Show that $a^Tz\ge b$ for $z\in K$ if $a^Tz>b$ for $z\in \operatorname{int}(K)$ with $K$ being a convex cone

Question

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.

Answer 1

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.

Answer 2

$\newcommand\int{\operatorname{int}}\newcommand\cl{\operatorname{cl}}$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, $a\in\cl(a,b]\subseteq\cl(\int(K_1))$.