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I'm trying to show the following :

Let $K,L$ two closed convexes of $\mathbb{R}^2,O=(0,0)$

  • If $O\notin K$ then there exists a straight line $D$ going through $O$ such that $K$ is in one of the half of plane defined by $D$

  • If $K$ is bounded and does not intersect with $L$ then there exists a straight line $D$ such that $K$,$L$ are in two distinct halves of plan defined by $D$

Those two properties are really easy to understand intuitively or with a graph, but I haven't been able to find a proper mathematical proof.

Hippalectryon
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  • Are the sets assumed open or closed? If not, you have to be careful: Consider the example where $K$ consists of points with $x^2+y^2<1$ and either $y>0$ or $y=0$ and $x>0$. Then let $L=-K$ (i.e., the reflection of $K$ through the origin). – Harald Hanche-Olsen Dec 15 '14 at 20:48
  • @HaraldHanche-Olsen They're closed, thanks, I forgot to add that. – Hippalectryon Dec 15 '14 at 20:49
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    By the way, it seems that this is a general situation not only for the plane, but for a locally convex space $X$, satisfying the Hahn-Banach Theorem. Since $K$ is closed, by Hahn-Banach separation Theorem there exist a real number $\varepsilon>0$ and a linear functional $f$ on $X$ such that $f(O)=0$ and $f|K>\varepsilon$. Geometrically it should mean that there is a hyperplane strictly separating the point $O$ from the set $K$. – Alex Ravsky Dec 15 '14 at 21:05
  • @AlexRavsky Absolutely. However, the Hahn–Banach separation theorem is much easier in Hilbert spaces! The proof outlined by me and elaborated by Mike Miller works equally well in that more general setting. – Harald Hanche-Olsen Dec 16 '14 at 08:50

2 Answers2

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Two hints for the price of one: For the first one, consider the point $P\in K$ closest to the origin, and take $D$ to be normal to the line from $O$ to $P$.

For the second one, apply the first one to the set $K-L=\{u-v\colon u\in K, v\in L\}$.

Harald Hanche-Olsen
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Here's a solution to the first problem. (For these sorts of problems I suggest drawing pictures - it's amazing how helpful blobs on a page can be.) Harald's gives you the second.

Because $K$ is closed, there's a point in $K$ of minimal distance to the origin; call it $v=(x,y)$. Then consider the normal line $D = \text{span}(-y,x)$. It suffices to show that $K \cap D = \varnothing$. Suppose not; say $(-ty,tx) \in K$. Because $v$ had minimal distance to the origin, $t \geq 1$. By convexity, every point of the form $s(-ty,tx) + (1-s)(x,y) \in K$ for $0 \leq s \leq 1$. Let's take the norm of this point. After some algebra, we obtain $$\sqrt{(1-s)^2+s^2t^2}\|v\|.$$ But for $s$ sufficiently small (pick $0 < s < \frac{2}{1+t}$), the first term is less than 1; this contradicts that $v$ was the point of $K$ of least norm. So $K \cap D = \varnothing$ as desired.