In mathematics, the Brunn–Minkowski theorem (or Brunn–Minkowski inequality) is an inequality relating the volumes (or more generally Lebesgue measures) of compact subsets of Euclidean space. The original version of the Brunn–Minkowski theorem (Hermann Brunn 1887; Hermann Minkowski 1896) applied to convex sets; the generalization to compact nonconvex sets stated here is due to Lazar Lyusternik (1935).
Statement
Let n ≥ 1 and let μ denote the Lebesgue measure on Rn. Let A and B be two nonempty compact subsets of Rn. Then the following inequality holds:
where A + B denotes the Minkowski sum:
The theorem is also true in the setting where are only assumed to be measurable and non-empty.
Multiplicative version
The multiplicative form of Brunn–Minkowski inequality states that for all .
The Brunn–Minkowski inequality is equivalent to the multiplicative version.
In one direction, use the inequality (exponential is convex), which holds for . In particular, .
Conversely, using the multiplicative form, we find
The right side is maximized at , which gives
.
The Prékopa–Leindler inequality is a functional generalization of this version of Brunn–Minkowski.
On the hypothesis
Measurability
It is possible for to be Lebesgue measurable and to not be; a counter example can be found in "Measure zero sets with non-measurable sum." On the other hand, if are Borel measurable, then is the continuous image of the Borel set , so analytic and thus measurable. See the discussion in Gardner's survey for more on this, as well as ways to avoid measurability hypothesis.
We note that in the case that A and B are compact, so is A + B, being the image of the compact set under the continuous addition map : , so the measurability conditions are easy to verify.
Non-emptiness
The condition that are both non-empty is clearly necessary. This condition is not part of the multiplicative versions of BM stated below.
Proofs
We give two well known proofs of Brunn–Minkowski.
We give a well-known argument that follows a general recipe of arguments in measure theory; namely, it establishes a simple case by direct analysis, uses induction to establish a finitary extension of that special case, and then uses general machinery to obtain the general case as a limit. A discussion of this history of this proof can be found in Theorem 4.1 in Gardner's survey on Brunn–Minkowski.
We prove the version of the Brunn–Minkowski theorem that only requires to be measurable and non-empty.
The case that A and B are axis aligned boxes:
By translation invariance of volumes, it suffices to take . Then . In this special case, the Brunn–Minkowski inequality asserts that . After dividing both sides by , this follows from the AM–GM inequality: .
The case where A and B are both disjoint unions of finitely many such boxes:
We will use induction on the total number of boxes, where the previous calculation establishes the base case of two boxes. First, we observe that there is an axis aligned hyperplane H that such that each side of H contains an entire box of A. To see this, it suffices to reduce to the case where A consists of two boxes, and then calculate that the negation of this statement implies that the two boxes have a point in common.
For a body X, we let denote the intersections of X with the "right" and "left" halfspaces defined by H. Noting again that the statement of Brunn–Minkowski is translation invariant, we then translate B so that ; such a translation exists by the intermediate value theorem because is a continuous function, if v is perpendicular to H has limiting values 0 and as , so takes on at some point.
We now have the pieces in place to complete the induction step. First, observe that and are disjoint subsets of , and so Now, both have one fewer box than A, while each have at most as many boxes as B. Thus, we can apply the induction hypothesis: and .
Elementary algebra shows that if , then also , so we can calculate:
The case that A and B are bounded open sets:
In this setting, both bodies can be approximated arbitrarily well by unions of disjoint axis aligned rectangles contained in their interior; this follows from general facts about the Lebesgue measure of open sets. That is, we have a sequence of bodies , which are disjoint unions of finitely many axis aligned rectangles, where , and likewise . Then we have that , so . The right hand side converges to as , establishing this special case.
The case that A and B are compact sets: