# Convex Optimization: Separation of Cones

I am trying to solve Exercise 2.39 at Boyd and Vandenberghe's Convex Optimization book. In one source, the answer is given as:

2.39 Separation of cones. Let $$K$$ and $$\tilde K$$ be two convex cones whose interiors are nonempty and disjoint. Show that there is a nonzero $$y$$ such that $$y\in K^*$$, $$-y\in\tilde K^*$$.

Solution. Let $$y\ne0$$ be the normal vector of a separating hyperplane separating the interiors: $$y^\top x\ge\alpha$$ for $$x\in\boldsymbol{\operatorname{int}}K_1$$ and $$y^\top x\le\alpha$$ for $$x\in\boldsymbol{\operatorname{int}}K_2$$. We must have $$\alpha=0$$ because $$K_1$$ and $$K_2$$ are cones, so if $$x\in\boldsymbol{\operatorname{int}}K_1$$, then $$tx\in\boldsymbol{\operatorname{int}}K_1$$ for all $$t>0$$. This means that $$y\in(\boldsymbol{\operatorname{int}}K_1)^*=K_1^*,\quad-y\in(\boldsymbol{\operatorname{int}}K_2)^*=K_2^*.$$

I don't get the final part $$(\boldsymbol{\operatorname{int}}K)^* = K^*$$. Why is this a valid equality? My idea is because if a boundary point of $$K$$ has $$y^\top x < 0$$ this would contradict $$\boldsymbol{\operatorname{int}}K$$ being an open set, but I can not formalize this.

Ok, after seeing the wrong attempt below which has been edited multiple times, I believe it is time to close this question. I will just leave my attempt:

Assume $$K^* \neq (\operatorname{int}K)^*$$, so $$\exists x_0 \in \operatorname{bd} K: \ x_0^\top y < 0$$. Because of the strict inequality, we know that we can take a very small ball around $$x_0$$, say $$B(x_0)$$ and all the points $$x' \in B(x_0)$$ will have $$x'^\top y < 0$$. By the definition of the boundary, we have $$(B(x_0) \cap \operatorname{int} K)\neq \emptyset$$ hence for some $$x \in \operatorname{int}K$$ we have $$x^\top y < 0$$, which is a contradiction. Hence $$K^* = \operatorname{int}K^*$$.

A new approach focusing only on $$(\boldsymbol{\operatorname{int}}K)^* = K^*$$, since that seems to be the biggest problem to you.

From section 2.6 of Convex Optimization (Boyd, Vanderberghe) we have that:

• (3) $$K^*$$ is closed and convex.
• (1) $$K^{**}$$ is the closure of the convex hull.
• (2) For $$K$$ convex and closed, $$K^{**} = K$$.

Now, we know that $$K$$ is the closure of the convex hull of $$\boldsymbol{\operatorname{int}}K$$, that is, $$(\boldsymbol{\operatorname{int}}K)^{**}=K$$ (1).
But, $$K$$ is convex and closed, such that $$K = K^{**} = (\boldsymbol{\operatorname{int}}K)^{**}$$ (2).
Both $$K^{*}$$ and $$(\boldsymbol{\operatorname{int}}K)^{*}$$ are closed and convex (3), therefore, $$K^{**} = (\boldsymbol{\operatorname{int}}K)^{**} \iff K^{*} = (\boldsymbol{\operatorname{int}}K)^{*}$$ (X).