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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.

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