The Minkowski sum of two sets $A$ and $B$ in the plane is defined as $$A+B = \{ a + b \mid a \in A, b \in B \}.$$ The Minkowski difference $A-B$ is defined similarly.

For any convex set $A$, is it always true that $$|A-A| \ge |A+A|?$$

For example, if $A$ is a triangle, then $|A - A| = \frac{3}{2} |A + A|$. If $A$ is symmetric about a point, then $|A-A| = |A+A|$.

  • $\begingroup$ I presume in the last paragraph you mean 'symmetric' in the sense that $A=(-A)$, i.e. $a\in A\Longleftrightarrow (-a)\in A$? $\endgroup$ Jul 2, 2014 at 17:54
  • $\begingroup$ Correct, edited. $\endgroup$
    – keej
    Jul 2, 2014 at 17:55
  • $\begingroup$ I asked a question related to this one: math.stackexchange.com/questions/855854/…. I don't know if it will help, but I think if the answer to my question is "yes," then the answer to your question is yes, because the set $\frac12(A-A)$ has equal diameter to $A$ but is symmetric, so it would have greater area than $A$, and we would have $|A-A|=4|\frac12(A-A)|\geq 4|A| =|A+A|$. Unless all of my speculation is wrong, that is. $\endgroup$ Jul 3, 2014 at 22:37

1 Answer 1


Yes. By the Brunn–Minkowski inequality, $|A-A|^{1/2}\geq|A|^{1/2}+|-A|^{1/2}$, so $|A-A|\geq 4|A|$. The hypotheses of this theorem do not include convexity, although they do include compactness.

On the other hand, under the assumption that $A$ is convex, $A+A = 2A$, and $|A+A|=4|A|$. Therefore $|A-A|\geq |A+A|$.


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