Quantitative stability for the Brunn-Minkowski inequality

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1 Quantitative stability for the Brunn-Minkowski inequality Alessio Figalli and David Jerison Abstract We prove a quantitative stability result for the Brunn-Minkowski inequality: if A = B = 1, t [τ, 1 τ] with τ > 0, and ta+1 tb 1/n 1+δ for some small δ, then, up to a translation, both A and B are quantitatively close in terms of δ to a convex set K. 1 Introduction Given two sets A, B R n, and c > 0, we define the set sum and scalar multiple by A + B := {a + b : a A, b B}, ca := {ca : a A} 1.1 Let E 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 states that, given A, B R n nonempty measurable sets, A + B 1/n A 1/n + B 1/n. 1.2 In addition, if A, B > 0, then equality holds if and only if there exist a convex set K R n, λ 1, λ 1 > 0, and v 1, v 2 R n, such that λ 1 A + v 1 K, λ 2 B + v 2 K, K \ λ 1 A + v 1 = K \ λ 2 B + v 2 = 0. Our aim is to investigate the stability of such a statement. When n = 1, the following sharp stability result holds as a consequence of classical theorems in additive combinatorics an elementary proof of this result can be given using Kemperman s theorem [C3, C4]: Theorem 1.1. Let A, B R be measurable sets. If A + B < A + B + δ for some δ min{ A, B }, then there exist two intervals I, J R such that A I, B J, I \ A δ, and J \ B δ. The University of Texas at Austin, Mathematics Dept. RLM 8.100, 2515 Speedway Stop C1200, Austin, TX USA. address: figalli@math.utexas.edu Department of Mathematics, Massachusetts Institute of Technology, 77 Massachusetts Ave, Cambridge, MA USA. address: jerison@math.mit.edu 1

2 Concerning the higher dimensional case, in [C1, C2] M. Christ proved a qualitative stability result for 1.2, namely, if A+B 1/n is close to A 1/n + B 1/n then A and B are close to homothetic convex sets. On the quantitative side, first V. I. Diskant [D] and then H. Groemer [G] obtained some stability results for convex sets in terms of the Hausdorff distance. More recently, sharp stability results in terms of the L 1 distance have been obtained by the first author together with F. Maggi and A. Pratelli [FMP1, FMP2]. Since this latter result will play a role in our proofs, we state it in detail. We begin by noticing that, after dilating A and B appropriately, we can assume A = B = 1 while replacing the sum A + B by a convex combination S := ta + 1 tb. It follows by 1.2 that S = 1 + δ for some δ 0. Theorem 1.2. see [FMP1, FMP2] There is a computable dimensional constant C 0 n such that if A, B R n are convex sets satisfying A = B = 1, ta + 1 tb = 1 + δ for some t [τ, 1 τ], then, up to a translation, A B C 0 nτ 1/2n δ 1/2 Here and in the sequel, E F denotes the symmetric difference between two sets E and F, that is E F = E \ F F \ E. Our main theorem here is a quantitative version of Christ s result. His result relies on compactness and, for that reason, does not yield any explicit information about the dependence on the parameter δ. Since our proof is by induction on the dimension, it will be convenient to allow the measures of A and B not to be exactly equal, but just close in terms of δ. Here is the main result of this paper, which shows that the measure of the difference between the sets A and B and their convex hull is bounded by a power δ ɛ, confirming a conjecture of Christ [C1]. Theorem 1.3. Let n 2, let A, B R n be measurable sets, and define S := ta + 1 tb for some t [τ, 1 τ], 0 < τ 1/2. There are computable dimensional constants N n and computable functions M n τ, ε n τ > 0 such that if A 1 + B 1 + S 1 δ 1.3 for some δ e Mnτ, then there exists a convex set K R n such that, up to a translation, Explicitly, we may take A, B K and K \ A + K \ B τ Nn δ εnτ. M n τ = 23n+2 n 3n log τ 3n τ 3n, ε n τ = τ 3n 2 3n+1 n 3n log τ 3n. It is interesting to make some comments on the above theorem: first of all, notice that the result holds only under the assumption that δ is sufficiently small, namely δ e Mnτ. A smallness assumption on δ is actually necessary, as can be easily seen from the following example: A = B := B ρ 0 {2L e 1 }, 2

3 where L 1, e 1 denotes the first vector of the canonical basis in R n, and ρ > 0 is chosen so that B ρ 0 = 1. Then it is easily checked that 1 2 A B = B ρ 0 B ρ/2 L e 1 {2L e 1 } = n, while coa L can be arbitrarily large, hence the result is false unless we assume that δ < 2 n. Concerning the exponent ε n τ, at the moment it is unclear to us whether a dimensional dependency is necessary. It is however worth to point out that there are stability results for functional inequalities where a dimensional dependent exponent is needed see for instance [BP, Theorem 3.5], so it would not be completely surprising if in this situation the optimal exponent does depend on n. We plan to investigate this very delicate question in future works. Another important direction to develop would be to understand the analytic counterpart of the Brunn-Minkowski inequality, namely the Prékopa-Leindler inequality. At the moment, some stability estimates are known only in one dimension or for some special class of functions [BB1, BB2], and a general stability result would be an important direction of future investigations. The paper is structured as follows. In the next section we introduce a few notations and give an outline of the proof along with some commentary on the techniques and ideas. Then, in Section 3 we collect most of the technical results we will use. Since the proofs of some of these technical results are delicate and involved, we postpone them to Section 5. Section 4 is devoted to the proof of Theorem 1.3. Acknowledgements: AF was partially supported by NSF Grant DMS DJ was partially supported by the Bergman Trust and NSF Grant DMS Notation and an outline of the proof Let H k denote k-dimensional Hausdorff measure on R n. Denote by x = y, s 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., πy, s := y and πy, s := s. Given a compact set E R n, y R n 1, and λ > 0, we use the notation E y := E π 1 y {y} R, Es := E π 1 s R n 1 {s}, 2.1 Following Christ [C2], we consider different symmetrizations. Eλ := { y R n 1 : H 1 E y > λ }. 2.2 Definition 2.1. Let E R n be a compact set. We define the Schwarz symmetrization E of E as follows. For each t R, - If H n 1 Es > 0, then E s is the closed disk centered at 0 R n 1 with the same measure. - If H n 1 Es = 0, then E s is empty. We define the Steiner symmetrization E of E so that for each y R n 1, the set E y is empty if H 1 E y = 0; otherwise it is the closed interval of length H 1 E y centered at 0 R. Finally, we define E := E. 3

4 Outline of the proof of Theorem 1.3 The proof of Theorem 1.3 is very elaborate, combining the techniques of M. Christ with those developed by the present authors in [FJ] where we proved Theorem 1.3 in the special case A = B and t = 1/2, as well as several new ideas. For that reason, we give detailed description of the argument. In Section 4.1 we prove the theorem in the special case A = A and B = B. In this case we have that A y = {y} [ ay, ay] and B y = {y} [ by, by], for some functions a, b : R n 1 R +, and it is easy to show that a and b satisfy the 3-point concavity inequality tay + 1 tby [ta + 1 tb]y + δ 1/4 2.3 whenever y, y, and y := ty + 1 ty belong to a large subset F of πa πb. From this 3-point inequality and an elementary argument Remark 4.1 we show that a satisfies the 4-point concavity inequality ay 1 + ay 2 ay 12 + ay t δ1/4 2.4 with y 12 := t y t y 2, y 12 := t y t y 2, t := 1 2 t, t := 1 t, provided all four points belong to F. The analogous inequality for b involves a different set of four points. Using this inequality and Lemma 3.6, we deduce that a is quantitatively close in L 1 to a concave function. The proof, in Section 5, of Lemma 3.6, although reminiscent of Step 4 in the proof of [FJ, Theorem 1.2], is delicate and involved. Once we know that a and analogously b is L 1 -close to a concave function, we deduce that both A and B are L 1 -close to convex sets K A and K B respectively, and we would like to say that these convex sets are nearly the same. This is demonstrated as part of Proposition 3.4, which is proved by first showing that S is close to tk A + 1 tk B, then applying Theorem 1.2 to deduce that K A and K B are almost homothetic, and then constructing a convex set K close to A and B and containing both of them. This concludes the proof of Theorem 1.3 in the case A = A and B = B. In Section 4.2 we consider the general case, which we prove in several steps, culminating in induction on dimension. Step 1. This first step is very close to the argument used by M. Christ in [C2], although our analysis is more elaborate since we have to quantify every estimate. Given A, B, and S, as in the theorem, we consider their symmetrizations A, B, and S, and apply the result from Section 4.1 to deduce that A and B are close to the same convex set. This information combined with Christ s Lemma 3.1 allows us to deduce that functions y H 1 A y and y H 1 B y are almost equipartitioned that is, the measure of their level sets Aλ and Bλ are very close. This fact combined with a Fubini argument yields that, for most levels λ, Aλ and Bλ are almost optimal for the n 1-dimensional Brunn-Minkowski inequality. Thus, by the inductive step, we can find a level λ δ ζ ζ > 0 such that we can apply the inductive hypothesis to A λ and B λ. Consequently, after removing sets of small measure both from A and B and 4

5 translating in y, we deduce that πa, πb R n 1 are close to the same convex set. Step 2. This step is elementary: we apply a Fubini argument and Theorem 1.1 to most of the sets A y and B y for y A λ B λ to deduce that they are close to their convex hulls. Note, however, that to apply Fubini and Theorem 1.1 it is crucial that, thanks to Step 1, we found a set in R n 1 onto which both A and B project almost fully. Indeed, in order to say that H 1 A y + B y H 1 A y + H 1 B y it is necessary to know that both A y and B y are nonempty, as otherwise the inequality would be false! Step 3. The argument here uses several ideas from our previous paper [FJ] to obtain a 3-point concavity inequality as in 2.3 above for the upper profile of A and B and an analogous inequality for the lower profile. This inequality allows us to say that the barycenter of A y satisfies the 4-point inequality 2.4 both from above and from below, and from this information we can deduce that, as a function of y, the barycenter of A y resp. B y is at bounded distance from a linear function see Lemma 5.1. It follows that the barycenters of S y are a bounded distance from a linear function for a set S which is almost of full measure inside S. Then a variation of [FJ, Proof of Theorem 1.2, Step 3] allows us to show that, after an affine measure preserving transformation, S is universally bounded, that is, bounded in diameter by a constant of the form C n τ Mn where C n and M n are dimensional constants. Step 4. By a relatively easy argument we find sets A and B of the form A = {y} [a A y, b A y] B = {y} [a B y, b B y] y F which are close to A and B, respectively, and are universally bounded. Step 5. This is a crucial step: we want to show that A and B are close to convex sets. As in the case A = A and B = B, we would like to apply Lemma 3.6 to deduce that b A and b B resp. a A and a B are L 1 -close to concave resp. convex functions. The main issue is that the hypothesis of the lemma, in addition to asking for boundedness and concavity of b A and b B at most points, also requires that the level sets of b A and b B be close to their convex hulls. To deduce this we wish to show that most slices of A and B are nearly optimal in the Brunn-Minkowski inequality in dimension n 1 and invoke the inductive hypothesis. We achieve this by an inductive proof of the Brunn-Minkowski inequality, based on combining the validity of Brunn-Minkowski in dimension n 1 with 1-dimensional optimal transport see Lemma 3.5. An examination of this proof of the Brunn-Minkowski inequality in the situation near equality shows that if A and B are almost optimal for the Brunn-Minkowski inequality in dimension n, then for most levels s, the slices As and BT s have comparable n 1-measure, where T is the 1-dimensional optimal transport map, and this pair of sets is almost optimal for the Brunn- Minkowski inequality in dimension n 1. In particular, we can apply the inductive hypothesis to deduce that most n 1-dimensional slices are close to their convex hulls. This nearly suffices to apply Lemma 3.6. But this lemma asks for control of the superlevel sets of the function b A, which a priori may be very different from the slices of A. To avoid this issue, 5 y F

6 we simply replace A and B by auxiliary sets A and B which consist of the top profile of A and B with a flat bottom, so that the slices coincide with the superlevel sets of b A and b B. We then show that Lemma 3.5 applies to A and B. In this way, we end up proving that A and B are close to convex sets, as desired. Step 6. Since A and B are close to A and B respectively, we simply apply Proposition 3.4 as before in 4.1 to conclude the proof of the theorem. Step 7. Tracking down the exponents in the proof, we provide an explicit lower resp. upper bound on ε n τ resp. M n τ. 3 Technical Results In this section we state most of the important lemmas we will need. The first three are due to M. Christ or are easy corollaries of his results. It is well-known that both the Schwarz and the Steiner symmetrization preserve the measure of sets, while they decrease the measure of the semi-sum see for instance [C2, Lemma 2.1]. Also, as shown in [C2, Lemma 2.2], the -symmetrization preserves the measure of the sets Eλ. We combine these results into one lemma, and refer to [C2, Section 2] for a proof. Lemma 3.1. Let A, B R n be compact sets. Then A = A = A = A, ta +1 tb ta+1 tb, ta +1 tb ta+1 tb, ta +1 tb ta+1 tb, and, with the notation in 2.2, A \ π 1 Aλ = A \ π 1 A λ and H n 1 Aλ = H n 1 A λ for almost every λ > 0. Another important fact is that a bound on the measure of ta+1 tb in terms of the measures of A and B implies bounds relating the sizes of sup H 1 A y, y sup H 1 B y, H n 1 πa, H n 1 πb. y Lemma 3.2. Let A, B R n be compact sets such that A, B 1/2 and ta + 1 tb 2 for some t 0, 1, and set τ := min{t, 1 t} There exists a dimensional constant M > 1 such that sup y H 1 A y τ n sup y H 1 B y M, M H n 1 πa τ τ n, H n 1 πb n M, M τ n, sup H 1 A y H n 1 πa 1 y M, M τ 2n, sup H 1 B y H n 1 πb 1 y M, M τ 2n. and, up a measure preserving affine transformation of the form y, s λy, λ 1 n t with λ > 0, we have H n 1 πa + H n 1 πb + sup y In this case, we say that A and B are M, τ-normalized. 6 H 1 A y + sup H 1 B y M. 3.1 y τ 2n

7 Proof. As observed in [C2, Lemma 3.1] and in the discussion immediately after that lemma, sup y H 1 A y H n 1 πb ta + 1 tb t1 t n 1 2 τ n, sup y H 1 A y H n 1 πa A 1/2. By exchanging the roles of A and B, the first part of the lemma follows. To prove the second part, it suffices to choose λ > 0 so that λ n 1 H n 1 πa = 1/τ n. The third lemma is a result of Christ [C1, Lemma 4.1] showing that sup s H n 1 As and sup s H n 1 Bs are close in terms of δ: Lemma 3.3. Let A, B R n be compact sets, define S := ta + 1 tb for some t [τ, 1 τ], and assume that 1.3 holds for some δ 1/2. Then there exists a numerical constant L > 0 such that sup s H n 1 As sup s H n 1 Bs 1 Lτ 1/2 δ 1/2, 1 + Lτ 1/2 δ 1/2. Proof. Set sups H n 1 As 1 t γ := sup s H n 1 Bs sups H n 1 Bs t, γ := sup s H n 1 As, and after possibly exchanging A and B, we may assume that γ 1. By the argument in the proof of [C1, Lemma 4.1] we get S tγ 1 A + 1 t γ 1 B, so, by 1.3, tγ tγ t/1 t 1 4δ. The function γ tγ tγ t/1 t is convex for γ 0, 1], attains its minimum at γ = 1, and its second derivative is bounded below by τ. It follows that 4δ tγ tγ t/1 t 1 τ 2 γ 1 2, which proves the result. There are several other important ingredients in the proof of Theorem 1.3, which are to our knowledge new. Because their proofs are long and involved, we postpone them to Section 5. The first of these results shows that if A and B are L 1 -close to convex sets K A and K B respectively, then A and B are close to each other, and we can find a convex set K which contains both A and B with a good control on the measure. As we shall see, the proof relies primarily on Theorem 1.2. Proposition 3.4. Let A, B R n be compact sets, define S := ta + 1 tb for some t [τ, 1 τ], and assume that 1.3 holds. Suppose A, B B R, for some R τ Nn with N n a dimensional constant N n > 1. Suppose further that we can find a convex sets K A, K B R n such that A K A + B K B ζ 3.2 7

8 for some ζ δ. Then there exists a dimensional constant L n > 1 such that after a translation, A B τ Ln ζ 1/2n and there exists a convex set K containing both A and B such that K \ A + K \ B τ Ln ζ 1/2n3. Our next result is a consequence of a proof of the Brunn-Minkowski inequality by induction, using horizontal n 1-dimensional slices. The lemma says that when A, B, and S satisfy 1.3, then most of their horizontal slices chosen at suitable levels satisfy near equality in the Brunn- Minkowski inequality and the ratio of their volumes of the slices is comparable to 1 on a large set. Lemma 3.5. Given compact sets A, B R n and S := ta + 1 tb, and recalling the notation Es R n 1 {s} in 2.1, we define the probability densities on the real line ρ A s := Hn 1 As, ρ B s := Hn 1 Bs, ρ S s := Hn 1 Ss. 3.3 A B S Let T : R R be the monotone rearrangement sending ρ A onto ρ B, that is T is an increasing map such that T ρ A = ρ B. 1 Then S t A 1/n + 1 t B 1/n n e n 1 s t + 1 tt s ds, 3.4 R where T t s := ts + 1 tt s and e n 1 s := H n 1 ST t s [ th n 1 As 1/n th n 1 BT s ] 1/n 1 n 1. Moreover, if t [τ, 1 τ] and 1.3 holds with δ/τ n is sufficiently small, then ρ A s ρ B T s 1 ρ As ds Cn τ n/2 δ1/ R Finally, we have a lemma saying that if a function ψ is nearly concave on a large set, and most of its level sets are close to their convex hulls, then it is L 1 -close to a concave function. Here and in the sequel, given a set E we will use coe to denote its convex hull. 1 T denotes the push-forward through the map T, that is, T ρ A = ρ B ρ Bs ds = ρ As ds E T 1 E E R Borel. An explicit formula for T can be given using the distribution functions of ρ A and ρ B: if we define G As := s ρ As ds, G Bs := and we set G 1 B r := inf{s R : GAs > t}, then T = G 1 B GA. 8 s ρ Bs ds,

9 Lemma 3.6. Let 0 < τ 1/2 and fix t such that 1/2 t 1 τ/2. Let t = 1 t, and for all y 1 and y 2 in R n 1 define y 12 := t y 1 + t y 2 ; y 12 := t y 1 + t y 2. Let σ, ς > 0, ˆM 1, F R n 1, and let ψ : F R be a function satisfying ψy 1 + ψy 2 ψy 12 + ψy 12 + σ y 1, y 2, y 12, y 12 F, 3.6 Also, we assume that there exists a set H R such that H n 1 co{ψ > s} \ {ψ > s} ds + H Ω := cof, H n 1 Ω \ F ς, 3.7 B r Ω B n 1r, 1/n < r < n, 3.8 ˆM ψy ˆM y F. 3.9 R\H H n 1 {ψ > s} ds ς Then there exist a concave function Ψ : Ω [ 2 ˆM, 2 ˆM] and a dimensional constant L n such that Ψy ψy dy τ L n ˆM σ + ς β n,τ, 3.11 where 4 Proof of Theorem 1.3 F β n,τ := τ 16n 1 log τ As explained in [FJ], by inner approximation 2 it suffices to prove the result when A, B are compact sets. Hence, let A and B be compact sets, define S := ta + 1 tb for some t [τ, 1 τ], and assume that 1.3 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 τ Nn δ εnτ. In order to simplify the notation, C will denote a generic constant, which may change from line to line, and that is bounded from above by τ Nn for some dimensional constant N n > 1 recall that by assumption τ 1/2. We will say that such a constant is universal. Observe that, since the statement and the conclusions are invariant under measure preserving affine transformations, by Lemma 3.2 we can assume that A and B are M, τ-normalized see The approximation of A and analogously for B is by a sequence of compact sets A k A such that A k A and coa k coa. One way to construct such sets is to define A k := A k V k, where A k A are compact sets satisfying A k A, and V k V k+1 A are finite sets satisfying cov k coa. 9

10 4.1 The case A = A and B = B Let A, B R n be compact sets satisfying A = A, B = B. Since π As π A0 = πa and π Bs π B0 = πb are disks centered at the origin, applying Lemma 3.3 we deduce that Hence, if we define H n 1 πa πb C δ 1/ S := y πa πb ta y + 1 tb y, then S y S y for all y R n 1. In addition, using 1.3, 3.1, and 4.1, we have 1 + δ S = H 1 S y dy H 1 S y dy H 1 S y dy R n 1 πa πb πa πb = S t H 1 A y dy + 1 t H 1 B y dy πa πb t A + 1 t B 2 which implies since S S πa πb C H n 1 πa πb 1 C δ 1/2, S \ S C δ 1/ Also, by Chebyshev s inequality we deduce that there exists a set F πa πb such that H n 1 πa πb \ F C δ 1/4, H 1 S y \ S y δ 1/4 y F. This implies that, if we write with a and b radial decreasing, then A y := {y} [ ay, ay] and B y := {y} [ by, by], tay + 1 tby [ta + 1 tb]y + δ 1/4 y = ty + 1 ty, y, y, y F. We show next that a three-point inequality for two functions f and g implies a four-point inequality for each of f and g separately. Remark 4.1. Let F R n 1, and f, g : F R be two bounded Borel functions satisfying tfy + 1 tgy [tf + 1 tg]y + σ y = ty + 1 ty, y, y, y F, for some σ 0. Let t [τ, 1 τ] and define t := 1 2 t, t := 1 t ; y 12 := t y t y 2, y 12 := t y t y

11 We claim that fy 1 + fy 2 fy 12 + fy σ. 4.4 t The analogous statement for g involves replacing t with 1 t, so gives different values of t and t. Notice that, if τ t 1/2, then 1/2 t 2/3 4.5 independent of τ, whereas if 1/2 t 1 τ, then To prove 4.4, note that the definitions above imply 2/3 t 1 τ/ y 12 = ty ty 12, y 12 = ty ty 2. Hence, assuming that y 1, y 2, y 12, y 12 F, we can add together the two inequalities to get 4.4. tfy tgy 12 [tf + 1 tg]y 12 + σ, tfy tgy 12 [tf + 1 tg]y 12 + σ, By the remark above and Lemma 3.6 notice that the level sets of a and b are both disks, so 3.10 holds with ς = 0, we obtain that both functions a and b are L 1 -close to concave functions Ψ A and Ψ B, both defined on πa πb. Hence, if we define the convex sets K A := { y, s R n : y πa πb, Ψ A y s Ψ A y }, K B := { y, s R n : y πa πb, Ψ B y s Ψ B y }, we deduce that A K A + B K B C δ βn,τ /4. Hence, it follows from Proposition 3.4 that, up to a translation, there exists a convex set K such that A B K and A B C δ βn,τ /8n, K \ A + K \ B C δ βn,τ /8n Notice that, because A = A and B = B, it is easy to check that the above properties still hold with K in place of K. Hence, in this case, without loss of generality one can assume that K = K. 4.2 The general case Since the result is true when n = 1 by Theorem 1.1, we assume that we already proved Theorem 1.3 through n 1, and we want to show its validity for n. 11

12 Step 1: There exist a dimensional constant ζ > 0 and λ δ ζ such that the inductive hypothesis applies to A λ and B λ. Let A and B be as in Definition 2.1. Thanks to Lemma 3.1, A and B still satisfy 1.3, so we can apply the result proved in Section 4.1 above to get see 4.7 H 1 A y H 1 B y dy H 1 A y B y dy = A B C δᾱ 4.8 R n 1 R n 1 and for some convex set K = K, where K A B, K \ A + K \ B C δᾱ/n2 4.9 ᾱ := β n,τ 8n In addition, because A and B are M, τ-normalized see 3.1, so are A and B, and by 4.9 we deduce that there exists a universal constant R > 0 such that K B R Also, by 4.8 and Chebyshev s inequality we obtain that, up to a set of measure C δᾱ/2, H 1 A y H 1 By δᾱ/2. Thus, recalling Lemma 3.1, for almost every λ > 0 H n 1 Aλ = H n 1 A λ H n 1 B λ δᾱ/2 + C δᾱ/2 = H n 1 Bλ δᾱ/2 + C δᾱ/2. Since, by 3.1, τ 2n M 0 H n 1 Bλ H n 1 Bλ + δᾱ/2 δ ᾱ/2 dλ = H n 1 Bλ dλ C δᾱ/2, 0 by Chebyshev s inequality we deduce that H n 1 Aλ H n 1 Bλ + C δᾱ/4 for all λ outside a set of measure δᾱ/4. Exchanging the roles of A and B we obtain that there exists a set G [0, τ 2n M] such that H 1 G C δᾱ/4, H n 1 Aλ H n 1 Bλ C δᾱ/4 λ [0, ] \ G Using the elementary inequality ta + 1 tb n 1 ta n tb n 1 C a b 2 0 a, b M τ 2n, and replacing a and b with a 1/n 1 and b 1/n 1, respectively, we get ta 1/n tb 1/n 1 n 1 ta + 1 tb C a b 2/n a, b M τ 2n 4.13

13 notice that a 1/n 1 b 1/n 1 a b 1/n 1. Finally, it is easy to check that taλ + 1 tbλ Sλ λ > 0. Hence, by the Brunn-Minkowski inequality 1.2 applied to Aλ and Bλ, using 1.3, 3.1, 4.13, and 4.12, we get 1 + δ S = τ 2n M 0 τ 2n M 0 τ 2n M 0 C H n 1 Sλ dλ th n 1 Aλ 1/n th n 1 Bλ 1/n 1 n 1 dλ th n 1 Aλ + 1 th n 1 Bλ dλ τ 2n M 0 = t A + 1 t B C δᾱ/[2n 1] 1 C δᾱ/[2n 1]. H n 1 Aλ H n 1 Bλ 2/n 1 dλ 4.14 We also observe that, since K = K, by Lemma 3.1, 4.11, and [C2, Lemma 4.3], for almost every λ > 0 we have A \ π 1 Aλ = A \ π 1 A λ K \ π 1 Kλ + M H n 1 A λ Kλ 4.15 and analogously for B. Also, by 4.9, τ 2n M 0 We set Cλ 2 + M H n 1 A λ Kλ, H n 1 A λ Kλ + H n 1 B λ Kλ dλ K \ A + K \ B C δᾱ/n { } and we notice that η min ᾱ 2n 1, ᾱ 4. Take ζ > 0 to be fixed later. inequality, we can find a level such that H n 1 S λ η := ᾱ n 2, 4.17 Then by 4.12, 4.14, 4.15, 4.16, and by Chebyshev s [ 10 δ ζ λ, τ ] 20 δζ τ 4.18 th n 1 A λ 1/n th n 1 B λ 1/n 1 n 1 + C δ η ζ,

14 In addition, from the properties A \ π 1 A λ + B \ π 1 B λ C δ 2ζ + δ η ζ, 4.20 H n 1 A λ H n 1 B λ C δ η H n 1 Aλ τ 2n M for any λ > 0 see 3.1, τ 2n M 0 H n 1 Aλ dλ = A 1 δ, λ H n 1 Aλ is a decreasing function, we deduce that τ 2n 2M Hn 1 Aλ M τ 2n λ 0, τ 2n 2M 1. The same holds for B and S, hence H n 1 S λ, H n 1 A λ, H n 1 B λ [ τ 2n 2M 1, τ 2n M ] provided δ τ N for some large dimensional constant N. Set ρ := 1/H n 1 A λ 1/n 1 [1/C, C], and define A := ρa λ, B := ρb λ, S := ρs λ. By 4.19 and 4.21 we get H n 1 A = 1, H n 1 B 1 C δ η, H n 1 S 1 + C δ η ζ. while, by 1.2, H n 1 S 1/n 1 th n 1 A 1/n th n 1 B 1/n 1 1 C δ η, therefore H n 1 A 1 + H n 1 B 1 + H n 1 S 1 C δ η ζ. Thus, by Theorem 1.3 applied with n 1, up to a translation there exists a n 1-dimensional convex set Ω such that Define ζ by Ω A B, H n 1 Ω \ A + H n 1 Ω \ B C δ η ζε n 1τ. ζ := ε n 1τ η, and set Ω := Ω /ρ. Then we obtain recall that 1/ρ C and that ε n 1 τ 1 Ω A λ B λ, H n 1 Ω \ A λ + H n 1 Ω \ B λ C δ 2ζ

15 Step 2: Theorem 1.1 applies to most of the sets A y and B y for y A λ B λ. Define C := A λ B λ S λ. By 4.20, 4.23, 3.1, and 4.22, 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 y dy + B λ\a λ H1 B y dy C δ 2ζ + δ η ζ + C H n 1 Ω \ A λ + H n 1 Ω \ B λ C δ 2ζ + δ η ζ C δ 2ζ 4.24 Hence, by 1.3 and 4.24, H 1 S y \ [ ta y + 1 tb y dy = H 1 S y H 1 ta y + 1 tb y ] dy C C [ H 1 S y th 1 A y 1 th 1 B y ] dy C = S π 1 C t A π 1 C 1 t B π 1 C S t A 1 t B + t A \ π 1 C + 1 t B \ π 1 C C δ 2ζ Write C as C 1 C 2, where C 1 := { y C : H 1 S y th 1 A y 1 th 1 B y δ ζ}, C 2 := C \ C 1. By Chebyshev s inequality while, recalling 4.18, H n 1 C 2 C δ ζ, 4.26 min { H 1 A y, H 1 B y } λ > δ ζ /2 y C 1. Hence, by Theorem 1.1 applied to A y, B y R for y C 1, we deduce that H 1 coa y \ A y + H 1 cob y \ B y C δ ζ 4.27 recall that coe denotes the convex hull of a set E. Let Ĉ1 C 1 denote the set of y C 1 such that H 1 S y \ ta y + 1 tb y δ ζ, 4.28 and notice that, by 4.25 and Chebyshev s inequality, H n 1 C 1 \ Ĉ1 C δ ζ. compact set C 1 Ĉ1 such that H n 1 Ĉ1 \ C 1 δ ζ to obtain Then choose a H n 1 C 1 \ C 1 C δ ζ In particular, it follows from 4.23 that H n 1 Ω \ C 1 C δ ζ

16 Step 3: There is S S so that S \ S is small and S is bounded. Define the compact sets Ā := y C 1 A y, B := Note that by 3.1, 4.24, 4.26, 4.29, y C 1 B y, S := y C 1 ta y + 1 tb y A Ā + B B = A \ π 1 C 1 + B \ π 1 C 1 Cδ ζ, 4.32 therefore S = H 1 [ ta y + 1 tb y dy th 1 A y + 1 th 1 B y ] dy = t Ā + 1 t B 1 Cδ ζ. C 1 C 1 Hence, by 1.3 and S S, S S Cδ ζ 4.33 Next we show that S is bounded. First recall that S y = ta y + 1 tb y, 4.34 y=ty +1 ty and by 4.28 we get H 1 ta y + 1 tb y \ ta y + 1 tb y δ ζ y=ty +1 ty y C Recalling that π : R n R is the orthogonal projection onto the last component that is, πy, s = s, we define the characteristic functions χ A y s := { 1 if s πtay 0 otherwise, { χ A, 1 if s πt coay y s := 0 otherwise, and analogously for B y with 1 t in place of t. Hence, by 4.27 we have the following estimate on the convolution of the functions χ y and χ y: χ A, y χ B, y χ A y χb y L R χ B, y χ B y L 1 R + χ A, y χ A y L 1 R H 1 cob y \ B y + H 1 coa y \ A y < 3 δ ζ y, y C Let us denote by [a, b] the interval π t coa y + 1 t cob y, and notice that, since by construction min { th 1 A y, 1 th 1 B y } min{τ, 1 τ} λ 10 δ ζ y C 1 see 4.18, this interval has length greater than 20 δ ζ. Also, it is easy to check that the function χ y χ y is supported on [a, b], has slope equal to 1 resp. 1 inside [a, a + 3 δ ζ ] resp. [b 3 δ ζ, b], 16

17 and it is greater than 3 δ ζ inside [a + 3 δ ζ, b 3 δ ζ ]. Hence, since π ta y + 1 tb y contains the set {χ y χ y > 0}, by 4.36 we deduce that which implies in particular that π ta y + 1 tb y [a + 3 δ ζ, b 3 δ ζ ], 4.37 H 1 t coa y + 1 t cob y H 1 ta y + 1 tb y + 6 δ ζ y, y C We claim that if y, y, y = ty + 1 ty C 1, then π t coa y + 1 t cob y [α y 16 δ ζ, β y + 16 δ ζ ], 4.39 where [α y, β y ] := π t coa y + 1 t cob y. Indeed, if this was false, since π t coa y + 1 t cob y = [a, b] is an interval of length at least 20δ ζ 16δ ζ, it follows that This implies that so, by 4.37, H 1 [a, b] \ [α y, β y ] 16δ ζ. H 1 [a + 3δ ζ, b 3δ ζ ] \ [α y, b y ] 10δ ζ, H 1 π ta y + 1 tb y \ [αy, β y ] 10δ ζ. However, since [α y, β y ] π ta y + 1 tb y, this contradicts 4.35 and proves the claim Now, if we write coa y = {y} [a A y, b A y], cob y = {y} [a B y, b B y], co S y = {y} [a Sy, b Sy], and we denote by c A y := aa y+b A y 2 the barycenter of coa y and analogously for B and S, then b S = tb A + 1 tb B and it follows from 4.39 that tb A y + 1 tb B y b Sy + 16 δ ζ y, y, y C 1, y = ty + 1 ty 4.40 and analogously for a. Hence, from the fact that H 1 coa y and H 1 cob y are universally bounded see 3.1 and 4.27 one easily deduces that tc A y + 1 tc B y c Sy C y, y, y C 1, y = ty + 1 ty. Hence, by Remark 4.1 in Section 4.1 we get that c A y 1 + c A y 2 c A t Ay t Ay 2 c A t Ay t Ay 2 C whenever y 1, y 2, t A y t A y 2, t A y t A y 2 C 1, with t A := 1 2 t and t A := 1 t A. As proved in Lemma 5.1, this estimate in one dimension implies that, along any segment [y 1, y 2 ] 17

18 on which C 1 has large measure and such that y 1, y 2, t A y t A y 2, t A y t A y 2 C 1, c A is at bounded distance from a linear function l A. Analogously, c B y 1 + c B y 2 c B t By t By 2 c B t By t By 2 C whenever y 1, y 2, t B y t B y 2, t B y t B y 2 C 1, now with t B := 1 1+t and t B := 1 t B recall again Remark 4.1, so c B is at bounded distance from a linear function l B along any segment [y 1, y 2 ] on which C 1 has large measure and such that y 1, y 2, t B y t B y 2, t B y t B y 2 C 1. Hence, along any segment [y 1, y 2 ] on which C 1 has large measure and such that y 1, y 2, t A y t A y 2, t A y t A y 2, t B y t B y 2, t B y t B y 2 C 1, c S is at bounded distance from the linear function ly := tl A y + 1 tl B y. We now use this information to deduce that, up to an affine transformation of the form R n 1 R y, s T y, t Ly + y 0, t with T : R n 1 R n 1, dett = 1, and y 0, t 0 R n, the set S is universally bounded, say S B R for some universal constant R. Indeed, first of all, since C 1 is almost of full measure inside the convex set Ω see 4.23, 4.26, and 4.29, by a simple Fubini argument see the analogous argument in [FJ, Proof of Theorem 1.2, Step 3-b] for more details we can choose n good points y 1,..., y n C 1 such that: a All points y 1,..., y n and t Ay t Ay 2, t Ay t Ay 2, t By t By 2, t By t By 2 belong to C 1. b Let Σ i, i = 1,..., n, denote the i 1-dimensional simplex generated by y 1,..., y i, and define Σ i := [ t AΣ i +1 t Ay i+1 ] [ t AΣ i +1 t Ay i+1 ] [ t B Σ i +1 t By i+1 ] [ t BΣ i +1 t By i+1 ], i = 1,..., n 1. Then i ii H i 1 Σ i c n, H i 1 Σ i C 1 H i 1 Σ i C 1 H i 1 Σ i 1 δ ζ/2, i = 2,..., n; H i 1 Σ i 1 δ ζ/2 i = 2,..., n 1. Then, thanks to John s Lemma [J], up to an affine transformation of the form 4.41 we can assume that B r Ω B n 1r, 1/C n < r < C n with C n dimensional, 4.42 and y k, 0 S, k = 1,..., n

19 We then prove that S is universally bounded as follows: first of all, thanks to 4.42 we only need to show that S is bounded in the last variable. Then, by 4.43 and a and b-i above, thanks to Lemma 5.1 we deduce that c S is universally bounded on Σ 2 C 1. One then iterates this construction: since c S is universally bounded on Σ 2 C 1 and at y 3, for any point z Σ 2 C 1 such that t A z+1 t A y 3, t A z+1 t A y 3, t B z+1 t B y 3, t B z+1 t B y 2 Σ 2 C 1 these are most of the points we can apply again Lemma 5.1 to deduce that c S is universally bounded on the set [z, y 3 ] C 1. An iteration of this argument as in [FJ, Proof of Theorem 1.2, Step 3-d] shows that c S is universal bounded on a set Σ n such that H n 1 Σ n \ Σ n C δ ζ/2. We now conclude as in [FJ, Proof of Theorem 1.2, Step 3-e]: fix a point ȳ 1 C 1. Then we can find another point ȳ 2 C 1 such that t Aȳ1+1 t A ȳ 2, t Aȳ1+1 t A ȳ 2, t Bȳ1+1 t B ȳ 2, t Bȳ1+1 t B ȳ 2 C 1, most of the points on the segment [ȳ 1, ȳ 2 ] belong to C 1, and H 1 [ȳ 1, ȳ 2 ] Σ n c n for some dimensional constant c n > 0. Hence, on this segment c S must be at some bounded distance from a linear function l, but at the same time we know that c S is universally bounded on [ȳ 1, ȳ 2 ] Σ n, so l is universally bounded there. Since this set has non-trivial measure, this implies that l has to be universally bounded on the whole segment [ȳ 1, ȳ 2 ] since l is a linear function. Thus c S is universally bounded on [ȳ 1, ȳ 2 ] C 1 as well, and this provides a universal bound for c Sȳ 1, concluding the proof. Step 4: There are uniformly bounded vertically convex sets A and B near A and B. Let Ā, B, and S be as in 4.31, and recall that by the previous step there exists a constant R such that S { x n R}. Let us apply opposite translations along the e n -axis to tā and 1 t B see 4.31, i.e., tā tā + µe n, 1 t B 1 t B µe n, for some µ R, so that Ā {x n R} and Ā {x n = R} recall that Ā is compact. This means that min a A y = R. y C 1 Notice that, thanks to 4.39, ta A y + 1 ta B y a Sy 16 δ ζ y, y, y C 1, y = ty + 1 ty Let ȳ C 1 be such that a A ȳ = R, and set C 1 := C 1 C 1 tȳ 1 t. Then, since a A ȳ = R and a S R, it follows from 4.44 that a B y R C δ ζ R 1 y C 1. Define A := {y} [a A y, b A y], B := {y} [a B y, b B y]. y C 1 y C 1 19

20 We have shown that A B {x n R 1}. It remains to prove the upper bounds. Note that because ȳ Ω and Ω is convex, it follows from 4.30 that H n 1 Ω \ C 1 tȳ = 1 t 1 n H n 1 1 tω + tȳ \ 1 t C 1 1 t 1 n H n 1 Ω \ C 1 C δ ζ. Therefore, using 4.30 again, we have H n 1 Ω \ C 1 C δζ We now claim that A B {x n CR} for some universal constant C. Indeed, if for instance b A ỹ CR for some ỹ C 1, then we could use 4.40 and the fact that bb a B R 1 on C 1 to get b Sy t CR 1 tr δ ζ τ CR R 2 y C 1 tỹ + 1 tc 1, and since the latter set is nonempty because of 4.30, 4.45 and the convexity of Ω this contradicts the fact that b S R provided C is large enough the case b B ỹ CR for some ỹ C 1 is completely analogous. Thus, A and B are universally bounded. Finally, note that 3.1, 4.32, 4.45, and 4.27 imply A A + B B Cδ ζ Step 5: The inductive hypothesis applies to horizontal sections and hence there are convex sets close to A and B. The main goal of this section is to show that the hypotheses of Lemma 3.6 apply to the function b A and similarly to b B, a A, and a B. The fact that A and B are close to convex sets will then follow easily. As explained in the outline of the proof in Section 2, to be able to apply Lemma 3.6 we will construct auxiliary sets A and B which consist of the top profile of A and B with a flat bottom, for which the slices coincide with the superlevel sets of b A and b B, and we will apply Lemma 3.5 to such sets. However, to be able to do this, we must show that A and B are almost optimal in the Brunn-Minkowski inequality. As we showed in Step 4, A and B are universally bounded, so we may choose universal constants M A 0 and M B 0 such that M A a A y, M B a B y y C 1, and such that the sets A := {y} [ M A, b A y], B := {y} [ M B, b B y], y C 1 20 y C 1

21 are universally bounded. We may also adjust the constants M A and M B so that A = B. Define S := ta + 1 tb ; C y := {y, y C C : ty + 1 ty = y}. We estimate the measure of S using 4.40 as follows: S = H 1 t[ M A, b A y ] + 1 t[ M B, b B y ] dy tc 1 +1 tc 1 tc 1 +1 tc 1 C 1 y,y C y b Sy + 16 δ ζ + tm A + 1 tm B dy b Sy + tm A + 1 tm B dy + C δ ζ, where, in the final inequality, we used that H n 1 tc1 + 1 tc 1 \ C 1 C δ ζ recall that C1 Ω and Ω is convex, thus tc tc 1 Ω and the bound follows from 4.45 and that b S is universally bounded on C1. Next, since b S = tb A + 1 tb B and A = B, it follows that S t b A y M A dy + 1 t b B y M B dy + C δ ζ C 1 On the other hand 1.2 implies Hence, in all, we find that = t A + 1 t B + C δ ζ = A + C δ ζ, S t A 1/n + 1 t B 1/n n = A. C 1 0 S A C δ ζ and A = B 4.47 We are now in a position to apply Lemma 3.5 to A and B to confirm that hypothesis 3.10 of Lemma 3.6 is valid for b A and b B. Let us recall the notation Es R n 1 {s} in 2.1. Since ca = cb = 1 for some universal constant c > 0, by applying 4.47 and Lemma 3.5 to the sets ca, cb, and cs, we find a monotone map T : R R such that T ρ A = ρ B, ρ A s := Hn 1 A s A, ρ B s := Hn 1 B s B, T s = Hn 1 A s B H n 1 B T s A ρ A -a.e., 4.48 e n 1 s [t + 1 tt s] ds C δ ζ, 4.49 R 21

22 and where T t s = ts + 1 tt s and R ρ A s ρ B T s 1 ρ A s ds C δζ/2, 4.50 e n 1 s := H n 1 S T t s [ th n 1 A s 1/n th n 1 B T s ] 1/n 1 n 1. Let us define the set G := {s R : e n 1 s δ ζ/2 }, 4.51 and observe that, thanks to 4.49, 4.51, and T 0, H 1 R \ G 1 [t + 1 tt s] ds C δ ζ/ τ R\G Next, note that the formula for T with A and B replaced by A and B given in the footnote in the statement of Lemma 3.5 implies that the distributional derivative of T has no singular part on T 1 {ρ B > 0}. Hence, the area formula gives H 1 R \ T G {ρ B > 0} = T s ds, R\G T 1 {ρ B >0} and it follows that H 1 R \ T G {ρ B > 0} T s ds 1 R\G τ R\G [t + 1 tt s] ds C δ ζ/ Also, we define and I A := { s R : H n 1 A s > δ ζ/4}, I B := Notice that, thanks to 4.50, I T := { s R : 2 3 ρ A s ρ B T s 3 }. 2 I T ρ A s ds 1 C δ ζ/2. { s R :, H n 1 B s > δ ζ/4}, Also, since ρ A and ρ B are probability densities supported inside some bounded interval being A and B universally bounded, we have ρ A s ds = ρ A s ds 1 C δ ζ/4 I A {ρ A >δ ζ/4 A } and using the condition T ρ A = ρ B ρ A s ds = ρ B s ds = T 1 I B I B 22 {ρ B >δ ζ/4 B } ρ B s ds 1 C δ ζ/4.

23 Therefore I ρ A s ds = T I ρ B s ds 1 C δ ζ/4, I := I A T 1 I B I T We now apply the inductive hypothesis to A s, B T s, S T t s: define Then, since 2 3 A s := S s := t s := A s H n 1 A s 1/n 1, Bs := B T s H n 1 B T s 1/n 1, S T t s th n 1 A s 1/n th n 1 B T s 1/n 1, th n 1 A s 1/n 1 th n 1 A s 1/n th n 1 B T s 1/n 1. ρ A s ρ B T s 3 2 for s I, and A = B, it follows that t s [ τ 2, 1 τ ] 2 s I. In addition, recalling the definition of G, for any s I G we also have Hence, if S s = t s A s + 1 t s B s, H n 1 A s = H n 1 B s = 1, H n 1 S s 1 + δ ζ/4. δ ζ/4 e M n 1τ/2, 4.55 then by the inductive hypothesis we deduce the existence of a convex set K s such that, up to a translation which may depend on s Thus, in particular, which implies that K s A s B s, H n 1 K s \ A s + H n 1 K s \ B s C δ ζ 4 ε n 1τ/2. H n 1 coa s \ A s + H n 1 cob s \ B s δ ζ 4 ε n 1τ/2, H n 1 co A s \ A s + H n 1 co B T s \ B T s δ ζ 4 ε n 1τ/2 s I G. Hence, integrating with respect to s I G and using that T C on I G as a consequence of 4.48 and the fact that I I T we obtain H n 1 co A s \ A s ds C δ ζ 4 εn 1τ/2, 4.56 I G 23

24 H n 1 co B s \ B s ds = T I G I G H n 1 co B T s \ B T s T s ds C δ ζ 4 ε n 1τ/2. Also, recalling the definition of ρ A and ρ B, it follows from 4.54, 4.52, 4.53, that H n 1 A s ds + H n 1 B s ds C δ ζ/4 R\I G R\T I G 4.57 notice that B s = on {ρ B = 0}. By the bound above, 4.40, 4.44, 4.54, 4.56, 4.57, and Remark 4.1 see Section 4.1, we can apply Lemma 3.6 to b A find a concave function Ψ + y defined on Ω such that b A y Ψ + y dy C δ ζ βn,τ ε 4 n 1 τ/2. C 1 Similarly, there is a convex function Ψ on Ω such that a A y Ψ y dy C δ ζ βn,τ ε 4 n 1 τ/2, so the convex set C 1 K A := { y, s : y Ω, Ψ y s Ψ + y } satisfies A K A C δ ζ βn,τ ε 4 n 1 τ/2. The same argument also applies to B so that, in all, we have A K A + B K B C δ ζ βn,τ ε 4 n 1 τ/ Step 6: Conclusion. By 4.58 and 4.46, we can apply Proposition 3.4 to deduce that, up to a translation, there exists a convex set K convex such that A B K and concluding the proof. Step 7: An explicit bound for ε n τ and M n τ. K \ A + K \ B C δ ζ βn,τ 8 n 3 ε n 1τ/2, 4.59 By 4.59 and 4.55 it follows that the recurrence for ε n τ and M n τ is given, respectively, by Recall that see 4.22 and Lemma 3.6 ε n τ = ζ β n,τ 8 n 2 ε n 1τ/2, M n τ = 4 ζ M n 1τ/2. ζ = ε n 1τ τ η, β n,τ = 3 16n 1 logτ. 24

25 For n = 1, Theorem 1.1 implies that if δ < τ/2, then K \ A + K \ B 8δ/τ. In other words, ε 1 τ = 1 and M 1 τ = logτ/3 are admissible choices. For n 2 we have see 4.10 and 4.17 thus which gives ε n τ = η = ᾱ n 2 = β n,τ 2 4 n 2 = τ 2 7 n 1n 2 log τ, ζ = τ 2 7 3n 1n 2 log τ ε n 1τ, τ n 1 2 n 4 log τ 2 ε n 1τε n 1 τ/2. In particular, we obtain and explicit lower bound for all n 2 which can be easily checked to hold by induction: τ 3n ε n τ. 2 3n+1 n 3n log τ 3n Concerning M n τ we have M n τ = 4 ζ M n 1τ/2 = 29 3n 1n 2 log τ τ ε n 1 τ M n 1 τ/2 from which we get M n τ 23n+2 n 3n log τ 3n τ 3n. 5 Proof of the technical results As in the previous section, we use C to denote a generic constant, which may change from line to line, and that is bounded from above by τ Nn for some dimensional constant N n > 1. Again, we will say that such a constant is universal. 5.1 Proof of Lemma 3.5: an inductive proof of the Brunn-Minkowski inequality In this section we show how to prove the Brunn-Minkowski inequality 1.2 by induction on dimension. 3 As a byproduct of our proof we obtain the bounds 3.4 and 3.5. Given compact sets A, B R n and S := ta+1 tb, we define the probability densities on the real line ρ A, ρ B, and ρ S as in 3.3, and we let T : R R be the monotone rearrangement sending 3 The one-dimensional case is elementary, and can be proved for instance as follows: given A, B R compact, after translation we can assume that A, 0], B [0, +, A B = {0}. Then A + B A B, hence A + B A B = A + B, as desired. 25

26 ρ A onto ρ B. Since monotone functions are differentiable almost everywhere, as a consequence of the Area Formula one has see for instance [AGS, Lemma 5.5.3] T s = ρ As ρ B T s ρ A -a.e. 5.1 Set T t s := ts + 1 tt s and observe that ST t s tas + 1 tbt s, so by the Brunn-Minkowski inequality in R n 1 we get H n 1 ST t s 1/n 1 th n 1 As 1/n th n 1 BT s 1/n We now write S = S = R R ρ S s ds S ρ S T t s T ts ds R H n 1 ST t s [t + 1 tt s] ds, Here, when we applied the change of variable s T t s, we used the fact that since T t is increasing, its pointwise derivative is bounded from above by its distributional derivative. Define µ 1 s =... = µ n 1 s := 1 t B 1/n 1 ρ B T s 1/n 1 t A 1/n 1 ρ A s 1/n 1, µ n s := 1 t t ρ A s ρ B T s 5.3 Using 5.2 and 5.1, we obtain S th n 1 As 1/n th n 1 BT s 1/n 1 n 1 [t + 1 tt s] ds R = t A 1/n 1 ρ A s 1/n t B 1/n 1 ρ B T s 1/n 1 n 1 ρ A s t + 1 t ds R ρ B T s n = A t n 1 + µ i s ρ A s ds. R i=1 We now use the following inequality, see [FMP2, Equation 22] and [FMP1, Lemma 2.5]: there exists a dimensional constant cn > 0 such that, for any choice of nonnegative numbers {µ i } i=1,...,n, n 1 + µ i i=1 Hence, we get n 1 + µ i s i=1 n t t i=1 1/n n 1 µ i + cn max i µ i B 1/n n A 1/n + cn 1 max i µ i s n n 1/n 2 µ j µ i. j=1 n j=1 i=1 µ j s 1 t t B 1/n 2 A 1/n, 26

27 which gives recall that ρ A = 1 S A t n t R t + cn A t n R B 1/n n A 1/n ρ A s ds n 1 max i µ i s t A 1/n + 1 t B 1/n n, j=1 µ j s 1 t t B 1/n 2 A 1/n ρ A sds which proves the validity of Brunn-Minkowski in dimension n. As a byproduct of this proof we will deduce 3.4 and 3.5. Indeed 3.4 is immediate from our proof. Moreover, we have S t A 1/n +1 t B 1/n n cn A t n R R 1 max i µ i s n j=1 µ j s 1 t t With the further assumption 1.3, 5.4 gives n 1 t 2 max i µ i s 1 t µ js B 1/n A 1/n ρ A s ds n R j=1 1 max i µ i s n j=1 Cn t n 2 1 t 2 δ Cn τ n δ, which, combined with the Schwarz inequality, leads to n t 1 t µ js B 1/n ρa A 1/n s ds R j=1 Cn δ1/2 τ n/2 R Cn δ1/2 τ n/2 Cn δ1/2 τ n/2 t 2 1 t µ js B 1/n A 1/n ρ A s ds max µ i s ρ A s ds i R t max i 1 t µ is B 1/n ρa A 1/n s ds + n R j=1 Hence, provided δ/τ n is sufficiently small we get n t 1 t µ js B 1/n R j=1 t 1 t µ js B 1/n ρa A 1/n s ds + 2 A 1/n ds Cn τ n/2 δ1/2. B 1/n 2 A 1/n ρ A s ds B 1/n A 1/n Recalling the definition of µ i see 5.3 and using that 1 4δ B / A 1 + 4δ, we deduce that 3.5 holds. 27

28 5.2 Proof of Lemma 3.6 We first remark that it suffices to prove the result in the case ˆM = 1, since the general case follows by applying the result to the function f/ ˆM. The proof of this result is rather involved and is divided into several steps. Step a: Making ψ uniformly concave at points that are well separated Let β 0, 1/3] to be fixed later, and define ϕ : Ω R as ϕy := { ψy σ + ς β y 2 y F, 0 y Ω \ F. 5.5 Notice that, y y 12 2 y 1 2 y 2 2 = 2t 1 t y 1 y 2 2 τ 2 y 1 y 2 2. Because of this, 3.9 and 3.6, we have 0 ϕ 3 and ϕy 1 + ϕy 2 ϕy 12 + ϕy 12 + σ 10τ σ + ς β y 1 y 2 2 y 1, y 2, y 12, y 12 F, which implies in particular that Also, since β 1/3, ϕy 1 + ϕy 2 ϕy 12 + ϕy 12 + σ y 1, y 2, y 12, y 12 F. 5.6 ϕy 1 + ϕy 2 < ϕy 12 + ϕy 12 τσ + ς β y 1 y 2 2 y 1, y 2, y 12, y 12 σ + ςβ F, y 1 y 2, τ that is ϕ is uniformly concave on points of F that are at least σ + ς β / τ-apart. Step b: Constructing a concave function that should be close to ϕ Let us take γ 0, β] to be fixed later, and define where h [0, 3] is given by Since 0 ϕ 3, we get [ϕy ϕy] dy = Ω ϕy := min{ϕy, h}, 5.7 h := inf { t > 0 : H n 1 {ϕ > t} σ + ς γ} M h H n 1 {ϕ > s} ds 3Mσ + ς γ. 5.9 Notice that whenever max{ϕy 12, ϕy 12 } h, ϕ satisfies 5.6 and 5.7. We define Φ : Ω [0, h] to be the concave envelope of ϕ, that is, the infimum among all linear functions that are above ϕ in Ω. Our goal is to show that Φ is L 1 -close to ϕ and hence to ϕ. 28

29 Step c: The set {Φ = ϕ} is Kσ + ς β dense in Ω \ co{ ϕ > h Kσ + ς β }. Let β 0, 1/3] be as in Step b. We claim that there exists a universal constant K > 0 such that the following holds, provided β is sufficiently small chosen later depending on τ and dimension: For any y Ω, either there is x {Φ = ϕ} Ω with y x Kσ + ς β ; or y belongs to the convex hull of the set { ϕ > h Kσ + ς β }. To prove this, we define Ω β := { y Ω : disty, Ω σ + ς β}. Of course, with a suitable value of K, it suffices to consider the case when y Ω β. So, let us fix y Ω β. Since Ω is a convex set comparable to a ball of unit size see 3.8 and Φ is a nonnegative concave function bounded by 3 inside Ω, there exists a dimensional constant C such that, for every linear function L Φ satisfying Ly = Φy, we have L C σ + ς β By [FJ, Step 4-c], there are m n points y 1,..., y m F such that y S := co{y 1,..., y m }, and all y j s are contact points: Φy j = Ly j = ϕy j, j = 1,..., m. If the diameter of S is less than Kσ + ς β, then its vertices are contact points within Kσ + ς β of y and we are done. Hence, let us assume that the diameter of S is at least Kσ + ς β. We claim that Observe that, if we can prove 5.11, then and we are done again. ϕy i > h Kσ β i = 1,..., m y S co{ ϕ > h Kσ + ς β }, It remains only to prove To begin the proof, given i {1,..., m}, take j {1,..., m} such that y i y j Kσ +ς β /2 such a j always exists because of the assumption on the diameter of S. We rename i = 1 and j = 2. Fix N N to be chosen later. For x, y Ω, define H N x, y := N k=0 j=0 k 1 t j t k j F [y] 1 t j t k j 1 x, where F [y] := F F t y 1 t F t y 1 t. 29

30 Observe that, since Ω is convex and by 3.7, H n 1 Ω \ F [y] CH n 1 Ω \ F C ς, H Ω n 1 1 \ t j t k j F [y] 1 t j t k j 1 x 1 = t jn 1 t k jn 1 Hn 1 t j t k j Ω + 1 t j t k j x \ F [y] 1 C t jn 1 t k jn 1 Hn 1 Ω \ F [y] t jn 1 t ς, k jn 1 and analogously for t, so H n 1 Ω \ H N x, y C C N k k=0 j=0 1 t 1 t jn 1 t k jn 1 ς Nn 1 ς, 5.12 where we used that t t. Choose N such that τ Nn 1 = C ς for some large dimensional constant C. In this way, from 3.8 and 5.12 we get H n 1 H N y 2, y 1 c n /2 > 0, and hence H N y 2, y 1 is nonempty. Now, choose w 0 H N y 2, y 1, and apply 5.6 iteratively in the following way: if we set w 1 := t w t y 2, w 1 := t w t y 2, then the fact that w 0 H N y 2, y 1 implies that w 1, w 1 F and also that t y t w 0, t y t w 0 F. Hence we can apply 5.6 to obtain Then define w 1 to be equal either to w 1 follows from the equation above that ϕw 1 + ϕw 1 ϕy 2 + ϕw 0 σ. or to w 1 so that ϕw 1 = max{ϕw 1, ϕw 1 }. Then, it ϕw 1 ϕy 2 + ϕw 0 2 σ 2. We now set w 2 := t w t y 2, w 2 := t w t y 2 F, and apply 5.6 again to get ϕw 2 + ϕw 2 ϕy 2 + ϕw 1 σ 1 1/4ϕy 2 + ϕw 0 / /2 σ. 30

31 Again we choose w 2 {w 2, w 2 } so that ϕw 2 = max{ϕw 2, ϕw 2 } and we keep iterating this construction, so that in N steps we get recall that 0 ϕ 3 Hence, In addition, since w 0 H N y 1, y 2, ϕw N 1 2 N ϕy N ϕw 0 2σ 1 2 N ϕy 2 2σ ϕy N 2σ ϕw N ϕy N 2σ y := t y 1 + t w N F, y := t y 1 + t w N F. Since the diameter of F is bounded see 3.7 and 3.8 it is easy to check that Therefore, by 5.10 we have w N y 2 C t N C1 τ/2 N. Ly + y Ly 1 + y 2 = Lw N y 2 C σ + ς β 1 τ/2 N. Hence, since y 1 and y 2 are contact points and L ϕ, using 5.14 and 5.13 we get ϕy 1 + ϕw N ϕy 1 + ϕy N 2σ = Ly 1 + y N 2σ Ly + y C 2 N + σ + σ + ς β 1 τ/2 N 5.15 ϕy + ϕy C σ + ς min{θ,κ β}, where recall 4.5 θ := 1 log2 n 1 logτ/2, κ := 1 n 1 Now assume by way of contradiction that We also have recalling that t = 1 t Hence, since Ly 1 = ϕy 1, ϕy 1 h Kσ + ς β. Lw N Lw N y 2 + Ly 2 = Lw N y 2 + ϕy 2 Cσ + ς β 1 τ/2 N + h h + Cσ + ς κ β Ly = t ϕy 1 + t Lw N t h K 2 σ + ςβ + t h + Cσ + ς κ β < h, 31 log1 τ/ logτ/2