Alessio FIGALLI David JERISON

(Dedicated to Professor Haim Brezis on the occasion of his 70th birthday)

1 Introduction

The Brunn-Minkowskiinequality is a very classical and powerful inequality in convex geometry that has found important applications in analysis,statistics,and information theory.We refer the reader to[14]for an extended exposition on the Brunn-Minkowski inequality and its relation to several other famous inequalities(see also[6–7]).

To state the inequality,we first need some basic notation.Given two subset A,B?Rn,and c>0,we define the set sum and scalar multiple by

We shall use|E|to denote the Lebesgue measure of a set E.(If E is not measurable,|E|denotes the outer Lebesgue measure of E.)The Brunn-Minkowski inequality says that,given A,B?Rnmeasurable sets,

In addition,if|A|,|B|>0,then equality holds if and only if there exists a convex set K?Rn,λA,λB>0,and vA,vB∈ Rn,such that

In other words,if equality holds in(1.2),then A and B are subsets of full measure in homothetic convex sets.

Because of the variety of applications of(1.2)as well as the fact the one can characterize the case of equality,a natural stability question that one would like to address is the following.

Let A,B be two sets for which equality in(1.2)almost holds.Is it true that,up to translations and dilations,A and B are close to the same convex set?

This question has a long history.First of all,when n=1 and A=B,inequality(1.2)reduces to|A+A|≥2|A|.If one approximates sets in R with finite unions of intervals,then one can translate the problem to Z,and in the discrete setting the question becomes a well studied problem in additive combinatorics.There are many results on this topic,usually called Freiman-type theorems.The precise statement in one dimension is the following.

Theorem 1.1Let A?R be a measurable set,and denote by co(A)its convex hull.Then

or,equivalently,if|A|>0,then

This theorem can be obtained as a corollary of a result of Freiman[12]about the structure of additive subsets of Z(see[13]or[17,Theorem 5.11]for a statement and a proof).However,it turns out that to prove Theorem 1.1,one only needs weaker results,and one can find an elementary self-contained proof of Theorem 1.1 in[8,Section 2].

In the case n=1 but AB,the following sharp stability result holds again as a consequence of classical theorems in additive combinatorics(an elementary proof of this result can be given using Kemperman’s theorem in[3–4].

Theorem 1.2Let A,B?R be measurable sets.If|A+B|<|A|+|B|+δfor some δ≤ min{|A|,|B|},then|co(A)A|≤ δand|co(B)B|≤ δ.

Concerning the higher dimensional case,in[1–2],Christ proved a qualitative stability result for(1.2),giving a positive answer to the stability question raised above.However,his results do not provide any quantitative control.

On the quantitative side,Diskant[5]and Groemer[15]obtained some stability results for convex sets in terms of the Hausdorff distance.More recently,in[10–11],the first author together with Maggi and Pratelli obtained a sharp stability result in terms of the L1distance,still on convex sets.Since this last result will be used later in our proofs,we state it in detail.(Here and from now on,E?F denotes the symmetric difference between sets E and F,that is,E?F=(EF)∪(FE).)

Theorem 1.3Let A,B?Rnbe convex sets,and define

There exists a computable dimensional constant C0(n)such that

Morerecently,in[8,Theorem 1.2 and Remark 3.2],the present authorsproved a quantitative stability result when A=B:Given a measurable set A?Rnwith|A|>0,set

Then,a power ofδ(A)dominates the measure of the difference between A and its convex hull co(A).

Theorem 1.4Let A?Rnbe a measurable set of positive measure.There exist computable dimensional constants δn,cn>0,such that ifδ(A)≤ δn,then

In addition,there exists a convex set K?Rnsuch that

After that,we investigated the general case AB.Notice that,after a dilation,one can always assume|A|=|B|=1 while replacing the sum A+B by a convex combination St:=tA+(1?t)B.It follows by(1.2)that|St|=1+δfor someδ≥0.The main theorem in[9]is a quantitative version of Christ’s result.Since the proof is by induction on the dimension,it is convenient to allow the measures of|A|and|B|not to be exactly equal,but just close in terms ofδ.Here is the main result of that paper.

Theorem 1.5Let n≥2,let A,B?Rnbe measurable sets,and define St:=tA+(1?t)B for someThere are computable dimensional constants Nnand computable functions Mn(τ),εn(τ)>0,such that if

for somethen there exists a convex set K ? Rnsuch that,up to a translation,

Explicitly,we may take

In particular,the measure of the difference between the sets A and B and their convex hull is bounded by a power δ?,confirming a conjecture of Christ[1].

The result above provides a general quantitative stability for the Brunn-Minkowski inequality in arbitrary dimension.However,the exponent degenerates very quickly as the dimension increases(much faster than in Theorem 1.4),and,in addition,the argument in[9]is very long and involved.The aim of this paper is to provide a shorter and more elementary proof when|A|=|B|>0,that we believe to be of independent interest.

After a dilation,one can assume with no loss of generality that|A|=|B|=1.In this case,it follows by(1.2)thatfor someδ≥0,and we want to show that a power ofδcontrols the closeness of A and B to the same convex set K.Again,as in the previous theorem,it will be convenient to allow the measures of|A|and|B|not to be exactly equal,but just close in terms ofδ.

Here is the main result of this paper.

Theorem 1.6Let A,B?Rnbe measurable sets,and define their semi-sumThere exist computable dimensional constantsδn,Cn>0,such that if

for someδ≤ δn,then there exists a convex set K ?Rnsuch that,up to a translation,

where

andαkis given by Theorem 1.4.(Recall that|S|is the outer measure of S if S is not measurable.)

The proof of this theorem is specific to the case|A|near|B|.It uses a symmetrization and other techniques introduced by Christ[2–3],Theorems 1.3–1.4,and two propositions of independent interest,Propositions 2.1–2.2 below.See Section 3 for further discussion of the strategy of the proof.

2 Notation and Preliminary Results

Let Hkdenote the k-dimensional Hausdorff measure on Rn.Denote by x=(y,t)∈ Rn?1×R a point in Rn,and letanddenote the canonical projections,i.e.,

Given a compact set E ?Rn,y∈Rn?1,andλ>0,we use the notation

Following Christ[2],we consider two symmetrizations and combine them.For our purposes(see the proof of Proposition 2.1),it is convenient to use a definition of Schwarz symmetrization that is slightly different from the classical one.(In the usual definition of Schwarz symmetrization,E?(t)= ? whenever Hd?1(E(t))=0.)

Definition 2.1Let E?Rnbe a compact set.We define the Schwarz symmetrization E?of E as follows.For each t∈R,

(1)If Hd?1(E(t))>0,then E?(t)is the closed disk centered at 0 ∈ Rn?1with the same measure.

(2)If Hd?1(E(t))=0 but E(t)is non-empty,then E?(t)={0}.

(3)If E(t)is empty,then E?(t)is empty as well.

We define the Steiner symmetrization E?of E so that for each y∈Rn?1,the setis empty if H1(Ey)=0;otherwise it is the closed interval of length H1(Ey)centered at 0∈R.Finally,we define E?:=(E?)?.

As for instance in[2,Section 2],both the Schwarz and the Steiner symmetrization preserve the measure of sets,and the ?-symmetrization preserves the measure of the sets E(λ).The following statement collects all these results.

Lemma 2.1Let A,B ? Rnbe compact sets.Then|A|=|A?|=|A?|=|A?|,

and,for almost everyλ>0,

where

Another important fact is that a bound on the measure of A+B in terms of the measures of A and B gives bounds relating the sizes of

We refer to[9,Lemma 3.2]for a proof.

Lemma 2.2Let A,B?Rnbe compact sets such thatand

There exists a dimensional constant M>1,such that

Thus,up a measure preserving affine transformation of the formwith τ>0,all the quantitiesare of order one.

In particular,

In this case,we say that A and B are M-normalized.

Lemma 2.3Let A,B?Rnbe compact sets,defineand assume that(1.5)holds for someAlso,suppose that A and B are M-normalized as defined in Lemma 2.2.

Then,there exists a dimensional constant C>0 such that

Two other key ingredients in our proof of Theorem 1.6 are the following propositions,whose proofs are postponed to Section 4.

Proposition 2.1Let A,B?Rnbe compact sets,defineand assume that(1.5)holds for someAlso,suppose that we can find a convex set K?Rnsuch that

for someα>0,where C>0 is a dimensional constant.Then there exists a dimensional constant C′>0 such that

Proposition 2.2Let A,B?Rnbe compact sets,define,and assume that(1.5)holds for someAlso,suppose that

for someβ>0,where C>0 is a dimensional constant.Then,up to a translation,

and there exists a convex set K containing both A and B such that

for some dimensional constant C′>0.

3 Proof of Theorem 1.6

As explained in[8],by inner approximation1The approximation of A(and analogously for B)is by a sequence of compact sets A k?A such thatandOne way to construct such sets is to definewhereare compact sets satisfyingand V k ? V k+1 ? A are finite sets satisfyingit suffices to prove the result when A,B are compact sets.Hence,let A and B be compact,define,and assume that(1.5)holds.We want to prove that there exists a convex set K such that,up to a translation,Moreover,since the statement and the conclusions are invariant under measure preserving affine transformations,by Lemma 2.2,we can assume that A and B are M-normalized(see(2.3)).

Ultimately,we wish to show that,up to translation,each of A,B,and S is of nearly full measure in the same convex set.The strategy of the proof is to show first that S is close to a convex set,and then apply Propositions 2.1–2.2.To obtain the closeness of S to a convex set,we would like prove thatis close to|S|and then apply Theorem 1.4.It is simpler,however,to construct a subsetsuch thatis small andis close to

To carry out our argument,one important ingredient will be to use the inductive hypothesis on the level sets A(λ)and B(λ)defined in(2.2).However,two difficulties arise here:First of all,to apply the inductive hypothesis,we need to know that Hn?1(A(λ))and Hn?1(B(λ))are close.In addition,the Brunn-Minkowski inequality does not have a natural proof by induction unless the measures of all the level sets Hn?1(A(λ))and Hn?1(B(λ))are the nearly same(see(3.11)below).Hence,it is important for us to have a preliminary quantitative estimate on the difference between Hn?1(A(λ))and Hn?1(B(λ))for most λ >0.For this,we follow an approach used first in[2]and readapted in[9],in which we begin by showing our theorem in the special case of symmetrized sets A=A?and B=B?(recall Definition 2.1).Thanks to Lemma 2.1,this will give us the desired closeness between Hn?1(A(λ))and Hn?1(B(λ))for mostλ>0,which allows us to apply the strategy described above and prove the theorem in the general case.

Throughout the proof,C will denote a generic constant depending only on the dimension,which may change from line to line.

3.1 The case A=A?and B=B?

Let A,B ? Rnbe compact sets satisfying A=A?,B=B?.Sinceandare disks centered at the origin,applying Lemma 2.3,we deduce that

Hence,if we define

thenfor all y∈Rn?1.In addition,using(1.5),(2.3),and(3.1),we have

which implies(since S?S)

Furthermore,since each section Syis an interval centered at 0 ∈ R,for all y′,y′′∈ π(A)∩ π(B)such that

which gives

Recalling(1.3),by(3.2)–(3.3),we obtain that.Hence,we can apply Theorem 1.4 toto find a convex setsuch that

Hence,by(3.3),

and using Propositions 2.1–2.2,we deduce that,up to a translation,there exists a convex set K such that A∪B?K and

Notice that,becauseandit is easy to check that the above properties still hold within place of K.Hence,in this case,without loss of generality,one can assume that

3.2 The general case

Since,by Theorem 1.2,the result is true when n=1,we may assume that we already proved Theorem 1.6 through n?1,and we want to show its validity for n.

Step 1There exist a dimensional constantζ>0 andsuch that we can apply the inductive hypothesis to

Let A?and B?be as in Definition 2.1 and denote

Thanks to Lemma 2.1,A?and B?still satisfy(1.5),so we can apply the result proved in Section 3.1 above to get(see(3.4))

and

for some convex set K=K?.

In addition,because A and B are M-normalized(see(2.3)),so are A?and B?,and by(3.7)we deduce that there exists a dimensional constant Rn>0 such that

Also,by(3.6)and Chebyshev’s inequality,we obtain that,except for a set of measure

Thus,recalling Lemma 2.1,for almost everyλ>0,

Since,by(2.3),

by Chebyshev’s inequality,we deduce that

for allλoutside a set of measureExchanging the roles of A and B,we obtain that there exists a set F?[0,M],such that

Using the elementary inequality

and replacing a and b withrespectively,we get

(notice thatFinally,it is easy to check that

Hence,by the Brunn-Minkowski inequality(1.2)applied to A(λ)and B(λ),using(1.5),(2.3)and(3.9)–(3.10),we get

We also observe that,since K=K?,by Lemma 2.1,(3.8),and[2,Lemma 4.3],for almost every λ>0,we have

and analogously for B.Also,by(3.7),

Define

and note that.Let ζ∈ (0,η)to be fixed later.Then by(3.9),(3.11)–(3.13),and by Chebyshev’s inequality,we can find a level

such that

In addition,from thepropertiesfor any λ >0(see(2.3)),=|A|≥ 1 ? δ,andis a decreasing function,we deduce that

The same holds for B and S,hence

By(3.16)–(3.17),we get

while,by(1.2),

therefore

Thus,by the inductive hypothesis of Theorem 1.6,up to a translation there exists an(n?1)-dimensional convex set?′,such that

and definingwe obtain(recall that

Step 2We apply Theorem 1.2 to the sets Ayand Byfor most

DefineBy(3.18)–(3.19)and(2.3),we have

provided that we choose

(recall thatβn?1≤ 1).Hence,by(1.5)and(3.20),

Write C as C1∪C2,where

By Chebyshev’s inequality and(3.22),

while,recalling(3.15),

Hence,by Theorem 1.2 applied to Ay,By?R for y∈C1,we deduce that

Letdenote the set ofsuch that

and notice that,by(3.22)and Chebyshev’s inequality,.Then choose a compact setsuch thatto obtain

Step 3We findso thatandare small.

Define the compact set

Observe,thanks to(3.20),(3.23),(3.26),(2.3)and(1.5),

So,since

Now,we want to estimate the measure ofFirst of all,since

by(3.25),we get

Also,if we define the characteristic functions

and analogously for By,by(3.24)we have the following estimate on their convolutions:

Recalling thatis the orthogonal projection onto the last component(that is,we denote by[a,b]the interval,and notice that,since by construction

(see(3.15)),this interval has length greater thanAlso,it is easy to check that the functionis supported on[a,b],has slope equal to 1(resp.?1)in[a,a+3δζ](resp.and it is greater than 3δζin[a+3δζ,b?3δζ].Hence,sincecontains the setby(3.30),we deduce that

which implies in particular that

Also,by the same argument as in[8,Step 2-a],if we denote by

using(3.25)and(3.31),we have

(Compare with[8,(3.25)].)

Hence,by(3.33),we deduce that each of the latter sets is contained inside the convex setso also their semi-sum is contained in the same set,and using(3.32)with y′=y′′=y,we get

(notice thatExchanging the role of A and B and adding up the two inequalities,we deduce that

As shown in[8,Step 3],this estimate combined with the fact thatis almost of full measure inside the convex set ?(see(3.19),(3.23)and(3.26))proves that,up to an affine transformation of the form

withandthe set S is universally bounded,sayfor some dimensional constant R.This implies that[?R,R],so

Hence,sinceby(3.34),(3.19)and(3.21),

that is,

Step 4Conclusion

By the previous step,we have thatHence,applying Theorem 1.4 towe find a convex setsuch that

so,by(3.27),

Using this estimate together with Propositions 2.1–2.2,we deduce that,up to a translation,there exists a convex set K convex such that A∪B?K and

Recalling the definition ofζ(see(3.5),(3.14),(3.21)),we see that

Sinceβ1=1(by Theorem 1.2),it is easy to check that

concluding the proof.

4 Technical Results

As in the previous section,we use C to denote a generic constant depending only on the dimension,which may change from line to line.

4.1 Proof of Proposition 2.1

Assume that

for some α ∈ (0,1].By John’s lemma(see[16]),after a volume preserving affine transformation,we can assume that,with rbounded above and below by positive dimensional constants.Note,however,that with this normalization,we will not be able to assume that A and B are M-normalized,since we have already chosen a different affine normalization.

We want to prove that

Letand setWithout loss of generality,we can assume thatfor someWe need to prove that

Let us consider the setsobtained from A,B,S,K performing a Schwarz symmetrization around the en-axis(see Definition 2.1).SetSince

and,by(1.5)(notice that S′? S?and that|S′|≥ 1?Cδby(1.2)),

we get thatIn addition,Hence,without loss of generality,we can assume from the beginning that A=A?,B=B?,S=and K=K?.

For a compact set E ? Rn,recall the notation E(t)? Rn?1×{t}in(2.1),and define E[s]?R by

Sincewe have

so,by(1.2),we deduce that

Hence

and integrating with respect to s,by(1.5),we get

Recall that K=K?,so that the canonical projection π(K)onto Rn?1is a ball.We denote it BR:= π(K),and note that R ≤ nrn,with rn=rn(K)given by John’s lemma at the beginning of this proof.Then,since|S?K|≤ Cδα,we have

so,by(4.3),

Hence,recalling that|A|and|B|are≥1?δ,we deduce that

and since R is universally bounded(being less than nrn)and both functions

are decreasing,there exists a small dimensional constant c′>0,such that

Also,by(4.4),

and since|S?K|≤ Cδαand K ?{xn≤ τ},

Hence,thanks to(4.6)–(4.8),we use Theorem 1.2 and Chebishev’s inequality to find a value

such that

(notice thatα≤1)and

Sincethis implies

Hence,after applying opposite translations along the en-axis to A and B,i.e.,

for some?∈R,we can assume that

Since the setsare decreasing,we deduce that

In addition,sinceπ(A)andπ(B)are(n?1)-dimensional disks centered on the en-axis,|S?K|≤Cδαand,we easily deduce that

provided thatδis small enough.Hence,combining(4.11)and(4.12),we deduce that Hn?1(π(B))is bounded from away from zero by a dimensional constant,thus

Hence,by(4.5),(4.10),(4.13)and(4.9),

and,analogously,

Now,given r≥0,let us define the sets

By(4.14)–(4.15),we know that

and it is immediate to check that

Also,since K is a convex set satisfying Brn?K?Bnrn,there exists a dimensional constant cn>0 such that

Hence

and by(1.2)applied towe get

which gives

(and analogously for B).

Since the pointbelongs to,there as to be a pointsuch thatWithout loss of generality,assume thatThen,by(4.16)applied with r=ρ,we get

so

which impliesproving(4.1).

Hence,from which the result follows immediately.

4.2 Proof of Proposition 2.2

Since

by(1.2),(2.4)–(1.5),we have

from which we deduce that

Also,by Theorem 1.3 and the fact that||co(A)|? |co(B)||≤ Cδβαn(see(4.17)),we obtain that,up to a translation,

This estimate combined with(4.17)implies that

In addition,if we define K:=co(A∪B),then we will conclude our argument by showing that

Indeed,by John’s lemma(see[16]),after a volume preserving affine transformation,we can assume that Br?co(A)?Bnrfor some radius r bounded above and below by positive dimensional constants.By(4.18)and a simple geometric argument,we easily deduce that

Thus and(4.19)follows by(4.17)–(4.18).

AcknowledgementsThis work started during Alessio Figalli’s visit at MIT during the fall 2012.Alessio Figalli wishes to thank the Mathematics Department at MIT for its warm hospitality.

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