Brunn–Minkowski theorem
In mathematics, the Brunn–Minkowski theorem is an inequality relating the volumes of compact subsets of Euclidean space. The original version of the Brunn–Minkowski theorem applied to convex sets; the generalization to compact nonconvex sets stated here is due to Lazar Lyusternik.
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.
To prove the equivalence one direction, apply the Brunn–Minkowski inequality to sets and, which gives
where we used the fact that is homogeneous. Using now the inequality, which holds for all, we find that, concluding the proof.
Conversely, using the multiplicative form, we find
The right side is maximized at, which gives
concluding the proof.
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 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.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 .
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:
- The case where A and B are both disjoint unions of finitely many such boxes:
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 less 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:
- The case that A and B are compact sets:
- The case of bounded measurable sets:
- The case of measurable sets:
We give a proof of the Brunn–Minkowski inequality as a corollary to the Prékopa–Leindler inequality, a functional version of the BM inequality. We will first prove PL, and then show that PL implies a multiplicative version of BM, then show that multiplicative BM implies additive BM. The argument here is simpler than the proof via cuboids, in particular, we only need to prove the BM inequality in one dimensions. This happens because the more general statement of the PL-inequality than the BM-inequality allows for an induction argument.
- The multiplicative form of the BM inequality
- Prékopa–Leindler inequality
Proof :
We will need the one dimensional version of BM, namely that if are measurable, then. First, assuming that are bounded, we shift so that. Thus,, whence by almost disjointedness we have that. We then pass to the unbounded case by filtering with the intervals
We first show the case of the PL inequality. Let . . Thus, by the one-dimensional version of Brunn–Minkowski, we have that. We recall that if is non-negative, then Fubini's theorem implies. Then, we have that, where in the last step we use the weighted AM–GM inequality, which asserts that for.
Now we prove the case. For, we pick and set. For any c, we define, that is, defining a new function on n-1 variables by setting the last variable to be . Applying the hypothesis and doing nothing but formal manipulation of the definitions, we have that.
Thus, by the inductive case applied to the functions, we obtain. We define and similarly. In this notation, the previous calculation can be rewritten as:. Since we have proven this for any fixed, this means that the function satisfy the hypothesis for the one dimensional version of the PL theorem. Thus, we have that, implying the claim by Fubini's theorem. QED
- PL implies multiplicative BM
- Multiplicative BM implies Additive BM
We assume that both A,B have positive volume, as otherwise the inequality is trivial, and normalize them to have volume 1 by setting. We define ;. With these definitions, and using that, we calculate using the multiplicative Brunn–Minkowski inequality that:
The additive form of Brunn–Minkowski now follows by pulling the scaling out of the leftmost volume calculation and rearranging.
Important corollaries
The Brunn–Minkowski inequality gives much insight into the geometry of high dimensional convex bodies. In this section we sketch a few of those insights.Concavity of the radius function (Brunn's theorem)
Consider a convex body. Let be vertical slices of K. Define to be the radius function; if the slices of K are discs, then r gives the radius of the disc K, up to a constant. For more general bodies this radius function does not appear to have a completely clear geometric interpretation beyond being the radius of the disc obtained by packing the volume of the slice as close to the origin as possible; in the case when K is not a disc, the example of a hypercube shows that the average distance to the center of mass can be much larger than r. Sometimes in the context of a convex geometry, the radius function has a different meaning, here we follow the terminology of .By convexity of K, we have that. Applying the Brunn–Minkowski inequality gives, provided. This shows that the radius function is concave on its support, matching the intuition that a convex body does not dip into itself along any direction. This result is sometimes known as Brunn's theorem.
Brunn–Minkowski symmetrization of a convex body
Again consider a convex body . Fix some line and for each let denote the affine hyperplane orthogonal to that passes through . Define, ; as discussed in the previous section, this function is concave. Now, let. That is, is obtained from by replacing each slice with a disc of the same -dimensional volume centered inside of . The concavity of the radius function defined in the previous section implies that that is convex. This construction is called the Brunn–Minkowski symmetrization.Grunbaum's theorem
Theorem : Consider a convex body. Let be any half-space containing the center of mass of ; that is, the expected location of a uniform point sampled from Then.Grunbaum's theorem can be proven using Brunn–Minkowski inequality, specifically the convexity of the Brunn–Minkowski symmetrization.
Grunbaum's inequality has the following fair cake cutting interpretation. Suppose two players are playing a game of cutting up an dimensional, convex cake. Player 1 chooses a point in the cake, and player two chooses a hyperplane to cut the cake along. Player 1 then receives the cut of the cake containing his point. Grunbaum's theorem implies that if player 1 chooses the center of mass, then the worst that an adversarial player 2 can do is give him a piece of cake with volume at least a fraction of the total. In dimensions 2 and 3, the most common dimensions for cakes, the bounds given by the theorem are approximately respectively. Note, however, that in dimensions, calculating the centroid is hard, limiting the usefulness of this cake cutting strategy for higher dimensional, but computationally bounded creatures.
Applications of Grunbaum's theorem also appear in convex optimization, specifically in analyzing the converge of the center of gravity method.
Isoperimetric inequality
Let denote the unit ball. For a convex body, K, let define its surface area. This agrees with the usual meaning of surface area by the Minkowski-Steiner formula. Consider the function. The isoperimetric inequality states that this is maximized on Euclidean balls.First, observe that Brunn–Minkowski implies where in the last inequality we used that for. We use this calculation to lower bound the surface area of via Next, we use the fact that, which follows from the Minkowski-Steiner formula, to calculate Rearranging this yields the isoperimetric inequality:
Applications to inequalities between mixed volumes
The Brunn–Minkowski inequality can be used to deduce the following inequality, where the term is a mixed-volume. Equality holds if and only if K,L are homothetic.We recall the following facts about mixed volumes :, so that in particular if, then.
Let . Brunn's theorem implies that this is concave for . Thus,, where denotes the right derivative. We also have that . From this we get, where we applied BM in the last inequality.
Concentration of measure on the sphere and other strictly convex surfaces
We prove the following theorem on concentration of measure, following and . See also Concentration of measure#Concentration on the sphere.Theorem: Let be the unit sphere in. Let. Define, where d refers to the Euclidean distance in. Let denote the surface area on the sphere. Then, for any we have that.
Proof: Let, and let . Then, for one can show, using and for, that . In particular, .
We let, and aim to show that . Let . The argument below will be symmetric in, so we assume without loss of generality that and set . Then,
This implies that .
Thus, we know that, so . We apply the multiplicative form of the Brunn–Minkowski inequality to lower bound the first term by, giving us .
. QED
Version of this result hold also for so-called strictly convex surfaces, where the result depends on the modulus of convexity. However, the notion of surface area requires modification, see:
Remarks
The proof of the Brunn–Minkowski theorem establishes that the functionis concave in the sense that, for every pair of nonempty compact subsets A and B of Rn and every 0 ≤ t ≤ 1,
For convex sets A and B of positive measure, the inequality in the theorem is strict
for 0 < t < 1 unless A and B are positive homothetic, i.e. are equal up to translation and dilation by a positive factor.
Examples
Rounded cubes
It is instructive to consider the case where an square in the plane, and a ball of radius . In this case, is a rounded square, and its volume can be accounted for as the four rounded quarter circles of radius, the four rectangles of dimensions along the sides, and the original square. Thus,This example also hints at the theory of mixed-volumes, since the terms that appear in the expansion of the volume of correspond to the differently dimensional pieces of A. In particular, if we rewrite Brunn–Minkowski as, we see that we can think of the cross terms of the binomial expansion of the latter as accounting, in some fashion, for the mixed volume representation of. This same phenomenon can also be seen for the sum of an n-dimensional box and a ball of radius, where the cross terms in, up to constants, account for the mixed volumes. This is made precise for the first mixed volume in the section above on the applications to mixed volumes.