Quantitative stability of the Brunn-Minkowski inequality for sets of equal volume

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1 Quantitative stabilit of the Brunn-Minkowski inequalit for sets of equal volume Alessio Figalli and David Jerison Abstract We prove a quantitative stabilit result for the Brunn-Minkowski inequalit on sets of equal volume: if A = B > and A + B 1/n = + δ A 1/n for some small δ, then, up to a translation, both A and B are close in terms of δ to a convex set K. Although this result was alread proved in our previous paper [9] even for sets of different volume, we provide here a more elementar proof that we believe has its own interest. Also, in terms of the stabilit exponent, this result provides a stronger estimate than the result in [9]. 1 Introduction The Brunn-Minkowski inequalit is a ver classical and powerful inequalit in convex geometr that has found important applications in analsis, statistics, and information theor. We refer the reader to [14] for an extended exposition on the Brunn-Minkowski inequalit and its relation to several other famous inequalities; see also [6, 7]. To state the inequalit, we first need some basic notation. Given two subset A, B R n, and c >, we define the set sum and scalar multiple b A + B := {a + b : a A, b B}, ca := {ca : a A} 1.1 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 inequalit sas that, given A, B R n measurable sets, A + B 1/n A 1/n + B 1/n. 1. In addition, if A, B >, then equalit holds if and onl if there exist a convex set K R n, λ A, λ B >, and v A, v B R n, such that A λ A K + v A, B λ B K + v B, λ A K + v A \ A = λ B K + v B \ B =. In other words, if equalit holds in 1., then A and B are subsets of full measure in homothetic convex sets. The Universit of Texas at Austin, Mathematics Dept. RLM 8.1, 515 Speedwa Stop C1, Austin, TX USA. address: figalli@math.utexas.edu Department of Mathematics, Massachusetts Institute of Technolog, 77 Massachusetts Ave, Cambridge, MA USA. address: jerison@math.mit.edu 1

2 Because of the variet of applications of 1. as well as the fact the one can characterize the case of equalit, a natural stabilit question that one would like to address is the following: Let A, B be two sets for which equalit in 1. 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 histor. First of all, when n = 1 and A = B, inequalit 1. reduces to A + A 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 man results on this topic, usuall called Freiman-tpe theorems. The precise statement in one dimension is the following. Theorem 1.1. Let A R be a measurable set, and denote b coa its convex hull. Then A + A A min{ coa \ A, A }, or, equivalentl, if A > then δa 1 { } coa \ A min, 1. A This theorem can be obtained as a corollar of a result of G. Freiman [1] 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 of Theorem 1.1 one onl needs weaker results, and one can find an elementar self-contained proof of Theorem 1.1 in [8, Section ]. In the case n = 1 but A B, the following sharp stabilit result holds again as a consequence of classical theorems in additive combinatorics an elementar proof of this result can be given using Kemperman s theorem [3, 4]: Theorem 1.. Let A, B R be measurable sets. If A + B < A + B + δ for some δ min{ A, B }, then coa \ A δ and cob \ B δ. Concerning the higher dimensional case, in [1, ] M. Christ proved a qualitative stabilit result for 1., giving a positive answer to the stabilit question raised above. However, his results do not provide an quantitative control. On the quantitative side, V. I. Diskant [5] and H. Groemer [15] obtained some stabilit results for convex sets in terms of the Hausdorff distance. More recentl, in [1, 11], the first author together with F. Maggi and A. Pratelli obtained a sharp stabilit result in terms of the L 1 distance, 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 smmetric difference between sets E and F, that is E F = E \ F F \ E. Theorem 1.3. Let A, B R n be convex sets, and define { } A x + τb A 1/n { A A A, B := inf : τ =, σa, B := max x R n A B B, B }. A

3 There exists a computable dimensional constant C n such that A + B 1/n A 1/n + B 1/n { A A, B } 1 + C n σa, B 1/n. More recentl, in [8, Theorem 1. and Remark 3.], the present authors proved a quantitative stabilit result when A = B: given a measurable set A R n with A >, set 1 δa := A + A A + A 1 = A A Then, a power of δa dominates the measure of the difference between A and its convex hull coa. Theorem 1.4. Let A R n be a measurable set of positive measure. dimensional constants δ n, c n > such that if δa δ n, then There exist computable coa \ A δa αn c n, α n := A In addition, there exists a convex set K R n such that K A δa nαn c n. A 1 8 n 1 n! [n 1!]. After that, we investigated the general case A B. Notice that, after a dilation, one can alwas assume A = B = 1 while replacing the sum A + B b a convex combination S t := ta + 1 tb. It follows b 1. that S t = 1 + δ for some δ. The main theorem in [9] is a quantitative version of Christ s result. Since the proof is b induction on the dimension, it is convenient to allow the measures of A and B not to be exactl equal, but just close in terms of δ. Here is the main result of that paper. Theorem 1.5. Let n, let A, B R n be measurable sets, and define S t := ta + 1 tb for some t [τ, 1 τ], < τ 1/. There are computable dimensional constants N n and computable functions M n τ, ε n τ > such that if A 1 + B 1 + S t 1 δ 1.4 for some δ e Mnτ, then there exists a convex set K R n such that, up to a translation, Explicitl, we ma take A, B K and K \ A + K \ B τ Nn δ εnτ. M n τ = 3n+ n 3n log τ 3n τ 3n, ε n τ = τ 3n 3n+1 n 3n log τ 3n. 3

4 In particular, the measure of the difference between the sets A and B and their convex hull is bounded b a power δ ɛ, confirming a conjecture of Christ [1]. The result above provides a general quantitative stabilit for the Brunn-Minkowski inequalit in arbitrar dimension. However the exponent degenerates ver quickl as the dimension increases much faster than in Theorem 1.4, and, in addition, the argument in [9] is ver long and involved. The aim of this paper is to provide a shorter and more elementar proof when A = B >, that we believe to be of independent interest. After a dilation, one can assume with no loss of generalit that A = B = 1. In this case, it follows b 1. that 1 A + B = 1 + δ for some δ, and we want to show that a power of δ controls 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 exactl equal, but just close in terms of δ. Here is the main result of this paper: Theorem 1.6. Let A, B R n be measurable sets, and define their semi-sum S := 1 A+B. There exist computable dimensional constants δ n, C n > such that if A 1 + B 1 + S 1 δ 1.5 for some δ δ n, then there exists a convex set K R n such that, up to a translation, where A, B K and K \ A + K \ B C n δ βn, β 1 := 1, β n := 1 6n 5 3 n 1 n!n 1! n αk n, and α k is given b 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 smmetrization and other techniques introduced b Christ [, 3], Theorems 1.3 and 1.4, and two propositions of independent interest, Propositions.5 and.6 below. See Section 3 for further discussion of the strateg of the proof. Acknowledgements: AF was partiall supported b NSF Grant DMS and NSF Grant DMS DJ was partiall supported b NSF Grant DMS-1695 and DMS This work started during AF s visit at MIT during the Fall 1. AF wishes to thank the Mathematics Department at MIT for its warm hospitalit. Notation and preliminar results Let H k denote the k-dimensional Hausdorff measure on R n. Denote b x =, t R n 1 R a point in R n, and let π : R n R n 1 and π : R n R denote the canonical projections, i.e., k=1 π, t := and π, t := t. 4

5 Given a compact set E R n, R n 1, and λ >, we use the notation E := E π 1 {} R, Et := E π 1 t R n 1 {t},.1 Eλ := { R n 1 : H 1 E > λ }.. Following Christ [], we consider two smmetrizations and combine them. For our purposes see the proof of Proposition.5, it is convenient to use a definition of Schwarz smmetrization that is slightl different from the classical one. In the usual definition of Schwarz smmetrization E t = whenever H d 1 Et =. Definition.1. Let E R n be a compact set. We define the Schwarz smmetrization E of E as follows. For each t R, - If H d 1 Et >, then E t is the closed disk centered at R n 1 with the same measure. - If H d 1 Et = but Et is non-empt, then E t = {}. - If Et is empt, then E t is empt as well. We define the Steiner smmetrization E of E so that for each R n 1, the set E is empt if H 1 E = ; otherwise it is the closed interval of length H 1 E centered at R. Finall, we define E := E. As for instance in [, Section ], both the Schwarz and the Steiner smmetrization preserve the measure of sets, and the -smmetrization preserves the measure of the sets Eλ. The following statement collects all these results. Lemma.. Let A, B R n be compact sets. Then A = A = A = A, A + B A + B, A + B A + B, A + B A + B, and, for almost ever λ >, A \ π 1 Aλ = A \ π 1 A λ and H n 1 Aλ = H n 1 A λ, where Aλ := { R n 1 : H 1 A > λ } and A λ := { R n 1 : H 1 A > λ }. 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 sup H 1 A, sup H 1 B, H n 1 πa, H n 1 πb. We refer to [9, Lemma 3.] for a proof. Lemma.3. Let A, B R n be compact sets such that A, B 1/ and 1 A + B. There exists a dimensional constant M > 1 such that sup H 1 A H n 1 πa sup H 1 1/M, M, B H n 1 1/M, M, πb 5

6 sup H 1 A H n 1 πa 1/M, M, sup H 1 B H n 1 πb 1/M, M. Thus, up a measure preserving affine transformation of the form, t τ, τ 1 n t with τ >, all the quantities sup H 1 A, sup H 1 B, H n 1 πa, H n 1 πb are of order one. In particular, H n 1 πa + H n 1 πb + sup In this case, we sa that A and B are M-normalized. H 1 A + sup H 1 B M..3 The following result of Christ [1, Lemma 4.1] shows that sup t H n 1 At and sup t H n 1 Bt are close in terms of δ: Lemma.4. Let A, B R n be compact sets, define S := 1 A + B, and assume that 1.5 holds for some δ 1/. Also, suppose that A and B are M-normalized as defined in Lemma.3. Then, there exists a dimensional constant C > such that sup t H n 1 At sup t H n 1 Bt 1 Cδ 1/, 1 + Cδ 1/. Two other ke ingredients in our proof of Theorem 1.6 are the following propositions, whose proofs are postponed to Section 4: Proposition.5. Let A, B R n be compact sets, define S := 1 A + B, and assume that 1.5 holds for some δ 1/. Also, suppose that we can find a convex set K R n such that S K Cδ α for some α >, where C > is a dimensional constant. Then there exists a dimensional constant C > such that cos \ S C δ α/n. Proposition.6. Let A, B R n be compact sets, define S := 1 A + B, and assume that 1.5 holds for some δ 1/. Also, suppose that cos \ S Cδ β.4 for some β >, where C > is a dimensional constant. Then, up to a translation, A B C δ β/ and there exists a convex set K containing both A and B such that for some dimensional constant C >. K \ A + K \ B C δ β/n, 6

7 3 Proof of Theorem 1.6 As explained in [8], b inner approximation 1 it suffices to prove the result when A, B are compact sets. Hence, let A and B be compact, define S := 1 A + B, and assume that 1.5 holds. We want to prove that there exists a convex set K such that, up to a translation, A, B K, K \ A + K \ B C n δ βn. Moreover, since the statement and the conclusions are invariant under measure preserving affine transformations, b Lemma.3 we can assume that A and B are M-normalized see.3. Ultimatel, we wish to show that, up to translation, each of A, B, and S is of nearl full measure in the same convex set. The strateg of the proof is to show first that S is close to a convex set, and then appl Propositions.5 and.6. To obtain the closeness of S to a convex set, we would like prove that 1 S + S is close to S and then appl Theorem 1.4. It is simpler, however, to construct a subset S S such that S \ S is small and 1 S + S is close to S. To carr out our argument, one important ingredient will be to use the inductive hpothesis on the level sets Aλ and Bλ defined in.. However, two difficulties arise here: first of all, to appl the inductive hpothesis, we need to know that H n 1 Aλ and H n 1 Bλ are close. In addition, the Brunn-Minkowski inequalit does not have a natural proof b induction unless the measures of all the level sets H n 1 Aλ and H n 1 Bλ are the nearl same. See 3.11 below. Hence, it is important for us to have a preliminar quantitative estimate on the difference between H n 1 Aλ and H n 1 Bλ for most λ >. For this we follow an approach used first in [] and readapted in [9], in which we begin b showing our theorem in the special case of smmetrized sets A = A and B = B recall Definition.1. Thanks to Lemma., this will give us the desired closeness between H n 1 Aλ and H n 1 Bλ for most λ >, which allows us to appl the strateg described above and prove the theorem in the general case. Throughout the proof, C will denote a generic constant depending onl on the dimension, which ma change from line to line. 3.1 The case A = A and B = B Let A, B R n be compact sets satisfing A = A, B = B. Since π At π A = πa and π Bt π B = πb are disks centered at the origin, appling Lemma.4 we deduce that H n 1 πa πb Cδ 1/. 3.1 Hence, if we define S := πa πb A + B, 1 The approximation of A and analogousl for B is b a sequence of compact sets A k A such that A k A and coa k coa. One wa to construct such sets is to define A k := A k V k, where A k A are compact sets satisfing A k A, and V k V k+1 A are finite sets satisfing cov k coa. 7

8 then S S for all R n 1. In addition, using 1.5,.3, and 3.1, we have 1 + δ S = H 1 S d H 1 S d H 1 S d R n 1 πa πb πa πb = S 1 H 1 A d + 1 H 1 B d πa πb πa πb A + B MH n 1 πa πb 1 Cδ 1/, which implies since S S S \ S Cδ 1/. 3. Furthermore, since each section S is an interval centered at R, for all, πa πb such that + =, which gives S + S = A + B + A + B = A + B S + S + A + B S + S = S, S. 3.3 Recalling 1.3, b 3. and 3.3 we obtain that δ S Cδ 1/. Hence, we can appl Theorem 1.4 to S to find a convex set K such that Hence, b 3.3, S K Cδ nαn/. S K Cδ nαn/, and using Propositions.5 and.6 we deduce that, up to a translation, there exists a convex set K such that A B K and A B Cδ αn/8, K \ A + K \ B Cδ αn/8n. 3.4 Notice that, because A = A and B = B, it is eas to check that the above properties still hold with K in place of K. Hence, in this case, without loss of generalit one can assume that K = K. 3. The general case Since, b Theorem 1., the result is true when n = 1, we ma assume that we alread proved Theorem 1.6 through n 1, and we want to show its validit for n. Step 1: There exist a dimensional constant ζ > and λ δ ζ such that we can appl the inductive hpothesis to A λ and B λ. Let A and B be as in Definition.1 and denote ᾱ := α n

9 Thanks to Lemma., A and B still satisf 1.5, so we can appl the result proved in Section 3.1 above to get see 3.4 H 1 A H 1 B d H 1 A B d = A B Cδᾱ 3.6 R n 1 R n 1 and K A B, K \ A + K \ B Cδᾱ/n 3.7 for some convex set K = K. In addition, because A and B are M-normalized see.3, so are A and B, and b 3.7 we deduce that there exists a dimensional constant R n > such that K B Rn. 3.8 Also, b 3.6 and Chebshev s inequalit we obtain that, except for a set of measure Cδᾱ/, H 1 A H 1 B δᾱ/. Thus, recalling Lemma., for almost ever λ > H n 1 Aλ = H n 1 A λ H n 1 B λ δᾱ/ + Cδᾱ/ = H n 1 Bλ δᾱ/ + Cδᾱ/. Since, b.3, M H n 1 Bλ H n 1 Bλ + δᾱ/ δ ᾱ/ dλ = H n 1 Bλ dλ Mδᾱ/, b Chebshev s inequalit we deduce that H n 1 Aλ H n 1 Bλ + Cδᾱ/4 for all λ outside a set of measure Cδᾱ/4. Exchanging the roles of A and B we obtain that there exists a set F [, M] such that H 1 F Cδᾱ/4, H n 1 Aλ H n 1 Bλ Cδᾱ/4 λ [, ] \ F. 3.9 Using the elementar inequalit a + b n 1 a n 1 + b n 1 C a b a, b M, and replacing a and b with a 1/n 1 and b 1/n 1, respectivel, we get a 1/n 1 + b 1/n 1 n 1 a + b C a b /n 1 a, b M 3.1 notice that a 1/n 1 b 1/n 1 a b 1/n 1. Finall, it is eas to check that Aλ + Bλ Sλ λ >. 9

10 Hence, b the Brunn-Minkowski inequalit 1. applied to Aλ and Bλ, using 1.5,.3, 3.1, and 3.9, we get 1 + δ S = M 1 1 n 1 M C H n 1 Sλ dλ M H n 1 Aλ 1/n 1 + H n 1 Bλ 1/n 1 n 1 dλ H n 1 Aλ + H n 1 Bλ dλ M A + B = 1 Cδᾱ/[n 1]. H n 1 Aλ H n 1 Bλ /n 1 dλ Cδᾱ/[n 1] 3.11 We also observe that, since K = K, b Lemma., 3.8, and [, Lemma 4.3], for almost ever λ > we have A \ π 1 Aλ = A \ π 1 A λ K \ π 1 Kλ + M H n 1 A λ Kλ 3.1 and analogousl for B. Also, b 3.7, M Define Cλ + M H n 1 A λ Kλ, H n 1 A λ Kλ + H n 1 B λ Kλ dλ K \ A + K \ B Cδᾱ/n { η := min ᾱ n 1, ᾱ 4 }, 3.14 and note that η ᾱ/n. Let ζ, η to be fixed later. Then b 3.9, 3.11, 3.1, 3.13, and b Chebshev s inequalit, we can find a level λ [1δ ζ, δ ζ ] 3.15 such that H n 1 A λ H n 1 B λ Cδ η n 1 H n 1 S λ H n 1 A λ 1/n 1 + H n 1 B λ 1/n 1 n 1 + Cδ η ζ, 3.17 A \ π 1 A λ + B \ π 1 B λ C δ ζ + δ η ζ, 3.18 In addition, from the properties H n 1 Aλ M for an λ > see.3, M H n 1 Aλ dλ = A 1 δ, and s H n 1 Aλ is a decreasing function, we deduce that 1 M Hn 1 Aλ M λ, M 1. 1

11 The same holds for B and S, hence H n 1 S λ, H n 1 A λ, H n 1 B λ [ M 1, M ] provided δ is small enough. Set ρ := 1/H n 1 A λ 1/n 1 [1/M 1/n 1, M 1/n 1 ], and define B 3.17 and 3.16 we get while, b 1., H n 1 A = 1, A := ρa λ, B := ρb λ, S := ρs λ. H n 1 B 1 Cδ η, H n 1 S 1 + Cδ η ζ. H n 1 S 1/n 1 Hn 1 A 1/n 1 + H n 1 B 1/n 1 1 Cδ η, therefore H n 1 A 1 + H n 1 B 1 + H n 1 S 1 Cδ η ζ. Thus, b the inductive hpothesis of Theorem 1.6, up to a translation there exists a n 1- dimensional convex set Ω such that Ω A B, H n 1 Ω \ A + H n 1 Ω \ B Cδ η ζβ n 1, and defining Ω := Ω /ρ we obtain recall that 1/ρ M 1/n 1 Ω A λ B λ, H n 1 Ω \ A λ + H n 1 Ω \ B λ Cδ η ζβ n Step : We appl Theorem 1. to the sets A and B for most A λ B λ. Define C := A λ B λ S λ. B 3.18, 3.19, and.3, we have A \ π 1 C + B \ π 1 C A \ π 1 A λ + B \ π 1 B λ + A λ\b λ H1 A d + B λ\a λ H1 B d C δ ζ + δ η ζ + M H n 1 Ω \ A λ + H n 1 Ω \ B λ C δ ζ + δ η ζ + δ η ζβ n 1 Cδ ζ provided we choose ζ := ηβ n 1 3 recall that β n 1 1. Hence, b 1.5 and 3., H 1 S \ A + B [ d H 1 S 1 H 1 A + H 1 B ] d C C = S π 1 C A π 1 C + B π 1 C A + B S + A \ π 1 C + B \ π 1 C Cδ ζ

12 Write C as C 1 C, where C 1 := { C : H 1 S H 1 A H 1 B δ ζ / }, C := C \ C 1. B Chebshev s inequalit and 3., while, recalling 3.15, H n 1 C Cδ ζ, 3.3 min { H 1 A, H 1 B } λ 1δ ζ > δ ζ / C 1. Hence, b Theorem 1. applied to A, B R for C 1, we deduce that Let Ĉ1 C 1 denote the set of C 1 such that H 1 coa \ A + H 1 cob \ B δ ζ. 3.4 H 1 S \ A + B δ ζ, 3.5 and notice that, b 3. and Chebshev s inequalit, H n 1 C 1 \Ĉ1 Cδ ζ. Then choose a compact set C 1 Ĉ1 such that H n 1 Ĉ1 \ C 1 δζ to obtain Step 3: We find S S so that S \ S and δ S are small. Define the compact set H n 1 C 1 \ C 1 Cδ ζ. 3.6 S := C 1 A + B Observe, thanks to 3., 3.3, 3.6,.3, and 1.5, S H 1 A d + H 1 B d So, since S S, C 1 C 1 R n. A + B A \ π 1 C B \ π 1 C M H n 1 C \ C 1 S Cδ ζ. Now, we want to estimate the measure of 1 S + S. First of all, since S S Cδ ζ. 3.7 S = A + B, 3.8 = + 1

13 b 3.5 we get H 1 A + B = + Also, if we define the characteristic functions { χ A 1 if, λ A λ := otherwise, \ A + B δ ζ C { χ A, 1 if, λ coa λ := otherwise, and analogousl for B, b 3.4 we have the following estimate on their convolutions: χ A, χ B, χ A χb L R χ B, χ B L 1 R + χ A, χ A L 1 R = H 1 cob \ B + H 1 coa \ A δ ζ < 3δ ζ, C Recalling that π : R n R is the orthogonal projection onto the last component that is, π, t = t, denote b [a, b] the interval π coa + cob, and notice that, since b construction min{h 1 A, H 1 B } λ 1δ ζ C 1 see 3.15, this interval has length greater than δ ζ. Also, it is eas to check that the function χ A, χ B, is supported on [a, b], has slope equal to 1 resp. 1 in [a, a+3δ ζ ] resp. [b 3δ ζ, b], and it is greater than 3δ ζ in [a+3δ ζ, b 3δ ζ ]. Hence, since πa +B contains the set {χ A χ B > }, b 3.3 we deduce that π A + B [a + 3δ ζ, b 3δ ζ ], 3.31 which implies in particular that H 1 coa + cob H 1 A + B + 6δ ζ Also, b the same argument as in [8, Step -a], if we denote b using 3.5 and 3.31 we have [α, β ] := π coa + cob, π coa + cob [α 16δ ζ, β + 16δ ζ ],, = + Compare with [8, Equation 3.5]. We now estimate the size of [ 1 S + S ]. Observe that, for all C 1, [ 1 S + S ] = 1 A + B + 1 A + B = +,, C 1 1 A + B + 1 = A + B 1 = +,, C 1 = +,, C 1 1 A + B + 13, C = +,, C 1 C A + B.

14 Hence, b 3.33 we deduce that each of the latter sets is contained inside the convex set {} [α 16δ ζ, β + 16δ ζ ], so also their semi-sum is contained in the same set, and using 3.3 with = = we get H 1 [ S + S/] H 1 coa + cob + 16δ ζ H 1 A + B + δ ζ 3.34 = H 1 S + δ ζ C 1. In order to estimate [ 1 S + S ] when C 1 +C 1 \ C 1 we argue as follows: b 3.33 and the fact that H 1 coa and H 1 cob are universall bounded see.3 and 3.4, the following holds: if we denote b c A the barcenter of coa and analogousl for B and S, we have c A + c B c S C,, C 1, = + notice that co S = coa + cob. Exchanging the role of A and B and adding up the two inequalities, we deduce that c S + c S c S C,, C 1, = +. As shown in [8, Step 3], this estimate combined with the fact that C 1 is almost of full measure inside the convex set Ω see 3.19, 3.3, and 3.6 proves that, up to an affine transformation of the form R n 1 R, t T, t L +, t 3.35 with T : R n 1 R n 1, L : R n 1 R, dett = 1, and, t R n, the set S is universall bounded, sa S B R for some dimensional constant R. This implies that [ 1 S + S ] [ R, R], so H 1[ 1 S + S ] R. Hence, since 1 C 1 + C 1 Ω, b 3.34, 3.19, and 3.1, S + S \ S = H 1 [ S + S/] H 1 S d [ 1 C 1 +C 1 ] C 1 + H 1 [ S + S/] d [ 1 C 1 +C 1 ]\C 1 δ ζ H n 1 Ω + R H n 1 Ω \ C 1 Cδ ζ, that is, δ S Cδ ζ. Step 4: Conclusion. B the previous step we have that δ S Cδ ζ. Hence, appling Theorem 1.4 to S we find a convex set K such that S K Cδ nαnζ, 14

15 so, b 3.7, S K Cδ nαnζ. Using this estimate together with Propositions.5 and.6 we deduce that, up to a translation, there exists a convex set K convex such that A B K and K \ A + K \ B Cδ αnζ/4n. Recalling the definition of ζ see 3.5, 3.14, 3.1, we see that β n := α { nζ 1 4n = min n 1, 1 } α n 3 6 n β n 1. Since β 1 = 1 b Theorem 1., it is eas to check that concluding the proof. β n = 1 6n 5 3 n 1 n!n 1! n αk n, k=1 4 Technical results As in the previous section, we use C to denote a generic constant depending onl on the dimension, which ma change from line to line. 4.1 Proof of Proposition.5 Assume that S K Cδ α for some α, 1]. B John s Lemma [16], after a volume preserving affine transformation, we can assume that B rn K B nrn, with r n = r n K > bounded above and below b 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 alread chosen a different affine normalization. We want to prove that S 1 + Cδ α/n K. 4.1 Let x S \ K, and set ρ := dist x, K = x x 1 with x 1 K. With no loss of generalit we can assume that x 1 = τe n, for some τ >, x = τ + ρe n, and K {x n τ}. We need to prove that ρ Cδ α/n. Let us consider the sets A, B, S, K obtained from A, B, S, K performing a Schwarz smmetrization around the e n -axis see Definition.1. Set S := 1 A + B. Since S K S K Cδ α, and, b 1.5 notice that S S and that S 1 Cδ b 1., S \ S = S S = S S Cδ, 15

16 we get that S K Cδ α. In addition, K {x n τ}, x 1 K, and x S. Hence, without loss of generalit we can assume from the beginning that A = A, B = B, S = 1 A + B, and K = K. For a compact set E R n, recall the notation Et R n 1 {t} in.1, and define E[s] R b E[s] := { t : H n 1 Et s } 4. Since S = 1 A + B we have At so, b 1. we deduce that Hence + Bt S[s] St t R, A[s] + B[s] and integrating with respect to s, b 1.5 we get s >. H 1 A[s] + H 1 B[s] H 1 S[s] s >, 4.3 4δ S A B = H 1 S[s] H 1 A[s] H 1 B[s] ds. 4.4 Recall that K = K, so that the canonical projection πk onto R n 1 is a ball. We denote it B R := πk, and note that R nr n, with r n = r n K given b John s lemma at the beginning of this proof. Then, since S K Cδ α we have so, b 4.3, Cδ α S \ π 1 B R = A \ π 1 B R + B \ π 1 B R = H n 1 B R H n 1 B R Hence, recalling that A and B are 1 δ, we deduce that H n 1 B R H 1 A[s] ds 1/, H 1 S[s] ds, H 1 A[s] + H 1 B[s] ds Cδ α. 4.5 H n 1 B R H 1 B[s] ds 1/, and since R is universall bounded being less than nr n and both functions s H 1 A[s], s H 1 B[s] are decreasing, there exists a small dimensional constant c > such that min { H 1 A[s], H 1 B[s] } c s, c. 4.6 Also, b 4.4, c H 1 S[s] H 1 A[s] H 1 B[s] ds 4δ,

17 and since S K Cδ α and K {x n τ} c H 1 S[s] \, τ] ds S \ {x n τ} Cδ α. 4.8 Hence, thanks to 4.6, 4.7, 4.8, we use Theorem 1. and Chebishev s inequalit to find a value such that notice that α 1 and s [δ α/, δ α/ ] 4.9 H 1 co A[ s] \ A[ s] + H 1 co B[ s] \ B[ s] Cδ 1 α/ Cδ α/ Since 1 A[ s] + B[ s] S[ s], this implies H 1 S[ s] \, τ] Cδ α/. co A[ s] + co B[ s], τ + Cδ α/ ]. Hence, after appling opposite translations along the e n -axis to A and B, i.e., for some l R, we can assume that A A + le n, B B le n, co A[ s], τ + Cδ α/ ], co B[ s], τ + Cδ α/ ]. Since the sets s A[s], B[s] are decreasing, we deduce that co A[s], co B[s], τ + Cδ α/ ], s s. 4.1 We now want to bound sup s> H 1 A[s]. Recall that we cannot assume that A and B are M-normalized, since we alread made an affine transformation to ensure that B rn K B nrn. Since A = A, we have sup s> H 1 A[s] = sup R n 1 H 1 A, so, b Lemma.3, sup H 1 A[s] s> M H n 1 πb, H n 1 πa H n 1 πb M 1, M In addition, since πa and πb are n 1-dimensional disks centered on the e n -axis, S K Cδ α, and B rn K B nrn, we easil deduce that πa + πb = πs B rn/, 4.1 provided δ is small enough. Hence, combining 4.11 and 4.1 we deduce that H n 1 πb is bounded from awa from zero b a dimensional constant, thus sup s> H 1 A[s] C

18 Hence, b 4.5, 4.1, 4.13, and 4.9, A \ {x n τ} A \ π 1 B R + π 1 B R {τ x n τ + Cδ α/ } + and, analogousl, Cδ α + Cδ α/ + C s Cδ α/, Now, given r, let us define the sets s H 1 A[s] ds 4.14 B \ {x n τ} Cδ α/ A r := A {x n τ r}, B r := B {x n τ r}, S r := S {x n τ r}. B 4.14 and 4.15 we know that and it is immediate to check that A + B r A, B 1 Cδ α/, S r/, A r + B S r/. Also, since K is a convex set satisfing B rn K B nrn, there exists a dimensional constant c n > such that K {τ r/ x n τ} c n min { r n, 1 }. Hence and b 1. applied to A r and B we get which gives S r/ S S {τ r/ x n τ} S + S K K {τ r/ x n τ} 1 + Cδ α c n min { r n, 1 }, 1 Cδ α/ C A {τ r x n τ} A r 1/n + B 1/n S r/ 1/n 1 + Cδ α c n min { r n, 1 }, C A {τ r x n τ} c n min { r n, 1 } Cδ α/ and analogousl for B Since the point x = τ + ρe n belongs to S = A + B/, there as to be a point x A B such that x e n τ + ρ. With no loss of generalit, assume that x B. Then, b 4.16 applied with r = ρ we get S {x n τ} x + A {τ ρ x n τ}, so Cδ α S {x n τ} A {τ ρ x n τ} n c n C min{ ρ n, 1 } Cδ α/, which implies ρ Cδ α/n, proving 4.1. Hence cos 1 + Cδ α/n K, from which the result follows immediatel. 18

19 4. Proof of Proposition.6 Since b 1.,.4, and 1.5 we have coa + cob = cos, from which we deduce that coa 1/n + cob 1/n coa + cob 1/n = cos 1/n S 1/n + Cδ β A 1/n + B 1/n + Cδ β coa 1/n + cob 1/n + Cδ β, coa \ A + cob \ B Cδ β Also, b Theorem 1.3 and the fact that coa cob Cδ βαn see 4.17 we obtain that, up to a translation, coa cob C δ β/ + δ β Cδ β/ This estimate combined with 4.17 implies that A B Cδ β/. In addition, if we define K := coa B, then we will conclude our argument b showing that K \ A + K \ B Cδ β/n Indeed, b John s Lemma [16], after a volume preserving affine transformation we can assume that B r coa B nr for some radius r bounded above and below b positive dimensional constants. B 4.18 and a simple geometric argument we easil deduce that Thus and 4.19 follows b 4.17 and References cob 1 + Cδ β/n coa. coa cob K 1 + Cδ β/n coa, [1] Christ M. Near equalit in the two-dimensional Brunn-Minkowski inequalit. Preprint, 1. Available at [] Christ M. Near equalit in the Brunn-Minkowski inequalit. Preprint, 1. Available at [3] Christ M. An approximate inverse Riesz-Sobolev inequalit. Preprint, 11. Available at 19

20 [4] Christ M. Personal communication. [5] Diskant, V. I. Stabilit of the solution of a Minkowski equation. Russian Sibirsk. Mat. Ž , , 696. [6] Figalli A. Stabilit results for the Brunn-Minkowski inequalit. Colloquium De Giorgi 13-14, to appear. [7] Figalli A. Quantitative stabilit results for the Brunn-Minkowski inequalit. Proceedings of the ICM 14, to appear. [8] Figalli A.; Jerison D. Quantitative stabilit for sumsets in R n. J. Eur. Math. Soc. JEMS, 17 15, no. 5, [9] Figalli A.; Jerison D. Quantitative stabilit for the Brunn-Minkowski inequalit. Preprint, 14. [1] Figalli, A.; Maggi, F.; Pratelli, A. A mass transportation approach to quantitative isoperimetric inequalities. Invent. Math. 18 1, no. 1, [11] Figalli, A.; Maggi, F.; Pratelli, A. A refined Brunn-Minkowski inequalit for convex sets. Ann. Inst. H. Poincaré Anal. Non Linéaire 6 9, no. 6, [1] Freiman, G. A. The addition of finite sets. I. Russian Izv. Vss. Ucebn. Zaved. Matematika, 1959, no. 6 13, -13. [13] Freiman, G. A. Foundations of a structural theor of set addition. Translated from the Russian. Translations of Mathematical Monographs, Vol 37. American Mathematical Societ, Providence, R. I., [14] Gardner, R. J., The Brunn-Minkowski inequalit. Bull. Amer. Math. Soc. N.S. 39, no. 3, [15] Groemer, H. On the Brunn-Minkowski theorem. Geom. Dedicata , no. 3, [16] John F. Extremum problems with inequalities as subsidiar conditions. In Studies and Essas Presented to R. Courant on his 6th Birthda, Januar 8, 1948, pages Interscience, New York, [17] Tao, T.; Vu, V. Additive combinatorics. Cambridge Studies in Advanced Mathematics, 15. Cambridge Universit Press, Cambridge, 6.