# Convexity cones and polar set.

Let $$\mathbb{K} \subseteq \mathbb{R^n}$$ be a closed convex cone.

Show that:

$$\mathbb{K}$$ is a linear subspace $$\iff$$ $$\mathbb{K}^º \cap (-\mathbb{K})=\{0_n\}$$

where $$\mathbb{K}^º:=\{x\in\mathbb{R}^n\,|\,\langle k,x\rangle\leq 0,\forall k\in \mathbb{K}\}$$

I've done this implication $$"\Rightarrow"$$ so easy but I can't get the another one.

We assume that $$K$$ is a closed convex cone in $$\mathbb{R}^n$$. For now, assume that $$K^º\cap -K = \{0_n\}$$ (thus $$K$$ and $$K^º$$ are nonempty). Since $$K$$ is a closed convex cone, so are the sets $$-K$$, $$(-K)^º$$, and their sum. We also have $$-K = ((-K)^º)^º$$ and the following property,
$$(K_1 + K_2)^º = K_1^º\cap K_2^º.$$
$$\{0_n\}=K^º \cap (-K) = -K^º\cap K = (K^º + (-K))^º\\ \iff \mathbb{R}^n = \{0_n\}^º = K^º + (-K)$$
In other words, we want to show that if $$K$$ is satisfies the above, then $$K$$ is a linear subspace. Since we are already assuming $$K$$ is a closed convex cone, being a linear subspace is equivalent to say $$K=-K$$.
For a proof by contrapositive, assume now that $$K$$ is not a subspace, i.e. $$K\neq -K$$. We will show that $$-K + K^º\neq \mathbb{R}^n$$. Let $$u\in K\backslash -K$$ and assume for the sake of contradiction that $$u = a+b$$ with $$a\in -K$$ and $$b\in K^º$$. Then, $$u-a\in K$$ but also $$u-a=b\in K^º$$. Thus $$u-a=0_n\implies u=a$$ but $$a\in -K$$ and $$u\not\in-K$$, a contradiction. Thus we cannot write $$u=a+b$$ with $$a\in K$$ and $$b\in K^º$$ and so the proof is complete.