THE STEINER FORMULA FOR EROSIONS. then one shows in integral geometry that the volume of A ρk (ρ 0) is a polynomial of degree n:

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1 THE STEINER FORMULA FOR EROSIONS. G. MATHERON Abstract. If A and K are compact convex sets in R n, a Steiner-type formula is valid for the erosion of A by K if and only if A is open with respect to K. The if part is proved in the general case. The only if part is conjectured, and proved only in the case n = 2. Keywords: erosion, Steiner formula. If A and K are compact and convex in R n, and if denotes Minowsi sum B C = {b + c, b B, c C}, then one shows in integral geometry that the volume of A ρk (ρ 0) is a polynomial of degree n: V (A ρk) = n W (A, K)ρ. (1) The coefficients of this polynomial are the values in (A, K) of the mixed volume functionals of Minowsi. In particular, W 0 (A, K) = V (A) and W n (A, K) = V (K). More generally, one also has n p p W p (A ρk, K) = W +p (A, K)ρ, p = 0, 1,..., n. (2) These mixed functionals (in the two variables A and K) are read as usual functionals of a single variable (Quermassintegrale) by the intermediary of passage to means of rotation: if Ω denotes the group of rotations of R n and ω(dω) is the unique invariant probability measure on Ω, one has, in effect, the classical relation W (ωa, K) ω(dω) = 1 W (A)W n (K) (3) b n Ω where b n denotes the volume of the unit ball in R n. In particular, the mean volume of the rotation of A ρk is given by V (ωa ρk) ω(dω) = 1 n W (A)W n (K)ρ. (4) b n Ω In contrast, one does not generally have such simple results for Minowsi subtraction. Rather than Minowsi subtraction, it is interesting to consider the erosion of A by ρk, in Translation by Erin P. J. Pearse, May 20, Full reference for the original article appears at the end. 1

2 other words, the difference A ρǩ where Ǩ = K is the symmetric of K in relation to the origin. (In a general manner, B Č = {x C. x B}, where C x denotes the translate of C by x.) My objective, in what follows, is to examine conditions under which the formulas (1), (2), and (4) remain valid for erosions: A must be open by K, i.e., A = A K, with A K = (A Ǩ) K. This first result can be improved. Suppose that all the functionals W p, p = 1, 2,..., n, satisfy the relation in question. I mae a conjecture stronger than the converse: it suffices only to suppose that the volume of the erosion satisfies the formula for which A is open by K. Similarly, if one passes to means of rotation, one deduces from the first result a general demonstration of a conjecture of [Miles] (demonstrated by this author in the case n = 2) according to which the Steiner formula is applied to the mean of rotation of the erosion for which A and all its rotational transforms are open by K. In my turn, I put forth a new conjecture for which condition is not just sufficient, but also necessary. I demonstrate these two new conjectures in paragraph 3, but only for 2 dimensions. In the last paragraph, I give (always in R 2 ) the necessary and sufficient condition for the volume of the erosion to be a polynomial of second degree. 1. The formula (2) for erosions. We first give the principal result. Theorem 1. Let A and K be compact and convex in R n. One has: n p W p (A ρǩ, K) = p ( 1) ρ W +p (A, K) (2 ) for p = 0, 1,..., n and all ρ (0, 1) if and only if A is open by K. Proof. ( ) Suppose that A is open by K, so A = A 1 K with A 1 = A Ǩ. So for 0 r 1, one also has A rǩ = A 1 (1 r)k and, according to (2): n p W p (A rǩ, K) = p (1 r) W +p (A 1, K) = Q p (r), is a polynomial in r. By the same formula, for r 0 one also finds n p p W p (A rk, K) = r W +p (A, K) = P p (r). But one has A rk = A 1 (1 + r)k and thus again: n p p W p (A rk, K) = (1 + r) W +p (A 1, K) = Q p ( r). Consequently, Q p ( r) = P p (r) for r 0: as it is a question of polynomials, this also gives rise to Q p (r) = P p ( r) for r (0, 1), i.e., W p (A rǩ, K) = P p( r). 2

3 ( ) Conversely, suppose: W p (A Ǩ, K) = P p( 1) for p = 0, 1,..., n. Applying the formula (1) to the opening A K = (A Ǩ) K gives n n V (A K ) = W (A Ǩ, K) = P ( 1). An elementary calculation thus gives ( n ) P ( 1) = V (A). Lie A K A, the equality V (A K ) = V (A) brings forth the equality A K = A: A is open by K. Corollary (1). If A is open by K, for all ρ (0, 1) one has: n V (A ρǩ) = ( 1) W (A, K)ρ. (1 ) This corollary is trivial, but its converse is not. Below, we state a form of this conjecture for R n, and we prove only the case n = 2. Conjecture 1. (Strong converse). If (1 ) is true, then A is open by K. We will see just below that the volume of the erosion is differentiable in ρ with derivative: d dρ V (A ρǩ) = nw 1(A ρǩ, K), ρ (0, 1) (5) (provided that A Ǩ ). It follows that (2 ) hence (1 ) is verified for p = 1. For two dimensions, this suffices (since W 2 (A ρǩ, K) = V (K) trivially verifies (2 )), and Conjecture 2 is a simple corollary of the theorem. But it is no longer the same when n 3. In this regard to the means of rotation we obtain the result of [Miles] named in the conjecture (and proved by him in the case n = 2). Corollary (2). If A and all its rotations are open by K, then for all ρ (0, 1), one has: V (ωa ρǩ) ω(dω) = 1 n ( 1) W (A)W n (K)ρ. (4 ) b n Ω In effect, it suffices to apply (1 ) to ωa and to use the relation (3). Remar. It is easy to see that our condition (ωa open by K for all rotations ω Ω) is equivalent to that given by [Miles] (the sup of the radii of the curvature of K must be bounded by the inf of the radii of the curvature of A) or rather: there exists a ball Br of the ray r 0 such that A is open by Br and Br is open by K. We prove further the converse of the corollary in the case n = 2. For n > 2, we give a conjecture. 3

4 Conjecture 2. If (4 ) is true, then A and all its rotations are open by K. 2. Differentiability and convexity of V (A ρǩ) In the following, A and K always designate compact convex sets for which the erosion A Ǩ is not empty. As the many functionals studied here are invariant under translation, we may assume 0 K A, which lets us use the support functions h K and h A of these convex bodies. Our first objective is to establish the differentiability of V (A ρǩ), and the relation (5) used to prove the conjecture in the case n = 2. We first note an inequality. From A A ρk = (A ρǩ) ρk and the relation (1), we have V (A) V (A ρǩ) + nρw 1(A ρǩ, K) +... (taing into account the continuity of W 1 ) and by the following, for ρ tending to 0: V (A) V (A ρǩ) lim inf nw 1 (A, K). (6) ρ We use the following lemma (which we will prove in a moment). Lemma 1 (Convexity Lemma). For all ρ (0, 1), one has V (A ρk) V (A) V (A) V (A ρǩ). (7) According to this lemma, one obtains for ρ tending to 0: lim sup V (A) V (A ρǩ) ρ lim V (A ρk) V (A) nw 1 (A, K). ρ Taing (6) into account, this indicates that V (A ρǩ) admits a right derivative at ρ = 0 equal to nw 1 (A, K). Upon replacing A by A ρǩ, on finds immediately that for 0 ρ < 1 d + V (A ρǩ) dρ = nw 1 (A ρǩ, K). For 0 < ρ 1, analogous reasons show that the left derivative exists and again taes the value nw 1 (A ρǩ, K). This establishes differentiability and the relation (5). Moreover, this derivative is increasing (because W 1 (C, K) is increasing function of the convex body C) so that the function ρ V (A ρǩ) is convex on (0, ). If we put { V (A ρǩ), ρ 0, Φ(ρ) = V (A ρǩ), ρ < 0, this function Φ admits the same left and right derivative nw 1 (A ρǩ, K) at ρ = 0. As it is convex on the positive half-line, and also, after (1), on the negative half-line, this function is convex on the entire interval (, ). 4

5 Proof of the convexity lemma. We initially consider the case where A is a polyhedron (with nonempty interior). Let F i, i = 1,..., N be the faces of A, and S i the (n 1)-volume of F i, and let u i be the exterior unit normal vector of F i. If h K is the support function of K, one has: N V (A ρk) V (A) + ρ S i h K (u i ) and nw 1 (A, K) = i=1 N S i h K (u i ). We show that V (A ρǩ) has right derivative nw 1(A, K) at ρ = 0. Let x A be a point of A. One has x A Ǩ iff x F i ρǩ for one or more faces F i. But it is easy to see that the volume of A (F i ρǩ) has an upper bound of the form i=1 V ((F i ρǩ) A) ρ(s ih K (u i ) + ε i ) with ε i 0 if ρ 0. (h K is the support function of K). Therefore: N V (A) V (A ρǩ) V [ N (F i ρǩ) A] ρ V (S i h K (u i ) + ε i ) and lim sup i=1 V (A) V (A ρǩ) ρ i=1 N S i h K (u i ) = nw 1 (A, K). Taing into account (6), this implies the existence of a right derivative at ρ = 0. Just as this derivative is nw 1 (A, K), the function { V (A ρǩ), ρ 0, Φ(ρ) = V (A ρk), ρ < 0, similarly admits a derivative i=1 Φ (0) = nw 1 (A, K) at ρ = 0. For ρ > 0, A ρǩ is again a polyhedron (as an intersection of translates of A), and one can apply the same reasoning replacing A by A ρǩ. Whence one sees that Φ has, for all ρ (, 1), the derivative { Φ nw 1 (A ρǩ, K), ρ 0, (ρ) = nw 1 (A ρk, K), ρ < 0. This derivative is increasing so that Φ is convex. In particular, A verifies the relation (7). Now if A is not a polyhedron, consider a decreasing sequence of convex polyhedrons {A n } such that A = A n : one has A ρk = (A n ρk), A ρǩ = (A n ρǩ), 5

6 so that the relation (7), which holds for each A n, passes to the limit. The surface measure. When A is a convex polyhedron, one has for all compact convex K: nw 1 (A, K) = h K (u)g A (du), (8) where G A is the measure on the unit sphere defined by G A = N i=1 S iδ ui. Because the support functions h K of the compact convex sets K form a dense subspace of the continuous functions on the unit sphere (under uniform convergence), this relation (8) passes to the limit: for all compact convex sets A, there is a measure G A on the unit sphere such that (8) is valid and the function A G A is continuous. (If A n A for the Hausdorff metric, then G An converges vaguely to G A.) Moreover, if A is a polyhedron, the (n 1)-volumes S i (ρ) of the faces of the erosion A ρǩ are bounded above by the (n 1)-volumes S i of the corresponding faces of A. Then one also has G A ρ Ǩ G A. This relation similarly passes to the limit: for all compact convex A, the function ρ G A ρ Ǩ is decreasing. In particular, we see the following result. Lemma 2. In 2 dimensions, A is open by A Ǩ whenever A Ǩ. Proof. In effect, in R 2, a measure G on the unit circle is a perimeter measure of a compact convex (defined up to a translation) iff e iθ G(dθ) = 0 (condition of the closure of the contour). If A Ǩ, one has G A Ǩ G A by the above. Thus, the measure G = G A G A Ǩ is positive, and as it satisfies the conditions of closure, G is the perimeter measure of a compact convex set K 1. But G A = G A Ǩ + G K1 is equivalent to A + (A Ǩ) K 1. Thus A is open by A Ǩ. Remar. Lemma 2 is not true for n 3. Counterexample: in R 3, let A be a cone with a circular base, and let K be a segment b parallel to the base of A. Then A Ǩ is the intersection A A b, and one easily sees that A is not open by A A b. In R n, we have A 1 = A Ǩ, K 1 = A Ǎ1 = A (Ǎ K). It is easy to see that K 1 is the closure of K by the complement A c of A (i.e., the intersection of the translates of A which contain K): K 1 = K Ac = (K Ǎc ) A c = {A b. A b K}. So A Ǩ1 = A 1 (because A Ǩ1 is the closure of A 1 by A c, but A 1 is already closed by A c insofar as the intersection of the translates of A) and one has A K1 = A A1 = A 1 K 1 A. But the inclusion is strict in general, and it s only in the case n = 2 that the equality A = A 1 K 1 is guaranteed. 6

7 In R n, if A is open by K, let A = A 1 K and one necessarily has K 1 = K. In effect, A 1 K = A A 1 K 1 A 1 K follows from the equalities A 1 K 1 = A 1 K and K 1 = K. For n = 2, the converse is true: in R 2, A is open by K iff K 1 = K. In effect, according to lemma 2, in R 2 A is open by A 1 = A Ǩ. One has therefore A = A 1 C for the compact convex C which satisfies C = A Ǎ1 = K 1. Thus A = A 1 K 1. Consequently, if K 1 = K, one has A = A 1 K and A is open by K. 3. Demonstration of the Conjectures 1 and 2 for n = 2 In the case n = 2, one has an important inequality. Theorem 2. If A and K are compact and convex in R 2 and A Ǩ, one has: V (A ρǩ) V (A) 2ρW 1(A, K) + ρ 2 V (K) (9) for all ρ (0, 1), with equality iff A is open by ρk. Proof. In effect, by (5) one immediately has d dρ V (A ρǩ) = 2W 1(A ρǩ, K) and A A ρk = (A ρǩ) ρk follows immediately from (2): W 1 (A, K) W 1 (A ρǩ, K) + ρv (K). Consequently: d dρ V (A ρǩ) 2ρV (K) 2W 1(A, K) and by integrating from 0 to ρ 0 1: V (A ρ 0 Ǩ) V (A) 2ρ 0 W 1 (A, K) + ρ 2 0 V (K). If the equality is attained, one has d dρ V (A ρǩ) = 2ρV (K) 2W 1(A, K) for 0 ρ ρ 0, i.e., W 1 (A ρǩ) = W 1(A, K) ρv(k). Upon transferring these results into V (A ρ0 K) = V (A ρ 0 Ǩ) + 2ρ 0 W 1 (A ρ 0 Ǩ, K) + ρ 2 0 V (K) one thus finds V (A ρ0 K) = V (A), and consequently A ρ0 K = A is open by ρ 0 K. Corollary. The conjectures 1 and 2 are true in R 2. 7

8 Proof. For conjecture 1, this is immediate. Suppose we have V (ωa ρǩ) ω(dω) = V (A) (ρ/2π)s(a)s(k) + ρ2 V (K) Ω where S(A) = 2W 1 (A) and S(K) = 2W 1 (K) are the perimeters of A and K. inequality (9), this indicates that one has: V (ωa ρǩ) = V (A) 2ρW 1(A, K) + ρ 2 V (K), By the thus ωa is open by K, for ω-almost every rotation ω Ω. As the collection of compact convex sets open by K is closed (under the Hausdorff metric), this relation occurs, as a matter of fact, for all ω Ω, and conjecture 2 follows. At this stage, we put forth a third conjecture (from which the first and second conjectures would be immediate consequences). Conjecture 3. In R n, one has for all ρ (0, 1): n V (A ρǩ) ( 1) W (A, K)ρ with equality iff A is open by ρk. 4. Conditions for which V (A ρǩ) will be a polynomial When A isn t open by K, the function ρ V (A ρǩ) may be very complicated (for example, if one erodes a ball by a cube), but may, in certain cases, be a polynomial of degree n (by example, if one erodes a cube by a ball). In the case of 2 dimensions, one can obtain a precise result in this regard (which does not generalize to higher dimensions). In effect, for n = 2 we can put By Lemma 2, one has One can also write A ρ = A ρǩ and K ρ = A Ǎρ, ρ (0, 1). A = A ρ K ρ, and ρk K ρ ρk 1. A 1 = A Ǩ = (A ρǩ) (1 ρ)ǩ = A ρ (1 ρ)ǩ. By the same Lemma 2, it follows that A ρ is open by A 1, so let: with X ρ = A ρ Ǎ1. Whence A ρ = A 1 X ρ (10) A = A ρ K ρ = A 1 K ρ X ρ. As one also has A = A 1 K 1, it further follows that K 1 = K ρ X ρ. (10 ) 8

9 Therefore, K 1 is open by K ρ = A Ǎρ for all ρ (0, 1). As one also has ρk ρk 1, it follows that A ρǩ = A ρ A ρǩ1, thus V (A ρǩ) V (A ρǩ1) = V (A) 2ρW 1 (A, K 1 ) + ρ 2 V (K 1 ) (11) with equality when ρ = 0 and ρ = 1 (because A Ǩ = A Ǩ1). For ρ other than 0 or 1, the inequality (11) becomes an inequality iff A ρǩ = A ρǩ1, that is to say, if: A 1 X ρ = (A 1 K 1 ) ρǩ1 = A 1 (1 ρ)k 1, so finally also if X ρ = (1 ρ)k 1. But X ρ = (1 ρ)k 1,, by (10 ) is equivalent to K ρ = ρk 1. Whence a first result. Lemma 3. In R 2, the inequality (11) is true, and the equality is true for ρ (0, 1) iff K ρ = ρk 1. More generally, we have the following theorem. Theorem 3. In R 2, V (A ρǩ) is a polynomial of second degree in ρ iff K ρ = ρk 1 for every ρ (0, 1), and in this case one has { V (A ρǩ) = V (A) 2ρW (A, K 1 ) + ρ 2 V (K 1 ) W 1 (A, K 1 ) = W 1 (A, K). Proof. If K ρ = ρk 1, these relation follow from Lemma 3 and formula (5). Conversely, we suppose that V (A ρ ) is a second degree polynomial, V (A ρ ) = V (A) 2aρ + Hρ 2. Upon application of (5), we find a = W 1 (A, K) and W 1 (A ρ, K) = W 1 (A, K) ρh. But A = A ρ K ρ and W 1 (A, K) = W 1 (A ρ, K) + W 1 (K ρ, K). Consequently, In particular, for ρ = 1 it becomes So one can write for ρ = 1 ρh = W 1 (K ρ, K), ρ (0, 1). H = W 1 (K 1, K). V (A 1 ) = V (A) 2W 1 (A, K) + W 1 (K 1, K) = V (A) 2W 1 (A 1, K) W 1 (K 1, K) (seeing that A = A 1 K 1 ), since (by K K 1 ) we have V (A 1 ) V (A) 2W 1 (A 1, K 1 ) V (K 1 ) = V (A) 2W 1 (A, K 1 ) + V (K 1 ) = V (A 1 ). 9

10 It follows that the preceding inequalities are, in fact, equalities: W 1 (K 1, K) = V (K 1 ) W 1 (A 1, K) = W 1 (A 1, K 1 ) W 1 (A, K) = W 1 (A, K 1 ). Thus one has a = W 1 (A, K 1 ) and H = V (K 1 ). By Lemma 3, this follows from A ρ = A ρǩ1, and K ρ = ρk 1 for all ρ (0, 1). Remar. Theorem 3 is not true for n 3. For example, tae the unit ball in R 3 for A, and let K be a straight segment of unit length. Then for ρ (0, 1), one finds: V (A ρǩ) = π ( ρ + 12 ) ρ3. This is a polynomial of third degree. But the relation K ρ = ρk 1 is not satisfied. References [Math] Matheron, G., La Formule de Steiner pour les érosions, J. Appl. Prob. 15, pp (1978) [Miles] Miles, R. E., On the elimination of edge efects in planar sampling., in Stochastic Geometry. E. F. Harding and D. G. Kendall, eds., Wiley, London (1974) 10