# On the dual face of a subset of a convex cone

Suppose $C$ is a convex cone of a finite dimensional normed space $X$. Suppose $F$ is a face of $C$, and $x_0$ is an interior point of $F$. The dual face of $F$ is then given by $$F' := \{ y \in C^\circ: \langle x,y \rangle = 0 \text{ for all } x \in F \}$$ why is this set equal to $G := \{ y \in C^\circ: \langle x_0,y \rangle = 0\}$? By definition we have $F' \subset G$, but what about the other inclusion?

And why is $F''' \subset F'$?

• What is $F'''$?
– gerw
Oct 29, 2015 at 20:22
• It's the third dual of $F$. And now that you mention it, I think it follows from the definition. My bad. Oct 29, 2015 at 20:31
• This is by definition. The other inclusion is also not hard.
– gerw
Oct 29, 2015 at 20:45

Let $v \in G$ be given. Then, $v \in C^\circ$ and $(v,x_0) = 0$. Now, let $f \in F$ be given. Since $x_0$ is a (relative) interior point of $F$, $x_0 \pm \varepsilon \, f \in F$ for some small $\varepsilon > 0$. Hence, $0 \ge (v, x_0 \pm \varepsilon \, f) = (v, \pm \varepsilon \, f)$. This yields $(v,f) = 0$. Now, the assertion follows from $v \in C^\circ \cap F^\perp = F'$.