aa r X i v : . [ m a t h . M G ] F e b RIGIDITY FOR MEASURABLE SETS
DORIN BUCUR, ILARIA FRAGAL `A
Abstract.
Let Ω ⊂ R d be a set with finite Lebesgue measure such that, for a fixedradius r >
0, the Lebesgue measure of Ω ∩ B r ( x ) is equal to a positive constant when x varies in the essential boundary of Ω. We prove that Ω is a ball (or a finite union ofequal balls) provided it satisfies a nondegeneracy condition, which holds in particularfor any set of diameter larger than r which is either open and connected, or of finiteperimeter and indecomposable. The proof requires reinventing each step of the movingplanes method by Alexandrov in the framework of measurable sets. Introduction
In this paper we study the following rigidity problem for measurable sets of the Euclideanspace R d : for a given radius r >
0, identify measurable sets Ω of finite Lebesgue measuresuch that, for a positive constant c ,(1) | Ω ∩ B r ( x ) | = c ∀ x ∈ ∂ ∗ Ω , where B r ( x ) is the ball of radius r centred at x , | · | denotes the Lebesgue measure, and ∂ ∗ Ω indicates the essential boundary of Ω ( i.e. the set of points x ∈ R d at which bothΩ and its complement Ω c have a strictly positive d -dimensional upper density).Such an easy-to-state geometric question conceals several relations with classical ques-tions in Differential Geometry, as well as with recent advances in Convex Geometry andGeometric Measure Theory. We outline them below, before stating the results. A quick historical overview.
The question about rigidity criteria obtainable bymeasuring intersections of a domain Ω in R d with balls rolling along its boundary datesback to almost one century ago. The idea is to look at the behaviour, for x ∈ ∂ Ω, ofsurface integrals H d − (Ω ∩ ∂B r ( x )), or of volume integrals | Ω ∩ B r ( x ) | . In its first grain,this idea can be found in a paper dating back to 1932 by Cimmino (see [14]), where heraised the following question: is it possible to characterize surfaces Γ which bisect the H -measure of the boundary of any ball which is centred on Γ and has a sufficientlysmall radius? Cimmino’s problem was solved more than sixty years later by Nitsche,who proved that the only (smooth) surfaces with this property are the plane and thehelicoid [31].The problem reemerged in the first 2000s under different garments: motivated by thestudy of isothermic surfaces in heat conduction, also in relation with the so-called Matzohball soup problem [26], Magnanini, Prajapat and Sakaguchi were led to consider B -dense Date : February 25, 2021.2010
Mathematics Subject Classification.
Key words and phrases.
Rigidity results, measurable sets, moving planes, Steiner symmetrization. domains , namely subsets Ω in R d such that, for any r >
0, there exists a positive constant c ( r ) such that | Ω ∩ B r ( x ) | = c ( r ) for every x ∈ ∂ Ω. In particular they proved that, ifthe boundary of a B -dense domain Ω is a complete embedded minimal surface of finitetotal curvature in R , Ω must be a plane [25]. Later in 2016, Magnanini and Mariniproved that, in any space dimension and for any given convex body K , if a set Ω ofpositive and finite Lebesgue measure is K -dense (meaning that | Ω ∩ ( x + rK ) | = c ( r )for every x ∈ ∂ Ω), then Ω and K are homothetic ellipsoids [24] (see also [23, 2]). Notethat, as long as it is assumed to hold for any sufficiently small r >
0, the constancy ofthe volume measure | Ω ∩ B r ( x ) | is actually equivalent to that of the surface measure H d − (Ω ∩ ∂B r ( x )); so the difference between B -dense domains and those considered byCimmino is just that the volume fraction is no longer fixed to (in other words, theconstant c ( r ) may differ from | B r ( x ) | ).All the rigidity results mentioned so far are naturally related to a central question inDifferential Geometry, namely the classification of hypersurfaces with constant meancurvature in R d . Indeed, in view of the asymptotic expansion(2) | Ω ∩ B r ( x ) | = 12 ω d r d − d − d + 1) ω d − H Ω ( x ) r d +1 + O ( r d +2 ) , where ω d is the volume of the unit ball in R d and H Ω is the mean curvature of ∂ Ω [21], theLebesgue measure of Ω ∩ B r ( x ) can be interpreted as an integral approximation of H Ω ( x );differentiating with respect to r , the same assertion is valid for the ( d − ∩ ∂B r ( x ).In this perspective, the rigidity criteria quoted above can be read as counterparts of somecornerstone results for hypersurfaces with constant mean curvature. Thus, Magnanini-Marini criterion reminds the celebrated theorem proved in 1958 by Alexandrov [1]: if Ωis a bounded connected domain of class C such that ∂ Ω has constant mean curvature,then Ω is a ball (a generalization has been proved very recently by Delgadino-Maggi [17],showing that any set with finite Lebesgue measure and finite perimeter with constantdistributional mean curvature is a finite union of equal balls). Likewise, Nitsche criterion,though not involving any topological constraint, reminds the much harder problem,settled only in 2005 by Meeks-Rosenberg [29], of classifying the plane and the helicoidas the unique simply connected minimal surfaces embedded in R .Comparing the constancy of the mean curvature with the constancy of one of the mea-sures H d − (Ω ∩ ∂B r ( x )) and | Ω ∩ B r ( x ) | for any sufficiently small r >
0, it is clear thatthe former is in principle weaker, as it concerns only one among the coefficients of theexpansion (2) in powers of r ; on the other hand, defining a notion of mean curvaturerequires some boundary regularity, even if done in distributional sense, while measuringintersections with balls requires no smoothness at all.All in all, at present no rigidity result seems to be available for arbitrary measurable setsunder a fairly weak condition such as the constancy of a single and well-defined quantity.Aim of this paper is to provide a first contribution in this direction, by considering setswhich satisfy condition (1). The fact that we work with one fixed radius makes theapproach completely new. We call sets satisfying (1) r -critical . Note that, since theessential boundary ∂ ∗ Ω is included in the topological boundary ∂ Ω, condition (1) isweaker than the constancy of | Ω ∩ B r ( x ) | along ∂ Ω. Incidentally let us also mention that, at least for convex domains, r -criticality can be rephrased by saying that ∂ Ω is alevel surface for the cross-covariogram function of Ω and B r (0). (The cross-covariogramfunction of two convex bodies K and K is defined as g ( K ,K ) ( x ) := | K ∩ ( x + K ) | ,and the investigation of its level lines has attracted some attention in the literature onConvex Geometry, also in connection with the floating body problem, see for instance[6, 30]).The reason for the terminology “ r -critical”’ is that, notably, this notion has still avariational interpretation. Actually, sets of constant mean curvature may be viewedas stationary sets for the perimeter functional under a volume preserving perturbation.From this point of view, Alexandrov result, along with its extension in [17], allows toidentify critical sets for the isoperimetric inequality proved in 1958 by De Giorgi [16, 18].An interpretation in the same vein can be given to r -critical sets, as soon as the isoperi-metric inequality is replaced by another classical one, which is even more ancient, namelythe rearrangement inequality proved in 1932 by Riesz [33]. In a simplified version it statesthat, for any radially symmetric, decreasing, non negative function h , balls maximize,under a constraint of prescribed Lebesgue measure, the integral functional J h (Ω) := Z Ω Z Ω h ( x − y ) dx dy. Given an integrand h as above, it is not difficult to check that balls maximize J h providedthey maximize J χ Br (0) for all r >
0. Hence the choice h = χ B r (0) is of special relevance,and for such kernel stationary domains are precisely sets satisfying condition (1). Equiv-alently, in view of the equality | Ω ∩ B r ( x ) | − | Ω c ∩ B r ( x ) | = 2 | Ω ∩ B r ( x ) | − ω d r d , r -criticalsets may be viewed as stationary domains, under volume preserving perturbations, forthe nonlocal perimeter r -Per(Ω) := Z Ω Z Ω c χ {| x − y |
Before stating our main result, in order to introduce the key conditionfor rigidity, companion to r -criticality, we set the following definition:we say that a measurable set Ω in R d is r -degenerate ifinf x ,x ∈ ∂ ∗ Ω (cid:12)(cid:12) Ω ∩ ( B r ( x )∆ B r ( x )) (cid:12)(cid:12) k x − x k = 0 . A discussion about this notion is postponed to Section 2.1. Therein we shall provide, inparticular, a measure theoretic condition sufficient for nondegeneracy, which permits toshow that bounded open connected sets, as well as bounded indecomposable sets withfinite perimeter, are not degenerate for any r smaller than their diameter. Theorem 1.
Let Ω be a measurable set with finite Lebesgue measure in R d , and let r > . Assume that Ω is r -critical and not r -degenerate. Then Ω is equivalent to a finiteunion of balls of the same radius R > r , at mutual distance larger than or equal to r .Remark . Rigidity may fail if the finite measure assumption is dropped: any halfspaceor any strip { x ∈ R d : a < x < b } is critical and not degenerate for any r >
0. As well,rigidity may fail for r -critical sets of finite measure which are r -degenerate; for someexamples in this respect, see Section 2.1. Remark . A result analogue to Theorem 1 can be immediately deduced, by using thearea formula, if balls are replaced by ellipsoids: if E is a given ellipsoid, any set withfinite measure which satisfies the criticality and degeneracy conditions with x + E inplace of B r ( x ), is a finite union of ellipsoids homotetic to E . Remark . We point out that the initial choice of the radius r produces a sort of tunedbubbling phenomenon , which may occur only with a precise lower threshold both onthe size of the balls and on their mutual distance. In accordance with the short-rangenonlocal nature of our kernel, this behaviour should be compared with the local resultin [17], where bubbling can occur at any scale, and the fractional results in [15, 8],where bubbling cannot occur at all. Let us also mention that a further motivation forcharacterising finite unions of equal balls is their appearance as optimal domains inspectral shape optimization problems (see [20, 7]).We now present some consequences of Theorem 1 for sets enjoing some kind of regularity.We begin by the case of open sets: Corollary 5.
Let Ω be an open set with finite Lebesgue measure in R d , and let r > .Assume that there exists a positive constant c such that (3) | Ω ∩ B r ( x ) | = c ∀ x ∈ ∂ Ω , where ∂ Ω denotes the topological boundary.If r < inf i { diam(Ω i ) } , where Ω i are the open connected components of Ω , then Ω is afinite union of balls of the same radius R > r , at mutual distance larger than or equalto r . In particular, if Ω is connected and r < diam(Ω) , then Ω is a ball. Next we turn to the case of sets with finite perimeter. Recall that, following [3], any setΩ with finite perimeter can be written as finite or countable family of indecomposablecomponents Ω i . This means that each Ω i is indecomposable in the sense that it doesnot admit a partition (Ω + i , Ω − i ), with | Ω ± i | > i ) = Per(Ω + i ) + Per(Ω − i ), andthat the Ω i ’s are maximal indecomposable sets. Recall also that the reduced boundary F Ω is the collection of points x ∈ supp( Dχ Ω ) such that the generalized normal ν Ω ( x ) :=lim ρ → Dχ Ω ( B ρ ( x )) / | Dχ Ω | ( B ρ ( x )) exists in R d and satisfies ν Ω ( x ) = 1. For sets of finiteperimeter, a sufficient condition for r -criticality is the validity of condition (4) below,because the closure of F Ω \ N , for any H d − -negligible set N , turns out to contain ∂ ∗ Ω. Corollary 6.
Let Ω be a set of finite perimeter and finite Lebesgue measure in R d , andlet r > . Assume there exists a positive constant c such that (4) | Ω ∩ B r ( x ) | = c for H d − -a.e. x ∈ F Ω . If r < inf i { diam(Ω i ) } , where Ω i are the indecomposable components of Ω , then Ω isequivalent to a finite union of balls of the same radius R > r , at mutual distance largerthan or equal to r . In particular, if Ω is indecomposable and r < diam(Ω) , Ω is a ball. About the proof of Theorem 1.
Alexandrov rigidity theorem was obtained by a veryelegant proof, based on what he called reflection principle , nowadays commonly knownas the moving planes method . His brilliant idea was destined to have a tremendous im-pact also in the field of Mathematical Analysis: its implications in PDEs were firstlyenhanced in the seventies by Serrin [34] to get rigidity results for overdetermined bound-ary value problems, and afterwards enlivened to get symmetry and monotonicity prop-erties of solutions to nonlinear elliptic equations, in particular by Gidas-Ni-Nirenberg[19], Berestycki-Nirenberg [4, 5], Caffarelli-Gidas-Spruck [10].The proof of Theorem 1 is based on a reinvention of the moving planes method in thecontext of measurable sets. Alexandrov idea is that, if Ω has constant mean curvature,it must have a hyperplane of symmetry in every direction; this can be obtained startingfrom an arbitrary hyperplane, moving it in a parallel way until an appropriate stoppingtime, and reflecting Ω about such hyperplane. The conclusion is then reached by thequalitative behaviour of the constant mean curvature equation, specifically using thestrong maximum principle and Hopf boundary point lemma. Our situation is completelydifferent, for many reasons. First, no connectedness assumption is made on Ω, so thatthe proof cannot be obtained just by observing that the choice of the initial hyperplaneis arbitrary, but requires a new argument allowing to single out each ball and “extract”it from Ω once enough symmetries are detected. Second, no smoothness information isavailable: since the essential boundary does not admit a normal vector and is not locallya graph, all the the steps of the method loose their meaning. Third, even in cases whenthe boundary is locally a graph and the contact with the reflected cap holds in classicalsense, no PDE holds around the contact point, but merely the r -criticality conditionon the essential boundary. Thus we need to conceive new arguments, in particular todecipher why the movement can start and especially when it has to stop. In the smoothsetting, this occurs either when the boundary and the reflected cap become tangent, orwhen the boundary and the moving plane meet orthogonally; in the measurable setting,these two situations must be abandoned in favour of suitable notions of away or close contact. They are defined and handled relying on the concept of Steiner-symmetric sets,which plays a crucial role similarly as in De Giorgi’s proof of the isoperimetric theorem.A more detailed outline of the proof is given at the beginning of Section 3.2. Preliminaries
In this section we discuss the main issues about degeneracy, and we prepare the proofof Theorem 1, by analyzing the structure of certain Steiner symmetric sets obtained byreflection.2.1.
About r -degeneracy. We start by showing some counterexamples of r -criticalsets which escape from rigidity since they are degenerate. Examples 7.
Different kinds of bounded sets which, for some r >
0, are critical butdegenerate: (i) Small sets: any measurable set Ω with diam(Ω) ≤ r . (ii) Unions of small sets at large mutual distance: any measurable set obtained as theunion of a finite number of measurable sets Ω j , having the same measure, diam(Ω j ) ≤ r ∀ j , and dist(Ω j , Ω l ) ≥ r ∀ j = l . (iii) Unions of spaced small sets at small mutual distance: for d = 2 and any fixed n ∈ N , given r > || − e i πjn | − r | ≥ ε > ∀ j = 1 , . . . , n , any set obtained asthe union of n measurable sets Ω j , having the same measure, such that Ω j ⊆ B ε ( e i πjn ) ∀ j = 1 , . . . , n. This last example shows in particular that the connectedness of a r -neighbourhood of Ω is not sufficient to avoid r -degeneracy.Next we establish a measure-theoretic sufficient condition for nondegeneracy (Proposi-tion 8) which is useful, in particular, to deal with open sets and sets with finite perimeter(Proposition 10), and hence to deduce Corollaries 5 and 6 from Theorem 1.Such condition is expressed in terms of the total variation measure | Dχ B r ( x ) | , which isgiven by (see for instance [22, page 117])(5) | Dχ B r ( x ) | ( E ) = H d − ( ∂B r ( x ) ∩ E ) for any measurable set E ;we are thus led back to handle the measure of spherical hypersurfaces considered byCimmino.Hereafter and in the sequel, we denote by Ω ( t ) the set of points x ∈ R d at which Ω has d -dimensional density equal t . Proposition 8.
Let Ω be a bounded measurable set, and let r > . Assume there exists ε > such that (6) inf x ∈U ε ( ∂ ∗ Ω) | Dχ B r ( x ) | (Ω (1) ) > where U ε ( ∂ ∗ Ω) is the set of points at distance smaller than ε from ∂ ∗ Ω . Then Ω is not r -degenerate. Remark . (i) Taking points of density 1 in (6) is relevant in order to make the conditionsatisfied, for instance, by open sets deprived of a spherical hypersurface centred at aboundary point.(ii) Condition (6) is not necessary for nondegeneracy. For instance, consider in thecomplex plane the union of the sets { z = ρe iθ : 0 < ρ < , θ ∈ ( − π , π ) } and { z = ρe iθ :1 < ρ < , θ ∈ ( − π , π ) } . For r = 1, the set is not degenerate. However, the infimum (6)vanishes, by taking points x arbitrarily close to 0 on the negative real axis. Proof of Proposition 8.
Set α := inf x ∈U ε ( ∂ ∗ Ω) | Dχ B r ( x ) | (Ω (1) ). Assume by contradictionthat α > r -degenerate. Since Ω is bounded, we can find two sequences { x n } , { y n } ⊂ ∂ ∗ Ω, with lim n k x n − y n k = 0, such that(7) lim n → + ∞ (cid:12)(cid:12) Ω ∩ ( B r ( x n )∆ B r ( y n )) (cid:12)(cid:12) k x n − y n k = 0 . Without loss of generality, we can assume that for every n it holds x n , y n ∈ U ε ( ∂ ∗ Ω) and k x n − y n k ≤ ε . This implies that, if [ x n , y n ] is the closed segment with endpoints x n and y n , we have(8) | Dχ B r ( x ) | (Ω (1) ) ≥ α ∀ x ∈ [ x n , y n ] . For every x ∈ R d , we denote x ′ = ( x , . . . , x d − ) ∈ R d − , and we write x = ( x ′ , x d ). Wefix some δ ∈ (0 , ε ∧ r ) such that(9) H d − ( { x ∈ ∂B r (0) : | x d | ≤ δ } ) ≤ α . For convenience, we position our system of coordinates so that x n = (0 , δ n ) and y n =(0 , − δ n ). Then, for n large enough so that δ n ≤ δ , we get the following estimate: (cid:12)(cid:12) Ω ∩ ( B r ( x n )∆ B r ( y n )) (cid:12)(cid:12) k x n − y n k = 12 δ n Z B r ( x n )∆ B r ( y n ) χ Ω ( x ) dx ≥ δ n Z δ n − δ n h Z k x ′ k≤ √ r − δ χ Ω (1) ( x ′ , p r − k x ′ k + s ) dx ′ i ds + 12 δ n Z δ n − δ n h Z k x ′ k≤ √ r − δ χ Ω (1) ( x ′ , − p r − k x ′ k + s ) dx ′ i ds ≥ δ n p r − δ r Z δ n − δ n h Z k x ′ k≤ √ r − δ χ Ω (1) ( x ′ , p r − k x ′ k + s ) r p r − k x ′ k dx ′ i ds + 12 δ n p r − δ r Z δ n − δ n h Z k x ′ k≤ √ r − δ χ Ω (1) ( x ′ , − p r − k x ′ k + s ) r p r − k x ′ k dx ′ i ds ≥ p r − δ r α , where the last inequality follows from (8) and (9). This contradicts (7) and achieves theproof. (cid:3) Proposition 10.
Let Ω be either a bounded open set or a bounded set of finite perimeter,and let { Ω i } i denote the family respectively of its connected or indecomposable compo-nents. Then Ω is not r -degenerate for any r < inf i { diam(Ω i ) } .Remark . While sets Ω with a single component are not r -degenerate if and onlyif r < diam(Ω), for multiply connected domains the condition r < inf i { diam(Ω i ) } issufficient but clearly not necessary to avoid r -degeneracy: for instance, the disjointunion of two balls of radius R can be not r -degenerate also for radii r ≥ R providedthe balls are close enough. Remark . In the light of Proposition 10, it is natural to ask if there exists some notionof “connectedness” avoiding r -degeneracy also for arbitrary measurable sets. To thebest of our knowledge, the unique kind of such a general notion for Borel sets Ω hasbeen proposed in [12] under the name of essential connectedness , and amounts to askthat H d − (Ω (1) ∩ ∂ ∗ Ω + ∩ ∂ ∗ Ω − ) > + , Ω − ) of Ω.However, if Ω has not finite perimeter, relying on the possible lack of semicontinuity ofthe map t
7→ H d − (Ω ∩ B r ( x t ))) as x t → x ∈ ∂ ∗ Ω, it is possible to construct examplesof Borel sets Ω which are essentially connected but degenerate for some r < diam(Ω).
Proof of Proposition 10 . We focus our attention on the case of finite perimeter sets,since for open sets the proof can be obtained in a similar way. Working componentby component, we are reduced to prove that, if Ω is indecomposable, then it is not r -degenerate for any r < diam(Ω). To that aim, it is enough to prove that it satisfies,for some ε >
0, condition (6) in Proposition 8. Assume by contradiction this is not thecase. Then, in view of (5), it would be possible to find a sequence { x n } converging to apoint x ∈ ∂ ∗ Ω such that(10) lim n H d − ( ∂B r ( x n ) ∩ Ω (1) ) = 0 . Up to subsequences, we denote by A and C the limit in L respectively of the charac-teristic functions of the sets A n := Ω ∩ B r ( x n ) and C n := Ω \ B r ( x n ). We are going toshow that they provide a nontrivial partition of Ω such that Per(Ω) = Per( A ) + Per( C ).We have Per( A n ) = Per(Ω , B r ( x n )) + H d − ( ∂B r ( x n ) ∩ ∂ ∗ A n ∩ Ω ( ) ) + H d − ( ∂B r ( x n ) ∩ Ω (1) )Per( C n ) = Per(Ω , R d \ B r ( x n )) + H d − ( ∂B r ( x n ) ∩ ∂ ∗ C n ∩ Ω ( ) ) + H d − ( ∂B r ( x n ) ∩ Ω (1) )Since perimeter is lower semicontinuous with respect to L -convergence, and we have H d − ( ∂ ∗ A n ∩ ∂ ∗ C n ∩ Ω ( ) ) = 0, by passing to the limit in the two relations above andsumming, we getPer( A ) + Per( C ) ≤ Per(Ω) + 2 lim n H d − ( ∂B r ( x n ) ∩ Ω (1) ) . The conclusion follows by condition (10). (cid:3)
About some Steiner symmetric sets obtained by reflection.
A measurableset ω is Steiner symmetric about a hyperplane H with unit normal ν if the followingequality holds as an equivalence between Lebesgue measurable sets: ω = n x ∈ R d : x = z + tν , z ∈ H , | t | < H (cid:0) ω ∩ (cid:8) z + tν : t ∈ R (cid:9)(cid:1)o . In Proposition 13 below, we focus our attention on a special kind of Steiner symmetricsets obtained by reflection, that we shall need to handle in the proof of Theorem 1.To that aim and in the sequel, we shall make repeatedly use of the following elementaryobservation: given a measurable subset ω of R d , it holds(11) ω (1) \ ∂ ∗ ω = int( ω (1) ) , ω (0) \ ∂ ∗ ω = int( ω (0) ) . In particular, R d can be decomposed as a disjoint union,(12) R d = int( ω (1) ) ⊔ int( ω (0) ) ⊔ ∂ ∗ ω . Let us prove the first equality in (11), the second one being analogous. The inclusion ⊇ isimmediate. Viceversa, let x ∈ ω (1) \ ∂ ∗ ω , and let U be an open neighbourhood of x whichdoes not meet ∂ ∗ ω . Let us prove that U ⊂ ω (1) . By Federer’s Theorem, ω is of finiteperimeter in U . Then, by the relative isoperimetric inequality, min {| ω c ∩ U | , | ω ∩ U |} = 0.But it cannot be | ω ∩ U | = 0, because x ∈ ω (1) . Hence | ω c ∩ U | = 0. Then, U cannotcontain any point of density 0 for ω , since such point would be of density 1 for ω c , against | U ∩ ω c | = 0. Recalling that U does not meet ∂ ∗ ω , we conclude that U ⊂ ω (1) . (cid:3) Proposition 13.
Let H be a hyperplane with unit normal ν , and let ω be a boundedmeasurable set contained into H − = { H + tν : t ≤ } such that (13) ∀ z ∈ H , ∂ ∗ ω ∩ (cid:8) z + tν : t < (cid:9) is empty or a singleton ,so that ω can be viewed as the subgraph of the function g : H → R − defined by g ( z ) = 0 if the intersection in (13) is empty and g ( z ) = t if such intersection is z + tν . Thefollowing properties hold: (i) | ∂ ∗ ω | = 0 ; (ii) ω is essentially open; (iii) the union of ω and its reflection about H is Steiner-symmetric about H ; (iv) the function g is continuous.Proof. Statement (i) is an immediate consequence of the assumption (13) and FubiniTheorem. To obtain statement (ii), it is enough to show that ω (1) is essentially open.Such property follows from statement (i), after applying (11). To prove statement (iii), itis enough to show that the union of ω (1) and its reflection about H is Steiner-symmetricabout H . To that aim, let us fix z ∈ H such that ∂ ∗ ω ∩ (cid:8) z + tν : t < (cid:9) = { p } , and letus show that the open segment ( p, z ) is contained into ω (1) . Recalling the decomposition(12) and assumption (13), we infer that the open segment ( p, z ) is entirely containedeither in int( ω (1) ) or in int( ω (0) ). In the first case we are done. It remains to excludethat it is entirely contained in int( ω (0) ). Assume by contradiction this is the case. Weobserve that, by the first equality in (11), since p ∈ ∂ ∗ ω , p is the limit of a sequence of points { p n } ⊂ ω (1) . Since we are assuming that the open segment ( p, z ) is entirelycontained in int( ω (0) ), the points p n do not belong to such segment, so that they belongto straight lines of the form { z n + tν : t ∈ R } , for a sequence of points { z n } ⊂ H converging to z . But then some of these straight lines would necessarily contain atleast two points of ∂ ∗ ω (otherwise the segments ( p n , z n ) would be entirely contained intoint( ω (1) ) and could not converge to ( p, z ) which is entirely contained into int( ω (0) )).Let g be the function defined via (13) as in the statement, so that ω can be viewed asthe subgraph of g . To show the continuity of g at a fixed point z ∈ H , we considerseparately the cases g ( z ) = 0 and g ( z ) <
0. If g ( z ) = 0, we have to prove that, forany sequence { z n } ⊂ H converging to z , the sequence { g ( z n ) } converges to 0. Up to asubsequence, we may assume g ( z n ) → λ , with λ ≤
0. If λ <
0, for n large enough wehave g ( z n ) <
0, which by definition of g means that z n + g ( z n ) ν ∈ ∂ ∗ ω ; passing to thelimit in the last relation, we get z + λν ∈ ∂ ∗ ω , against g ( z ) = 0.Assume now g ( z ) <
0, and let { z n } ⊂ H be any sequence converging to z . We mayassume that g ( z n ) → λ , with λ ≤
0. If λ <
0, we get as above z + λν ∈ ∂ ∗ ω ; by (13), weconclude that g ( z ) = λ . It remains to show that the case λ = 0 cannot occur. Assume λ = 0. We consider the open segment S := ( z + g ( z ) ν, z ). By (13), S ∩ ∂ ∗ ω = ∅ , andhence S is entirely contained either into int( ω (1) ) or into int( ω (0) ). If S ⊆ int( ω (1) ), wepick a point x ∈ S and a small ball B ε ( x ) ⊂ ω (1) . For every x ∈ B ε ( x ), denoting by z x its projection onto H (in particular, z x = z ), by statement (iii) we have that thesegment ( x, z x ) lies into ω (1) . This property leads to a contradiction, as it implies on onehand that z ∈ int( ω (1) ) and on the other hand that that g ( z n ) < n large enough,which in turn gives z ∈ ∂ ∗ ω (passing to the limit in the relation z n + g ( z n ) ν ∈ ∂ ∗ ω ).If S ⊆ int( ω (0) ), we can pick a point x ∈ S and a small ball B ε ( x ) ⊂ ω (0) . Thiscontradicts statement (iii) and the fact that, since the point z + g ( z ) ν belongs to ∂ ∗ ω ,it is the limit of a sequence of points of density 1 for ω . (cid:3) Proof of Theorem 1
Outline of the proof.
We observe first of all that the equality (1) continues to holdat every point x ∈ ∂ ∗ Ω. Then we fix a direction ν ∈ S d − , and we consider an initialhyperplane H with unit normal ν , not intersecting ∂ ∗ Ω. Such an initial hyperplaneexists because, since Ω has finite measure and is r -critical, it is necessarily bounded. Westart moving H in the direction of its normal ν to new positions, so that at a certainmoment of the process it starts intersecting ∂ ∗ Ω. We continue the movement in direction ν , and we denote by H t the hyperplanes thus obtained. We set: H − t := the closed halfspace determined by H t containing H H + t := the closed halfspace determined by H t not containing H Ω t := Ω ∩ H − t R t := the reflection of Ω t about H t . • We say that symmetric inclusion holds at t if(14) R t ⊂ Ω and Ω t ∪ R t is Steiner symmetric about H t . • We say that symmetric inclusion occurs at t if with away contact if (14) holds andthere exists an “away contact point”, namely a point(15) p ′ ∈ (cid:2) ∂ ∗ R t ∩ ∂ ∗ Ω (cid:3) \ H t . when (14) holds but (15) is false, we say that symmetric inclusion at t holds withoutaway contact . • We say that symmetric inclusion occurs at t with close contact if (14) holds and thereexists a “close contact point”, namely a point(16) H t ∋ q = lim n q ,n = lim n q ,n , q i,n ∈ ∂ ∗ Ω ∩ { q + tν : t ∈ R } , q ,n = q ,n , (where one among q ,n and q ,n will always happen to belong to H + t , while the otherone may fall in H + t as well as in H − t .) Notice that symmetric inclusion can occur at thesame t with both away contact and close contact.The statement will be obtained in the following steps, which are carried over separatelyin the next subsections. Step 1 (start)
There exists ε > t ∈ [0 , ε ), symmetric inclusion holds. Step 2 (the stopping time: no close contact without away contact)
Setting T := sup n t > s ∈ [0 , t ), symmetric inclusion occurs without away contact o , we have T < + ∞ , and symmetric inclusion occurs at T with away or with close contact.But we are able to rule out the case of close contact without away contact, so necessarilyat t = T we are in the situation of away contact. Step 3 (decomposition of Ω into symmetric and non-symmetric part) We show that Ω can be decomposed asΩ = Ω s ⊔ Ω ns , where Ω s is an open set representing the Steiner symmetric part of Ω, given byΩ s := [ n ( p, p ′ ) : p ′ is an away contact point, p is its symmetric about H T o , ( p, p ′ ) being the open segment with endpoints p and p ′ , and Ω ns := Ω \ Ω s represents thenon-symmetric part. Moreover, denoting by Ω si the open connected components of Ω s ,we prove that: ∂ ∗ Ω si ∩ ( H ± T \ H T ) are connected sets;(17) ∂ ∗ Ω s ∩ ∂ ∗ Ω ns ⊂ H T . (18) Step 4 (conclusion)
We show that the open connected components of Ω s are balls ofthe same radius R > r/
2, lying at distance larger than or equal to r , while the set Ω ns is Lebesgue negligible. Proof of Step 1.
The proof is based on the following lemma.
Lemma 14 (no converging pairs) . Let Ω ⊂ R d be a measurable set which is r -criticaland not r -degenerate. Assume that Ω is contained into H +0 := { z + tν : z ∈ H , t ≥ } , H being a hyperplane with unit normal ν . Then there cannot exist two sequences ofpoints { p ,n } , { p ,n } in ∂ ∗ Ω ∩ H +0 which for every fixed n are distinct, with the sameprojection onto H , and at infinitesimal distance from H as n → + ∞ .Proof. We argue by contradiction. Setting t i,n := dist( p i,n , H ), we can assume up to asubsequence that t ,n > t ,n for every n . We are going to show that(19) lim inf n → + ∞ | Ω ∩ B r ( p ,n ) | − | Ω ∩ B r ( p ,n ) | t ,n − t ,n > , against the fact that Ω is r -critical.We have(20) | Ω ∩ B r ( p ,n ) |−| Ω ∩ B r ( p ,n ) | = | Ω ∩ ( B r ( p ,n ) \ B r ( p ,n )) |−| Ω ∩ ( B r ( p ,n ) \ B r ( p ,n )) | . Since Ω is not r -degenerate, there exists a positive constant C such that(21) | Ω ∩ ( B r ( p ,n ) \ B r ( p ,n )) | + | Ω ∩ ( B r ( p ,n ) \ B r ( p ,n )) | t ,n − t ,n ≥ C .
In view of (20) and (21), the inequality (19) holds true provided(22) | Ω ∩ ( B r ( p ,n ) \ B r ( p ,n )) | t ,n − t ,n ≤ C . In turn, by the inclusion Ω ⊂ H +0 , the inequality (23) is satisfied as soon as(23) | H +0 ∩ ( B r ( p ,n ) \ B r ( p ,n )) | t ,n − t ,n ≤ C . Such inequality follows from elementary geometric arguments. Indeed, for every fixed n , the set H +0 ∩ ( B r ( p ,n ) \ B r ( p ,n )) has volume not larger than the region D n obtainedas the difference between two right cylinders having the same axis, given by the straightline orthogonal to H through the common projection z n of p ,n and p ,n onto H , thesame height equal to t ,n + (1 / t ,n − t ,n ), and as bases the ( d − H , with center at z n and radii respectively equal to ( r − t ,n ) / and( r − t ,n ) / . Hence, to get (23) it is enough to show that | D n | = o ( t ,n − t ,n ). This isreadily checked since, setting γ n := ( t ,n − t ,n ) /
2, we have | D n | = ω d − (cid:0) ( r − t ,n ) n − − ( r − t ,n ) n − (cid:1)(cid:0) t ,n + 12 ( t ,n − t ,n ) (cid:1) ∼ d − ω d − r d − ( t ,n γ n + γ n ) (cid:0) t ,n + γ n (cid:1) . (cid:3) Assume the claim in Step 1 false. Then, at least one of the following assertions holds:(i) ∃{ t n } → ∀ n Ω t n ∪ R t n is not Steiner symmetric about H t n ;(ii) ∃{ t n } → ∀ n |R t n \ Ω | > In case (i), for every n we can apply Proposition 13 with H = H t n and ω = Ω − t n to inferthat, for some z n ∈ H t n , the set ∂ ∗ Ω t n ∩ { z n + tν : t < } contains at least two distinctpoints. Then ∂ ∗ Ω ∩ H +0 contains two sequences of points { p ,n } , { p ,n } which for every n are distinct, with the same projection onto H , and at infinitesimal distance from H as n → + ∞ , against Lemma 14.In case (ii), we may assume that Ω t n ∪ R t n is Steiner symmetric about H t n For every n let y ′ n ∈ Ω (0) ∩ R (1) t n , and let z n be the orthogonal projection of y ′ n on H t n .If on the segment ( z n , y ′ n ] there is some point in ∂ ∗ Ω, since Ω t n ∪R t n is Steiner symmetricabout H t n , we would have a pair of distinct points belonging to ∂ ∗ Ω ∩ H +0 , with thesame projection on H , and infinitesimal distance from H , against Lemma 14.If on the segment ( z n , y ′ n ] there is no point in ∂ ∗ Ω, invoking (12) and recalling that y ′ n ∈ Ω (0) , we infer that the whole segment ( z n , y ′ n ] is contained into int(Ω (0) ). Onthe other hand, since y ′ n ∈ R (1) t n , denoting by y n the reflection of y ′ n about H t n , wehave that the whole segment [ y n , z n ) is contained into int(Ω (1) ). We conclude that thepoint z n belongs to ∂ ∗ Ω. Then, by arguing in the same way as in the last part ofthe proof of Proposition 13, it would be possible to find some straight line of the form { e z n + tν : t ∈ R } , with e z n ∈ H t n arbitrarily close to z n , containing at least two pointsof ∂ ∗ Ω ∩ ( H ) + . Again, this would contradict Lemma 14.3.2. Proof of Step 2.
Since Ω is bounded, we have
T < + ∞ . Then the proof of Step2 is obtained by showing the following claims: • Claim 2a. Symmetric inclusion holds at T with away contact or with close contact. • Claim 2b. Symmetric inclusion cannot hold with close contact and no away contact.Proof of Claim 2a . Symmetric inclusion clearly continues to hold at T . Moreover, bydefinition of T , at least one of the following assertions is true:(i) ∃{ t n } → T + such that ∀ n symmetric inclusion with away contact holds at t n ;(ii) ∃{ t n } → T + such that ∀ n symmetric inclusion does not hold at t n .In case (i), for every n there exists an away contact point at t n , namely a point p ′ n ∈ (cid:2) ∂ ∗ R t n ∩ ∂ ∗ Ω (cid:3) \ H t n . Up to a subsequence, denote by p ′ the limit of p ′ n . Two cases mayoccur. If p ′ H T , then p ′ is an away contact point at T . If p ′ ∈ H T , denoting by p n thesymmetric of p ′ n about H t n , taking q ,n = p n and q ,n = p ′ n in (16), we see that p ′ is aclose contact point.To deal with case (ii), we point out the validity of the following Away inclusion property: If symmetric inclusion occurs without away contact at T , forevery δ > , there exists s δ > such that, for every s ∈ [0 , s δ ] the set U sT − δ := n x + (2 δ + 2 s ) ν : x ∈ R T − δ o is contained into Ω . The away inclusion property can be easily proved by contradiction. If it was false, wecould find an infinitesimal sequence { s n } of positive numbers, and a sequence of points { x ′ n } of density 1 for U s n T − δ but of density 0 for Ω. Up to a subsequence, there exists x ′ := lim n x ′ n . By construction, we have x ′ ∈ { x + 2 δν : x ∈ ∂ ∗ R T − δ } ⊂ ∂ ∗ R T . But,since we are assuming that symmetric inclusion occurs without away contact at T , it isreadily checked that ∂ ∗ R T ⊆ int(Ω (1) ). Then x ′ ∈ int(Ω (1) ), against the fact that x ′ n arepoints of density 0 for Ω.Now, going back to case (ii), we can assume that symmetric inclusion occurs at T withoutaway contact (otherwise Claim 2a. holds for free). Then, in view of the away inclusionproperty, the failure of symmetric inclusion at t n implies that, for every n and every δ >
0, there exist at least two distinct points in ∂ ∗ Ω, say q ,δ,n and q ,δ,n , which havethe same orthogonal projection onto H T and have distance less than δ from H T . By thearbitrariness of δ >
0, we can choose a diagonal sequence, and passing to the limit weget a close contact point at T according to definition (16). Proof of Claim 2b.
Assume by contradiction that symmetric inclusion holds at T withclose contact. We are going to contradict (4) by showing that, if { q ,n } and { q ,n } aresequences converging to a point q ∈ H T as in (16), it holds(24) lim inf n → + ∞ | Ω ∩ B r ( q ,n ) | − | Ω ∩ B r ( q ,n ) |k q ,n − q ,n k > . We have(25) | Ω ∩ B r ( q ,n ) | − | Ω ∩ B r ( q ,n ) | = | Ω ∩ ( B r ( q ,n ) \ B r ( q ,n )) | − | Ω ∩ ( B r ( q ,n ) \ B r ( q ,n )) | . In order to estimate the two terms at the r.h.s. of (25), we fix δ > s n > H T + s n contains the midpoint of the segment ( q ,n , q ,n ).Up to working with n large enough, since q ,n and q ,n converge to a point of H T , thanksto the away inclusion property we can assume that(26) U s n T − δ ⊂ Ω . Hence,(27) | Ω ∩ ( B r ( q ,n ) \ B r ( q ,n )) | = | U s n T − δ ∩ ( B r ( q ,n ) \ B r ( q ,h )) | + | (Ω \ U s n T − δ ) ∩ ( B r ( q ,n ) \ B r ( q ,h )) | . On the other hand,(28) | Ω ∩ ( B r ( q ,n ) \ B r ( q ,n )) | = | Ω T − δ ∩ ( B r ( q ,n ) \ B r ( q ,n )) | + | Ω ∩ ( H T + s n ⊕ B δ + s n ) ∩ ( B r ( q ,n ) \ B r ( q ,n )) || Here and below, we denote by H ⊕ B R (0) the strip given by points of R d with distanceless than R from a hyperplane H .The two sets Ω T − δ ∩ ( B r ( q ,n ) \ B r ( q ,n )) and U s n T − δ ∩ ( B r ( q ,n ) \ B r ( q ,n )) have the samemeasure as they are symmetric about the hyperplane H T + s n . Therefore, by subtracting(28) from (27), and recalling (25), we obtain(29) | Ω ∩ B r ( q ,n ) | − | Ω ∩ B r ( q ,n ) | = I n − J n , with I n := | (Ω \ U s n T − δ ) ∩ ( B r ( q ,n ) \ B r ( q ,n )) | J n := | Ω ∩ ( H T + s n ⊕ B δ + s n ) ∩ ( B r ( q ,n ) \ B r ( q ,n )) | . We are now going to estimate J n from above and I n from below. We have J n ≤ | ( H T + s n ⊕ B δ + s n ) ∩ ( B r ( q ,n ) \ B r ( q ,n ) | . In turn, the right hand side of the above inequality does not exceed the measure of theregion D n obtained as the difference between two right cylinders having both as axisthe straight line containing q ,n and q ,n , as height δ + s n , and as bases the ( d − H T − δ respectively with B r ( q ,n ) and B r ( q ,n ).The measure of such region D n satisfies ( cf. the proof of Lemma 14) | D n | ∼ d − ω d − r d − ( t ,n γ n + γ n ) (cid:0) t ,n + γ n (cid:1) , where γ n is the distance of q ,n and q ,n from H T + s n (or equivalently, 2 γ n is the distancebetween q ,n and q ,n ), and t ,n = δ + s n − γ n is the distance of q ,n from H T − δ . Weinfer that(30) J n ≤ d − ω d − r d − δ γ n , where the last inequality holds because s n ≤ δ for n large enough.We now turn to estimate I n . Let us begin by proving that: ∃ δ > n | Ω ∩ H + T + δ +2 s n ∩ ( B r ( q ,n ) \ B r ( q ,n )) | > , (31) ∃ γ > n dist ( U s n T − δ , ∂ ∗ Ω ∩ ( B r ( q ,n ) \ B r ( q ,n )) ≥ γ, ∀ δ ∈ (0 , δ ] . (32)If the infimum in (31) is zero, by virtue of (26) we obtaininf n (cid:12)(cid:12) Ω ∩ ( H T + s n ⊕ B δ (0)) ∩ ( B r ( q ,n )∆ B r ( q ,n )) (cid:12)(cid:12) k q ,n − q ,n k = inf n (cid:12)(cid:12) Ω ∩ ( B r ( q ,n )∆ B r ( q ,n )) (cid:12)(cid:12) k q ,n − q ,n k . This is not possible provided δ is small enough, because the left hand side of the aboveequality is infinitesimal as δ → + , while the right hand side is controlled from belowby a positive constant thanks to the nondegeneracy assumption.To prove (32), it is enough to take therein δ = δ , since, once proved the lower bound for δ = δ , it will hold a fortiori for δ ∈ (0 , δ ]. Now, if (32) was false for δ = δ , we couldfind a sequence { x n } ⊂ ∂ ∗ Ω ∩ ( B r ( q ,n ) \ B r ( q ,n )) such that lim n dist ( U s n T − δ , x n ) = 0.This is not possible because, up to a subsequence, the limit of { x n } would provide anaway contact point at T , against our assumption.Now, thanks to (32), we can consider a geodesic curve minimizing the distance between U s n T − δ and ∂ ∗ Ω ∩ ( B r ( q ,n ) \ B r ( q ,n ) inside B r ( q ,n ) \ B r ( q ,n ). We take the mid-point,say y n , and we consider the ball B γ ( y n ). The set B γ ( y n ) ∩ ( B r ( q ,n ) \ B r ( q ,n )) doesnot intersect U s n T − δ and, by (26) and (31), it is contained into Ω ∩ ( B r ( q ,n ) \ B r ( q ,n )).Since γ is fixed, we infer that there exists a positive constant K such that(33) I n ≥ Kγ n . By (29), (30), and (33) we conclude that that, up to taking δ smaller, it holdslim inf n → + ∞ | Ω ∩ B r ( q ,n ) | − | Ω ∩ B r ( q ,n ) |k q ,n − q ,n k > K . Thus (24) holds true and the proof of Step 2 is achieved.3.3.
Proof of Step 3.
By Step 2, we know that at t = T symmetric inclusion occurswith away contact. The proof of Step 3 is obtained by showing the following claims: • Claim 3a. If p ′ is an away contact point and p is its symmetric about about H T , (cid:12)(cid:12)(cid:2) B r ( p ′ ) \ B r ( p ) (cid:3) ∩ (cid:2) Ω \ R T (cid:3)(cid:12)(cid:12) = 0 , and hence | ( B r ( p ′ ) \ B r ( p )) ∩ Ω ns | = 0;(34) ∃ ε > (cid:12)(cid:12) B ε ( p ′ ) ∩ (Ω \ R T ) (cid:12)(cid:12) = 0 , and hence Ω s is open. (35) • Claim 3b. Properties (17) and (18) hold.
Proof of Claim 3a.
We have: (cid:12)(cid:12)(cid:2) B r ( p ′ ) \ B r ( p ) (cid:3) ∩ (cid:0) Ω \ R T (cid:1)(cid:12)(cid:12) = (cid:12)(cid:12) B r ( p ′ ) ∩ (cid:2) Ω \ (Ω T ∪ R T ) (cid:3)(cid:12)(cid:12) − (cid:12)(cid:12) B r ( p ) ∩ (cid:2) Ω \ (Ω T ∪ R T ) (cid:3)(cid:12)(cid:12) = (cid:12)(cid:12) B r ( p ′ ) ∩ Ω (cid:12)(cid:12) − (cid:12)(cid:12) B r ( p ′ ) ∩ (cid:0) Ω T ∪ R T (cid:1)(cid:12)(cid:12) − (cid:12)(cid:12) B r ( p ) ∩ Ω (cid:12)(cid:12) + (cid:12)(cid:12) B r ( p ) ∩ (cid:0) Ω T ∪ R T (cid:1)(cid:12)(cid:12) = 0 , where the first equality holds since B r ( p ′ ) \ B r ( p ) does not intersect Ω T , while in thelast one we have used r -criticality and the fact that the two sets B r ( p ′ ) ∩ (Ω T ∪ R T ) and B r ( p ) ∩ (Ω T ∪ R T ) are reflected of each other about H T . We have thus proved (34).In view of (34), the equality (35) is immediate in case p ′ B r ( p ). Therefore, we mayprove it having in mind that p ′ ∈ B r ( p ). We claim that(36) 0 < (cid:12)(cid:12)(cid:2) B r ( p ′ ) \ B r ( p ) (cid:3) ∩ Ω (cid:12)(cid:12) < (cid:12)(cid:12) B r ( p ′ ) \ B r ( p ) (cid:12)(cid:12) . Indeed, let us exclude both the equalities (cid:12)(cid:12)(cid:2) B r ( p ′ ) \ B r ( p ) (cid:3) ∩ Ω (cid:12)(cid:12) = 0 and (cid:12)(cid:12)(cid:2) B r ( p ′ ) \ B r ( p ) (cid:3) ∩ Ω (cid:12)(cid:12) = (cid:12)(cid:12) B r ( p ′ ) \ B r ( p ) (cid:12)(cid:12) . The former cannot hold since Ω is not r -degenerate. The latter, in view of (34), wouldimply that B r ( p ′ ) \ B r ( p ) is contained into R T , and hence B r ( p ) \ B r ( p ′ ) is containedinto Ω T . Since Ω T ∪ R T is Steiner-symmetric about H T , this would give (via FubiniTheorem) that p and p ′ belong to int(Ω (1) ), contradicting the fact that they belong to ∂ ∗ Ω.As a consequence of (36), we observe that(37) ∃ y ′ ∈ (cid:2) B r ( p ′ ) \ B r ( p ) (cid:3) ∩ ∂ ∗ Ω . Indeed, if (37) was false, B r ( p ′ ) \ B r ( p ) would be contained either into int(Ω (1) ) or intoint(Ω (0) ), against (36). Next we observe that, in view of (34), the two sets Ω and R T have the same density at every point of B r ( p ′ ) \ B r ( p ), and hence (cid:2) B r ( p ′ ) \ B r ( p ) (cid:3) ∩ ∂ ∗ Ω = (cid:2) B r ( p ′ ) \ B r ( p ) (cid:3) ∩ ∂ ∗ R T ;consequently, since the set B r ( p ′ ) \ B r ( p ) is open, we have(38) (cid:2) B r ( p ′ ) \ B r ( p ) (cid:3) ∩ ∂ ∗ Ω = (cid:2) B r ( p ′ ) \ B r ( p ) (cid:3) ∩ ∂ ∗ R T . By (37) and (38), it turns out that y ′ is itself an away contact point. Therefore, denotingby y its symmetric about H T , in the same way as we obtained (34), replacing the pair p, p ′ by the pair y, y ′ , we get(39) (cid:12)(cid:12)(cid:2) B r ( y ′ ) \ B r ( y ) (cid:3) ∩ (cid:0) Ω \ R T (cid:1)(cid:12)(cid:12) = 0 . Moreover, since the set B r ( p ′ ) \ B r ( p ) is open, for ε > B ε ( y ′ )is contained into B r ( p ′ ) \ B r ( p ), and hence(40) ∃ ε > B ε ( p ′ ) ⊂ (cid:2) B r ( y ′ ) \ B r ( y ) (cid:3) . By (39) and (40), (35) is proved.
Proof of Claim 3b.
In order to prove (17)-(18), we consider the subsets of H T defined by C T := n m ( p,p ′ ) : p ′ is an away contact point, p is its symmetric about H T o ,A T := n z ∈ H T : g ( z ) < o , where m ( p,p ′ ) ∈ H T denotes the mid-point of the segment ( p, p ′ ), and the function g isdefined as in Proposition 13 (iv) (applied with ω := Ω T and H := H T ).By Proposition 13 (iv), we know that g is continuous and hence the set A T turns out tobe open. Since Claim 3a. implies that C T is open in H T , we infer that C T is relativelyopen in A T . On the other hand, since ∂ ∗ Ω is a closed set, it easy to check that C T is also relatively closed in A T . Hence, C T consists in a non-empty union of connectedcomponents of A T . Accordingly, ∂ ∗ Ω s ∩ ( H − T \ H T ) (resp., ∂ ∗ Ω s ∩ ( H + T \ H T )) is a union ofconnected sets, which are the images of the open connected components of C T throughthe continuous function g (resp., the reflections of such images about H T ). Each of theseconnected sets corresponds to ∂ ∗ Ω si ∩ ( H − T \ H T ) (resp., ∂ ∗ Ω si ∩ ( H + T \ H T ) ) for someopen connected component Ω si of Ω s . This proves (17). Since by (35) none of the sets ∂ ∗ Ω si ∩ ( H ± T \ H T ) can intersect ∂ ∗ Ω ns , (18) follows.3.4. Proof of Step 4.
Relying on decomposition Ω = Ω s ⊔ Ω ns made in Step 3, we aregoing to analyze in detail the behaviour of the open connected components Ω si of Ω s . Tothat aim, we need to set up some additional definitions and notation.Given two two different open connected components Ω si , Ω sj of Ω s , we say that Ω si isin r -contact with Ω sj if there exists an away contact point p ′ ∈ ∂ ∗ Ω si \ H T such that,denoting by p its symmetric about H T , it holds (cid:12)(cid:12)(cid:0) B r ( p )∆ B r ( p ′ ) (cid:1) ∩ Ω sj (cid:12)(cid:12) > . It is not difficult to check that, if Ω si is in r -contact with Ω sj , Ω sj is in r -contact with Ω si .If Ω si is not in contact with any other component of Ω s , we say that Ω si is r -isolated .Since our strategy will require to let the initial hyperplane vary, we will writeΩ = Ω ν,s ⊔ Ω ν,ns , where the additional superscript ν indicates the direction of the parallel movement,namely the normal to the initial hyperplane H (and the decomposition is always meantwith respect to the parallel hyperplane H T at the stopping time T defined in Step 2).The proof of Step 4 is achieved by showing the following claims: • Claim 4a. Given ν ∈ S d − , let Ω ♭ be a r -isolated open connected component of Ω ν,s .Then Ω ♭ is a ball of radius at least r/ , and Ω \ Ω ♭ is r -critical and not r -degenerate,unless it has measure zero. • Claim 4b. The following family is empty: F := [ ν ∈ S d − n open connected components not r -isolated of Ω ν,s o . • Claim 4c (conclusion). Ω is equivalent to a finite union of balls of radius R > r/ , atmutual distance larger than or equal to r .Proof of claim 4a . Given ν ∈ S d − , let Ω ♭ be a r -isolated open connected component ofΩ ν,s . Assume by a moment to know that(41) Ω ♭ is r -critical and not r -degenerate.In this case, we can restart our proof, with Ω ♭ in place of Ω. Given an arbitrary direction e ν ∈ S d − , we make the decompositionΩ ♭ = Ω ˜ ν,s♭ ⊔ Ω ˜ ν,ns♭ . We are going to show that, unless Ω ˜ ν,ns♭ is empty, this decomposition splits Ω ♭ into twoopen sets, contradicting the connectedness of Ω ♭ . Hence Ω ♭ is Steiner symmetric abouta hyperplane with unit normal e ν . By the arbitrariness of e ν , we deduce that Ω ♭ is aball. (Indeed, since Ω ♭ is a compact set, following [35], there exists a sequence of Steinersymmetrizations of it converging to a ball; but since Ω ♭ is already Steiner symmetric inevery direction, it must coincide with such ball). Since Ω ♭ is not r -degenerate, the radiusof the ball is strictly larger than r/ e ν,ns♭ is not empty, let us show that every point of Ω ♭ is in the interior ofone among the two sets Ω e ν,s♭ and Ω e ν,ns♭ . Let us denote by e T the stopping time definedas in Step 2 for the parallel movement with normal e ν . Recall from (18) that(42) ∂ ∗ Ω e ν,s♭ ∩ ∂ ∗ Ω e ν,ns♭ ⊆ H e T , Let us consider separately the cases when x ∈ Ω ♭ \ H e T and when x ∈ Ω ♭ ∩ H e T .Let x ∈ Ω ♭ \ e H . Since Ω ♭ is open, there exists a ball B ε ( x ) contained into Ω ♭ \ e H . Itcannot be 0 < | Ω e ν,s♭ ∩ B ε ( x ) | < | B ε ( x ) | . Otherwise, by Federer’s Theorem, B ε ( x ) wouldcontain points of ∂ ∗ Ω e ν,s♭ ∩ ∂ ∗ Ω e ν,ns♭ , against (42). We deduce that B ε ( x ) is contained eitherinto Ω e ν,s♭ or into Ω e ν,ns♭ , namely x is an interior point for one among Ω e ν,s♭ and Ω e ν,ns♭ .Let now x ∈ Ω ♭ ∩ e H , and let B ε ( x ) be a ball contained into Ω ♭ . By the same argumentsas above, each of the two sets B ε ( x ) ∩ ( H − e T \ H e T ) and B ε ( x ) ∩ ( H + e T \ H e T ) must be entirelycontained either into Ω e ν,s♭ or into Ω e ν,ns♭ . Recalling that Ω e ν,s♭ is Steiner symmetric about H e T , we infer that either both sets are contained into Ω e ν,s♭ , or both sets are containedinto Ω e ν,ns♭ . Then, also in this case x is an interior point for one among Ω e ν,s♭ and Ω e ν,ns♭ . To conclude the proof of Claim 4a., it remains to show that (41) holds true and that thesame property is valid for Ω \ Ω ♭ , unless it has measure zero. For the sake of clearness,this will be obtained as the final product of three consecutive lemmas. Lemma 15.
Given ν ∈ S d − , let Ω ♭ be a r -isolated open connected component of Ω ν,s .Then inf x ,x ∈ ∂ ∗ Ω ♭ | Ω ν,s ∩ (cid:0) B r ( x )∆ B r ( x ) (cid:1) |k x − x k > . Proof.
Assume by contradiction that(43) inf x ,x ∈ ∂ ∗ Ω ♭ | Ω ν,s ∩ (cid:0) B r ( x )∆ B r ( x ) (cid:1) |k x − x k = 0 . Then there exist sequences of distinct points { x ,n } , { x ,n } ⊂ ∂ ∗ Ω ♭ , with k x ,n − x ,n k →
0, such that | Ω ν,s ∩ (cid:0) B r ( x ,n )∆ B r ( x ,n ) (cid:1) |k x ,n − x ,n k → . Up to subsequences, we may assume that k x ,n − x ,n k converges to 0 decreasingly, andthat { x ,n } and { x ,n } converge to some point x ∈ ∂ ∗ Ω ♭ , which may belong or not to H T , being as usual T the stopping time defined as in Step 2 for the parallel movementwith normal ν . Let us examine the two cases separately.In case x H T , we may assume without loss of generality that { x ,n } , { x ,n } ⊂ H + T \ H T .Recall that, by (17), the set ∂ ∗ Ω ♭ ∩ ( H + T \ H T ) is connected. Hence for every n ≥ x ,n to x ,n +1 by a continuous arc γ ,n ( s ) contained into ∂ ∗ Ω ♭ ∩ ( H + T \ H T ).We can repeat the same procedure for the second sequence, constructing a family ofcontinuous arcs γ ,n ( s ) joining x ,n to x ,n +1 for every n ≥ r whose centre moves along γ ,n ( s ) and γ ,n ( s ). Clearly these balls tends to superpose in the limit as n → + ∞ , since k x ,n − x ,n k decreases to 0. Moreover, we know from (34) that, during the continuous movement oftheir centre along along γ ,n ( s ) and γ ,n ( s ), the boundary of these balls cannot crosspoints of density 1 for Ω ν,ns . This property will give us the required contradiction. Moreprecisely, we argue as follows. Since Ω is not r -degenerate, (43) impliesinf x ,x ∈ ∂ ∗ Ω ♭ | Ω ν,ns ∩ (cid:0) B r ( x )∆ B r ( x ) (cid:1) |k x − x k > . In particular, for n = 1, we have | Ω ν,ns ∩ (cid:0) B r ( x , )∆ B r ( x , ) (cid:1) | >
0. Hence we can picka point p ∈ int( B r ( x , )∆ B r ( x , )) of density 1 for Ω ν,ns , and a radius ε > | B ε ( p ) ∩ Ω ν,ns | ≥ | B ε ( p ) | . Possibly reducing ε we can also assume that B ε ( p ) ⊆ (cid:0) B r ( x , )∆ B r ( x , ) (cid:1) . Recallingthat the boundaries of the balls of radius r whose centre moves along the continuousarcs γ ,n ( s ) and γ ,n ( s ) cannot meet Ω ns , we infer that, for n large, B ε ( p ) ∩ Ω ν,ns ⊆ B r ( x ,n )∆ B r ( x ,n ) ; hence, still for n sufficiently large, | B ε ( p ) ∩ Ω ν,ns | ≤ | B r ( x ,n )∆ B r ( x ,n ) | < | B ε ( p ) | , against (44).In case x ∈ H T , we proceed in the same way, except that we cannot ensure any morethat both sequences { x ,n } and { x ,n } belong to the same halfspace H + T or H − T . Thus,when we construct the continuous arcs γ ,n and γ ,n , they may belong indistinctly to ∂ ∗ Ω ♭ ∩ ( H − T \ H T ) or to ∂ ∗ Ω ♭ ∩ ( H + T \ H T ), but this does not affect the validity of theproof since the contradiction follows as soon as x ,n and x ,n are close enough. (cid:3) Lemma 16.
Given ν ∈ S d − , let Ω ♭ be a r -isolated open connected component of Ω ν,s .There exists a constant c ♭ > such that (45) | Ω ν,s ∩ B r ( x ) | = c ♭ ∀ x ∈ ∂ ∗ Ω ♭ . Moreover, the constant is the same for any other open connected component of Ω ν,s suchthat the closure of its essential boundary intersects ∂ ∗ Ω ♭ .Proof. We argue in a similar way as in the proof of the previous lemma. Given x , x ∈ ∂ ∗ Ω ♭ ∩ ( H + T \ H T ), by (17), they can be joined by a continuous arc γ ( s ) contained into ∂ ∗ Ω ♭ ∩ ( H + T \ H T ). By (34), the boundary of the ball of radius r centred at any point along γ ( s ) cannot cross points of density 1 for Ω ν,ns . We deduce that B r ( x )∆ B r ( x ) cannotcontain points of density 1 for Ω ν,ns . Since Ω is r -critical, it follows that | Ω ν,s ∩ B r ( x ) | = | Ω ν,s ∩ B r ( x ) | . By the arbitrariness of x , x , we infer that there exists a constant c + ♭ > | Ω ν,s ∩ B r ( x ) | = c + ♭ for every x ∈ ∂ ∗ Ω ♭ ∩ ( H + T \ H T ). In the same way, weobtain that there exists a constant c − ♭ > | Ω ν,s ∩ B r ( x ) | = c − ♭ for every x ∈ ∂ ∗ Ω ♭ ∩ ( H − T \ H T ). Since the two sets ∂ ∗ Ω ♭ ∩ H ± T have common points on H T ,we conclude that c + ♭ = c − ♭ , The same argument proves also the last assertion of thelemma. (cid:3) Lemma 17.
Given ν ∈ S d − , let Ω ♭ be a r -isolated open connected component of Ω ν,s .Then Ω ♭ is and r -critical and not r -degenerate. The same assertions hold true for itscomplement Ω \ Ω ♭ , unless it is of measure zero.Proof. The fact that Ω ♭ is not r -degenerate follows from Lemma 15 and the assumptionthat Ω ♭ is r -isolated. From equality (45) in Lemma 16, combined with (34) and theassumption that Ω ♭ is r -isolated, we infer that there exists a positive constant c ♭ suchthat | Ω ♭ ∩ B r ( x ) | = c ♭ for every x ∈ ∂ ∗ Ω ♭ , namely Ω ♭ is r -critical. Let us now considerthe complement Ω \ Ω ♭ . Assume it is of positive measure, and hence that ∂ ∗ (Ω \ Ω ♭ ) is notempty. The fact that Ω \ Ω ♭ is not r -degenerate follows from the assumption that Ω itselfit is not, combined with the fact that points of Ω ♭ and of Ω \ Ω ♭ cannot lie at distancesmaller than r (again by (34) and the assumption that Ω ♭ is r -isolated). Finally, it holds | (Ω \ Ω ♭ ) ∩ B r ( x ) | = c − c ♭ for every x ∈ ∂ ∗ (Ω \ Ω ♭ ), namely Ω \ Ω ♭ is r -critical. (cid:3) Proof of Claim 4b.
As a preliminary remark, we observe that the family F is at mostcountable. This is an immediate consequence of the fact that any open set of R d has at most countable connected components, and of the fact that, for two different directions ν and ν , it is not possible that a connected component of Ω ν ,s intersects a connectedcomponent of Ω ν ,s without being equal.We now prove Claim 4b. by contradiction.First of all let us show that, if the family F is not empty, it contains an element Ω ♯ whichis Steiner symmetric about d hyperplanes with linearly independent normals ν , . . . , ν d .Indeed, let S k denote the family of linear subspaces of dimension k in R d . For every k = d − , d − , . . .
1, we are going to associate with a given subspace V ∈ S k an elementof F , which will be denoted by Ω kV . These mappings(46) S k ∋ V −→ Ω kV ∈ F are constructed as follows.For k = d −
2, given V ∈ S d − , we consider all the subspaces e V ∈ S d − containing V . Forevery such e V , denoting by e ν the normal direction to e V , we perform the decompositionΩ e ν,s ∪ Ω e ν,ns . Since F is at most countable, there exist two distinct subspaces e V and e V in S d − such that the corresponding symmetric parts Ω e ν ,s and Ω e ν ,s share some openconnected component. We pick one among such shared connected components and weassociate it with V , denoting it by Ω d − V . Notice that neither the spaces e V , e V nor theshared connected component are unique, so the definition is made by choice.For k = d −
3, given V ∈ S d − , we consider all subspaces e V ∈ S d − containing V . Sincethe image of the mapping in (46) previously defined for k = d − e V and e V in S d − such that Ω d − e V = Ω d − e V . We set(again by choice) Ω d − V := Ω d − e V = Ω d − e V . We continue the process until we define the map in (46) for k = 1. Arguing as above,we find two distinct e V and e V in S such that Ω e V = Ω e V . We setΩ ♯ := Ω e V = Ω e V . By construction Ω ♯ is Steiner symmetric with respect to d hyperplanes with independentnormals ν , . . . , ν d .Next we consider any other element Ω ♯♯ of F which is in r -contact with Ω ♯ in thedecomposition with respect to one among the directions ν , . . . , ν d , say ν . If T is thestopping time for the parallel movement with normal ν , there exist p, p ′ ∈ ∂ ∗ Ω ♯ \ H T ,symmetric about H T , such that (cid:12)(cid:12)(cid:0) B r ( p )∆ B r ( p ′ ) (cid:1) ∩ Ω ♯♯ (cid:12)(cid:12) > . Since we are assuming that Ω ♯♯ is Steiner symmetric with respect to H T , the aboveinequality implies that ∂B r ( p ) contains points of density 1 for Ω ♯♯ . In particular, thisimplies that Ω ♯♯ is itself Steiner symmetric about the same hyperplanes as Ω ♯ is.Then, Lemma 18 below implies that the set Ω ♯ ∪ Ω ♯♯ is connected, yielding a contradiction. Lemma 18.
Assume that ω ⊆ R d is a bounded open set, Steiner symmetric about d hy-perplanes whose normals are linearly independent. Then ω is a connected set containingits centre of mass. Proof.
Let us denote the hyperplanes by H , . . . , H d , and by Π H k ( x ) the orthogonalprojection from R d onto H k , for k = 1 , . . . , d . Starting from a fixed point x ∈ ω , let usconsider the sequence of points defined by x n := Π H k ( x n − ) if n = k [mod d ]. It is easyto check that { x n } converges to the centre of mass G of ω . In fact, let us assume withoutlosing generality that G is at the origin. If α n ∈ (0 , π ] is the angle between the normalsto the hyperplanes H n and H n +1 (obtained by cyclically repeating H , . . . , H d ), and d n is the distance of x n to H n ∩ H n +1 , we have that k x n +1 k = k x n k − d n sin ( α n ). Then k x n k is decreasing and { d n } converges to 0, since { α n } is a periodic sequence of strictlypositive numbers. This readily implies that dist( x n , H k ) →
0, for every k = 1 , . . . , d ,and hence x n → ω is open, there exists ε > B ε ( x ) ⊆ ω . Fromthe assumption that ω is Steiner symmetric about H , . . . H d , we get that B ε ( x n ) ⊆ ω for every n and, more in general, that(47) [ n ≥ ([ x n − , x n ] ⊕ B ε (0)) ⊆ ω. By (47), it turns out that B ε ( G ) is contained into ω . Moreover, ω is connected becausethe initial point x was arbitrarily chosen, and by (47) it can be joined to G by acontinuous path contained into ω . (cid:3) Proof of Claim 4c.
We start the procedure by choosing a direction ν ∈ S d − . By Claim4b., we can pick a r -isolated open connected component of Ω ν,s , which by Claim 4a.turns out to be a ball of radius R > r/
2. We remove this ball from Ω. By Claim 4a., weare left with a set Ω ′ which is still r -critical and not r -degenerate (unless it has measurezero). So we can restart the process with Ω ′ in place of Ω. Again, by Claim 4b., we canpick a r -isolated open connected component of (Ω ′ ) ν,s , which by Claim 4a. turns outto be a ball of radius R > r/
2. We remove this ball from Ω ′ . We observe that, sincethe two balls of radii R and R that we have extracted from Ω are are r -isolated and r -critical, necessarily R = R =: R , and the balls lie at distance larger than or equalto r from each other. Since Ω has finite measure, we can repeat this process a finitenumber of times, until when we are left with a set of measure zero. (cid:3) Remark . A generalized version of Theorem 1 is expected to hold when the kernel χ B r (0) is replaced by a radially symmetric, decreasing, non negative function h satis-fying suitable assumptions: any set Ω with finite measure satisfying the criticality andnondegeneracy conditions, meant as Z Ω h ( x − y ) dy = c ∀ x ∈ ∂ ∗ Ω and inf x ,x ∈ ∂ ∗ Ω R Ω | h ( x − y ) − h ( x − y ) | dy , k x − x k > , will be a finite union of balls or a single ball, depending on the boundedness of thesupport of h . Acknowledgments.
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