Convex Optimization: Separation of Cones

Question

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.

Answer 1

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 $K^* = \operatorname{int}K^*$.

Answer 2

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, $K^{**} = K$.

I will add a conjecture:

• (X) The dual cone of a closed and convex cone is unique.

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).