Geometric and isoperimetric properties of sets of positive reach in E d

Geometric and isoperimetric properties of sets of positive reach in E d Andrea Colesanti and Paolo Manselli Abstract Some geometric facts concerning sets of reach R > 0 in the n dimensional Euclidean space are proved, and in particular we look for geometric properties that, as R, parallel corresponding properties of convex sets. Moreover, in the two dimensional case, an isoperimetric problem for sets as above is studied. AMS 2000 Subject Classification: 52A30. 1 Introduction Sets of positive reach were introduced by Federer in [2]. This class of sets can be viewed as an extension of that of convex sets. It is well known that every point x external to a closed convex set C in E d admits a unique projection on C, i.e. a point which minimizes the distance from x among all points in C. Sets of positive reach are those for which the projection is unique for the points of a parallel neighborhood of the set (and not necessarily for all external points). Along with their definition, Federer provided the main fundamental properties of sets of positive reach. Namely, the validity of global and local Steiner formulas and consequently the existence of curvature measures and many relevant properties of such measures. The study of properties of sets with positive reach has been continued by several authors and along various directions. Let us mention the contributions given by Zähle [9] and Rataj and Zähle [8] on integral representation of curvature measures, the results by Hug [4], and Hug and the first author [1] on singular points of sets with positive reach and the extensions of Steiner type formulas by Hug, Last and Weil [5]. Moreover, in [3] Fu proved several interesting connections between sets of positive reach and semi-convex functions. Further contributions are contained in [6] and [7]. As stated by Federer, closed convex sets represent a limit case of sets of positive reach, as the reach tends to. The following question was at the origin of the research carried out in this paper. Is it possible to see (at least some of) the geometric properties of convex sets as limit case of suitable geometric properties of sets of positive reach? The first property that we analyse is the very definition of convex set: if x 1 and x 2 belong to a convex set C, then the segment joining them is entirely contained in C. In 3 we prove a possible counterpart of this fact for sets of positive reach. For two points x 1 and x 2 in E d and R > 0 we denote by H(x 1, x 2, R) the intersection of all closed balls of 1

radius R containing x 1 and x 2. The set H(x 1, x 2, R) is a rugby ball-shaped set with cusps in x 1 and x 2 ; moreover for R, H(x 1, x 2, R) tends to the segment with endpoints x 1 and x 2. Theorem 3.8 states that reach(k) R if and only if for every x 1, x 2 K such that x 1 x 2 < 2R, H(x 1, x 2, R) K is connected. The proof of this result is geometric, in fact it is based only on the very definiton of reach of a set. As a corollary (see Theorem 3.10) we have the following fact: if reach(k) R > 0 and D is a closed ball of radius less than or equal to R, intersecting K, then reach(k D) R. The latter property can be seen as a counterpart, for sets with positive reach, of the well-known fact that the intersection of a convex set with an half-space is convex (if it is non-empty). We point out that Theorems 3.8 and 3.10 have been proved also by Rataj in [6] and [7], in a different way. Next, we consider the following problem: given a set A and a number R > 0 is it possible to find the minimal set (with respect to inclusion) containing A and having reach greater than or equal to R? The corresponding problem in the context of convexity (R = ) has an affirmative answer: every set admits a least convex cover, i.e. its convex hull. We will see through simple examples that this is not the case for arbitrary A and R and we will find necessary and sufficient conditions so that A admits a minimal cover of reach greater than or equal to R. In the final part of the paper we prove a result of isoperimetric type for sets with positive reach in the plane. More precisely, given a set K E 2 with reach(k) R > 0 and contained in a disk of radius g > 0, we find an explicit upper bound for the boundary measure of K in terms of R and g. The bound is sharp when g R. The paper is organized as follows: in 2 we introduce some notations; in 3 we prove Theorem 3.8 and some related results; in 4 we deal with the least cover with prescribed reach of a given set. In 5 we prove the isoperimetric inequality in the planar case. 2 Notations Let E d be the d-dimensional Euclidean space; for a, b E d, let b a be their distance and let (, ) denote the usual scalar product. If A is a subset of E d, then int(a), cl(a) and A c will denote the interior, the closure and the complement set of A, respectively. For x 0 E d and r > 0 we set B(x 0, r) = {x E d : x x 0 < r}, and D(x 0, r) = cl(b(x 0, r)). For A E d and a E d, the distance of a from A is given by δ A (a) = inf{ a x : x A}. Let us recall the definition of set of positive reach, introduced in [2]. Let K E d be closed; let Unp(K) be the set of points having a unique projection (or foot point) on K: Unp(K) := {a E d :! x K s.t. δ K (x) = a x }. This definition implies the existence of a projection mapping ξ K : Unp(K) K which assigns to x Unp(K) the unique point ξ K (x) K such that δ K (x) = x ξ K (x). For a point a K we set: reach(k, a) = sup{r > 0 : B(a, r) Unp(K)}. 2

The reach of K is then defined by: reach(k) = inf reach(k, a), a K and K is said to be of positive reach if reach(k) > 0. If K E d is compact and x K, the tangent and the normal spaces to K at a are: { Tan(K, a) = {0} u : ǫ > 0 b K s.t. 0 < b a < ǫ, b a b a u } u < ǫ, Nor(K, a) = {v : (u, v) 0, u Tan(K, a)}. Notice in particular that Nor(K, a) is a closed convex cone. Let reach(k) > 0; for a K we set: P a = {v : ξ K (a + v) = a}, Q a = {v : δ K (a + v) = v }. 3 Characterization and geometrical properties of sets with positive reach The following definition will be useful later. Definition 3.1 Let a, b E d, a b, R > 0, a b < 2R. Let We set D(a, b, R) = {D(x, R) : a x, b x R}. H(a, b, R) = D D(a,b,R) It is clear from the definition that H(a, b, R) is a compact convex set, containing a and b. The boundary of H(a, b, R) is obtained rotating an arc of circle of radius R joining a and b, about the line through a and b. Lemma 3.2 Let a, b E d be such that 0 < b a < 2R where R > 0. If c, d H(a, b, R), then H(c, d, R) H(a, b, R). Proof. If D D(a, b, R), then c, d D so that D D(c, d, R). The conclusion follows from Definition 3.1. A set is convex if and only if given any two points belonging to it, it contains the line segment joining them. In this section we prove (see Theorem 3.8) a characterization of sets of positive reach that somehow resembles the above characterization of convex sets. The proof of this result requires various lemmas. The next proposition is Theorem 4.8 (7) of [2]. Proposition 3.3 Let K E d be closed, x Unp(K) and reach(k, ξ K (x)) > 0. Then, for every b K (x ξ K (x), ξ K (x) b) ξ k(x) b 2 x ξ K (x). (1) 2 reach(k, ξ K (x)) 3 D.

Let R > 0 and a, b E d be such that 0 < a b < 2R. We define the cone { ( ) } v C(a, b, R) = v 0 : v, b a b a >. b a 2R A geometric version of the above proposition follows. Corollary 3.4 Let K be a closed subset of E d such that reach(k) R > 0. Let x Unp(K) \ K, a = ξ K (x) K and b K such that 0 < a b < 2R. Then x a / C(a, b, R). We proceed with some geometric considerations in the plane. Given v and w vectors in E 2, v, w 0, we set S(v, w) = {z : z = tv + τw, t, τ > 0}. Remark 3.5 Let R > 0 and z 1, z 2, z 3, z 4 E 2 be such that z 1 z 2 = z 2 z 3 = z 3 z 4 = z 4 z 1 = R, 0 < z 1 z 3 < 2R. We have C(z 1, z 3, R) = S(z 2 z 1, z 4 z 1 ). Lemma 3.6 Let R > 0, b 1, b 2 E 2 with 0 < b 1 b 2 < 2R, Γ j = B(b j, R), j = 1, 2, b 3, b 4 E 2 such that {b 3, b 4 } = Γ 1 Γ 2. Let (i) Σ Γ 1 be the closed arc joining b 3 and b 4, of smaller length; (ii) Σ Γ 1 be the closed arc having length πr and such that Σ Σ = {b 4 }. For every a B(b 4, R) \ D(b 3, R) there exist c Σ, c b 3, c b 4, and c Σ, uniquely determined, such that b 1 c = c a = a c = c b 1 = R. Σ c a b 4 b 1 c b 2 Σ b 5 b 3 Figure 1 4

Proof. We have a b 3 > R and a b 4 < R. Let us notice that b 3 and b 4 are the endpoints of Σ. By continuity, there exists c Σ such that a c = R. Let b 5 be the endpoint of Σ which does not coincide with b 4 ; we have b 4 b 5 = 2R and a b 5 + a b 4 b 4 b 5 = 2R; thus a b 5 2R a b 4 > R. By continuity, there exists c Σ such that a c = R. The points c and c are uniquely determined as intersection of Γ 1 and B(a, R). Lemma 3.7 Let R > 0, b 1, b 2 E 2, 0 < b 1 b 2 < 2R, B i = B(b i, R), Γ i = B i, i = 1, 2. Let b 3, b 4 be such that {b 3, b 4 } = Γ 1 Γ 2, B i = B(b i, R), i = 3, 4. Assume that a B 3 B 4 \ H(b 1, b 2, R) and c i, c i are such that b i c i = c i a = a c i = c i b i = R, for i = 1, 2, and let S i = S(c i a, c i a), for i = 1, 2. Then: S 1 S 2 S(b 2 a, b 1 a). (2) In particular 1 2 (b 1 + b 2 ) int(s 1 S 2 ). (3) c 1 a b 4 c 2 c 2 b 1 b 2 c 1 a b 3 Figure 2 Proof. S 1, S 2 and S(b 2 a, b 1 a) are open convex cones with apex in a; moreover b i a S i for i = 1, 2 so that {b 1 a, b 2 a} S 1 S 2. Let Σ 1 = Γ 1 D(b 2, R) and Σ 2 = Γ 2 D(b 1, R). By Lemma 3.6 we may assume that c i Σ i \ {b 3, b 4 } for i = 1, 2. This in turn implies c 1 c 2 < 2R (as c 1, c 2 H(b 3, b 4, R)). Hence it is uniquely determined a a such that {a, a } = B(c 1, R) B(c 2, R). The straight line through a and a bounds two open half-planes such that b 2 and c 1 (resp. b 1 and c 2 ) are in the same half-plane. Thus a a S 1 S 2. (4) This implies that S 1 S 2 is a convex cone and, since it contains b 1 and b 2, (2) follows. Theorem 3.8 If K E d is closed then reach(k) R > 0 if and only if for every b 1, b 2 K, b 1 b 2 < 2R, K H(b 1, b 2, R) is connected. 5

Proof. Let us assume that reach(k) R > 0. By contradiction, assume that K := K H(b 1, b 2, R) is not connected; then there exist K 1, K 2 K, closed, such that K = K 1 K 2 and K 1 K 2 =. By compactness, there exist c i K i for i = 1, 2 such that ρ := c 1 c 2 = inf{ x y : x K 1, y K 2 } > 0. As c 1, c 2 H(b 1, b 2, R), ρ R. We have B(c 1, ρ) B(c 2, ρ) K =. On the other hand it is easy to check that H(c 1, c 2, R) [B(c 1, ρ) B(c 2, ρ)] {c 1, c 2 }. By Lemma 3.2, H(c 1, c 2, R) H(b 1, b 2, R), so that H(c 1, c 2, R) K = {c 1, c 2 }. (5) In particular, c := c 1 + c 2 / K; as δ K (c) < R, c Unp(K)\K. Let c 3 = ξ K (c) K. Notice 2 that if c 3 H(c 1, c 2, R) then either c 3 = c 1 or c 3 = c 2 so that δ K (c) = c c 1 = c c 2 in contradiction with c Unp(K). Consequently, c 3 K \ H(c 1, c 2, R). We also observe that, for i = 1, 2, c i c 3 c i c + c c 3 < 2R as c c 3 = δ K (c) < R. We recall the definitions of the cones: { ( ) v C i (c 3, c i, R) = v 0 : v, c i c 3 > c } i c 3, i = 1, 2. c i c 3 2R By Corollary 3.4 we have that c c 3 / C 1 C 2. (6) Apply Remark 3.5 and Lemma 3.7 to the (uniquely determined) 2-dimensional plane containing c, c 1, c 2, c 3 to obtain a contradiction with (6). Vice versa, assume that for every b 1, b 2 K, b 1 b 2 < 2R, the set K H(b 1, b 2, R) is connected. If, by contradiction, reach(k) < R, then there exists x K c such that δ K (x) = r < R and x b 1 = x b 2 = r for some b 1, b 2 K, b 1 b 2. As b 1 b 2 < 2R, H(b 1, b 2, R) K is connected. On the other hand, r < R implies that H(b 1, b 2, R) B(x, r) {b 1, b 2 } so that there exists b K B(x, r) i.e. a contradiction. Remark 3.9 If reach(k) R > 0 and b 1, b 2 K are such that b 1 b 2 = 2R, then K H(b 1, b 2, R) is not necessarily connected. Any set consisting of two points at distance 2R is an example. Theorem 3.10 Let K be a closed set such that reach(k) R > 0. If D is a closed set such that for every b 1, b 2 D, H(b 1, b 2, R) D, then reach(k D) R. Proof. The argument is similar to the one used in the second part of the proof of Theorem 3.8. Let a (K D) c such that r = δ K D (a) < R; let us show that a Unp(K D). Assume by contradiction that there exist b 1, b 2 (K D) such that b 1 b 2 and a b 1 = a b 2 = r. 6

In particular b 1 b 2 < 2R. Clearly, H(b 1, b 2, R) D; consequently, by Theorem 3.8, H(b 1, b 2, R) (K D) is connected. Also, notice that (H(b 1, b 2, R) \ {b 1, b 2 }) B(a, r). Then there exists b K D such that a b < r, i.e. a contradiction. Corollary 3.11 If reach(k) R > 0, a, b K, a b 2R, then reach(k H(a, b, R)) R. It is well known that, if K is a closed convex set in E d and H is an open half space, satisfying H K =, then H K is either empty or a convex subset of H. Let us show that a similar property holds for sets of reach R > 0. Definition 3.12 Let S be a sphere of radius R > 0 in E d ; let K be a closed subset of S. We say that K is convex in S if x 1 K, x 2 K, dist(x 1, x 2 ) < 2R imply that the arc of great circle of S joining x 1 and x 2, and having smaller length, is contained in K. Theorem 3.13 Let K be a closed set in E d and reach(k) R > 0. Let B be an open ball of radius R satisfying B K =. Then B K is either empty or a convex subset of B. Proof. Theorem 3.10 implies that (B B) K = B K has reach R. Then, by theorem 3.8, if b 1, b 2 K B, b 1 b 2 < 2R, then K B H(b 1, b 2, R) is connected. Now K B H(b 1, b 2, R) is exactly the arc of great circle of B, joining b 1 and b 2 and having smaller length. 4 On the R-hull of a set Let A be a subset of E d and let R > 0. In this section we analyze the problem of finding K such that reach(k) R, K A and K is the minimal set (with respect to inclusion) having these properties. In other words we look for a sort of hull of reach R of A. Intuitively, when R = we are dealing with the convex hull of A which exists for every A. On the other hand, for finite R > 0 not every set A admits a hull of reach R (see the examples below). Our aim is to give necessary and sufficient conditions for A to have this property (see Theorems 4.4 and 4.6). Definition 4.1 Let A E d, R > 0. We say that A admits a R-hull if there exists  Ed such that: (i) A Â; (ii) reach(â) R; (iii) if reach(k) R and A K, then  K. If such a set exists, we call it the R-hull of A. 7

Example 1. For an arbitrary R > 0 we may construct an example of set which does not admit a R-hull. Let n = 2 and A = {a, b} with a b = R/2. Assume by contradiction that there exists the R-hull of A, and denote it by Â. Let Â1 be the closed line segment joining a and b: reach(â1) = so that Â1 Â. Let Γ be a circle of radius R passing through a and b and let Â2 Γ be the closed arc of smaller length joining a and b. We have reach(â2) = R so that Â2 Â. As Â1 Â2 = A, we must have  = A; on the other hand reach(a) = R/2 so we have a contradiction. Example 2. In E d consider a half-line L with endpoint in the origin. For every i = 1, 2,..., let a i be the point of L such that a i = 1/i. The set A = {a 1, a 2,... } does not admit a R-hull for any R (0, ). For an arbitrary set A E d and R > 0, we set A R = {x Ed : δ A (x) R}. The proof of the following proposition is an easy application of Theorem 3.8. Proposition 4.2 Let A E d, R > 0; reach(a R ) R if and only if for every a and b such that δ A (a), δ A (b) R and B(a, R) B(b, R), there exists a continuous arc Γ joining a and b, Γ H(a, b, R), such that δ A (x) R for every x Γ. Lemma 4.3 Let K E d, then (i) K (K R ) R {z Ed : δ K (z) < R}, (ii) if reach(k) R > 0 then reach(k R ) R and K = (K R ) R. Proof. If x K, then x y R for every y K R so that δ K R (x) R and x (K R ) R. On the other hand, if z (K R ) R then z / K R so that δ K(z) < R. Claim (i) is proved. For s 0 set K s = {x E d : δ K (x) s}. Corollary 4.9 in [2] implies that reach(k R 1/i ) R 1/i for every i = 1, 2,.... Moreover, the sequence K R 1/i converges to K R in the Hausdorff metric. On the other hand, the by Remark 4.14 in [2], for every ǫ > 0 the family {A E d : reach(a) R ǫ} is closed with respect to the Hausdorff metric. Then reach(k R ) R ǫ for every ǫ > 0. Now let us prove that if reach(k) R then (K R ) R \ K is empty. Let z (K R ) R \ K; (i) implies that z Unp(K). Let x = ξ K (z) and y t = x + t z x, t 0. Note that z x Nor(K, x) z x z x so that, by claim (12) of Theorem 4.8 of [2], if 0 < t < R, then δ K (y t ) = t and by continuity δ K (y R ) = R. Then y R K R and z y R < R, i.e. a contradiction. Theorem 4.4 Let A E d and R > 0. If reach(a R ) R then A admits R-hull  and  = (A R) R. 8

Proof. Let A 1 = (A R ) R ; we prove that A 1 is the R-hull of A. The inclusion A A 1 is part (i) in Lemma 4.3. By the same lemma, as reach(a R ) R we have reach(a 1) R. It remains to show that A 1 satisfies (iii) in Definition 4.1. Let K be such that K A and reach(k) R. Then K R A R and, by Lemma 4.3, K = (K r ) R (A R ) R = A 1. Corollary 4.5 Let A E d and R > 0. If for every a and b such that δ A (a), δ A (b) R and a b < 2R, there exists a continuous arc Γ, joining a and b such that δ A (x) R for every x Γ, Γ H(a, b, R), then A admits R-hull  and  = (A R) R. Theorem 4.6 Let K E d and R > 0. Assume that K admits R-hull ˆK. Then reach(k R ) R. Proof. We argue by contradiction: assume that reach(k r ) < R. By using Theorem 3.8, there exist b 1 and b 2 K R satisfying b 1 b 2 < 2R and such that H(b 1, b 2, R) K R is not connected. Then, as we saw in the proof of Theorem 3.8, there exist c 1 and c 2 K R such that c 1 c 2 < 2R and H(c 1, c 2, R) K R = {c 1, c 2 }. (7) For j = 1, 2 we have reach(b(c j, R) c ) = R and B(c j, R) c K thus B(c j, R) c ˆK. This implies in particular that c 1, c 2 ( ˆK) R. As reach( ˆK) R, by Lemma 4.3, reach[( ˆK) R ] R, then H(c 1, c 2, R) ˆK R is connected. Let a [H(c 1, c 2, R) \ {c 1, c 2 }] ( ˆK) R. We have B a (R) K B a (R) ˆK = then a K R which contradicts (7). From the above theorem another connection between convex sets and sets of positive reach can be deduced. The convex hull of a closed set C is the intersection of all the closed half-spaces containing C. Let us prove that if K admits R-hull ˆK, then ˆK is the intersection of the complement sets of all open balls that do not meet K. Note that for an arbitrary, non-empty, subset K of E d we have (K R) R = B x (R) c. δ K (x) R This remark and Theorem 4.6 lead to the following result. Corollary 4.7 Let K E d, R 0. Assume that K admits an R-hull ˆK. Then ˆK = B x (R) c. δ K (x) R The following result is a sufficient condition for a set to admit a R-hull. Theorem 4.8 Let A E 2 be a connected subset of B x0 (R), R > 0. Then A admits R-hull. Proof. We argue by contradiction. As in Theorem 4.6, there exists c 1 and c 2 A R such that: H(c 1, c 2, R) A R = {c 1, c 2 }. 9

Thus every x H(c 1, c 2, R) \ {c 1, c 2 } satisfies δ A (x) < R. Let {c 3, c 4 } = D R (c 1 ) D R (c 2 ); the above property of H(c 1, c 2, R) \ {c 1, c 2 } implies that there must be a j A (D R (c 1 ) D R (c 2 )) c close to c j+2 (j = 1, 2). On the other hand no connected subset of E 2 exists containing a 1 and a 2 and contained in an open disk of radius R. Remark 4.9 If A E 2 is a connected subset of D R (x 0 ), it may not admit an R-hull. Let 0 < b 1 b 2 > 2R, A = D R (b 1 ) \ B R (b 2 ). Assume that A admits R-hull Â. A is closed and connected, and A D R (B 1 ) and the sets A 1 = D R (b 1 ) \ B R (b 2 ) and A 2 = D R (b 1 ) have reach greater than or equal to R. Thus A 1, A 2 Â. We deduce that A = Â but this contradicts the fact that reach(a) < R. 5 An isoperimetric inequality for sets with positive reach in the plane For R, g > 0 let us introduce the set K (R) g = {K E 2 : reach(k) R, K D g }, where D g is a closed disk of radius g. We consider the following variational problem S (R) g Here M(K) is the Minkowski content of K, i.e. = sup{m(k) : K K (R) g }. M(K) = lim r 0 + K r K r and stands for the two-dimensional Lebesgue measure. The condition that K has positive reach implies that M(K) is well defined, as K obeys a Steiner formula (see [2]). Note that by virtue of Remark 5.10 in [2], S g (R) <. The aim of this section is to provide an explicit upper bound for S g (R), which is in fact its exact value for g R, in terms of R and g. Theorem 5.1 If, in the above notations, 0 < g R, then S (R) g = max{m(k) : K K (R) g } = 2πg. In particular, if g < R then D g is the unique maximizer, while if R = g there are infinitely many maximizers (and D g is one of them). In the general case the situation is less neat and we can only prove upper and lower bounds for S (R) g. Theorem 5.2 In the above notations we have π 2 R ( g 2 3) 3 R 2 S (R) g 4π2 R ( g ) 2 3 3 R + 3. (8) 10

The proof of Theorem 5.1 requires some preliminary steps. We start with a simple observation. Remark 5.3 Let a 1, a 2 E 2 be such that a 2 a 1 < 2R. For arbitrary x, y H(a 1, a 2, R) denote by Γ x,y the length of the shortest arc of a circle of radius R joining x and y. Then for every a H(a 1, a 2, R) Γ a1,a 2 Γ a1,a + Γ a,a2. (9) This inequality can be equivalently written in the form: ( ) ( ) ( ) a1 a 2 a1 a a a2 2R arcsin 2R arcsin + 2R arcsin. (10) 2R 2R 2R Lemma 5.4 Let K E 2 with reach(k) R > 0. Consider x 1, x 2 K such that x 1 x 2 = 2l < 2R and assume that K is contained in one of the half planes bounded by the straight line r through x 1 and x 2. Set K 1 = K H(x 1, x 2, R). Then every straight line which intersects orthogonally the segment joining x 1 and x 2, meets K 1 in at most two points. Proof. Let s be a straight line straight line which intersects orthogonally the segment joining x 1 and x 2 and let y 1, y 2, y 3 s K 1 ; assume that y 3 is contained in the segment with endpoints y 1 and y 2. Then H(x i, y j, R) K 1 is connected, for every i, j = 1, 2, moreover, y 3 / H(x i, y j, R) for i, j = 1, 2. Consequently, the connected component of the complement set of 2 i,j=1(h(x i, y j, R) K 1 ) which contains y 3 is contained in K 1. Then y 3 is an interior point of K 1 and this is a contradiction. In the notation of the previous lemma, let us fix an orthogonal coordinates frame Oξη with origin O in x 1 + x 2 and ξ-axes coinciding with the line through x 1 and x 2, oriented 2 toward x 2. Let Θ = {(ξ, η) : ξ l, η = max{y : (ξ, y) K 1 }}. (11) Lemma 5.5 In the assumptions of Lemma 5.4 and with the notation introduced before, if K is of class C 1, Θ is the graph of a C 1 function defined in [ l, l] and ( ) H 1 x1 x 2 (Θ) 2 arcsin. (12) 2R Proof. From Lemma 5.4 we have that Θ is a subarc of K 1, and consequently is of class C 1. Set K 2 = {(ξ, y) : ξ l, y η for every η such that (ξ, η) Θ}. (13) Assume by contradiction that (12) is not true; then for some Q > 1 we have l ( ) x1 b 1 + (η (ξ)) 2 2 dξ 2Q arcsin. 2R l Using Remark 5.3 and a standard bisection argument, we can find two sequences of points, k N, such that for every k ξ k and ξ k ξ k ξ k = 1 ( ξ k 1 2 ) ξ k 1 11

and ξ k ξ k ( ) bk a 1 + (η (ξ)) 2 k dξ 2QR arcsin, (14) 2R where a k = (ξ k, η(ξ k )), b k = (ξ k, η(ξ k )). Notice that in particular this implies a k, b k H(a k 1, b k 1, R) H(a k 2, b k 2, R) H(a 1, b 1, R), k. Without loss of generality, we may assume that ξ k converges to some ξ as k tends to infinity, so that we also have ξ k ξ and a k, b k c = ( ξ, η( ξ)) H(a 1, b 1, R). Now it is easy to see that ( ) 1 bk a k lim k ξ k 2R arcsin = 1 + (η ξ k 2R ( ξ)) 2. Hence, dividing both terms of (14) by (ξ k ξ k ) and letting k tend to infinity we obtain a contradiction. Consider a compact convex set C, with non-empty interior and with boundary of class C 1 and assume that C is contained in the ball D g centered at the origin and with radius g. For each point p C the normal cone Nor(C, x) consists of a half-line. In particular it is uniquely determined the point α(x) = (x + Nor(C, x)) D g. The map α : C D g is a homeomorphism. Moreover, its inverse is the restriction to D g of the metric projection on C. As the metric projection is non-expansive, i.e. is Lipschitzian, with Lipschitz constant smaller than or equal to one, this implies in particular the following fact. Lemma 5.6 In the above notations, for every Borel subset η of C we have that H 1 (η) H 1 (α(η)). Proof of Theorem 5.1. We first prove that we can reduce to sets with C 1 boundary. Let ǫ > 0, K (ǫ) = K D g ǫ (0) and (K (ǫ) ) ǫ/2 = {x R 2 : δ K (ǫ) ǫ/2}, where δ K (ǫ) is the distance from K (ǫ). We have that and consequently, using [2, Corollary 4.9], reach(k (ǫ) ) R reach((k (ǫ) ) ǫ/2 ) R ǫ 2. Moreover ((K (ǫ) ) ǫ/2 ) converges to K in the Hausdorff metric as ǫ 0 + and, for every fixed ǫ > 0, its boundary is of class C 1,1 (see [2]). Consequently, by Theorem 4.10 in [2], lim ǫ 0 + M((K(ǫ) ) ǫ/2 ) = M(K). 12

Hence, if the claim of the present theorem holds for ((K (ǫ) ) ǫ/2 ), i.e. ( M((K (ǫ) ) ǫ/2 ) 2π R ǫ ), 2 then letting ǫ tend to 0 + we get that the same is valid for K. From now on we assume that K is of class C 1 ; moreover let C be the convex hull of K and α be the map defined (on C) as in Lemma 5.6. The boundary of C can be decomposed as follows: ( ) C = ( C K) γ i, where {γ i } is a countable family of segments with end-points belonging to K. The above union is disjoint. Associated to each γ i there is a sub-arc Γ i of K having the same end-points as γ i and such that ( ) K = ( C K) Γ i. In particular H 1 ( K) H 1 ( K C) + i For each i, let γ i = α(γ i ) B g. By Lemma 5.5, H 1 (Γ i ) H 1 ( γ i ). On the other hand, by elementary geometric considerations we have H 1 ( γ i ) H 1 (α(γ i )). This bound is sharp if g < R. Hence by Lemma 5.6 i i H 1 (Γ i ). (15) H 1 ( K) H 1 (α( K C)) + i H 1 (α(γ i )) = H 1 (α( C)) = H 1 ( B g ) = 2πg. When g = R it is easy to check that there are infinitely many maximizers. For instance, let t (0, R) and let D t be a closed disk with radius R and center at distance t from the center of D g. Then D g \ D t is a maximizer. Remark 5.7 Let R > 0; recall that K (g) R is the collection of sets of reach at least R, contained in a closed disk of radius g. By Theorem 5.1 we have However for every ǫ > 0 S (R) R = max{m(k) : K K(R) R } = 2πR. S (R) R+ǫ 4π(R + ǫ). Indeed, let K η = D R+ǫ η \ B R+ǫ 2η where 0 < η < ǫ/2; we have that reach(k η ) R and M(K η ) = 4π(R + ǫ) 6πη. The claim follows as η can be arbitrarily small. 13

Proof of Theorem 5.2. Let z 1 = ( 3R, 0), z 2 = (3R/2, 3R/2) so that (0, 0), z 1 and z 2 are the vertices of an equilateral triangle of side length 3R. Consider the set of points Z = {hz 1 + kz 2 h, k Z} and let D = {D R (z) z Z}. The union of the elements of D covers the entire plane; in particular we have that K = {K D R (z) z D D g }. Moreover, if D D then reach(d K) R and, by Theorem 5.1, M(K D) 2πR. Thus, by the subadditivity of the Minkowski content M(K) 2πf(g), where f(g) is the cardinality of Z D g. In order to get an upper bound for f(g), notice that each z D D g is the lower left corner of a parallelogram, with sides of length 3R parallel to the straight lines though the origin and z 1 and z 2 respectively. These parallelograms have disjoint interiors; moreover each of them is contained in D g+3r. From the these facts it follows the inequality f(g) 2π ( g ) 2 3 3 R + 3, which concludes the proof of the right inequality in (8). In order to prove the left inequality we proceed as follows. Let b = (2R, 0), c = (R, 3R) and let T be the equilateral triangle with vertices a = (0, 0), b and c. We set C = T \ (B R (a) B R (b) B R (c)). Note that reach(c) = R and M(C) = πr. Now let K = {C + hb + kc h, k Z, C + hb + kc D g }. We have K D g, reach(k) R and M(K) 2πRf 1 (g) where f 1 (g) is the cardinality of the set {hb + kc h, k Z, hb + kc 2 3R}. Arguing as above we find the following lower bound f 1 (g) π ( g 2 3) 2 3 R 2 which concludes the proof. References [1] A. Colesanti and D. Hug, Steiner type formulae and weighted measures of singularities for semi-convex functions, Trans. Amer. Math. Soc. 352 (2000), 3239 3263 (electronic). [2] H. Federer, Curvature measures, Trans. Amer. Math. Soc. 93 (1959), 418 481. 14

[3] J. H. Fu, Tubular neighborhoods in Euclidean spaces, Duke Math. J. 52 (1985), 1025 1046. [4] D. Hug, Generalized curvature measures and singularities of sets with positive reach, Forum Math. 10 (1998), 699 728.. [5] D. Hug, G. Last and W. Weil, A local Steiner-type formula for general closed sets and applications, Math. Z. 246 (2004), 237 272. [6] J. Rataj Determination of spherical area measures by means of dilation volumes, Math. Nach. 235 (2002), 143 162. [7] J. Rataj On estimation of the Euler number by projections of thin slabs, Adv. Appl. Prob. 36 (2004), 715 724. [8] J. Rataj and M. Zähle, Mixed curvature measures for sets of positive reach and a translative integral formula, Geom. Dedicata 57 (1995), 259 283. [9] M. Zḧale, Integral and current representation of Federer s curvature measures, Arch. Math. (Basel) 46 (1986), 557 567. Andrea Colesanti, Dipartimento di Matematica U. Dini, Viale Morgagni 67/a, 50134 Firenze, Italy. E-mail: colesant@math.unifi.it Paolo Manselli, Dipartimento di Matematica e Applicazioni per l Architettura, via dell Agnolo 14, 50122 Firenze, Italy. E-mail: manselli@unifi.it 15