Home > Geometry, Inequalities, Measure Theory, shape optimization > The Brunn-Minkowski Inequality

## The Brunn-Minkowski Inequality

The Brunn-Minkowski Inequality gives an estimate for the measure of the set ${A+B=\{a+b : a \in A,b \in B\}}$ in terms of the Lebesgue measures of ${A}$ and ${B}$ in ${\Bbb{R}^d}$ (of course, we can only speak of such an inequality if all the sets ${A,B,A+B}$ are Lebesgue measurable). The inequality has the form

$\displaystyle |A+B|^{1/d} \geq |A|^{1/d}+|B|^{1/d}.$

First here are some facts that show that this is one of the only types of estimates we could expect. Since there are examples of sets ${A,B \subset \Bbb{R}^d}$ with ${|A|=|B|=0}$ and ${|A+B|>0}$ we cannot bound above ${|A+B|}$ in terms of ${|A|,|B|}$. For examples of such sets we have ${A=\mathcal{C},\ B=\mathcal{C}}$ in ${\Bbb{R}}$ (${\mathcal{C}}$ is the Cantor set) and in more dimensions we can consider sets of the form ${A= \{0\}\times X, B=Y \times\{0\}}$ such that ${|A+B|=|X \times Y|>0}$.

Secondly, if we search for inequalities of the form

$\displaystyle |A+B|^\alpha \geq |A|^\alpha + |B|^\alpha$

then by choosing ${A}$ convex and ${B= \lambda A}$ we would obtain the inequality

$\displaystyle (1+\lambda)^{d\alpha} \geq 1 + \lambda^{d \alpha}.$

Since we have

$\displaystyle (a+b)^\gamma \geq (\leq) a^\gamma+b^\gamma, a,b \geq 0,\ \gamma \geq (\leq) 1$

we see that the best ${\alpha}$ is ${\alpha=1/d}$.

Now let’s pass to the proof of the inequality. We consider ${A,B}$ of finite measure, or otherwise the inequality is just ${\infty \geq \infty}$. If ${A,B}$ are ${d}$-dimensional boxes then the inequality becomes

$\displaystyle \left(\prod_{i=1}^d (a_i+b_i) \right)^{1/d} \geq \left( \prod_{i=1}^d a_i \right)^{1/d}+\left(\prod_{i=1}^d b_i\right)^{1/d}$

Since this inequality does not change if we change ${(a_i,b_i)}$ to ${(\lambda_ia_i,\lambda_ib_i)}$ we can choose the factor ${\lambda_i}$ such that ${a_i+b_i=1,\ i=1,d}$. Then it becomes

$\displaystyle 1 \geq \left( \prod_{i=1}^d a_i \right)^{1/d}+\left(\prod_{i=1}^d b_i\right)^{1/d}$

which is an immediate consequence of the AM-GM inequality.

Consider now ${A}$ and ${B}$ as finite union of boxes as above with disjoint interiors. We would like to prove the inequality by induction after the number ${n}$ of boxes in ${A}$ and ${B}$. Note that we can translate ${A}$ and ${B}$ independently, and the inequality is the same, by using the invariance to translations of the Lebesgue measure. So we translate ${A}$ such that some hyperplane ${\{x_i=0 \}}$ separates two boxes in ${A}$, and denote ${A_+}$ and ${A_-}$ the intersection of ${A}$ with ${\{x_i \leq 0\}}$ and ${\{x_i \geq 0\}}$, respectively. Now translate ${B}$ such that

$\displaystyle \frac{ |A_\pm|}{|A|}=\frac{|B_\pm|}{|B|},$

where ${B_\pm}$ are defined in the same way as ${A_\pm}$. Then we have ${(A_+ +B_+) \cup (A_-+B_-) \subset A+B}$ and the number of boxes in ${A_+}$ and ${B_+}$ is at most ${n-1}$, and the same inequality holds for the number of boxes in ${A_-}$ and ${B_-}$. This allows us to apply the induction hypothesis and get

$\displaystyle |A+B| \geq |A_++B_+|+ |A_-+B_-| \geq$

$\displaystyle \geq (|A_+|^{1/d}+|B_+|^{1/d})^d+(|A_-|^{1/d}+|B_-|^{1/d})^d =$

$\displaystyle =|A_+|\left[1+ \left(\frac{|B|}{|A|}\right)^{1/d}\right]^d+|A_-|\left[1+ \left(\frac{|B|}{|A|}\right)^{1/d}\right]^d=$

$\displaystyle =(|A|^{1/d}+|B|^{1/d})^d.$

Consider now ${A,B}$ sets of finite measures and ${\varepsilon>0}$. Then we can find ${X,Y}$ which are finite union of boxes such that ${X \subset A,Y \subset B}$ and ${|A| \leq |X|+\varepsilon,\ |B| \leq |Y|+\varepsilon}$. Since ${X+Y \subset A+B}$, the inequality follows letting ${\varepsilon \rightarrow 0}$.

The case where ${A,B}$ are compact sets is proved using the previous case (note that ${A,B}$ compact implies ${A+B}$ compact). Denote ${A^\varepsilon=\{ x : d(x,A)<\varepsilon \}}$ and the same definition for ${B^\varepsilon}$, which make ${A^\varepsilon, B^\varepsilon}$ open sets. Since ${A+B \subset A^\varepsilon+B^\varepsilon \subset (A+B)^{2\varepsilon}}$, the inequality for ${A^\varepsilon, B^\varepsilon}$ implies the inequality for ${A,B}$ as ${\varepsilon \rightarrow 0}$.

When ${A,B}$ are arbitrary measurable sets with finite measures and ${A+B}$ is measurable, then approximate from inside ${A,B}$ with compact sets, and taking the limit, the inequality follows.