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I'm working on a proof and got stuck with the following situation. Given are two convex cones $K_1, K_2\in\mathbb{R}^n$ with non-empty interior such that $\operatorname{int}(K_1)\cap \operatorname{int}(K_2)=\emptyset$. With the hyperplane separation theorem we can find $a\in\mathbb{R}^n$ and $b\in\mathbb{R}$ such that $$a^Tx_1>b,\quad a^Tx_2<b\quad\forall x_1\in \operatorname{int}(K_1),\ x_2\in \operatorname{int}(K_2).$$ Is there an elegant way to show that $a^Tz\ge b$ for $z\in K_1\setminus \operatorname{int}(K_1)$ without invoking any hyperplane separation theorems for cones that are not open?

I tried to proof this by contradiction, assuming the existance of $z\in K_1\setminus\operatorname{int}(K_1)$ such that $a^Tz<b$, but I did not manage to do so.

Felix Crazzolara
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  • to understand you correctly: you don't want to use any hyperplane separation theorem at all or just not one specific for closed cones and using the one for open cones like you did can be part of the proof? – Targon Mar 08 '20 at 19:06
  • Isn't this true for all "nice enough" sets $K$ since the inner product is continuous ? I mean just construct a sequence of points $z_n\in \operatorname{int}(K)$ going to $z$ (such a sequence must exist so $K$ must be reasonably nice) and the inequality $a^\top z_n=\langle a, z_n\rangle\geq b$ will be preserved in the limit – Maximilian Janisch Mar 08 '20 at 19:07
  • @Targon I did use a hyperplane separation theorem for open convex sets to show that the interiors can be strictly separated. However I don't want to use any hyperplane separation theorems to show that $a^Tz\ge b$ for $z$ in $K_1\setminus\operatorname{int}(K_1)$, but rather show this using more elemental tools so to say. – Felix Crazzolara Mar 08 '20 at 19:14
  • @MaximilianJanisch is correct. "reasonably nice" could for example mean $\partial K = \overline{K} \setminus \text{int} K$. – Targon Mar 08 '20 at 19:17
  • @Targon I think rather something like $K=\overline{\operatorname{int} K}$ since $\partial K=\overline K\setminus\operatorname{int}K$ is always true – Maximilian Janisch Mar 08 '20 at 19:18
  • @MaximilianJanisch I don't want to make such kind of assumptions. I also don't think that we need to since I'm quite convinced that the above holds due to the convexity of $K_1$, but I don't know how to write it down. For example consider the projection of the interior of $K_1$ onto the separating hyperplane. Any $z$ on the "wrong" side must have rays going to the bounds of the projection due to the convexity of $K_1$ this in turn would imply that $\operatorname{int}(K_1)\cap{x:a^Tx<b}\ne\emptyset$ and thus lead to a contradiction, but I don't know how to write that properly. – Felix Crazzolara Mar 08 '20 at 19:22
  • @FelixCrazzolara It is easy to prove that if $K$ is a convex closed set with non-empty interior then $K=\overline{\operatorname{int} K}$ (see here) so I think you can even drop the cone assumption PS: ETH – Maximilian Janisch Mar 08 '20 at 19:24
  • @MaximilianJanisch I never assumed $K$ to be closed. – Felix Crazzolara Mar 08 '20 at 19:27
  • @FelixCrazzolara But your statement will hold also for $\overline K$ which is even better – Maximilian Janisch Mar 08 '20 at 19:27
  • If you want I can write a more explicit answer – Maximilian Janisch Mar 08 '20 at 19:29
  • Ok, if we can show what I asked for, for any $z$ in the closure of $K_1$ than my question is answered, if that's what you're saying. That would certainly satisfy me. It don't understand though how we properly show this and would appreciate to see a proper proof of this. – Felix Crazzolara Mar 08 '20 at 19:36
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    The set ${z\in\Bbb R^n:a^Tz\geq b}$ is closed and contains $\operatorname{int}(K_1)$, hence contains $K_1$ as well. – Fabio Lucchini Mar 08 '20 at 21:40
  • Have you decided to accept an answer to this question @FelixCrazzolara ? – Maximilian Janisch Mar 10 '20 at 14:58

2 Answers2

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$\newcommand\int{\operatorname{int}}\newcommand\cl{\operatorname{cl}}$The set $\{z\in\Bbb R^n:a^Tz\geq b\}$ is closed and contains $\int(K_1)$, hence contains $\cl(\int(K_1))$ as well.

We claim that $K_1\subseteq\cl(\int(K_1))$. Let $a\in K_1$ and for $t\in\Bbb R$ let \begin{align} &\lambda_t:\Bbb R^n\to\Bbb R^n&x\mapsto (1-t)a+tx \end{align} If $0\leq t\leq 1$, then $\lambda_t$ maps $K_1$ into itself. If $t\neq 0$, then $\lambda_t$ is an homeomorphism.

Let $b\in\int(K_1)$ and let $U$ be an open neighbourhood of $b$ contained in $K_1$. Then \begin{align} (a,b] &=\{\lambda_t(b):0<t\leq 1\}\\ &\subseteq\bigcup_{0<t\leq 1}\lambda_t[U]\\ &\subseteq K_1 \end{align} and $\bigcup_t\lambda_t[U]$ is open because union of open subsets. Thus $(a,b]\subseteq\int(K_1)$. Consequently, $a\in\cl(a,b]\subseteq\cl(\int(K_1))$.

Fabio Lucchini
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Proposition. Let $K\subset \mathbb R^n$ be a convex set such that for some $a\in\mathbb R^n$ and $b\in\mathbb R$ we have $$\langle z,a\rangle\geq b$$ for all $z\in \overset{\circ}K$. Then $\langle z,a\rangle\geq b$ for all $z\in K$.

Proof. I will prove the result for all $z\in\overline K$, which is stronger. Define $C=\overline K$. Then $C$ is convex and closed. Also, we have $\overset{\circ}C=\overset{\circ}K$ (see here). Since $C$ is convex and closed, we have $\overline{\overset\circ C}=C$ (see here).

Now let $z\in C$. By the previous fact, there is a sequence $z_n\in \overset\circ C=\overset\circ K$ converging to $z$. For each of the $z_n$, we have $\langle z_n,a\rangle\geq b$ by assumption so that by continuity of the inner product

$$ \langle z,a\rangle=\lim_{n\to\infty} \langle z_n,a\rangle\geq b. \square $$ A remark about this proof: You could probably shorten it a bit by directly applying the topological arguments that I just referenced above.

Maximilian Janisch
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