{R}_{n,2d} \mid \nabla f(x)=0 \}$, which implies that the codimension is at least $n$. Note that $\nabla f(x)=0$ already implies that $f(x)=0$ (Euler's identity). Left to show: Equality.