Let $F$ be a face of a convex cone $C$. Define the dual face $F^\Delta$ of $C^\ast$ by

$F^\Delta = \{ l \in C^\ast | l(x) = 0, \forall x \in F\}$.

(a) $F^\Delta$ is an exposed face of $C^\ast$.

(b) Part (a) implies that any face $F$ of $C$ is contained in an exposed face.

SOLUTION

(a)Fix $x$ in the relative interior of $F$. We will show that $F^\Delta =\operatorname{face}_x(C^\ast)=\{l \in C^\ast | l(x) \leq y(x), \forall y \in C^\ast\}= \{l \in C^\ast | l(x) = 0\}$.

From the definitions it is clear that $F^\Delta \subset \operatorname{face}_x(C^\ast)$. Suppose that $l \in \operatorname{face}_x(C^\ast)$ and $y \in F$. Since $x$ belongs to the relative interior of $F$, there exists $\epsilon > 0$ such that $x-\epsilon y \in F$. Then we have $l(y) = \frac{1}{\epsilon}l(\epsilon y)\leq \frac{1}{\epsilon}l(x)=0$. On the other hand, $l(y) \geq 0$. We have shown that for any $y \in F$, $l(y)=0$. In other words, $l \in F^\Delta$. Hence $F^\Delta$ is as claimed.

(b) It is true in general that $F \subset (F^\Delta)^\Delta$.-MH