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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.

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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,

$$(K_1 + K_2)^º = K_1^º\cap K_2^º.$$

Applying this to our assumption gives,

$$\{0_n\}=K^º \cap (-K) = -K^º\cap K = (K^º + (-K))^º\\ \iff \mathbb{R}^n = \{0_n\}^º = K^º + (-K)$$

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.

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