1 Introduction. A SHARP STABILITY RESULT FOR THE RELATIVE ISOPERIMETRIC INEQUALITY INSIDE CONVEX CONES A. Figalli and E. Indrei. 1.

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1 A SHARP STABILITY RESULT FOR THE RELATIVE ISOPERIMETRIC INEQUALITY INSIDE CONVEX CONES A. Figalli and E. Indrei Abstract The relative isoperimetric inequality inside an open, convex cone C states that, at fixed volume, B r C minimizes the perimeter inside C. Starting from the observation that this result can be recovered as a corollary of the anisotropic isoperimetric inequality, we exploit a variant of Gromov s proof of the classical isoperimetric inequality to prove a sharp stability result for the relative isoperimetric inequality inside C. Our proof follows the line of reasoning in [16], though several new ideas are needed in order to deal with the lack of translation invariance in our problem. 1 Introduction 1.1 Overview Recently, there has been a lot of interest in quantitative estimates for isoperimetric [4, 16, 18, 13, 3], Sobolev [10, 9, 3, 19], and Brunn-Minkowski [17, 35] type inequalities. The aim of all of these results is to show that if a set/function almost attains the equality in one of these inequalities, then it is close (in a quantitative way) to a minimizer. These results have natural applications in the study of the stability for minimizers of perturbed variational problems, see for instance [3, 14, 15]. Our goal is to investigate stability for the relative isoperimetric inequality inside convex cones. This inequality has been used, for instance, to characterize isoperimetric regions inside convex polytopes for small volumes [, Corollary 3]. Hence, as in [14, 15], one may use our stability result to prove quantitative closeness to such isoperimetric regions in perturbed situations. Let n, and C R n be an open, convex cone. We denote the unit ball in R n centered at the origin by B 1 (similar notation is used for a generic ball) and the De Giorgi perimeter of E relative to C by { } P (E C) := sup div ψdx : ψ C0 (C; R n ), ψ 1. (1.1) E 1

2 The relative isoperimetric inequality for convex cones states that if E C is a Borel set with finite Lebesgue measure E, then n B 1 C 1 n E n 1 n P (E C). (1.) If E has a smooth boundary, the perimeter of E is simply the (n 1)- Hausdorff measure of the boundary of E inside the cone (i.e. P (E C) = H n 1 ( E C)). We also note that if one replaces C by R n, then the above inequality reduces to the classical isoperimetric inequality for which there are many different proofs and formulations (see e.g. [33], [36], [8], [3], [7], [4]). However, (1.) is ultimately due to Lions and Pacella [7] (see also [34] for a different proof using secord order variations). Their proof is based on the Brunn-Minkowski inequality which states that if A, B R n are measurable, then A + B n A n + B n. (1.3) As we will show below, (1.) can be seen as an immediate corollary of the anisotropic isoperimetric inequality (1.7). This fact suggested to us that there should also be a direct proof of (1.) using optimal transport theory (see Theorem.), as is the case for the anisotropic isoperimetric inequality [10, 16]. 1 The aim of this paper (in the spirit of [16]) is to exploit such a proof in order to establish a quantitative version of (1.). To make this precise, we need some more notation. We define the relative isoperimetric deficit of a Borel set E by µ(e) := P (E C) n B 1 C 1 n E n 1 n 1. (1.4) Note that (1.) implies µ(e) 0. The equality cases were considered in [7] for the special case when C \ {0} is smooth (see also [34]). We will work out the general case in Theorem. with a self-contained proof. However, the (nontrivial) equality case is not needed in proving the following theorem (which is in any case a much stronger statement): 1 After completition of this work, we discovered that Frank Morgan had already observed that Gromov s argument may be used to prove the relative isoperimetric inequality inside convex cones [1, Remark after Theorem 10.6], though he was thinking about using the Knothe map instead of the Brenier map. However, as observed in [16, Section 1.5], the Brenier map is much more powerful than the Knothe map when dealing with stability estimates.

3 Theorem 1.1. Let C R n be an open, convex cone containing no lines, K = B 1 C, and E C a set of finite perimeter with 0 < E <. Suppose s > 0 satisfies E = sk. Then there exists a constant C(n, K) > 0 such that E (sk) C(n, K) µ(e). E The assumption that C contains no lines is crucial. To see this, consider the extreme case when C = R n. Let ν S n 1 be any unit vector and set E = ν + B 1 so that E B 1 = B 1 > 0. However, µ(e) = 0 so that in this case Theorem 1.1 can only be true up to a translation, and this is precisely the main result in [16] and [4]. Similar reasoning can be applied to the case when C is a proper convex cone containing a line (e.g. a half space). Indeed, if C contains a line, then by convexity one can show that (up to a change of coordinates) it is of the form R C, with C R n 1 an open, convex cone. Therefore, by taking E to be a translated version of K along the first coordinate, the symmetric difference will be positive, whereas the relative deficit will remain 0. In general (up to a change of coordinates), every convex cone is of the form C = R k C, where C R n k is a convex cone containing no lines. Indeed, Theorem 1.1 follows from our main result: Theorem 1.. Let C = R k C, where k {0,..., n} and C R n k is an open, convex cone containing no lines. Set K = B 1 C, and let E C be a set of finite perimeter with 0 < E <. Suppose s > 0 satisfies E = sk. Then there exists a constant C(n, K) > 0 such that { } E (sk + x) inf : x = (x 1,..., x k, 0,..., 0) C(n, K) µ(e). E Let us remark that if k = n, then C = R n and the theorem reduces to the main result of [16], the only difference being that here we do not attempt to find any explicit upper bound on the constant C(n, K). However, since all of our arguments are constructive, it is possible to find explicit upper bounds on C(n, K) in terms on n and the geometry of C (see also Section 1.4). 1. The anisotropic isoperimetric inequality As we will show below, our result is strictly related to the quantitative version of the anisotropic isoperimetric inequality proved in [16]. To show this link, 3

4 we first introduce some more notation. convex set, and let Suppose K is an open, bounded, ν K := sup{ν z : z K}. (1.5) The anisotropic perimeter of a set E of finite perimeter (i.e. P (E R n ) < ) is defined as P K (E) := ν E (x) K dh n 1 (x), (1.6) FE where FE is the reduced boundary of E, and ν E : FE S n 1 is the measure theoretic outer unit normal (see Section ). Note that for λ > 0, P K (λe) = λ n 1 P K (E) and P K (E) = P K (E + x 0 ) for all x 0 R n. If E has a smooth boundary, FE = E so that for K = B 1 we have P K (E) = H n 1 ( E). In general, one can think of K as a weight function on unit vectors. Indeed, P K has been used to model surface tensions in the study of equilibrium configurations of solid crystals with sufficiently small grains (see e.g. [39], [6], [37]) and also in modeling surface energies in phase transitions (see [5]). The anisotropic isoperimetric inequality states n K 1 n E n 1 n P K (E). (1.7) This estimate (including equality cases) is well known in the literature (see e.g. [1, 38, 11, 37, 0, 6, 31]). In particular, Gromov [31] uses certain properties of the Knothe map from E to K in order to establish (1.7). However, as pointed out in [10] and [16], the argument may be repeated verbatim if one uses the Brenier map instead. This approach leads to certain estimates which are helpful in proving a sharp stability theorem for (1.7) (see [16, Theorem 1.1]). Using the anisotropic perimeter, we now introduce the isoperimetric deficit of E P K (E) δ K (E) := n K 1 n E n 1 n 1. (1.8) Note that δ K (λe) = δ K (E) and δ K (E+x 0 ) = δ K (E) for all λ > 0 and x 0 R n. Thanks to (1.7) and the associated equality cases, we have δ K (E) 0 with equality if and only if E is equal to K (up to a scaling and translation). Note also the similarity between µ and δ K. Indeed, they are both scaling Usually in the definition of K, K is assumed to contain the origin. However, this is not needed (see Lemma.1). 4

5 invariant; however, µ may not be translation invariant (depending on C). We denote the asymmetry index of E by { } E (x0 + rk) A(E) := inf : x 0 R n, rk = E. (1.9) E The general stability problem consists of proving an estimate of the form A(E) Cδ K (E) 1 β, (1.10) where C = C(n, K) and β = β(n, K). In the Euclidean case (i.e. K = B 1 ), Hall conjectured that (1.10) should hold with β =, and this was confirmed by Fusco, Maggi, Pratelli [4]. Indeed, the 1 exponent is sharp (see e.g. [8, Figure 4]). Their proof depends heavily on the full symmetry of the Euclidean ball. For the general case when K is a generic convex set, non-sharp results were obtained by Esposito, Fusco, Trombetti [13], while the sharp estimate was recently obtained by Figalli, Maggi, Pratelli [16]. Their proof uses a technique based on optimal transport theory. For more information about the history of (1.10), we refer the reader to [16] and [8]. 1.3 Sketch of the proof of Theorem 1. We now provide a short sketch of the proof of Theorem 1. for the case when E = K and E has a smooth boundary. The first key observation is that the relative isoperimetric inequality inside a convex cone is a direct consequence of the anisotropic isoperimetric inequality with K = B 1 C. Indeed, as follows from the argument in Section., P K (E) H n 1 ( E C), so (1.) follows immediately from (1.7). This observation suggests that one may exploit Gromov s argument in a similar way as in the proof of [16, Theorem 1.1] to obtain additional information on E. Indeed, we can show that there exists a vector α = α(e) R n such that 3 1 x α dh n 1 C(n, K) δ K (E), (1.11) E C E (α + K) C(n, K) δ K (E). (1.1) 3 Let us point out that the existence of a vector α such that (1.1) holds is exactly the main result in [16]. However, here we need to show that we can find a vector such that both (1.11) and (1.1) hold simultaneously. 5

6 Let us write α = (α 1, α ), with α 1 R k and α R n k. Moreover, let Ẽ := E (α 1, 0). Then using that C = R k C, we obtain E C (α 1, 0) = Ẽ C; therefore, Ẽ C 1 x (0, α ) dh n 1 C(n, K) δ K (E), (1.13) Ẽ ((0, α ) + K) C(n, K) δ K (E). (1.14) Since δ K (E) µ(e) (see Corollary.3), (1.13) and (1.14) hold with µ(e) in place of δ K (E) (see Lemmas 3.5 & 3.6). Thanks to (1.14), we see that our result would readily follow if we can show α C(n, K) µ(e). (1.15) Indeed, since ((0, α ) + K) K α (see Lemmas 3.1 & 3.), Ẽ K E 1 ( Ẽ ((0, α ) + K) + ((0, α ) + K) K ) K C(n, K) µ(e), which, of course, implies Theorem 1.. Therefore, we are left with proving (1.15). Firstly, assume that µ(e) and α are sufficiently small (i.e. smaller than a constant depending only on n and K). By (1.13) and the fact that (see Section ) H n 1 ( Ẽ (B 3 ((0, α )) C)) = P (Ẽ B 3 ((0, α )) C), 4 4 we have C(n, K) µ(e) 1 x (0, α ) dh n 1 (x) Ẽ C 1 x (0, α ) dh n 1 (x) Ẽ C { 1 x (0,α ) 1 4 } 1 4 Hn 1( Ẽ C { 1 x (0, α ) Hn 1 ( Ẽ (B 3 4 ((0, α )) C)) = 1 4 P (Ẽ B 3 4 ((0, α )) C). }) 6

7 But since α is small, B 1 (0) C B 3 ((0, α )) C; hence, 4 P (Ẽ B 3 ((0, α )) C) P (Ẽ B 1 (0) C). 4 Moreover, thanks to the relative isoperimetric inequality inside B 1 (0) C (see e.g. [, Inequality (3.43)]), we have that for µ(e) small enough, C(n, K) µ(e) 1 4 c(n, K) min { Ẽ (B 1 (0) C) n 1 n, (B 1 (0) C) \ 1 4 c(n, K) min { Ẽ (B 1 (0) C), (B 1 (0) C) \ Ẽ } n 1 } Ẽ n = 1 4 c(n, K) (B 1 (0) C) \ Ẽ, (1.16) where in the last step we used that Ẽ is close to (0, α ) + K (see (1.14)) and α is small. Therefore, using (1.14) and (1.16), (B 1 (0) C) \ ((0, α ) + K) (B 1 (0) C) \ E + E \ ((0, α ) + K) µ(e) + C(n, K) µ(e) 4C(n, K) c(n, K) C(n, K) µ(e). (1.17) Since C contains no lines, by some simple geometric considerations one may reduce the problem to the case when α {(x k+1,..., x n ) R n k : x n 0} (see Lemma 3.7), and then it is not difficult to prove c(n, K) α (B 1 (0) C) \ ((0, α ) + K), which combined with (1.17) establishes (1.15), and hence, the theorem. Next, we briefly discuss the assumptions for the sketch of the proof above. Indeed, one may remove the size assumption on α by showing that if µ(e) is small enough, then α will be automatically small (see Proposition 3.11). 4 Furthermore, we may freely assume that µ(e) is small since if 4 Let us point out that this is a nontrivial fact. Indeed, in our case we want to prove in an explicit, quantitative way that µ(e) controls α (E); hence, we want to avoid any compactness argument. However, even using compactness, we do not know any simple argument which shows that α (E) 0 as µ(e) 0. 7

8 µ(e) c(n, K) > 0, then the theorem is trivial: E (sk) E c(n, K) µ(e). The regularity of E was used in order to apply the Sobolev-Poincaré type estimate [16, Lemma 3.1] which yields (1.13) (see Lemma 3.5). If E is a general set of finite perimeter in C with finite mass and small relative deficit, then Lemma 3.4 tells us that E has a sufficiently regular subset G so that E \ G and µ(g) are controlled by µ(e). Combining this fact with the argument above yields the theorem for general sets of finite perimeter (see Proposition 3.8). Lastly, the assumption on the mass of E (i.e. E = K ) can be removed by a simple scaling argument. 1.4 Sharpness of the result We now discuss the sharpness of the estimate in Theorem 1.. Indeed, it is well known that there exists a sequence of ellipsoids {E h } h N, symmetric with respect to the origin and converging to the ball B 1, such that ( where s h = lim sup h E h B 1 ) 1 n δb1 (E h ) E h (s h B 1 ) <, lim h δ K 0 (E h ) = 0, (see e.g. [8, pg. 38]). Consider the cone C = {x R n : x 1,..., x n > 0} and set Ẽh := E h C. By symmetry, it follows that δ B1 (Ẽh) = 1 δ n B1 (E h ) and Ẽh (s h K) = 1 E n h (s h B 1 ). We also note that P (Ẽh C) = H n 1 ( Ẽh C) = 1 n Hn 1 ( E h ) = 1 n P B 1 (E h ), Therefore, Ẽh = 1 n E h, B 1 C = 1 n B 1. µ(ẽh) = = P (Ẽh C) n B 1 C 1 n Ẽh n 1 n P B1 (E h ) n B 1 C 1 n E h n 1 n 1 = 1 H n 1 ( E n h ) n( 1 B n 1 ) 1 n ( 1 E ) n 1 n n 1 = δ B1 (E h ), 1 8

9 and we have lim sup µ(ẽh) h Ẽh (s h K) <. This example shows that the 1 exponent in the theorem cannot, in general, be replaced by something larger. One may wonder whether it is possible for Theorem 1. to hold with a constant depending only on the dimension and not on the cone. Indeed, this was a surprising feature of [16, Theorem 1.1]. However, this is not so in our case. To see this, consider a sequence of open, symmetric cones in R indexed by their opening θ. Let E θ be a unit half-ball along the boundary of ( ) 1 the cone C θ disjoint from s θ K θ (see Figure 1), where s θ =. Note that µ(e θ ) = ( θ π ) 1 ( π ) 1 1. Therefore, E θ (s θ K θ ) lim θ π E θ µ(e θ ) = lim θ π π θπ 1 =. E θ B 1 C θ Figure 1: An example which shows that the constant in Theorem 1. cannot be replaced by a constant depending only on the dimension. Preliminaries.1 Initial setup Endow the space R n n of n n tensors with the metric A = trace(a T A), where A T denotes the transpose of A. Let T L 1 loc (Rn ; R n ) and denote the 9

10 distributional derivative of T by DT. If C R n is a Borel set, then DT (C) is the total variation of DT on C with respect to the metric defined above, i.e. { DT (C) = sup DT (C h ) : C h C k =, } C h C. h N h N Let BV (R n ; R n ) be the set of all T L 1 (R n ; R n ) with DT (R n ) <. For such a T, decompose DT = T dx + D s T, where T is the density with respect to the Lebesgue measure and D s T is the corresponding singular part. Denote the distributional divergence of T by Div T := trace(dt ), and let div(t ) := trace( T (x)). Then we have Div T = div T dx + trace(d s T ). If DT is symmetric and positive definite, note that trace(d s T ) 0. (.1) If E is a set of finite perimeter in R n, then the reduced boundary of E, which we denote by FE, consists of all points x R n such that D1 E (B r (x)) < for all r > 0 and the following limit exists and belongs to S n 1 : lim r 0 + D1 E (B r (x)) D1 E (B r (x)) =: ν E(x). We call ν E the measure theoretic outer unit normal to E. The reduced boundary of a set of finite perimeter is closely related to the De Giorgi perimeter of the set relative to C defined in (1.1): P (E C) = H n 1 (FE C) (see e.g. [, Theorem 3.61] and [, Equation (3.6)]). This fact along with one of the equality cases in (1.) n B 1 C = H n 1 ( B 1 C) yields the following useful representation of the relative deficit (recall that s > 0 satisfies E = sk ): µ(e) = Hn 1 (FE C) H n 1 ( B s C). (.) H n 1 ( B s C) Next, if T BV (R n ; R n ), then for H n 1 - a.e. x FE there exists an inner trace vector tr E (T )(x) R n (see [, Theorem 3.77]) which satisfies 1 lim T (y) tr r 0 + r n E (T )(x) dy = 0. B r(x) {y:(y x) ν E (x)<0} Furthermore, E (1) denotes the set of points in R n having density 1 with respect to E; i.e. x E (1) means E B r (x) lim r 0 + B r (x) 10 = 1.

11 Having developed the necessary notation, we are ready to state the following general version of the divergence theorem (see e.g. [, Theorem 3.84]) which will help us prove the isoperimetric inequality for convex cones (i.e. Theorem.): Div T (E (1) ) = tr E (T )(x) ν E (x)dh n 1 (x). (.3) FE Now we develop a few more tools that will be used throughout the paper. Fix K := B 1 C, and let D := {E C : P (E C) <, E < }. To apply the techniques in [16], we need a convex set that contains the origin. Therefore, let us translate K by the vector x 0 K which minimizes the ratio M K 0 m K0, where K 0 = K + x 0, m K0 := inf{ ν K0 : ν S n 1 } > 0, M K0 := sup{ ν K0 : ν S n 1 } > 0, (.4) and ν K0 is defined as in (1.5). Next, we will define an important geometric quantity: { z K0 := inf λ > 0 : z } λ K 0. (.5) Note that the convexity of K 0 implies the triangle inequality for K0 so that it behaves sort of like a norm; however, if K 0 is not symmetric with respect to the origin, x K0 x K0. Hence, this norm is in general not a true norm. Nevertheless, the following estimates relate this quantity with the standard Euclidean norm (see [16, Equations (3.) and (3.9)]): x M K0 x K0 x m K0, (.6) y K0 M K 0 m K0 y K0. (.7) Recall that the isoperimetric deficit δ K ( ) is scaling and translation invariant in its argument. The next lemma states that it is also translation invariant in K (observe that if z 0 + K does not contain the origin, then z0 +K can also be negative in some direction). 11

12 Lemma.1. Let E D. Then δ z0 +K(E) = δ K (E) for all z 0 R n. Proof. It suffices to prove P z0 +K(E) = P K (E). P z0 +K(E) = = = = FE FE FE FE sup{ν E (x) z : z z 0 + K}dH n 1 (x) sup{ν E (x) (z 0 + z) : z K}dH n 1 (x) (ν E (x) z 0 + sup{ν E (x) z : z K})dH n 1 (x) ν E (x) z 0 dh n 1 (x) + P K (E). By using the divergence theorem for sets of finite perimeter [, Equation (3.47)], we obtain ν E (x) z 0 dh n 1 (x) = div(z 0 )dx = 0, FE which proves the result.. Isoperimetric inequality inside a convex cone Here we show how to use Gromov s argument to prove the relative isoperimetric inequality for convex cones. As discussed in the introduction, the first general proof of the inequality was due to Lions and Pacella [7] and is based on the Brunn-Minkowski inequality. The equality cases were considered in [7] for the special case when C \ {0} is smooth. Our proof of the inequality closely follows the proof of [16, Theorem.3] with some minor modifications. Theorem.. Let C be an open, convex cone and E <. Then n E n 1 n K 1 n E H n 1 (FE C). Moreover, if C contains no lines, then equality holds if and only if E = sk. Proof. Proof of the inequality. By rescaling, if necessary, we may assume that K = E (i.e. s = 1). Define the probability densities dµ + (x) = 1 E 1 E(x)dx 1

13 and dµ (y) = 1 K 1 K(y)dy. By classical results in optimal transport theory, it is well known that there exists an a.e. unique map T : E K (which we call the Brenier map) such that T = φ where φ is convex and det( T (x)) = 1 for a.e. x E (see e.g. [5, 30, 9]). Moreover, since T (x) is symmetric and nonnegative definite, its eigenvalues λ k (x) are nonnegative for a.e. x R n and T BV (R n ; K) (see [1, Proposition 5.1]). As a result, we may apply the arithmetic-geometric mean inequality to conclude that for a.e. x E, n = n ( det T (x) ) 1 n = n ( n λ k (x) ) 1 n n λ k (x) = div T (x). (.8) k=1 k=1 Therefore, n E n 1 n K 1 n = n E = n det( T (x)) 1 n dx E div T (x)dx = div T (x)dx, E E (1) where we recall that E (1) denotes the set of points with density 1 (see Section.1). Next, we use (.1) and (.3): div T (x)dx div T (x)dx + (Div T ) s (E (1) ) E (1) E (1) = Div T (E (1) ) = tr E (T )(x) ν E (x)dh n 1 (x). (.9) By the convexity of K and the fact that T (x) K for a.e. x E, it follows that tr E (T )(x) K, so by the definition of K, Hence, FE FE tr E (T )(x) ν E (x) ν E (x) K. tr E (T )(x) ν E (x)dh n 1 (x) Furthermore, note that if z K, then z 1; therefore, FE ν E (x) K dh n 1 (x) = P K (E). ν E (x) K = sup{ν E (x) z : z K} (.10)

14 Moreover, observe that by the definition of K, it follows easily that ν C (x) K = 0 for H n 1 -a.e. x C \ {0}; therefore, ν E (x) K = 0 for H n 1 -a.e. x FE C. Thus, ν E (x) K dh n 1 (x) = ν E (x) K dh n 1 (x) H n 1 (FE C), FE FE C and this proves the inequality. Equality case. If E = K, then T (x) = x and it is easy to check that equality holds in each of the inequalities above. Conversely, suppose there is equality. In particular, n K = P K (E). By [0] (see also [16, Theorem A.1]), we obtain that E = K + a with a C. Suppose a 0. Then since C + a C, we have K C K (C a). But H n 1 ( K C) = n K = H n 1 ( (K + a) C) = H n 1 ( K (C a)). Therefore, K C = K (C a). (.11) Now, note that for t 1, we also have C a C ta. In fact, we claim K (C a) = K (C ta). (.1) Indeed, if not, then there exists x K such that x (C ta) \ (C a). In particular, t > 1 and thanks to (.11) we have that x / C; therefore, x K C (C ta) and we may write x = c ta for some c C, so that x = 1 t (c ta) + ( 1 1 t ) x = ( c t + ( 1 1 t ) x ) a. But since x C, c C, and t > 1, it follows that ( c t + (1 1 t )x) C; thus, x C a, a contradiction. Therefore, if we let C := t 1(C ta), then thanks to (.11) and (.1) we obtain This implies K C = K C = ( ( B 1 C) C ) ( (B1 C) C ). B 1 C C =. (.13) Since C is a convex cone (as it is easy to check), (.13) gives C C =. (.14) 14

15 Note that C C. If x C \C, then pick a point y C. Since λx+(1 λ)y C for all λ [0, 1], we obtain that C C, contradicting (.14). Therefore, C = C, but this is a contradiction since C contains the line {ta} t R. Hence, a = 0 and we are done. Corollary.3. If E D, then δ K (E) µ(e). Proof. Since the inequality is scaling invariant, we may assume that E = K. From (.10) and the fact that n K = H n 1 ( B 1 C) we obtain P K (E) n K H n 1 (FE C) n K = H n 1 (FE C) H n 1 ( B 1 C). Dividing by n K and using the representation of µ(e) given by (.) yields the result. Corollary.4. Let E D with E = K, and let T 0 : E K 0 be the Brenier map from E to K 0. Then ( 1 tre (T 0 x 0 )(x) ) dh n 1 (x) n K µ(e). FE C Proof. Let T : E K be the Brenier map from E to K so that T is the a.e. unique gradient of a convex function φ. Then T 0 (x) = T (x) + x 0 (this follows easily from the fact that T (x) + x 0 = φ(x) + x 0 = (φ(x) + x 0 x) and φ(x) + x 0 x is still convex). Therefore, by (.9), n E FE tr E (T 0 x 0 )(x) ν E (x)dh n 1 (x) (.15) Next, we recall from the proof of Theorem. that tr E (T 0 x 0 )(x) K. Hence, tr E (T 0 x 0 )(x) ν E (x) 0 for H n 1 a.e. x FE C and tr E (T 0 x 0 )(x) 1 for H n 1 a.e. x FE C. Therefore, using (.15), n E FE C FE C tr E (T 0 x 0 )(x) ν E (x)dh n 1 (x) tr E (T 0 x 0 )(x) dh n 1 (x) H n 1 (FE C). The fact that n E = n K = H n 1 ( B 1 C) finishes the proof. 15

16 3 Proof of Theorem 1. We split the proof in several steps. In Section 3.1, we collect some useful technical tools. Then in Section 3., we prove Theorem 1. under the additional assumption that E is close to K (up to a translation in the first k coordinates). Finally, we remove this assumption in Section 3.3 to conclude the proof of the theorem. Let {e k } n k=1 be the standard orthonormal basis for Rn. Recall that C = R k C, where C R n k is an open, convex cone containing no lines. Hence, up to a change of coordinates, we may assume without loss generality that C {x n = 0} = {0}. With this in mind and a simple compactness argument, we note that b = b(n, K) := inf{t > 0 : B 1 (0) C {x n < t} } > 0, (3.1) where B 1 (0) is the ball in R n k. Next, we introduce the trace constant of a set of finite perimeter. Recall the definition of K 0 given in Section.1, so that (.6) and (.7) hold. Given a set E D, let τ(e) denote the trace constant of E, where { } P K0 (F ) E τ(e) := inf ν : F E, F. (3.) n 1 FF FE E K0 dh Note that τ is scaling invariant, and in general τ(e) 1. The trace constant contains valuable information about the geometry of E. For example, if E has multiple connected components or outward cusps, then τ(e) = 1. In general, sets for which τ(e) > 1 enjoy a nontrivial Sobolov-Poincaré type inequality (see [16, Lemma 3.1]). 3.1 Main tools In what follows, we list all the technical tools needed in order to prove Theorem 1.. We decided to move some of the proofs to the appendix in order to make this section more accessible. The following two lemmas are facts about general convex sets. Lemma 3.1. Let A R n be an open, bounded, convex set. Then there exists C 3.1 (n, A) > 0 such that for any y R n, (y + A) A C 3.1 (n, A) y. 16

17 Figure : d b = 1/ b. Lemma 3.. Let A R n be an open, bounded, convex set. Then there exist C 3. (n, A), c 3. (n, A) > 0 such that if y R n, then min{c 3. (n, A), C 3. (n, A) y } (y + A) A. Lemma 3.3. There exists a bounded, convex set K B 1 (0) C so that for all y = (0,..., 0, y k+1,..., y n ) with y n 0, we have Furthermore, if y n 0, then K \ (y + K) = K \ (y + C). (3.3) (y + K) \ K = (y + K) \ C. (3.4) Proof. We will show that one may pick b = b(n, K) > 0 small enough so that K := B 1 (0) C ( k i=1 { x i < b} ) {x n < b} has the desired properties. We will establish (3.3) first. Since y + K y + C, it suffices to prove K \ (y + K) K \ (y + C). If (for contradiction) there exists x K (y + K) c (y + C), then x K and x y C \ K. Since x K and y n 0, it follows that x n y n < b. Also, x i y i = x i < b for i {1,..., k}. Now, x y C = R k C, hence, (x k+1 y k+1,..., x n y n ) C. Let b = b(n, K) be the constant from (3.1), and assume without loss of generality that b < b. If z {x n = b} C is such that z = d, where d = sup{ v : v C, v n = b}, then z = 1/ t := b x n y n (see Figure ). Let γ := b 1, b b > 1, and recall that (x k+1 y k+1,..., x n y n ) C. Since C is a cone, we have w := t(x k+1 y k+1,..., x n y n ) C with w n = b. Hence, w z = γ b, but since t > 1 we obtain (x k+1 y k+1,..., x n y n ) B γ b(0), where B γ b(0) denotes the ball in dimension n k. Therefore, x y k b + (γ b). 17

18 Next, pick M = M(n, K) N so that (k + γ ) ( b M ) < 1. Thus, by letting 4 b := b, we obtain x y B M 1 (0). Therefore, we conclude x y K, a contradiction. Hence, (3.3) is established. Since (y + K) \ K = y + ( K \ ( y + K) ) and (y + K) \ C = y + ( ) K \ ( y + C), (3.4) follows from (3.3). The next lemma (whose proof is postponed to the appendix) tells us that a set with finite mass, perimeter, and small relative deficit has a subset with almost the same mass, good trace constant, and small relative deficit (compare with [16, Theorem 3.4]). Lemma 3.4. Let E D with E = K. Then there exists a set of finite perimeter G E and constants k(n), c 3.4 (n), C 3.4 (n, K) > 0 such that if µ(e) c 3.4 (n), then E \ G µ(e) E, (3.5) k(n) τ(g) 1 + m K 0 M K0 k(n), (3.6) µ(g) C 3.4 (n, K)µ(E). (3.7) The big advantage of using G in place of E is that (3.6) implies a nontrivial trace inequality for G which allows us to exploit Gromov s proof in order to prove (1.11) with G in place of E. Indeed, if E is smooth with a uniform Lipschitz bound on E, one may take G = E. Lemma 3.5. Let E D, E = K, and assume µ(e) c 3.4 (n), with G E and c 3.4 (n) as in Lemma 3.4. Moreover, let r > 0 satisfy rg = K. Then there exists ˆα = ˆα(E) R n and a constant C 3.5 (n, K) > 0 such that F(rG) C 1 x ˆα dhn 1 C 3.5 (n, K) µ(e). Proof. Let T 0 : rg K 0 be the Brenier map from rg to K 0, and denote by S i the i th component of S(x) = T 0 (x) x. For all i, we apply [16, Lemma 3.1] to the function S i and the set rg to obtain a vector a = a(e) = (a 1,..., a n ) R n such that 18

19 F(rG) C tr (rg) ( S i (x) + a i ) ) ν (rg) K0 dh n 1 (x) M K0 m K0 (τ(rg) 1) DS i K0 ((rg) (1) ) MK 0 m K0 (τ(rg) 1) DS i ((rg) (1) ) MK 0 m K0 (τ(rg) 1) DS ((rg)(1) ), where we have used (.6) in the second inequality. scaling invariant. Hence, using (3.6) we have Next, recall that τ is F(rG) C tr (rg) ( S i (x) + a i ) ) ν (rg) K0 dh n 1 (x) M 3 K 0 m K 0 k(n) DS ((rg)(1) ). But by [16, Corollary.4] and Corollary.3, (3.8) DS ((rg) (1) ) 9n K δ K0 (rg) 9n K µ(rg) = 9n K µ(g). Therefore, by summing over i = 1,,..., n we obtain F(rG) C tr (rg) ( S(x)+a) ) ν (rg) K0 dh n 1 (x) 9n3 K MK 3 0 µ(g). (3.9) m K 0 k(n) Let ˆα = ˆα(E) := a + x 0, with x 0 as in the definition of K 0 (see (.4)). The triangle inequality implies 1 x ˆα = 1 x (a + x 0 ) 1 tr (rg) ( T 0 (x) x 0 ) + tr (rg) ( T 0 (x) x 0 ) (x (a + x 0 )) = 1 tr (rg) ( T 0 (x) x 0 ) + tr (rg) ( T 0 (x) x + a ). 19

20 Hence, by Corollary.4, (3.9), and (3.7) we have m K0 1 x ˆα dh n 1 (x) F(rG) C 1 x ˆα ν(rg) K0 dh n 1 (x) F(rG) C 1 tr(rg) ( T 0 (x) x 0 ) ν(rg) K0 dh n 1 (x) F(rG) C + tr (rg) ( S(x)) + a ) ν (rg) K0 dh n 1 (x) F(rG) C M K0 n K µ(g) + 9n 3 M 3 K K 0 µ(g) m K 0 k(n) M K0 n K C 3.4 (n, K)µ(E) + 9n3 K MK 3 0 C3.4 (n, K) µ(e). m K 0 k(n) As µ(e) 1, the result follows. The translation ˆα from Lemma 3.5 can be scaled so that it enjoys some nice properties which we list in the next lemma. The proof is essentially the same as that of [16, Theorem 1.1], adapted slightly in order to accommodate our setup. However, for the sake of completion, we include it in the appendix. Lemma 3.6. Suppose E D with E = K. Let ˆα = ˆα(E), G, and r be as in Lemma 3.5. Define α = α(e) := ˆα. Then there exists a positive constant r C 3.6 (n, K) such that for µ(e) c 3.4 (n), with c 3.4 (n) as in Lemma 3.4, we have E (α + K) C 3.6 (n, K) µ(e), (3.10) (rg) (ˆα + K) C 3.6 (n, K) µ(e), (3.11) and r 1 + µ(e) k(n). (3.1) Next, define R n + := {(x 1, x,..., x n ) R n : x n 0} (R n is defined in a similar manner). In the case α R n, the following lemma tells us that the last (n k) components of α are controlled by the relative deficit. 0

21 Lemma 3.7. Let E D with E = K, and let α = α(e) = (α 1, α ) R k R n k be as in Lemma 3.6. There exist positive constants c 3.7 (n, K), C 3.7 (n, K) such that if α R n and µ(e) c 3.7 (n, K), then α C 3.7 (n, K) µ(e). Proof. Let K C be the bounded, convex set given by Lemma 3.3. application of Lemma 3. and (3.4) yields An 1 min{c 3.(n, K),C 3. (n, K) α } 1 ((0, α ) + K) K = ((0, α ) + K) \ K = ((0, α ) + K) \ C Now, note that E (α 1, 0) C = R k C; hence, by using this fact and (3.10) we obtain ((0, α ) + K) \ C ((0, α ) + K) \ (E (α 1, 0)) (α + K) \ E C 3.6 (n, K) µ(e). Therefore, there exists c 3.7 (n, K) > 0 such that for µ(e) c 3.7 (n, K), 1 C 3.(n, K) α C 3.6 (n, K) µ(e). Thus, the result follows with C 3.7 (n, K) = C 3.6(n,K) C 3. (n, K) determines K). (note that K completely 3. Proof of the result when α (E) is small Proposition 3.8. Let E D with E = K, and let α = α(e) = (α 1, α ) R k R n k be as in Lemma 3.6. Then there exist positive constants c 3.8 (n, K), c 3.8 (n, K), and C 3.8 (n, K) such that if µ(e) c 3.8 (n, K) and α c 3.8 (n, K), then α C 3.8 (n, K) µ(e). Proof. Thanks to Lemma 3.7, we may assume without loss of generality that α R n +. Let G := rg (ˆα 1, 0), with G as in Lemma 3.4 and r > 0 such that 1

22 rg = K. By Lemma 3.5 and the fact that C = R k C, C 3.5 (n, K) µ(e) 1 x ˆα dh n 1 (x) F(rG) C = 1 x (0, ˆα ) dh n 1 (x) F G C 1 x (0, ˆα ) dh n 1 (x) F G C { 1 x (0,ˆα ) 1 4 } However, thanks to (3.1), small we have B Hn 1( F G C { 1 x (0, ˆα ) 1 4 }) 1 4 Hn 1 (F G (B 3 ((0, ˆα )) C)) = P ( G B 3 ((0, ˆα )) C). 4 ˆα 1+ (0) C B 3 4 α, so for α k(n) µ(e) and µ(e) sufficiently ((0, ˆα )) C, and this implies P ( G B 3 ((0, ˆα )) C) P ( G B 1 (0) C). 4 Next, by using the relative isoperimetric inequality (apply [, Inequality (3.41)] to 1 (rg) and the set B 1 (0) C), we have that for µ(e) small enough, C 3.5 (n, K) µ(e) Furthermore, 1 4 c(n, K) min { G (B 1 (0) C) n 1 n, (B 1 (0) C) \ n 1 } G n 1 4 c(n, K) min { G (B 1 (0) C), (B 1 (0) C) \ G }. (3.13) (B 1 (0) C) \ G K \ G G K ( (rg (ˆα1, 0) ) ( K + (0, ˆα ) )) ( (K + (0, ˆα ) ) K and by using (3.11), Lemma 3.1, and (3.1), (B 1 (0) C) \ G C 3.6 (n, K) µ(e) + C 3.1 (n, K) ˆα C 3.6 (n, K) ( µ(e) + C 3.1 (n, K) 1 + ) k(n) µ(e) α. ),

23 Therefore, we can select c 3.8 (n, K), c 3.8 (n, K) > 0 such that if µ(e) c 3.8 (n, K) and α c 3.8 (n, K), then min { G (B 1 (0) C), (B 1 Thus, using (3.13) we obtain (0) C) \ G } = (B 1 (0) C) \ G. 1 4 c(n, K) (B 1 (0) C) \ G C 3.5 (n, K) µ(e). (3.14) Hence, by (3.14), (3.11), and Lemma 3.1 it follows that ( B 1 (0) C ) \ ( (0, α ) + K ) (B 1 (0) C) \ G + ( G \ (0, α ) + K ) (B 1 (0) C) \ G + ( G (0, ˆα ) + K ) + ( (0, ˆα ) + K ) ( (0, α ) + K ) 4C 3.5(n, K) µ(e) + (rg) (ˆα + K) + C3.1 (n, K) α ˆα c(n, K) 4C 3.5(n, K) µ(e) + C3.6 (n, K) µ(e) + C 3.1 (n, K) α ˆα. (3.15) c(n, K) But α ˆα = α (r 1), and from (3.1) it readily follows that α ˆα α µ(e) c k(n) 3.8(n, K) µ(e). Combining this fact with (3.15) yields a k(n) positive constant C(n, K) such that (B 1 (0) C) \ ( (0, α ) + K ) C(n, K) µ(e). (3.16) Next, let K B 1 (0) C be the bounded, convex set given by Lemma 3.3. We note that since α R n +, (3.3) implies K \ ( (0, α ) + K ) = K \ ( (0, α ) + K ). Therefore, using Lemma 3. and (3.16) we have min{c 3. (n, K), C 3. (n, K) α } ( (0, α ) + K ) K = K \ ( (0, α ) + K ) = K \ ( (0, α ) + K ) (B 1 (0) C) \ ( (0, α ) + K ) C(n, K) µ(e). 3

24 Thus, for c 3.8 (n, K) sufficiently small we can take C 3.8 (n, K) = C(n,K) to C 3. (n, K) conclude the proof. Corollary 3.9. Let E D with E = K, c 3.8 (n, K) and c 3.8 (n, K) be as in Proposition 3.8, and α = α(e) = (α 1, α ) R k R n k be as in Lemma 3.6. Then there exists a positive constant C 3.9 (n, K) such that if µ(e) c 3.8 (n, K) and α c 3.8 (n, K), then (E (α 1, 0)) K C 3.9 (n, K) µ(e). Proof. Note that by Proposition 3.8 we obtain α C 3.8 (n, K) µ(e). Next, by applying Lemma 3.1 and (3.10) we have (E (α 1, 0)) K E (α + K) + ((0, α ) + K) K C 3.6 (n, K) µ(e) + C 3.1 (n, K) α (C 3.6 (n, K) + C 3.1 (n, K)C 3.8 (n, K)) µ(e). Therefore, we may take C 3.9 (n, K) = C 3.6 (n, K) + C 3.1 (n, K)C 3.8 (n, K) to conclude the proof. 3.3 Reduction step In Proposition 3.11 below, we refine Corollary 3.9. Namely, we show that if µ(e) is small enough, then the assumption on the size of α is superfluous. However, to prove Proposition 3.11 we need to reduce the problem to the case when α C R n k (recall C = R k C). This is the content of Lemma For arbitrary y R n k + \ C, decompose y as y = y c + y p, (3.17) where y c C is the closest point on the boundary of the cone C to y and y p := y y c (see Figure 3). Note that y p is perpendicular to y c. Lemma Let E D with E = K, and let α = α(e) = (α 1, α ) R k R n k be as in Lemma 3.6. There exist constants c 3.10 (n, K), C 3.10 (n, K) > 0 such that if µ(e) c 3.10 (n, K) and α R n k + \ C, then α p C 3.10 (n, K) µ(e). Proof. Firstly, observe that ((0, α c ) + K) \ (C (0, α p )) = ((0, α ) + K) \ C (α + K) \ (C + (α 1, 0)) = (α + K) \ C (α + K) \ E C 3.6 (n, K) µ(e). 4

25 Figure 3: Control of α p. Since (0, α c ) C, it follows that ((0, α c ) + K) has a nontrivial intersection with C. Let z := 1 ( (0, α c ) + ( 0, α c sup{t > 0 : (0, tα c ) ((0, α c ) + K)} )), and note that, by convexity, z ((0, α) c + K) C. Next, pick r = α. p Observe that r is the smallest radius for which B r (z) (C (0, α)) p so that it contains some w R n (see Figure 3). Since C is convex, there exists a constant c 0 (n, K) > 0 such that B r (z) ((0, α) c + K) c 0 (n, K)r n. But B r (z) ((0, α) c + K) ((0, α) c + K) \ (C (0, α)). p Therefore, r n µ(e), and since r = α p ) 1 n µ(e) 1 C 3.6 (n,k) c 0 (n,k) we have that α p ( C 3.6 (n) n c 0. (n,k) Therefore, by taking µ(e) small enough we can make α p as small as we wish. This fact and the convexity of C yield the existence of a radius r 0 = r 0 (n, K) > 0 and a constant c 3.10 (n, K) > 0 such that a spherical cone centered at w with base radius r 0 and height α p is contained in ((0, α) c + K) \ (C (0, α)), p provided that µ(e) c 3.10 (n, K) (see Figure 4). This 5

26 implies, ω n 1 r0 n 1 1 n αp ((0, α) c + K) \ (C (0, α)) p C 3.6 (n, K) µ(e), and we take C 3.10 (n, K) = C 3.6(n, K)n ω n 1 r0 n 1. Figure 4: Convexity allows us to squeeze in a spherical cone in ((0, α c )+K)\ (C (0, α p )). Proposition Let E D with E = K, and let α = α(e) = (α 1, α ) R k R n k be as in Lemma 3.6. Then there exist c 3.11 (n, K) > 0 such that if µ(e) c 3.11 (n, K), then α c 3.8 (n, K) with c 3.8 (n, K) as in Proposition 3.8. Proof. If α R n, then the result follows from Lemma 3.7. If α R n +, then write α = α p + α c as in (3.17) with the understanding that α C if and only if α p = 0. In the case where α p > 0 (i.e. α R n k + \ C), thanks to Lemma 3.10, we have α p C 3.10 (n, K) µ(e). Therefore, it suffices to 6

27 prove that for µ(e) sufficiently small, α c 1 c 3.8(n, K). We split the proof into three steps. The idea is as follows: firstly, we assume (for contradiction) that α c > 1 c 3.8(n, K). This allows us to translate E by a suitable vector β so that (E β) C is a distance 1 4 c 3.8(n, K) from the origin (see Figure 4). The second step consists of showing that up to a small mass adjustment, Figure 5: If E has small relative deficit but is far away from the origin, we can translate it a little bit and show that the resulting set thanks to Proposition 3.8 should in fact be a lot closer to the origin. the relative deficit of this new set is controlled by µ(e). Lastly, we show that the new set satisfies the hypotheses of Proposition 3.8; therefore, we conclude that it should be a lot closer to the origin than it actually is. Step 1. Assume for contradiction that α c > 1 c 3.8(n, K). Select γ (0, 1) such that for β := (0, γα c ) C we have (0, α c ) β = (1 γ) α c = 1 4 c 3.8(n, K). By (3.10), Lemma 3.1, and Lemma 3.10, E ((α 1, α c ) + K) E (α + K) + (α + K) ((α 1, α c ) + K) C 3.6 (n, K) µ(e) + C 3.1 (n, K) α p C 3.6 (n, K) µ(e) + C 3.1 (n, K)C 3.10 (n, K) µ(e). Now set Ẽ := E (α 1, 0) so that (0, α c ) satisfies an inequality as in (3.10) with constant C(n, K) = C 3.6 (n, K) + C 3.1 (n, K)C 3.10 (n, K); namely, Ẽ ((0, αc ) + K) C(n, K) µ(e). (3.18) Let F = t((ẽ β) C)) where t 1 is chosen so that F = Ẽ. Therefore, F = Ẽ = Ẽ β = (Ẽ β) C + (Ẽ β) \ C. (3.19) 7

28 Next, let us focus on the second term on the right side of (3.19): using (3.18), (Ẽ β) \ C = Ẽ \ (C + β) Ẽ \ ((0, αc ) + K) + ((0, α c ) + K) \ (C + β) C(n, K) µ(e) + ((0, α c ) β) + K) \ C. (3.0) But, (0, α c ) β = (0, (1 γ)α c ) C, therefore, ((0, α c ) β) + K C, and hence, (((0, α c ) β) + K) \ C = 0. Thus, combining the previous fact with (3.19) and (3.0), F (Ẽ β) C C(n, K) µ(e). (3.1) Step. From the definition of F and (3.1), we deduce (t n 1) (Ẽ β) C C(n, K) µ(e), so that for µ(e) K C(n,K), by (3.1) again and the fact that F = K, ( t 1 + C(n, K) K µ(e) ) 1 n. (3.) Since C is a convex cone, it follows that 1 C = C and β + C C. Thus, t P (F C) = t n 1 P ( Ẽ β + C ) t n 1 P (Ẽ C) = tn 1 P (E C + (α 1, 0)) ( 1 + C(n, ) n 1 K) n µ(e) P (E C) K ( 1 + C(n, ) K) µ(e) P (E C). (3.3) K Recall that P (F C) = H n 1 (FF C) and P (E C) = H n 1 (FE C) (see Section ). Upon subtracting P (B C) from both sides of (3.3), dividing by n K (recall n K = H n 1 ( B 1 C)), and using that P (E C) = n K µ(e) + n K we have µ(f ) µ(e) + C(n, K) µ(e)p (E C) n K = µ(e) + C(n, K) K µ(e) 3 C(n, K) + µ(e). K 8

29 Let w(n, K) := C(n,K). Then, assuming without loss of generality that K µ(e) 1, µ(f ) w(n, K) µ(e). (3.4) Step 3. Using Lemma 3.1 and (3.), for µ(e) small enough we have F (((0, α c ) β) + K) F t(((0, α c ) β) + K) + t(((0, α c ) β) + K) (((0, α c ) β) + K) t n (Ẽ β) C (((0, αc ) β) + K) + t(((0, α c ) β) + K) (t((0, α c ) β) + K) + (t((0, α c ) β) + K) (((0, α c ) β) + K) (Ẽ β) C (((0, αc ) β) + K) + (tk) K + C 3.1 (n, K) (0, α c ) β (t 1) (Ẽ β) C (((0, αc ) β) + K) + C(n, K) µ(e). (3.5) Next, we claim ((Ẽ β) C) (((0, αc ) β) + K) C(n, K) µ(e). (3.6) Indeed, from (3.18) we deduce that ((Ẽ β) C) (((0, αc ) β) + K) But since ((Ẽ β) C) + β Ẽ, = ( ((Ẽ β) C) + β) ((0, α c ) + K) ( ((Ẽ β) C) + β) Ẽ + Ẽ ((0, α c ) + K) ( ((Ẽ β) C) + β) Ẽ + C(n, K) µ(e). (3.7) ( ((Ẽ β) C) + β) Ẽ = Ẽ \ ( (( Ẽ β) C) + β ) = ( Ẽ β) \ (Ẽ β) C = (Ẽ β) \ C = Ẽ \ (β + C) Ẽ \ ((0, αc ) + K) + ((0, α c ) + K) \ (β + C) C(n, K) µ(e) + (((0, α c ) β) + K) \ C. (3.8) 9

30 As before, (((0, α c ) β)+k)\c = 0 (since ((0, α c ) β)+k C). Therefore, (3.7) and (3.8) imply the claim (i.e. (3.6)). Furthermore, by using (3.5) and (3.6), it follows that for some constant w(n, K), F (((0, α c ) β) + K) w(n, K) µ(e). (3.9) Next, let α(f ) be the translation as in Lemma 3.6 for the set F C, so that F (α(f ) + K) C 3.6 (n, K) µ(f ). By Lemma 3. and (3.9), min{c 3. (n, K),C 3. (n, K) ((0, α c ) β) α(f ) } (((0, α c ) β) + K) (α(f ) + K) (((0, α c ) β) + K) F + F (α(f ) + K) w(n, K) µ(e) + C 3.6 (n, K) µ(f ). (3.30) Moreover, (3.4) and (3.30) imply that if µ(e) is sufficiently small, then there exists a constant w (n, K) so that and α (F ) α(f ) (0, α c ) β + w (n, K)µ(E) 1 4 = 1 4 c 3.8(n, K) + w (n, K)µ(E) 1 4, (3.31) α 1 (F ) w (n, K)µ(E) 1 4 (3.3) (since α 1 (F ) ((0, α c ) β) α(f ) ). Furthermore, using (3.31) and (3.4), we deduce that for µ(e) small enough α (F ) c 3.8 (n, K), µ(f ) c 3.8 (n, K), where c 3.8 is as in Proposition 3.8. Thus, by applying Proposition 3.8 to F and using (3.4) again, it follows that α (F ) C 3.8 (n, K) µ(f ) C 3.8 (n, K) w(n, K)µ(E) 1 4. (3.33) Combining (3.30), (3.3), and (3.33) we obtain 1 4 c 3.8(n, K) = (0, α c ) β α(f ) + w (n, K)µ(E) 1 4 α (F ) + w (n, K)µ(E) 1 4 ( C 3.8 (n, K) ) w(n, K) + w (n, K) µ(e) 1 4, which is impossible if µ(e) is sufficiently small. This concludes the proof. 30

31 We are now in a position to prove Theorem 1.. Firstly, we assume that E = K. Indeed, let c 3.8 and c 3.11 be the constants given in Proposition 3.8 and 3.11, respectively, and set c(n, K) := min{c 3.8 (n, K), c 3.11 (n, K)}. If µ(e) c(n, K), then it follows from Proposition 3.11 and Corollary 3.9 that (E (α 1, 0)) K K C 3.9(n, K) µ(e). K Let C(n, K) := C 3.9(n,K) and suppose now that E K. Pick t > 0 such K that te = K and apply the previous estimate to the set te to obtain (te (α 1 (te), 0)) K) te C(n, K) µ(te) = C(n, K) µ(e), and this implies (E ( α 1(tE), 0)) ( 1K) t t E C(n, K) µ(e). Since s = 1, this yields the theorem for the case when µ(e) c(n, K). If t µ(e) > c(n, K), then E (sk) E c(n, K) µ(e). { Therefore, we obtain the theorem with C(n, K) = min C(n, K), }. c(n,k) 4 Appendix Proof of Lemma 3.1. If y > diam(a), then (y + A) A = A, and (y + A) A A y. Next, suppose y diam(a). Note that for any diam(a) z 0 R n, (y + A) A = (y + (A + z 0 )) (A + z 0 ), therefore we may assume without loss of generality that A contains the origin. Then the assumptions on A imply the triangle inequality for A. Hence, if x A, by (.6) we have y + x A < y A m A y + 1. Therefore, it is 31

32 sufficient to dilate A by 1 m A y + 1 in order for y + A (1 + 1 m A y )A. Since y + A = A and y diam(a), we have that (y + A) A = (y + A) \ A (1 + 1 m A y )A \ A = ( 1 (1 + y )A 1 A ) = ((1 + y ) n 1) A m A m A C(n, A) y. Note that it is easy to find an explicit upper bound on C(n, A) by using the binomial expansion (since y diam(a)). Therefore, we may take C 3.1 (n, A) = max{ A, C(n, A)}. diam(a) Proof of Lemma 3.. For t > 0 and w S n 1 let f w (t) := (A + tw) A. Note that f w (t) = 1 R n (A+tw) (x) 1 A (x) dx = 1 R n A (x + tw) 1 A (x) dx. Furthermore, for t > 0, let g w (t) := ν A (x + tw), w dh n 1 (x), (A tw) A where ν A is the outer unit normal to the boundary of A. Our strategy is as follows: first we show that g w (t) has a positive lower bound for all t > 0 sufficiently small that is also uniform in w S n 1. Then we prove f w(t) = g w (t) for almost every t. Since f w is Lipschitz, this will conclude the proof. The idea of the proof for the first part is simple though the precise argument is a bit technical: for fixed w and t, by convexity, it follows that there is a set M (w,t) A so that M (w,t) tw (A tw) A and on which the angle between normal vectors and w is uniformly bounded away from π (see Figure 6). However, by compactness of A, one can show that H n 1 (M (w,t) ) is uniformly bounded from below by a positive constant independent of w and t and this yields a uniform lower bound on g w (t). To begin, we claim that for s > 0, if 0 < t < s, then (A sw) A + (s t)w (A tw) A. (4.1) 3

33 Figure 6: Controlling g w (t) from below. Indeed, let x = y + (s t)w with y (A sw) A. Thus, y = a sw, with a A and so x = a tw. Since y A, convexity tells us that for all λ (0, 1), λy + (1 λ)a A. But λy + (1 λ)a = λ(a sw) + (1 λ)a = a λsw. By letting λ = t we see s a tw = x A, and this completes the proof of the claim. Note that by the convexity of A, we also have ν A (x + tw), w 0 for all x ( A tw) A (indeed, if y A and ν A (y) w < 0, then y tw {z : z y, ν A (y) > 0}, and the latter is disjoint from A). Therefore, by the change of variable x = y + (s t)w and (4.1), it follows that g w (s) = ν A (y + sw), w dh n 1 (y) (A sw) A = ν A (x + tw), w dh n 1 (x) (A sw) A+(s t)w g w (t), provided 0 < t < s. Next, we claim that for a particular value of s = s(n, A) > 0 (which we will specify later), inf g w(s) > 0. (4.) w S n 1 Indeed, let w S n 1, and for each point y A find r y > 0 so that B ry (y) A is the graph of a concave function u y : R n 1 R. Upon a possible 33

34 relabeling and reorientation of the coordinate axes, we have A B ry (y) = {x B ry (y) : x n < u y (x 1,..., x n 1 )}. In what follows, Bβ (α) denotes the open ball in R n 1 centered at α with radius β. Since A B ry/(y), y A by compactness of A there exists N N and {y k } N k=1 A such that A N k=1 B r k /(y k ). Denote the corresponding concave functions by {u k } N k=1, with the understanding that A B rk (y k ) is the graph of u k. Next, let y A be such that the normal to one of the supporting hyperplanes at y is w. Then y A B rj /(y j ) for some j {1,,..., N}. Moreover, u j : B rj (ˆx j ) R, where (ˆx j, u j (ˆx)) = y j, and A B rj (y j ) = {x B rj (y) : x n < u(x 1,..., x n 1 )}. (4.3) Next, let ˆx B rj /(ˆx j ) be such that (ˆx, u j (ˆx )) = y, r = r(n, A) := min r k /4, and s = s(n, A) := r/4. Since ˆx B rj /(ˆx j ), we have B r (ˆx ) k B 3rj /4(ˆx j ). Let us denote the superdifferential of a concave function φ : R n 1 R at a point ˆx in its domain by + φ(ˆx) := {ŷ R n 1 : ẑ R n 1, φ(ẑ) φ(ˆx) + ŷ, ẑ ˆx }. Since w is the normal to a supporting hyperplane of A at (ˆx, u j (ˆx )), there exists + u j (ˆx ) + u j (ˆx ) such that w = ( + u j (ˆx ),1) =: (ŵ, w ). Up to 1+ + u j (ˆx ) an infinitesimal rotation, we can assume without loss of generality + u j (ˆx ) > 0. Let e 1 = + u j (ˆx ) + u j (ˆx ) and {e,..., e n 1 } be a corresponding orthonormal basis for R n 1. We define { N ɛ := ẑ B n 1 r (ˆx ) : ẑ ˆx = γ 1 e 1 + γ i e i, i= n 1 γ i e i ɛ, i= } r 4 γ 1, and { Ñ ɛ := ẑ B n 1 r (ˆx ) : ẑ ˆx = γ 1 e 1 + γ i e i, 34 i= n 1 γ i e i ɛ, i= } r γ 1.

35 Let ẑ Ñɛ, and note that u j (ẑ) u j (ẑ sŵ) + + u j (ẑ sŵ), sŵ s = u j (ẑ sŵ) u j (ˆx ) + u j (ẑ sŵ), + u j (ˆx ). (4.4) Furthermore, if ẑ B r (ˆx ), then there exists + u j (ẑ) + u j (ẑ) such that ν A ((ẑ, u j (ẑ))) = ( + u j (ẑ),1). Thus, 1+ + u j (ẑ) ν A ((ẑ, u j (ẑ))), w = = ( + u j (ẑ), 1) u j (ẑ), ( + u j (ˆx ), 1) u j (ˆx ) u j (ẑ) + u j (ˆx ) u j (ẑ) u j (ˆx ) (4.5) If C = max k {Lip 3 4 (u k )} > 0, where Lip 3 (u k ) denotes the Lipschitz constant 4 k (y k)), then since sup + u j (ẑ) C, of u k on B 3 4 r k (ˆx k) (recall ˆx k := u 1 we have ẑ B r(ˆx ) u j (ẑ) u j (ˆx ) C =: C 0. (4.6) Next, the monotonicity formula for the superdifferential of the concave function u j tells us that + u j (ẑ) + u j (ˆx ), ẑ ˆx 0, and since + u j (ˆx ), ẑ ˆx 0 for ẑ N ɛ, we have n u j (ẑ), ẑ ˆx = + u j (ẑ), γ 1 e 1 + γ i e i As 0 < r 4 γ 1 we obtain γ 1 + u j (ˆx ) + u j (ẑ), + u j (ˆx ) C 0 ɛ. i= + u j (ẑ), + u j (ˆx ) 4C 0ɛ + u j (ˆx ) r 4C Cɛ 0. (4.7) r 35

36 Note that if ẑ Ñɛ, then ẑ sŵ N ɛ so by (4.7) Combining this with (4.4) yields + u j (ẑ sŵ), + u j (ˆx ) 4C Cɛ 0. r 4C 0 Cɛ u j (ẑ sŵ) u j (ẑ) s r u j (ˆx ) u j(ẑ) s 4C Cɛ 0. (4.8) r Now pick ɛ ( 0, min{ r 4C 0 C 1+C, r } ). Note that ɛ = ɛ(n, A) and with 8C 0 C this choice of ɛ, (4.5), (4.6), and (4.7) imply that for ẑ Ñɛ, whereas (4.8) implies ν A ((ẑ, u j (ẑ))), w 1 C 0, (4.9) u j (ẑ sŵ) > u j (ẑ) s 1 + C u j(ẑ) sw. (4.10) From (4.10), we obtain (ẑ sŵ, u j (ẑ) sw ) (A sw) A. Moreover, let U s : R n 1 R n be given by U s ( ˆd) = ( ˆd, u j ( ˆd + sŵ) sw ). Then U s (Ñɛ sŵ) (A sw) A. (4.11) Therefore, if y U s (Ñɛ sŵ) so that y = (ẑ sŵ, u j (ẑ) sw ) for some ẑ Ñɛ, then ν (A sw) (y) = ν A ((ẑ, u j (ẑ))); hence, (4.9) implies ν (A sw) (y), w = ν A ((ẑ, u j (ẑ)), w 1 C 0. This fact and (4.11) yields g s (w) = ν A sw (y), w dh n 1 (y) (A sw) A 1 U C 0dH n 1 (y) = 1 C 0H n 1 (U s (Ñɛ sŵ)) s(ñɛ sŵ) 1 C 0H n 1 (Ñɛ sŵ) 1 C 0γ(n, ɛ)r n 1 = 1 C 0γ(n, ɛ) min k ( ) n 1 rk, 4 36

37 and this proves (4.). Next, we claim f w(t) = g w (t) almost everywhere. To see this, let h(y) := A \ (A y) = 1 A\(A y) (x)dx, R n and ψ Cc (R n ; R n ). By Lemma 3.1, h(y) is Lipschitz; therefore, using integration by parts, Fubini, and that ψ is divergence free we obtain ψ h(y)dy = div ψ(y)h(y)dy R n R n ( ) = div ψ(y)1 A\(A y) (x)dy dx R n R ( n ) = div ψ(y)1 A\(A y) (x)dy dx A R ( n ) = div ψ(y)(1 1 (A y) (x))dy dx A R ( n ) = div ψ(y)1 (A y) (x)dy dx. (4.1) A R n Next, note that 1 (A y) (x) = 1 (A x) (y) and by applying the divergence theorem, it follows that div ψ(y)1 (A x) (y)dy = ψ(y), ν (A x) (y) dh n 1 (A x)(y) R n R n = ψ(z x), ν A (z) dh n 1 A(z). (4.13) R n Hence, by (4.1), (4.13), and Fubini we obtain ( ) ψ h(y)dy = ψ(z x), ν A (z) dh n 1 A(z) dx R n A R ( n ) = ψ(z x)1 A (x)dx ν A (z)dh n 1 A(z) R n R ( n ) = ψ(y)1 A (z y)dy ν A (z)dh n 1 A(z) R n R n ( ) = ψ(y) 1 A (z y)ν A (z)dh n 1 A(z) dy R n R ( n ) = ψ(y) 1 A (x)ν A (x + y)dh n 1 (x) dy. R n (A y) 37