2

Hi my question is quite straightforward, if we have two disjoint compact convex sets A and B, is their minkowski difference A-B then convex again?

Thanks!

1233023
  • 543
  • Do you know, for instance, that the Minkowksi sum of two convex sets is convex? The proof here is fairly trivial given that lemma. (And that the map $f(x)=-x$ preserves convexity - or, as happens to be the case, any linear map) – Milo Brandt Aug 04 '16 at 14:13
  • Minkowski difference, as it turns out, is not all that different from Minkowksi sum. – Ivan Neretin Aug 04 '16 at 14:15
  • Okay, yeah I knew that the sums were convex, but I was a little in doubt about the difference. I have a further question, is it then also true, that if two convex sets are disjoint, then they're also separated? – 1233023 Aug 04 '16 at 14:17

1 Answers1

5

Each point in $A-B$ is of the form $a-b$, where $a\in A$ and $b\in B$.

Letting $a-b$ and $a^{\prime}-b^{\prime}$ be two points in $A-B$, we note that for any $\theta\in[0,1]$, $$ \theta\left(a-b\right)+\left(1-\theta\right)\left(a^{\prime}-b^{\prime}\right)=\left[\theta a+\left(1-\theta\right)a^{\prime}\right]-\left[\theta b+\left(1-\theta\right)b^{\prime}\right]. $$ What does this imply?

parsiad
  • 25,154
  • 1
    Okay Cool, then I have a further question! If two convex sets are disjoint, that implies that they are separated right? – 1233023 Aug 04 '16 at 14:12
  • You should ask a new question, as this is unrelated to your current question. Also, please consider upvoting/accepting correct/helpful answers. – parsiad Aug 04 '16 at 15:06
  • I did, but my reputation is so low, that it doesn't let me upvote :) – 1233023 Aug 04 '16 at 15:53