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RANDOM POLYTOPES AND AFFINE SURFACE AREA

arXiv:math/9302210v1 [math.MG] 4 Feb 1993RANDOM POLYTOPES AND AFFINE SURFACE AREACarsten Sch¨uttAbstract. Let K be a convex body in Rd.

A random polytope is the convex hull[x1, ..., xn] of finitely many points chosen at random in K. E(K, n) is the expectationof the volume of a random polytope of n randomly chosen points. I. B´ar´any showedthat we have for convex bodies with C3 boundary and everywhere positive curvaturec(d) limn→∞vold(K) −E(K, n)( vold(K)n)2d+1=Z∂Kκ(x)1d+1 dµ(x)where κ(x) denotes the Gauß-Kronecker curvature.

We show that the same for-mula holds for all convex bodies if κ(x) denotes the generalized Gauß-Kroneckercurvature.1991 Mathematics Subject Classification. 52A22.Supported by NSF-grant DMS-9301506Tt bAMS T X

2CARSTEN SCH¨UTT1. IntroductionLet K be a convex body in Rd.

A random polytope in K is the convex hull offinitely many points in K that are chosen at random with respect to a probabilitymeasure on K. Here we consider the normalized Lebesgue measure on K. For a fixednumber n of points we are interested in the expectation of the volume of that partof K that is not contained in the convex hull [x1, ....., xn] of the chosen points. WedenoteE(K, n) =ZK×···×Kvold([x1, ..., xn])dP(x1, ...xn)where P is the n-fold product of the normalized Lebesgue measure on K. We areinterested in the asymptotic behavior ofvold(K) −E(K, n) =ZK×···×Kvold(K \ [x1, ...., xn])dP(x1, ..., xn)In [R-S1, R-S2] the asymptotic behavior of this expression has been determinedfor polygons and smooth convex bodies in R2.Theorem 1.

Let K be a convex body in Rd. Then we havec(d) limn→∞vold(K) −E(K, n)( vold(K)n)2d+1=Z∂Kκ(x)1d+1 dµ(x)where κ(x) is the generalized Gauß-Kronecker curvature andc(d) = 2(vold−1(Bd−12)d + 1)2d+1(d + 3)(d + 1)!

(d2 + d + 2)(d2 + 1)Γ( d2+1d+1 )This problem was posed by Schneider and Wieacker [Schn-W] and Schneider[Schn]. It has been solved by B´ar´any [B] for convex bodies with C3 boundary andeverywhere positive curvature.

Our result holds for arbitrary convex bodies.The main ingredients of the proof are taken from [B-L],[B], and [Sch¨u-W 1].We introduce the notion of generalized Gauß-Kronecker curvature. A convexfunction f : X →R, X ⊆Rd is called twice differentiable at x0 in a generalizedsense if there are a linear map d2f(x0) ∈L(Rd) and a neighborhood U(x0) so thatwe have for all x ∈U(x0) and all subdifferentials df(x)∥df(x) −df(x0) −(d2f(x0))(x −x0) ∥≤Θ(∥x −x0 ∥) ∥x −x0 ∥where limt→0 Θ(t) = Θ(0) = 0 and where Θ is a montone function.

d2f(x0) issymmetric and positive semidefinite. If f(0)=0 and df(0)=0 then the ellipsoid orelliptical cylinder

RANDOM POLYTOPES AND AFFINE SURFACE AREA3xtd2f(0)x = 1is called the indicatrix of Dupin at 0. The general case is reduced to the casef(0)=0 and df(0)=0 by an affine transform.

The eigenvalues of d2f(0) are called thepricipal curvatures and their product the Gauß-Kronecker curvature κ(0). Alek-sandrov [A, Ba] proved that a convex surface is almost everywhere differentiablein the generalized sense.

As surface measure on ∂K we take the restriction of the(d-1)-dimensional Hausdorffmeasure to ∂K. For x ∈∂K the normal at x to ∂K isdenoted by N(x).

N(x) is almost everywhere unique. We denoteKt = {x ∈K|vold((−x + K) ∩(x −K)) ≥t}for t ∈[0, T] withT = maxy∈K vold((−y + K) ∩(y −K))Kt is a convex body and was studied and used in [St, F-R, K, Schm1] and wascalled convolution body in [K,Schm1].

It was shown in [St, F-R] that KT consistsof one point only. Therefore we may also interpret KT - in abuse of notation - as apoint.

For a given x ∈∂K there is a unique point xt ∈∂Kt that is the intersectionof the interval [KT , x] and ∂Kt.Pξ denotes the orthogonal projection onto the hyperplane orthogonal to ξ andpassing through the origin. Bd2(x, r) is the Euclidean ball in Rd with center x andradius r. Bd2 is the ball with center 0 and radius 1.

H(x, ξ) denotes the hyperplanethrough x and orthogonal to ξ. For a given hyperplane H the closed halfspaces aredenoted by H+ and H−.

Usually H+ is the halfspace containing KT if we considera convex body K.2. Outline Of The ProofWe outline the proof of Theorem 1.

We have thatvold(K) −E(K, n) =ZKP{(x1, ..., xn)|x /∈[x1, ..., xn]}dx= −Z T0ddtZKtP{x1, ..., xn)|x /∈[x1, ..., xn]}dxdtThe derivative can be computed and we getZ T0Z∂KtP{(x1, ..., xn) | x /∈[x1, ..., xn]}vold−1(PN(x)((−x + K) ∩(x −K)))dµt(x)dtwhere µt is the surface measure on ∂Kt. We pass to an integral on ∂K insteadof ∂Kt.ZZ TP{(x1, ...., xn) | xt /∈[x1, ...., xn]}vol(P(( x + K) ∩(xK)))∥xt∥d∥x∥d< x, N(x) >< xN(x ) >dtdµ(x)

4CARSTEN SCH¨UTTwhere xt is the unique element on ∂Kt that is on the line through KT and x.Since P{(x1, ..., xn)|xt /∈[x1, ..., xn]} is concentrated for large n near the boundary∂K we getlimn→∞vold(K) −E(K, n)( vold(K)n)2d+1=limn→∞(nvold(K))2d+1Z∂KZlog nn0P{(x1, ...., xn) | xt /∈[x1, ...., xn]}vold−1(PN(xt)((−xt + K) ∩(xt −K)))∥xt∥d∥x∥d< x, N(x) >< xt, N(xt) >dtdµThen we apply Lebesgue’s convergence theorem and obtainZ∂Klimn→∞(nvold(K))2d+1Zlog nn0P{(x1, ...., xn) | xt /∈[x1, ...., xn]}vold−1(PN(xt)((−xt + K) ∩(xt −K)))∥xt∥d∥x∥d< x, N(x) >< xt, N(xt) >dtdµThe hypothesis of Lebesgue’s convergence theorem is fulfilled since the followingfunction dominates the functions under the integral: For every x ∈∂K let r(x) bethe largest radius so that(1)Bd2(x −r(x)N(x), r(x)) ⊆Kr(x) may be 0, e.g. if N(x) is not unique.

The functions under the integral areuniformly smaller than a constant timesr(x)−d−1d+1which is integrable on ∂K. Then we show that the expression under the integralconverges to r(x)1/(d+1) times an appropriate constant.3.

Proof Of Theorem 1Lemma 2. Let K be a convex body in Rd, f a continuous function on K , andt ∈(0, T).

Then we haveddtZKtf(x)dx = −Z∂Ktf(x)vold−1(PN(x)((−x + K) ∩(x −K)))dµt(x)where µt is the surface measure on ∂Kt.For f(x) identical to 1 this is an unpublished result of Schmuckenschl¨ager [Schm2]. A similar result for convex floating bodies instead of convolution bodies can befound in [Sch¨u-W2].

RANDOM POLYTOPES AND AFFINE SURFACE AREA5Lemma 3. Let K be a convex body in Rd and Kt, t ∈[0, T], the convolution bodies.Then we have for all t0 ∈[0, T]vold(K) −E(K, n) =Z∂KZ t00P{(x1, ...., xn) | xt /∈[x1, ...., xn]}vold−1(PN(xt)((−xt + K) ∩(xt −K)))∥xt∥d∥x∥d< x, N(x) >< xt, N(xt) >dtdµ(x)+Z Tt0Z∂KtP{(x1, ..., xn) | x /∈[x1, ..., xn]}vold−1(PN(x)((−x + K) ∩(x −K)))dµt(x)dtwhere µ and µt are the surface measures on ∂K and ∂Kt respectively and {xt} =∂K ∩[KT , x].Proof.vold(K) −E(K, n) =ZK×...×Kvold(K \ [x1, ..., xn])dP=ZK×...×KZKχK\[x1,...,xn]dxdP=ZKZK×...×KχK\[x1,...,xn]dPdx=ZKP{(x1, ..., xn)|x /∈[x1, ..., xn]}dxSinceZKtP{(x1, ..., xn)|x /∈[x1, ..., xn]}dxis a bounded, continuous, decreasing function on [0, T] it is absolutely continuous.We getvold(K) −E(K, n) = −Z T0ddtZKtP{(x1, ..., xn)|x /∈[x1, ..., xn]}dxdtSince P{(x1, ..., xn)|x /∈[x1, ..., xn]} is a continuous function of x we get byLemma 2vold(K) −E(K, n) =Z T0Z∂KtP{(x1, ..., xn)|x /∈[x1, ..., xn]}vold−1(PN(x)((−x + K) ∩(x −K)))dµt(x)dt□

6CARSTEN SCH¨UTTLemma 4. Let cap(r,∆) be a cap of height ∆of a d-dimensional Euclidean spherewith radius r. Then we have2(2 −∆r )d−12 vold−1(Bd−12)d + 1∆d+12 rd−12≤vold(cap(r, ∆)) ≤2d+12 vold−1(Bd−12)d + 1∆d+12 rd−12Lemma 5.

Let K be a convex body in Rd. Then there are constants c, c′ > 0 so thatwe have for all x ∈∂K and for all r > 0 with Bd2(x −rN(x), r) ⊆Kvold−1(PN(xt)((−xt + K) ∩(xt −K))) ≥(c(tr)d−1d+1if 0 ≤t ≤c′rdctd−1dif c′rd ≤t ≤Twhere {xt} = ∂Kt ∩[x, KT].Proof.

We may assume that KT coincides with the origin. By convexity and thefact that KT is an interior point we get that there is c1 > 0 so that we have for allx ∈∂K(2)≥c1Now we choose(3)c =vold−1(Bd−12)(d + 1)vold(Bd2)min{(1 −q1 −c21)d+12 , (1 −r34)d+12 }Then we have for all t ∈[0, crdvold(Bd2)](4)∥x −xt ∥≤r (5)1 ≤| < N(x), N(xt) > |

RANDOM POLYTOPES AND AFFINE SURFACE AREA7Figure 1Geometrically (4) means the following: Let z be the midpoint of the interval[KT , x] ∩Bd2(x −rN(x), r). Then xt ∈[z, x] (see figure 1).We verify (4).

Since Bd2(x −rN(x), r) ⊆K we havevold((−z +K) ∩(z −K)) ≥vold((−z +Bd2(x −rN(x), r)) ∩(z −Bd2(x −rN(x), r)))The last expression equals twice the volume of a cap whose height is greater than∆= r(1 −q1 −c21)By Lemma 4 and (3) we getvold((−z + K) ∩(z −K)) ≥4(2 −∆r )d−12 vold−1(Bd−12)d + 1∆d+12 rd−12≥4crdvol (Bd) ≥4t > t

8CARSTEN SCH¨UTTThus z is an interior point of Kt and xt ∈[z, x]. Moreover, ∥z −x ∥= r .Now we check (5).

We get by (4) and figure 1 that xt has to be in the shadedarea of figure 2.Figure 2Assume that (5) is not true.Then it follows that the radius of the sphereH(xt, N(xt)) ∩Bd2(x −rN(x), r) is greater than r and that H(xt, N(xt)) cuts offa cap of height greater than (r(1 −q34)). By Lemma 4 we get that the volume ofthis cap is greater than2(2 −∆r )d−12 vold−1(Bd−12)d + 1rd(1 −r34)d+12Therefore we have for the center w of the sphere Bd2(x−rN(x), r)∩H(xt, N(xt))thatvold((−w + K) ∩(w −K)) ≥2vold(Bd2(x −rN(x), r) ∩H−(xt, N(xt)))≥4vold−1(Bd−12)d + 1(1 −r34)d+12 rd ≥4t > tThis means that w is an interior point of Kt.

On the other hand, z is an elementof the supporting hyperplane H(xt, N(xt)) to Kt. This gives a contradiction andwe conclude that (5) is valid

RANDOM POLYTOPES AND AFFINE SURFACE AREA9Figure 3We denote Θ = arccos().¿From figure 3 it follows that the distance of xt to the boundary of Bd2(x −rN(x), r) equals(cos(Θ) −sin(Θ) cot(π2 −α2 )) ∥x −xt ∥By figure 1 we have α ≤π2 −Θ so that the above expression is larger than(6)(cos(Θ) −sin(Θ) cot(π4 + Θ2 )) ∥x −xt ∥Please note that by (2) there is ǫ > 0 so that we have for all x ∈∂K(7)cos(Θ) −sin(Θ) cot(π + Θ) ≥ǫ

10CARSTEN SCH¨UTTWe assume now that(8)∥x −xt ∥≥1ǫt2d+1rd−1d+1 (d + 1vold−1(Bd−12))2d+1Then(−xt + K) ∩(xt −K) ⊇(−xt + Bd2(x −rN(x), r)) ∩(xt −Bd2(x −rN(x), r))has a volume greater than twice the volume of a cap of a Euclidean ball of radiusr and height (6). By Lemma 4, (6), and (7) we get as abovevold((−xt + K) ∩(xt −K)) ≥4twhich cannot be true.Therefore (8) does not hold.We deduce that the distance between the twoparallel hyperplanes H(x, N(x)) and H(2xt −x, −N(x)) is less than twice the righthand expression of (8).

Moreover,(−xt + K) ∩(xt −K) ⊆H+(x, N(x)) ∩H+(2xt −x, −N(x))where both half spaces are chosen so that xt is contained in them. Thereforethere must be a hyperplane H parallel to H(x, N(x)) so thatvold−1((−xt + K) ∩(xt −K) ∩H)≥ǫ2(rt)d−1d+1 (vold−1(Bd−12)d + 1)2d+1By (5) we get that the same inequality holds forPN(xt)((−xt + K) ∩(xt −K))with another constant.Now we consider the casecrdvold(Bd2) ≤t ≤TSince K is compact there are r1, r2 > 0 so that(9)Bd2(KT , r1) ⊆K ⊆Bd2(KT , r2)Suppose that(10)vold−1(PN(xt)((−xt + K) ∩(xt −K)))) ≤ctd−1dwith c

RANDOM POLYTOPES AND AFFINE SURFACE AREA11vold−1(((−xt + K) ∩(xt −K)) ∩H(xt, N(xt)) ≤ctd−1dand since (−xt + K) ∩(xt −K) is symmetric with respect to xt all other d −1dimensional sections of (−xt +K)∩(xt −K) that are parallel to H(xt, N(xt)) havea smaller d −1 dimensional volume. Therefore there must be a non-empty sectionwhose distance to H(xt, N(xt)) is at least12ct1d .

This means that there is z ∈Kwith (see figure 4)(11)d(z, H(xt, N(xt)) ≥12ct1dLet y be the unique point in the intersection[KT , z] ∩H(xt, N(xt))We show that y is an interior point of Kt which contradicts the fact that y ∈H(xt, N(xt)). The sphere[z, Bd2(KT , r1)] ∩H(y,z −y∥z −y ∥)has a radius larger than12cr1r2t1dThis follows from (11).

Thus we find that[z, Bd2(KT , r1)] ⊇Bd2(y, 14cr1r2t1d )and as above that y is an interior point of Kt.□Lemma 6. [Sch¨u-W1] Let K be a convex body in Rd, α ∈(0, 1) and r(x) as definedby (1).

Then we haveZ∂Kr(x)−αdµ(x) < ∞where µ is the surface measure on ∂K.Lemma 7. [B-L] Let K be a convex body in Rd and let x ∈∂Kt.

Then we haveP{(x1, ..., xn)|xt /∈[x1, ..., xn]} ≤2d−1Xi=0ni(t2vold(K))i(1 −t2vold(K))n−i

12CARSTEN SCH¨UTTFigure 4Lemma 8. Let K be a convex body in Rd, t0 ∈(0, T] and µt the surface measureon ∂Kt.

Then we havelimn→∞n2d+1Z Tt0Z∂KtP{(x1, ..., xn)|x /∈[x1, ..., xn]}vold−1(PN(x)((−x + K) ∩(x −K)))dµt(x)dt = 0Proof. Since t ∈[t0, T] and t0 > 0 there is a constant c > 0 so that we have forall x ∈∂K

RANDOM POLYTOPES AND AFFINE SURFACE AREA13vold−1(PN(x)((−x + K) ∩(x −K)) ≥cBy Lemma 7 we getn2d+1Z Tt0Z∂KtP{(x1, ..., xn)|x /∈[x1, ..., xn]}vold−1(PN(x)((−x + K) ∩(x −K)))dµt(x)dt≤2c n2d+1d−1Xi=0ni Z Tt0Z∂Kt(t2vold(K))i(1 −t2vold(K))n−idµt(x)dtSince vold−1(∂Kt) ≤vold−1(∂K) we get that the last expression is smaller than2c n2d+1 vold−1(∂K)d−1Xi=0ni Z Tt0(t2vold(K))i(1 −t2vold(K))n−idt≤4c n2d+1 vold−1(∂K)vold(K)d−1Xi=0niΓ(i + 1)Γ(n −i + 1)Γ(n + 2)= 4c vold−1(∂K)vold(K)n2d+1dn + 1≤4c vold−1(∂K)vold(K)dn−d−1d+1□Lemma 9. Let K be a convex body in Rd, let KT be the origin, and 0 ≤t1 ≤t2 < T.Then there is a constant c > 0 so that we have for all n ∈NZ t2t1P{(x1, ...., xn) | xt /∈[x1, ...., xn]}vold−1(PN(xt)((−xt + K) ∩(xt −K)))∥xt∥d∥x∥d< x, N(x) >< xt, N(xt) >dt(12)≤cr(x)−d−1d+1Zt22vold(K)t12vold(K)d−1Xi=0nisi−d−1d+1 (1 −s)n−idswhere r(x) is defined by (1).Proof.

We have ∥xt ∥≤∥x ∥and we have a constant c > 0 so that we have forall x ∈∂K < xt, N(xt) >≥c because t2 < T. Thus it is enough to estimateZ t2t1P{(x1, ...., xn) | xt /∈[x1, ...., xn]}vold−1(PN(xt)((−xt + K) ∩(xt −K)))dtBy Lemma 5 this is smaller than

14CARSTEN SCH¨UTTZ c′rdt1c(r(x)t)−d−1d+1 P{(x1, ..., xn)|xt /∈[x1, ..., xn]}dt+Z t2c′rd ct−d−1d P{(x1, ..., xn)|xt /∈[x1, ..., xn]}dtSince we have for t ∈[c′rd, t2] thatt−d−1d= t−d−1d(d+1) t−d−1d+1 ≤(c′r(x))−d−1d(d+1) t−d−1d+1we can estimate the above expression bycr(x)−d−1d+1Z t2t1t−d−1d+1 P{(x1, ..., xn)|xt /∈[x1, ..., xn]}dtwhere c is a new constant. Now it is left to apply Lemma 7.□Lemma 10.

Let K be a convex body in Rd, let KT be the origin, and let t1 < T.Then there is a constant c so that we have for all x ∈∂K and all n ∈NZ t10P{(x1, ...., xn) | xt /∈[x1, ...., xn]}vold−1(PN(xt)((−xt + K) ∩(xt −K)))∥xt∥d∥x∥d< x, N(x) >< xt, N(xt) >dt ≤cr(x)−d−1d+1Proof. By Lemma 9 we getZ t10P{(x1, ...., xn) | xt /∈[x1, ...., xn]}vold−1(PN(xt)((−xt + K) ∩(xt −K)))∥xt∥d∥x∥d< x, N(x) >< xt, N(xt) >dt≤cr(x)−d−1d+1 n2d+1d−1Xi=0niΓ(i + 1 −d−1d+1)Γ(n + 1 −i)Γ(n + 2 −d−1d+1)Sincelimk→∞Γ(k +2d+1)Γ(k)k−2d+1 = 1we can estimate the last expression bycr(x)−d−1d+1where c is a new constant that does not depend on n and x.□

RANDOM POLYTOPES AND AFFINE SURFACE AREA15Lemma 11. We havelimn→∞n2d+1Z 1log nnd−1Xi=0nisi−d−1d+1 (1 −s)n−ids = 0Lemma 12.

Let K be a convex body in Rd, let KT be the origin, and let t1 < T.Then we have for all x ∈∂K with r(x) > 0limn→∞n2d+1Z t1log nnP{(x1, ...., xn) | xt /∈[x1, ...., xn]}vold−1(PN(xt)((−xt + K) ∩(xt −K)))∥xt∥d∥x∥d< x, N(x) >< xt, N(xt) >dt = 0Proof. The result follows from Lemmata 9 and 11.□Lemma 13.

[Wie]limn→∞n2d+1 (vold(Bd2(0, r)) −E(Bd2(0, r), n)) =(d2 + d + 2)(d2 + 1)2(d + 3)(d + 1)! ((d + 1)vold(Bd2)vold−1(Bd−12))2d+1 Γ(d2 + 1d + 1 )vold−1(∂Bd2(0, r))By an affine transform we can change the indicatrix of Dupin into a Euclideansphere or a cylinder with a sphere as its base.Lemma 14.

Let K be a convex body in Rd with 0 ∈∂K and N(0) = (0, ..., 0, −1).Suppose that ∂K is twice differentiable in the generalized sense at 0. (i) If the indicatrix of Dupin at 0 is a d-2 dimensional sphere with radius√ρ, thenthere is a t0 > 0 and a monotone , increasing function ψ on R+ with limt→0 ψ(t) =ψ(0) = 1 so that we have for all t ∈(0, t0]{( x1ψ(t), ..., xd−1ψ(t) , t)|x ∈Bd2((0, ..., 0, ρ), ρ) and xn = t}⊆K ∩H(−tN(0), N(0))⊆{(ψ(t)x1, ..., ψ(t)xd−1, t)|x ∈Bd2((0, ..., 0, ρ), ρ) and xn = t}(ii) If the indicatrix of Dupin at 0 is a d-2 dimensional cylinder with radius√ρ,i e

16CARSTEN SCH¨UTTRk−1 × ∂Bd−k2(0,√ρ)then there is a function Φ on R+ so that for every ǫ > 0 there is a t0 > 0 so thatlimt→0√tΦ(t) = 0 and√tΦ(t) is increasing on R+ and so that we have for all t ∈(0, t0]{(y, x, t)|(x, t) ∈Bd−k+12((0, ..., 0, ρ −ǫ), ρ −ǫ) and y ∈[−Φ(t), Φ(t)]k−1}⊆K ∩H(−tN(0), N(0))Lemma 15. Let K be a convex body in Rd, c > 2, and x ∈∂K such that κ(x) > 0.Then there is tc > 0 so that we have for all t ∈(0, tc]: We have for the hyperplaneH whose normal coincides with N(x) and that satisfies vold(K ∩H−) = cd+1t andfor all n ∈N with n ≥vold(K)ct|P{(x1, ..., xn)|xt /∈[{x1, ..., xn} ∩H−]} −P{(x1, ..., xn)|xt /∈[x1, ..., xn]}|< 2d−1e−c1cwhere c1 is a constant that depends on d only.Proof.

We show thatP{(x1, ..., xn)|xt /∈[{x1, ..., xn} ∩H−]} and xt ∈[x1, ..., xn]} ≤2d−1e−c1cIf we havext /∈[{x1, ..., xn} ∩H−]} and xt ∈[x1, ..., xn]then there is y ∈H+ ∩K so that(13)[y, xt] ∩[{x1, ..., xn} ∩H−] = ∅For the following argument let us assume that N(x) = e1 and xt = 0. Moreover,since κ(x) > 0 we may assume that the indicatrix of Dupin at x is a Euclideansphere and by Lemma 14 ∂K can be approximated arbitrary well at x by a sphereif we choose the height of the cap or correspondingly tc sufficiently small.Weassume for the following arguments that K ∩H−is a cap of a sphere.

Later weshall see that we control the error by choosing tc sufficiently small. We considerthe following sets (figure 5)

RANDOM POLYTOPES AND AFFINE SURFACE AREA17Figure 5cornΘ = K ∩H−∩H+(xt, N(x)) ∩{d\i=2H−(xt, e1 + Θiλei)}where Θ2, ..., Θd = ±1. We have 2d−1 sets and they are best described as cornersets.

λ is chosen so thatH−(xt, e1 + Θiλei) ∩H+consists of exactly one point.By the height h1 of cornΘ we understand the minimal distance of H(xt, N(x))and a parallel hyperplane so that cornΘ lies between them. We getvold(cornΘ) ≥2−d+1 h1d vold−1(K ∩H(xt, N(x))Let h2 denote the height of the cap K ∩H−(xt, N(x)) and h3 the height ofK ∩H−.

By Lemma 4 we get that2cd+1 =vold(K ∩H−)vold(K ∩H−(xt, N(x))) ≤2d+12 (h3h2)d+12orh2c2 ≤2h3The height h1 is of the order h3d . Altogether we get(14)vol (corn ) ≥2−d+1c c2h vol(K ∩H(xN(x)) ≥c c2t

18CARSTEN SCH¨UTTwhere c2 is a constant depending only on d. We have by (13){(x1, ..., xn)|xt /∈[{x1, ..., xn} ∩H−] and xt ∈[x1, ..., xn]} ⊆{(x1, ..., xn)|∃Hxt : xt ∈Hxt, H−xt ∩K ∩H+ ̸= ∅and H+xt ⊃{x1, ..., xn} ∩H−}Indeed, by the theorem of Hahn-Banach there is a hyperplane Hxt separatingthe convex sets [{x1, ..., xn} ∩H−] and the ray {xt + λ(y −xt)|λ ∈R}. By (13)they are disjoint.

We may assume that at least one point of the ray is an elementof Hxt. So xt is also an element of Hxt.

Let H−xt be the halfspace containing y,then H−xt ∩K ∩H+ contains y and is not empty.Such a halfspace always contains one of the corner sets cornΘ. This follows sincewe have in Rd for a hyperplane passing through the origin: The correspondinghalfspaces contain at least one 2d-tant.

Therefore we get{(x1, ..., xn)|∃Hxt : H−xt ∩K ∩H+ ̸= ∅and H+xt ⊃{x1, ..., xn} ∩H−}⊆[Θ{(x1, ..., xn)|{x1, ..., xn} ⊆K \ cornΘ}And consequently we get by (14)P{(x1, ..., xn)|xt /∈[{x1, ..., xn} ∩H−] and xt ∈[x1, ..., xn]}≤2d−1P{(x1, ..., xn)|{x1, ..., xn} ⊂K \ cornΘ}≤2d−1(1 −c2c2tvold(K))n ≤2d−1 exp(−n c2c2tvold(K))By the assumption on n we get that the last expression is smaller than2d−1 exp(−c2c)This argument also works if the volumes of the considered sets differ by a smallerror. Therefore the proof also goes through if K is not a sphere but can be ap-proximated arbitrarily well by a sphere at the point x.□Lemma 17.

Let K be a convex body in Rd and B a Euclidean ball of the samevolume. Let x ∈∂K and z ∈∂B and assume that κ(x) > 0.

Then for every ǫ > 0there is tǫ > 0 so that we have for all t ∈(0, tǫ] and all n ∈N with n ≥2d(15)|P {(xx )|x ∈[xx ]}P {(zz )|z ∈[zz ]}| < ǫ

RANDOM POLYTOPES AND AFFINE SURFACE AREA19Proof. We show first that there is c > 1 so that we have (15) whenever n ≤vold(K)ctor n ≥c vold(K)t. As c we can choose a number satisfyingc ≥max{1ǫ , d}(16)(d4cd+2)d+1e−c2 < ǫ2d−1e−c1c < ǫwhere c1 is the constant introduced in Lemma 15.We consider the case n ≤vold(K)ct.

Since the curvature at x is strictly positivewe may assume that the indicatrix of Dupin is a sphere.If we choose tǫ smallenough then by Lemma 14 there is a hyperplane H through xt,0 < t ≤tǫ , so thatvold(K ∩H−) ≤t. Therefore we get1 ≥PK{(x1, ..., xn)|xt /∈[x1, ..., xn]} ≥PK{(x1, ..., xn)|{x1, ..., xn} ⊆K ∩H+}≥(1 −tvold(K))n ≥(1 −1cn)n ≥1 −1cThe same estimate holds for PB and we get (15).

Now we consider the casen ≥2c vold(K)t. By Lemma 7 we get0 ≤P{(x1, ..., xn)|xt /∈[x1, ..., xn]}≤2d−1Xi=0ni(t2vold(K))i(1 −t2vold(K))n−iThe function si(1 −s)n−i attains its maximum atin. Since i < d, d ≤c, and2d ≤n we get that the latter expression is less thand−1Xi=0ni( cn)i(1 −cn)n−i ≤dcdexp(−c2) ≤ǫThe same holds for PB and we get (15) again.Now we consider the case vold(K)ct≤n ≤c vold(K)t. By triangle inequality we get|PK{(x1, ..., xn)|xt /∈[x1, ..., xn]} −PB{(z1, ..., zn)|zt /∈[z1, ..., zn]}| ≤|P {(xx )|x /∈[xx ]}P {(xx )|x /∈[{xx } ∩H−]}|

20CARSTEN SCH¨UTT+|PK{(x1, ..., xn)|xt /∈[{x1, ..., xn}∩H−]}−PB{(z1, ..., zn)|zt /∈[{z1, ..., zn}∩˜H−]}|+|PB{(z1, ..., zn)|zt /∈[z1, ..., zn]} −PB{(z1, ..., zn)|zt /∈[{z1, ..., zn} ∩˜H−]}|where H and ˜H are hyperplanes whose normals coincide with N(x) and N(z)respectively and vold(K ∩H−) = vold(B ∩˜H−) = cd+1t.The first and thirdsummands of the latter expression can be estimated by Lemma 15. We estimatenow the second summand.

Again, we may assume that the indicatrix of Dupinat x ∈∂K is a Euclidean sphere. Moreover, we may assume that the radius ofthe indicatrix equals the radius of B.

This is done by a volume preserving, affinetransform. We havePK{(x1, ..., xn)|xt /∈[{x1, ..., xn} ∩H−]} =(17)nXk=onk(vold(K ∩H−)vold(K))k(1−vold(K ∩H−)vold(K))n−kPK∩H−{(x1, ..., xk)|xt /∈[x1, ..., xk]}and the same for PB.

We get by Lemma 7 for k ≥4dcd+2PK∩H−{(x1, ..., xk)|xt /∈[x1, ..., xk]}≤2d−1Xi=0ki(12cd+1 )i(1 −12cd+1 )k−i ≤2dkde−14 kc−d−1The function sde−as attains its maximum atda.Therefore, and because of4dcd+2 ≤k the last expression is smaller than2d(d4cd+2)de−dc < ǫWe get the same for PB. Therefore we have|PK{(x1, ..., xn)|xt /∈[{x1, ..., xn} ∩H−]} −PB{(z1, ..., zn)|zt /∈[{z1, ..., zn} ∩˜H−]}|≤X0≤k≤4dcd+2nk(vold(K ∩H−)vold(K))k(1 −vold(K ∩H−)vold(K))n−k|P{(xx )|x /∈[xx ]}P{(zz )|z /∈[zz ]}| + 2ǫ

RANDOM POLYTOPES AND AFFINE SURFACE AREA21On the other hand, if we choose tǫ sufficiently small we have for all t ∈(0, tǫ]and all k, 0 ≤k ≤4dcd+2(18)|PK∩H−{(x1, ..., xk)|xt /∈[x1, ..., xk]} −PB∩˜H−{(z1, ..., zk)|zt /∈[z1, ..., zk]}| < ǫThis finishes the proof. We establish now (18).

For k = 0, ..., d the difference istrivially 0. Now we assume that x = z and N(x) = N(z).

We have that(19)PK∩H−{(x1, ..., xk)|xt /∈[{x1, ..., xk} ∩B ∩˜H−]} =kXm=0 km(vold(K ∩H−∩B ∩˜H−)vold(K ∩H−))m(1 −vold(K ∩H−∩B ∩˜H−)vold(K ∩H−))k−mPK∩H−∩B∩˜H−{(x1, ..., xm)|xt /∈[x1, ..., xm]}If we choose tǫ small enough we have by Lemma 14 for all t ∈(0, tǫ] that1 −vold(K ∩H−∩B ∩˜H−)vold(K ∩H−)is so small that we get by (19) and k ≤4dcd+2(20)|PK∩H−{(x1, ..., xk)|xt /∈[{x1, ..., xk} ∩B ∩˜H−]} ≤PK∩H−∩B∩˜H−{(x1, ..., xk)|xt /∈[x1, ..., xk]} + e−16dcMoreover, we havePK∩H−∩B∩˜H−{(x1, ..., xk)|xt /∈[x1, ..., xk]}(21)≥PK∩H−∩B∩˜H−{(x1, ..., xk)|{x1, ..., xn} ⊂K ∩H+(xt, N(x))}≥(1 −c−d−1)4dcd+2 ≥e−8dcThe last inequality holds because we have 1 −1s ≥e−2s for s ≥2. By (20) and(21) we get nowPK∩H−{(x1, ..., xk)|xt /∈[{x1, ..., xk} ∩B ∩˜H−]}≤(1 + exp(−c))PK∩H−∩B∩˜H−{(x1, ..., xk)|xt /∈[x1, ..., xk]}Therefore we get now

22CARSTEN SCH¨UTTPK∩H−{(x1, ..., xk)|xt /∈[x1, ..., xk]} ≤PK∩H−{(x1, ..., xk)|xt /∈[{x1, ..., xk} ∩B ∩˜H−]}≤(1 + e−c)PK∩H−∩B∩˜H−{(x1, ..., xk)|xt /∈[x1, ..., xk]}≤(1 + e−c)(vold(K ∩H−)vold(K ∩H−∩B ∩˜H−))kPK∩H−{(x1, ..., xk)|xt /∈[x1, ..., xk] and {x1, ..., xk} ⊂B ∩˜H−}≤(1 + e−c)(vold(K ∩H−)vold(K ∩H−∩B ∩˜H−))kPK∩H−{(x1, ..., xk)|xt /∈[x1, ..., xk]}Thus we get that|PK∩H−{(x1, ..., xk)|xt /∈[x1, ..., xk]}−PK∩H−∩B∩˜H−{(x1, ..., xk)|xt /∈[x1, ..., xk]}| < ǫif we choose c sufficiently big. We have the same inequality for PB∩˜H−.

Thisimplies (18).□Lemma 18. Let K be a convex body in Rd and x ∈∂K.

Suppose that ∂K is twicedifferentiable at x in the generalized sense. Then we have(i) limt→0< x, N(x) >< xt, N(xt) > = 1(ii)limt→0td−1d+1vold−1(PN(xt)((−xt + K) ∩(xt −K))) = κ(x)1d+1 (2d + 1)d−1d+1 vold−1(Bd−12)−2d+1Proof.

(i) The same arguments as in the proof of Lemma 5 are applied. We justsketch the argument.

Suppose (i) is not true. Then we find a supporting hyperplaneH(xt, N(xt)) so that xt is very close to x but N(xt) is not close to N(x).

By theassumption we have that all the points in the set H(xt, N(xt)) ∩K do not belongto the interior of Kt. On the other hand, the volume vold(K ∩H−(xt, N(xt)) is sobig that we can single out a point in H(xt, N(xt)) ∩K that is in the interior of Kt.

(ii) We consider the case κ(x) > 0. The case κ(x) = 0 is treated in an analogousway By Lemma 14 K can be approximated by an ellipsoid We may assume it is

RANDOM POLYTOPES AND AFFINE SURFACE AREA23a Euclidean sphere. By (i) < xt, N(xt) > is as close to < x, N(x) > as we chooseit to be for small t. Altogether we have thatvold−1(PN(xt((−xt + K) ∩(xt −K)))is up to some error equal tovold−1(H(xt, N(x)) ∩K)orvold−1(H(xt, N(x)) ∩Bd2(x −κ(x)−1d−1 N(x), κ(x)−1d−1 )It is left to apply Lemma 4.□Proof of Theorem 1.

We may assume that KT coincides with the origin. ByLemma 3 and 8 we havelimn→∞vold(K) −E(K, n)( 1n)2d+1=limn→∞n2d+1Z∂KZT20P{(x1, ...., xn) | xt /∈[x1, ...., xn]}vold−1(PN(xt)((−xt + K) ∩(xt −K)))∥xt∥d∥x∥d< x, N(x) >< xt, N(xt) >dtdµprovided the limit exists.

We apply now Lebesgue’s convergence theorem in orderto change limit and integration over ∂K. The hypothesis of Lebesgue’s theorem isfulfilled because of Lemma 6 and 10.

By Lemma 12 we get that the latter expressionequalsZ∂Klimn→∞n2d+1Zlog nn0P{(x1, ...., xn) | xt /∈[x1, ...., xn]}vold−1(PN(xt)((−xt + K) ∩(xt −K)))∥xt∥d∥x∥d< x, N(x) >< xt, N(xt) >dtdµBy Lemma 18 this expression equalsZ∂Klimn→∞n2d+1Zlog nn0P{(x1, ...., xn) | xt /∈[x1, ...., xn]}vold−1(PN(xt)((−xt + K) ∩(xt −K)))dtdµBy Lemma 18 (ii) we getZ∂Kκ(x)1d+1 dµ limn→∞n2d+1(2d+1)d−1d+1vold−1(Bd−12)2d+1Zlog nn0P{(x1, ..., xn)|xt /∈[x1, ..., xn]}td−1d+1dtBy Lemma 17 we have for x ∈∂K with κ(x) > 0lim n2d+1Zlog nnt−d−1d+1 PK{(x1, ..., xn)|xt /∈[x1, ..., xn]}dt

24CARSTEN SCH¨UTT= limn→∞n2d+1Zlog nn0t−d−1d+1 PB{(z1, ..., zn)|zt /∈[z1, ..., zn]}dtwhere B is a Euclidean ball whose volume is the same as that of K. The limitfor B exists by Lemma 13. Thus we getlimn→∞vold(K) −E(K, n)( 1n)2d+1=Z∂Kκ(x)1d+1 dµ limn→∞n2d+1(2d+1)d−1d+1vold−1(Bd−12)2d+1Zlog nn0PB{(z1, ..., zn)|zt /∈[z1, ..., zn]}td−1d+1dtSince this formula holds for all convex bodies it holds in particular for the Eu-clidean ball.

By Lemma 13 we determine the coefficient.□References[A]A.D. Aleksandrov, Almost everywhere existence of the second differential of a convexfunction and some properties of convex surfaces connected with it, Uchenye ZapiskiLeningrad Gos. Univ., Math.

Ser. 6 (1939), 3–35.[Ba]V.

Bangert, Analytische Eigenschaften konvexer Funktionen auf Riemannschen Man-nigfaltigkeiten, J. Reine Angew. Math.

307 (1979), 309–324.[B]I. B´ar´any, Pandom polytopes in smooth convex bodies, Mathematika 39 (1992), 81–92.[B-L]I.

B´ar´any and D.G. Larman, Convex bodies, economic cap covering, random polytopes,Mathematika 35 (1988), 274–291.[F-R]I.

F´ary and L. R´edei, Der zentralsymmetrische Kern und die zentralsymmetrischeH¨ulle von konvexen K¨orpern, Math. Ann.

122 (1950), 205–220.[G]B. Gr¨unbaum, Measures of symmetry for convex sets, Convexity, Proceedings of Sym-posion in Pure Mathematics (V.L.

Klee, ed. ), AMS, 1963, pp.

233–270.[K]K. Kiener, Extremalit¨at von Ellipsoiden und die Faltungsungleichung von Sobolev,Arch.

Math. 46 (1986), 162–168.[R-S1]A.

R´enyi and R. Sulanke, ¨Uber die konvexe H¨ulle von n zuf¨allig gew¨ahlten Punkten,Zeitschr. Wahrschein.

verwandte Geb. 2 (1963), 75–84.[R-S2]A.

R´enyi and R. Sulanke, ¨Uber die konvexe H¨ulle vo n zuf¨allig gew¨ahlten Punkten II,Zeitschr. Wahrschein.

verwandte Geb. 3 (1964), 138–147.[Schm1]M.

Schmuckenschl¨ager, The distribution function of the convolution square of a convexsymmetric body in Rn, Israel J. Math.

78 (1992), 309–334.[Schm2]M. Schmuckenschl¨ager, Notes.[Schn]R.

Schneider, Random approximation of convex sets, J. Microscopy 151 (1988), 211–227. [Schn-Wie] R. Schneider and J.A.

Wieacker, Random polytopes in a convex body, Z. Wahrschein-lichkeitstheorie verw. Gebiete 52 (1980), 69–73.

[Sch¨u-W 1] C. Sch¨utt and E. Werner, The convex floating body, Math. Scand.

66 (1990), 275–290. [Sch¨u-W 2] C. Sch¨utt and E. Werner, The convex floating body of almost polygonal bodies, Ge-ometriae Dedicata 44 (1992), 169–188.[St]S.

Stein, The symmetry function in a convex body, Pacific J. Math.

6 (1956), 145–148.[Wie]J.A. Wieacker, Einige Probleme der polyedrischen Approximation, Diplomarbeit, Freiburg im Breisgau, 1978.Oklahoma State University, Department of Mathematics, Stillwater, Oklahoma74078Christian Albrechts Universit¨at Mathematisches Seminar 24118 Kiel Germany

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