Specifically, I have the following in mind:

  • Consider the positive quadrant as the domain (i.e. $(x,y)$ with $x,y\geq 0 $).
  • Consider a downward sloping line that divides this domain into two sets (the set that lies below the line and the set that lies above)

Both of these sets are convex.

However, suppose that we introduce a kink in this line. That is, suppose that at some point $x$, the absolute value of the slope increases (so it becomes more downward sloping).

Now the set of points below this line is still convex, but the set of points above the line is no longer convex

  • What is the difference between the two sets that causes one to stop being convex and the other remains convex?

I am asking because in the example above we had a straight line. However, I feel like the result would hold even if the boundary was a curve (i.e., if a downward sloping boundary of a convex set suddenly becomes steeper at some point, the set will stop being convex). I have no idea how one would go about proving this, though. (Maybe choose 2 points and find a convex combination not in the set?)

(sorry if kink is not the correct term)

  • $\begingroup$ If the boundary between two sets is any kind of curve other than a single straight line (segment), then the two sets can't both be convex. $\endgroup$ – Gerry Myerson Apr 29 at 3:54
  • $\begingroup$ @GerryMyerson Sorry, I was not clear enough in my question. If if it is the upper set that is convex, and the boundary between them is downward sloping, I feel like making it (the boundary) steeper at some point will make the upper set non-convex as well. As an aside, if you could point me towards a proof of your claim (I believe it but I want to see a proof), I would appreciate it. I feel like such a proof is similar to what a proof would look like for the case I described in my question. $\endgroup$ – user106860 Apr 29 at 14:08
  • 1
    $\begingroup$ Assume the boundary is not a single straight line (segment). Let $A,B,C$ be three points on the boundary that don't all lie on one straight line, with $B$ between $A$ and $C$. Consider the line segment joining $A$ and $C$. It can't lie entirely in both sets, since to do that it would have to form part of the boundary, and thus would have to go through $B$, which it doesn't by our hypothesis. So the two sets can't both be convex. $\endgroup$ – Gerry Myerson Apr 29 at 22:57
  • $\begingroup$ That is pretty elegant. Thanks. I guess the difference with my question is that my question pertains to which set is not convex when the boundary is not a single straight line (segment) $\endgroup$ – user106860 Apr 30 at 14:15

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.