Separating hyperplane

Question

Let $K_1,K_2$ be disjoint convex sets in $\mathbb C$. Let $z_1\in\partial K_1,z_2\in\partial K_2$ be the minimizers of $\mbox{dist}(K_1,K_2)=\inf{|x-y|}$ where $x\in K_1$ and $y\in K_2$. Is it true that we can pick the separating hyperplane in this case to be orthogonal to the line connecting $z_1$ and $z_2$?

Answer 1

Edit: never mind, your sets are disjoint, so they can be separated properly. Then you can use the theorem stating that: if $C,D$ are non-empty sets in $\mathbb{R}^n$ then there exists a hyperplane separating $C,D$ properly if and only if there exists a vector v such that: $$\inf\{x\cdot v, x\in C\}\geq \sup\{y\cdot v, y\in D\} \wedge \sup\{x\cdot v, x\in C\}>\inf\{y\cdot v, y\in D\}$$ Then the hyperplane separating them has $v$ as normal.

So yes you can pick it to be orthogonal to that line, but you can also pick a hyperplane that is not orthogonal and it still may separate them porperly as long as it fulfills the above requirements in the theorem (basically what you require is unnecessarily strong).

To see why your assumption is correct, you can use the projection theorem. Then $(z_2-z_1) \cdot (x-z_1)\leq 0,\forall x\in K_1$ and $(z_1-z_2) \cdot (y-z_2)\leq 0,\forall y\in K_2$. Thus if you pick any point $z = (1-\lambda)z_1+\lambda z_2$, and use $z_1-z_2$ as the hyperplane normal, then $(y-z)\cdot(z_1-z_2)\leq 0$ and $(x-z)\cdot(z_1-z_2)\geq 0$ and the inequality is strict for $\lambda \in (0,1)$ since the sets are disjoint.

Edit2: Fixed proper/strong separation mistake.

Edit3: Proof clarification: You have the minimizers $z_1 \in \partial K_1 ,z_2 \in \partial K_2$, but from their definition and the definition of projection, you have that the projection of $z_1$ onto $cl(K_2)$ is $z_2$ and the projection of $z_2$ onto $cl(K_1)$ is $z_1$. Applying the projection theorem to $z_1,z_2$ we get: $$(z_2-z_1)\cdot(x - z_1)\leq 0, \forall x \in cl(K_1)$$ $$(z_1-z_2)\cdot(y - z_2)\leq 0, \forall y \in cl(K_2)$$

Having these results we want to show that $(z_2-z_1)\cdot(x-z) \leq 0, \forall x \in cl(K_1)$: $$(z_2-z_1)\cdot(x-z) = (z_2-z_1)\cdot(x-(1-\lambda)z_1 - \lambda z_2) = $$ $$(z_2-z_1)\cdot(x-z_1) + \lambda(z_2-z_1)\cdot(z_1-z_2) = $$ $$(z_2-z_1)\cdot(x-z_1) - \lambda||z_2-z_1||^2 \leq 0 $$ Where I have used in the last line that $(z_2-z_1)\cdot(x-z_1) \leq 0$ and the norm being non-negative and $\lambda \geq 0$. Similarly $(z_1-z_2)\cdot(y-z) \leq 0, \forall y \in cl(K_2)$: $$(z_1-z_2)\cdot(y-z) = (z_1-z_2)\cdot(y-(1-\lambda)z_1 - \lambda z_2) =$$ $$(z_1-z_2)\cdot(y-z_2) + (1-\lambda)(z_1-z_2)\cdot(z_2-z_1) = $$ $$(z_1-z_2)\cdot(y-z_2) - (1-\lambda)||z_1-z_2||^2 \leq 0$$ Where I have used in the last line that $(z_1-z_2)\cdot(y-z_2) \leq 0$ and the norm being non-negative and $\lambda \geq 0$. Thus $(z_1-z_2)\cdot(y-z) \leq 0, \forall y \in cl(K_2)$ and $(z_2-z_1)\cdot(x-z) \leq 0, \forall x \in cl(K_1)$ or equivalently $(z_1-z_2)\cdot(x-z) \geq 0, \forall x \in cl(K_1)$. However these two conditions simply mean that the plane defined as $(z_1-z_2)\cdot w = (z_1-z_2)\cdot z$ separates $cl(K_1)$ and $cl(K_2)$. Since the plane shares a point with $cl(K_1)$ or $cl(K_2)$ only when $\lambda$ is zero or one, and the or is exclusive, then it separates the two sets properly.