Convex Geometric Analysis MSRI Publications Volume 34, 998 Floating Boy, Illumination Boy, an Polytopal Approximation CARSTEN SCHÜTT Abstract. Let K be a convex boy inr an K t its floating boies. There is a polytope that satisfies K t P n K an has at most n vertices, where n e 6 vol (K \ K t t vol (B. Let K t be the illumination boies of K an Q n a polytope that contains K an has at most n ( -imensional faces. Then vol (K t \ K c 4 vol (Q n \ K, where n c t vol (K t \ K.. Introuction We investigate the approximation of a convex boy K in R by a polytope. We measure the approximation by the symmetric ifference metric. The symmetric ifference metric between two convex boies K an C is S (C, K = vol ((C \ K (K \ C. We stuy in particular two questions: How well can a convex boy K be approximate by a polytope P n that is containe in K an has at most n vertices an how well can K be approximate by a polytope Q n that contains K an has at most n ( -imensional faces. Macbeath [Mac] showe that the Eucliean Ball B 2 is an extremal case: The approximation for any other convex boy is better. We have for the Eucliean ball c vol (B 2n 2 S (P n, B 2 c 2 vol (B 2n 2, (. 99 Mathematics Subject Classification. 52A22. This paper was written while the author was visiting MSRI at Berkeley in the spring of 996. 203
204 CARSTEN SCHÜTT provie that n (c 3 ( /2. The right han inequality was first establishe by Bronshtein an Ivanov [BI] an Duley [D,D 2 ]. Goron, Meyer, an Reisner [GMR,GMR 2 ] gave a constructive proof for the same inequality. Müller [Mü] showe that ranom approximation gives the same estimate. Goron, Reisner, an Schütt [GRS] establishe the left han inequality. Gruber [Gr 2 ] obtaine an asymptotic formula. If a convex boy K in R has a C 2 -bounary with everywhere positive curvature, then inf { S (K, P n P n K an P n has at most n vertices} is asymptotically the same as ( + ( 2 2 el κ(x + µ(x, K n where el is a constant that is connecte with Delone triangulations. In this paper we are not concerne with asymptotic estimates, but with uniform. Int(M enotes the interior of a set M. H(x, ξ enotes the hyperplane that contains x an is orthogonal to ξ. H + (x, ξ enotes the halfspace that contains the vector x ξ, an H (x, ξ the halfspace containing x + ξ. e i, i =,..., enotes the unit vector basis in R. [A, B] is the convex hull of the sets A an B. The convex floating boy K t of a convex boy K is the intersection of all halfspaces whose efining hyperplanes cut off a set of volume t from K. The illumination boy K t of a convex boy K is [W] {x R vol ([x, K] \ K t }. K t is a convex boy. It is enough to show this for polytopes. Let F i enote the faces of a polytope P, ξ i the outer normal an x i an element of F i. Then vol ([x, P ] \ P = n max{0, ξ i, x x i } vol (F i. i= The right-han sie is a convex function. 2. The Floating Boy Theorem 2.. Let K be a convex boy in R. Then, for every t satisfying 0 t 4 e 5 vol (K, there exist n N with n e 6 vol (K \ K t t vol (B an a polytope P n that has n vertices an such that K t P n K.
FLOATING BODY, ILLUMINATION BODY, AND APPROXIMATION 205 We want to see what kin of asymptotic estimate we get for boies with smooth bounary from Theorem 2.. We have [SW] Since we get Thus we get vol (K \ K t t 2 ( + 2 t 2 + + vol (B K n 2 ( vol (K \ K t 2 vol (K \ K t + 2 n 2 + κ(x + µ(x. t vol (K \ K t, n vol (K \ K t K 2 + K κ(x + µ(x 2 + K κ(x κ(x + µ(x. ( vol (K \ P n vol (K \ K t 2 n 2 When K is the Eucliean ball we get K vol (B 2 \ P n c 2 n 2 vol (B 2, + µ(x, κ(x + µ(x +. where c is an absolute constant. If one compares this to the optimal result (. one sees that there is an aitional factor. The volume ifference vol (P vol (P t for a polytope P is of a much smaller orer than for a convex boy with smooth bounary. In fact, we have [S] that it is of the orer t ln t. In [S] this has been use to get estimates for approximation of convex boies by polytopes. The same result as in Theorem 2. hols if we fix the number of (-- imensional faces instea of the number of vertices. This follows from the same proof as for Theorem 2. an also from the economic cap covering for floating boies [BL, Theorem 6]. I. Bárány showe us a proof for Theorem 2. using the economic cap covering. The constants are not as goo as in Theorem 2.. The following lemmata are not new. They have usually been formulate for symmetric, convex boies [B,H,MP]. Lemma 2.2 is ue to Grünbaum [Grü]. Lemma 2.2. Let K be a convex boy in R an let H(cg(K, ξ be the hyperplane passing through the center of gravity cg(k of K an being orthogonal to ξ. Then we have, for all ξ B 2: (i ( + vol (K vol (K H + (cg(k, ξ ( ( + vol (K.
206 CARSTEN SCHÜTT (ii For all hyperplanes H in R that are parallel to H(cg(K, ξ, ( vol (K H vol (K H(cg(K, ξ. + The sequence ( +, = 2, 3,... is monotonely ecreasing. Inee, by Bernoulli s inequality we have (, or 2 ( 2. Therefore 2 we get ( + (, which implies ( + (. Therefore we get for the inequalities (i e vol (K vol (K H + (cg(k, ξ ( e vol (K. (2. By the preceing calculations, ( + is a monotonely increasing sequence. Thus ( + < e. For (ii we get vol (K H e vol (K H(cg(K, ξ. (2.2 Proof. (i We can reuce the inequality to the case that K is a cone with a Eucliean ball of imension as base. To see this we perform a Schwarz symmetrization parallel to H(cg(K, ξ an enote the symmetrize boy by S(K. The Schwarz symmetrization replaces a section parallel to H(cg(K, ξ by a ( -imensional Eucliean sphere of the same ( -imensional volume. This oes not change the volume of K an K H + (cg(k, ξ an the center of gravity cg(k is still an element of H(cg(K, ξ. Now we consier the cone such that [z, S(K H(cg(K, ξ] vol ([z, S(K H(cg(K, ξ] = vol (K H (cg(k, ξ an such that z lies on the axis of symmetry of S(K an in H (cg(k, ξ. See Figure. The set K = (K H + (cg(k, ξ [z, S(K H(cg(K, ξ] is a convex set such that vol (K = vol ( K an such that the center of gravity cg( K of K is containe in [z, S(K H(cg(K, ξ]. Thus vol ( K H + (cg( K, ξ vol ( K H + (cg(k, ξ = vol (K H + (cg(k, ξ. We apply a similar argument to the set S(K H + (cg(k, ξ an show that we may assume that S(K is a cone with z as its vertex. Thus we may assume that K = [ (0,..., 0,, { (x,..., x, 0 i= x i 2 }] an ξ = (0,..., 0,. Then vol (K = vol (B
FLOATING BODY, ILLUMINATION BODY, AND APPROXIMATION 207 H(cg(K, ξ z S(K an x x = vol (K K We obtain 0 Figure. t( t t = vol (K H (cg(k, (0,..., 0, = 0 ( ss s = +. ( vol (K. + (ii Let H be a hyperplane parallel to H(cg(K, ξ an such that vol (K H > vol (K H(cg(K, ξ. Otherwise there is nothing to prove. We apply a Schwarz symmetrization parallel to H(cg(K, ξ to K. The symmetrize boy is enote by S(K. Let z be the element of the axis of symmetry of S(K such that [z, S(K H] H(cg(K, ξ = S(K H(cg(K, ξ. Since vol (K H > vol (K H(cg(K, ξ there is such a z. assume that H + (cg(k, ξ is the half-space containing z. Then We may [z, S(K H] H (cg(k, ξ S(K H (cg(k, ξ, [z, S(K H] H + (cg(k, ξ S(K H + (cg(k, ξ. Therefore cg([z, S(K H] H + (cg(k, ξ.
208 CARSTEN SCHÜTT Therefore, if h cg enotes the istance of z to H(cg(K, ξ an h the istance of z to H, we get as in the proof of (i that ( h cg h. + Thus we get vol (K H(cg(K, ξ = vol (S(K H(cg(K, ξ ( + vol (S(K H = ( + vol (K H. Lemma 2.3. Let K be a convex boy in R an let Θ(ξ be the infimum of all positive numbers t such that Then vol (K H(cg(K, ξ e vol (K H(cg(K + tξ, ξ. 2e 3 vol (K Θ(ξ vol (K H(cg(K, ξ e vol (K. Proof. The right han inequality follows from Fubini s theorem an Brunn Minkowski s theorem. Now we verify the left han inequality. We consier first the case in which, for all t such that t > Θ(ξ, Then, by (2. an (2.2, K H(cg(K + tξ, ξ =. e vol (K vol (K H + (cg(k, ξ = Θ(ξ 0 vol (K H(cg(K + tξ, ξ t e Θ(ξ vol (K H(cg(K, ξ. If, for some t such that t > Θ(ξ, we have K H(cg(K + tξ, ξ, then, by continuity, vol (K H(cg(K, ξ = e vol (K H(cg(K + Θ(ξξ, ξ. We perform a Schwarz symmetrization parallel to H(cg(K, ξ. We consier the cone [z, S(K H(cg(K, ξ] such that z is an element of the axis of symmetry of S(K an such that [z, S(K H(cg(K, ξ] H(cg(K + Θ(ξξ, ξ = S(K H(cg(K + Θ(ξξ, ξ.
FLOATING BODY, ILLUMINATION BODY, AND APPROXIMATION 209 H(cg(K, ξ z S(K H(cg(K + Θ(ξξ, ξ Figure 2. Let H + (cg(k, ξ an H + (cg(k+θ(ξξ, ξ be the half-spaces that contain z. Then, by convexity, [z, S(K H(cg(K, ξ] H + (cg(k + Θ(ξξ, ξ We get by (2. e vol (K vol (K H + (cg(k, ξ S(K H + (cg(k + Θ(ξξ, ξ. (2.3 = vol (K H + (cg(k, ξ H (cg(k + Θ(ξξ, ξ + vol (K H + (cg(k + Θ(ξξ, ξ = vol (S(K H + (cg(k, ξ H (cg(k + Θ(ξξ, ξ + vol (S(K H + (cg(k + Θ(ξξ, ξ. By the hypothesis of the lemma we have, for all s with 0 s Θ(ξ, vol (K H(cg(K, ξ e vol (K H(cg(K + sξ, ξ. Using this an (2.2 we estimate the first summan. The secon summan is estimate by using (2.3. Thus the above expression is not greater than e 2 vol ([z, S(K H(cg(K, ξ] H (cg(k + Θ(ξξ, ξ + vol ([z, S(K H(cg(K, ξ] H + (cg(k + Θ(ξξ, ξ.
20 CARSTEN SCHÜTT This is the volume of a cone with the base S(K H(cg(K, ξ. By an elementary computation for the volume of a cone we get that the latter expression is smaller than 2e 2 vol ([z, S(K H(cg(K, ξ] H (cg(k + Θ(ξξ, ξ. Since in a cone the base has the greatest surface area, the above expression is smaller than 2e 2 Θ(ξ vol (K H(cg(K, ξ. Lemma 2.4. Let K be a convex boy in R. Then there is a linear transform T with et(t = so that, for all ξ B 2, T (K x, ξ 2 x = T (K i= x, e i 2 x. We say that a convex boy is in an isotropic position if the linear transform T in Lemma 2.4 can be chosen to be the ientity. See [B,H]. Proof. We claim that there is a orthogonal transform U such that, for all i, j =,..., with i j, x, e i x, e j x = 0. Clearly, the matrix U(K ( x, e i x, e j x K i,j= is symmetric. Therefore there is an orthogonal -matrix U so that ( U x, e i x, e j x U t K is a iagonal matrix. We have ( U x, e i x, e j x K i,j= U t = = = ( ( ( K i,j= i,j= u l,i x, e i x, e j u k,j x x, U t (e l x, U t (e k x K y, e l y, e k y U(K l,k= l,k= l,k= So the latter matrix is a iagonal matrix. All the iagonal elements are strictly positive. This argument is repeate with a iagonal matrix so that the iagonal.
FLOATING BODY, ILLUMINATION BODY, AND APPROXIMATION 2 elements turn out to be equal. Therefore there is a matrix T with et T = such that 0 if i j, x, e i x, e j x = T (K x, e j 2 x if i = j. T (K j= From this the lemma follows. Lemma 2.5. Let K be a convex boy in R that is in an isotropic position an whose center of gravity is at the origin. Then, for all ξ B 2, 24e 0 vol (K 3 vol (K H(cg(K, ξ 2 K 6 e 3 vol (K 3. Proof. By Lemma 2.4 we have, for all ξ B 2, K x, e i 2 x = i= By Fubini s theorem, this equals K x, ξ 2 x. x, e i 2 x i= t 2 vol (K H(tξ, ξ t Θ(ξ t 2 0 vol (K H(tξ, ξ t, where Θ(ξ is as efine in Lemma 2.3. expression is greater than By the efinition of Θ(ξ the above e vol (K H(cg(K, ξ Θ(ξ By Lemma 2.3 this is greater than 0 t 2 t 3e Θ(ξ3 vol (K H(cg(K, ξ. vol (K 3 24e 0 vol (K H(cg(K, ξ 2. Now we show the right han inequality. By Lemma 2.4 we have K x, e i 2 x = i= Θ(ξ = + 0 0 Θ( ξ K x, ξ 2 x = t 2 vol (K H(tξ, ξ t + t 2 vol (K H(tξ, ξ t + t 2 t 2 Θ(ξ Θ( ξ t 2 vol (K H(tξ, ξ t vol (K H(tξ, ξ t vol (K H(tξ, ξ t.
22 CARSTEN SCHÜTT By (2.2 this is not greater than e 3 Θ(ξ3 vol (K H(cg(K, ξ + t 2 Θ(ξ + e 3 Θ( ξ3 vol (K H(cg(K, ξ + The integrals can be estimate by vol (K H(tξ, ξ t Θ( ξ t 2 vol (K H(tξ, ξ t. 2 Θ(ξ 3 vol (K H(cg(K, ξ an 2 Θ( ξ 3 vol (K H(cg(K, ξ, respectively. We treat here only the case ξ; the case ξ is treate in the same way. If the integral equals 0, there is nothing to show. If the integral oes not equal 0, we have vol (K H(cg(K, ξ = e vol (K H(cg(K + Θ(ξξ, ξ. We consier the Schwarz symmetrization S(K of K with respect to the plane H(cg(K, ξ. We consier the cone C that is generate by the Eucliean spheres S(K H(cg(K, ξ an S(K H(cg(K + Θ(ξξ, ξ. We an the height of C equals S(K H + (cg(k + Θ(ξξ, ξ C Θ(ξ e Since ( + < e, we have e >. Thus the height of the cone C is less than Θ(ξ. Thus, for all t with Θ(ξ t Θ(ξ, ( vol (K H(cg(K + tξ, ξ t vol (K H(cg(K, ξ. Θ(ξ Now we get Θ(ξ t 2 vol (K H(tξ, ξ t Therefore x, e i 2 x K i= Θ(ξ Θ(ξ t 2 (. t Θ(ξ vol (K H(cg(K, ξ t vol (K H(cg(K, ξ( Θ(ξ 3 s 2 ( s s = vol (K H(cg(K, ξ( Θ(ξ 3 2 ( + ( + 2 2 vol (K H(cg(K, ξθ(ξ 3. ( e 3 + (Θ(ξ 3 + Θ( ξ 3 vol (K H(cg(K, ξ. 0
FLOATING BODY, ILLUMINATION BODY, AND APPROXIMATION 23 Now we apply Lemma 2.3 an get 2( e 3 + 2e3 vol (K 3 vol (K H(cg(K, ξ 2. Lemma 2.6. Let K be a convex boy in R such that the origin is an element of K. Then K x, e i 2 x 2 + 2 vol ( B2 2 vol (K +2. i= Proof. Let r(ξ be the istance of the origin to the bounary of K in irection ξ. By passing to spherical coorinates we get K x, e i 2 x = i= B 2 r(ξ 0 ρ + ρ ξ = By Höler s inequality, this expression is greater than ( vol ( B2 +2 ( + 2 vol ( B2 r(ξ ξ B 2 r(ξ +2 ξ ( + 2 B2 = 2 + 2 vol ( B 2 2 vol (K +2. The following lemma can be foun in [MP]. It is formulate there for the case of symmetric convex boies. Lemma 2.7. Let K be a convex boy in R such that the origin coincies with the center of gravity of K an such that K is in an isotropic position. Then B 2(cg(K, 24e 5 π vol (K K 4e 4 vol (K. An affine transform can put a convex boy into this position. Proof. As in Lemma 2.3, let Θ(ξ be the infimum of all numbers t such that vol (K H(cg(K, ξ e vol (K H(cg(K + tξ, ξ. By Lemma 2.3, Θ(ξ vol (K 2e 3 vol (K H(cg(K, ξ. By Lemma 2.5 we get ( Θ(ξ 2e 3 x, e 6e 3 i 2 x 2 vol (K We have π 2 K i= vol (B2 = Γ( 2 + π( /2 (2e 2 + 2, 2.
24 CARSTEN SCHÜTT an thus Therefore, by Lemma 2.6, Θ(ξ 2e 3 6e 3 2 On the other han, ( + 2 vol (B 2 2πe. vol (K vol ( B2 vol (K H (cg(k + 2 Θ(ξξ, ξ Θ(ξ 2e 5 π vol (K. vol (K H(cg(K + tξ, ξ t, 2 Θ(ξ where H (cg(k+ 2Θ(ξξ, ξ is the half-space not containing the origin. By the efinition of Θ(ξ this expression is greater than Θ(ξ 2e vol (K H(cg(K, ξ. By Lemma 2.3 we get that this is greater than 4e 4 vol (K. Therefore, every hyperplane that has istance 24e 5 π vol (K from the center of gravity cuts off a set of volume greater than 4e 4 vol (K. Proof of Theorem 2.. We are choosing the vertices x,..., x n K of the polytope P n. N(x k enotes the normal to K at x k. x is chosen arbitrarily. Having chosen x,..., x k we choose x k such that {x,..., x k } Int(K H (x k k N(x k, N(x k =, where k is etermine by vol (K H (x k k N(x k, N(x k = t. If the normal at x k is not unique it suffices that just one of the normals satisfies the conition. It coul be that the hyperplane H(x k k N(x k, N(x k is not tangential to the floating boy K t, but this oes not affect the computation. We claim that this process terminates for some n with n e 6 vol (K \ K t t vol (B. (2.4 This claim proves the theorem: If we cannot choose another x n+, then there is no cap of volume t that oes not contain an element of the polytope P n = [x,..., x n ]. By the theorem of Hahn Banach we get K t P n. We show now
FLOATING BODY, ILLUMINATION BODY, AND APPROXIMATION 25 x k H(x k k N(x k, N(x k K t K Figure 3. the claim. We assume that we manage to choose points x,..., x n where n is to big that (2.4 oes not hol. We put an S k = K ( n i=k+ S n = K H (x n n N(x n, N(x n (2.5 H + (x i N(x i, N(x i for k =,..., n. For k l, we have Let k < l < n. Then ( n S k S l = K K i=k+ ( n vol (S k S l = 0. H + (x i N(x i, N(x i i=l+ H + (x i N(x i, N(x i H + (x l l N(x l, N(x l H (x l l N(x l, N(x l = H(x l l N(x l, N(x l. Thus we have H (x k k N(x k, N(x k H (x k k N(x k, N(x k H (x l l N(x l, N(x l vol (S k S l vol (H(x l l N(x l, N(x l = 0. (2.6
26 CARSTEN SCHÜTT The case k < l = n is shown in the same way. We have, for k =,..., n, ( n S k = K H + (x i N(x i, N(x i H (x k k N(x k, N(x k i=k+ [x k, K t ] H (x k k N(x k, N(x k [x k, (K H (x k k N(x k, N(x k t ] H (x k k N(x k, N(x k, where k is etermine by vol (K H (x k k N(x k, N(x k = 4e 4 t. By Lemma 2.7 there is an ellipsoi E containe in (K H (x k k N(x k, N(x k t whose center is cg(k H (x k k N(x k, N(x k an that has volume vol (E = 4e 4 (24e 5 π t vol (B 2 Since (K H (x k k N(x k, N(x k t is containe in K t, E is containe in K t. Thus S k [x k, E] H (x k k N(x k, N(x k. We claim now that [x k, E] H (x k k N(x k, N(x k contains an ellipsoi Ẽ such that vol (Ẽ = 4e 4 (24e 5 π (4e 5 t vol (B2, an consequently vol (S k 4e 4 (24e 5 π (4e 5 t vol (B2 4e 4 = (96e 0 π t vol (B2. (2.7 For this we have to see that k 4e 5 k. By the assumption t 4 e 5 vol (K we get vol (K H (x k k N(x k, N(x k e vol (K. Therefore, by (2., cg(k H + (x k k N(x k, N(x k. We consier two cases. If vol (K H(x k k N(x k, N(x k vol (K H(x k k N(x k, N(x k, the theorem of Brunn Minkowski implies that, for all s in the range k s k, we have vol (K H(cg(K, N(x k vol (K H(x k k N(x k, N(x k vol (K H(x k sn(x k, N(x k. (2.8
FLOATING BODY, ILLUMINATION BODY, AND APPROXIMATION 27 We get, by (2.2, By (2.8, k ( k k vol (K H(cg(K, N(x k t e vol (K H(cg(K, N(x k. vol (K H (x k k N(x k, N(x k vol (K H (x k k N(x k, N(x k. This implies Therefore k k k (4e 4 t vol (K H(cg(K, N(x k. (4e 4 t vol (K H(cg(K, N(x k + k 4e 5 k. If vol (K H(x k k N(x k, N(x k vol (K H(x k k N(x k, N(x k, the theorem of Brunn Minkowski implies that, for all u in the range 0 u k, an all s in the range k s k, we have vol (K H(x k un(x k, N(x k vol (K H(x k k N(x k, N(x k We get an Therefore k k k k vol (K H(x k sn(x k, N(x k. t vol (K H(x k k N(x k, N(x k (4e 4 t vol (K H(x k k N(x k, N(x k. (4e 4 t vol (K H(x k k N(x k, N(x k + k 4e 4 k. We have verifie (2.7. From (2.6 an (2.7 we get vol (K \ K t vol ( n S k = k= n 4e 4 vol (S k n (96e 0 π t vol (B2. k= Thus we get the esire equation (2.4: vol (K \ K t e 6 n t vol (B 2.
28 CARSTEN SCHÜTT 3. The Illumination Boy Theorem 3.. Let K be a convex boy in R such that c B 2 K c 2 B 2. Let 0 t (5c c vol (K an let n N be such that ( 28 7 π( /2 n 32 et vol (K t \ K. Then we have, for every polytope P n that contains K an has at most n ( - imensional faces, vol (K t \ K 0 7 2 (c c 2+ vol (P n \ K. We want to see what this result means for boies with a smooth bounary. We have the asymptotic formula [W] Thus lim t 0 vol (K t vol (K = 2 t 2 + An by the theorem we have vol (K t vol (K t 2 + ( 2 ( + + vol (B2 κ(x + µ(x. K K κ(x + µ(x. Thus or n t vol (K t \ K. ( 2 vol (K t vol (K n vol (K t + \ K κ(x + µ(x, K ( vol (K t \ K + n ( 2 vol (K t \ K n By Theorem 3. we now get vol (P n \ K ( c c 2 + + ( n 2 + ( K K 2 ( By a theorem of F. John [J] we have c c 2. κ(x + µ(x, κ(x + µ(x +. K κ(x + µ(x +.
FLOATING BODY, ILLUMINATION BODY, AND APPROXIMATION 29 The following lemma is ue to Bronshtein an Ivanov [BI] an Duley [D, D 2 ]. It can also be foun in [GRS]. Lemma 3.2. For all imensions, 2, an all natural numbers n, n 2, there is a polytope Q n that has n vertices an is containe in the Eucliean ball B2 such that H (Q n, B2 6 ( vol ( B2 2 7 vol (B2 n 2. We have π 2 vol ( B2 = vol (B2 = Γ( 2 + Since 2 = Γ( 2 + π Γ( 2 + vol (B2 π vol (B2. (3. 4 an ( t t, (3. yiels H (B2, Q n 6 ( π 7 n 2 64 7 π n 2. (3.2 Proof of Theorem 3.. We enote the ( -imensional faces of P n by F i, for i =,..., n, an the cones generate by the origin an a face F i by C i, for i =,..., n. Take x i F i an let ξ i, with ξ i 2 =, be orthogonal to F i an pointing to the outsie of P n. Then H(x i, ξ i is the hyperplane containing F i an H + (x i, ξ i the halfspace containing P n. See Figure 4. We may assume that the hyperplanes H(x i, ξ i, i =,..., n, are supporting hyperplanes of K. Otherwise we can choose a polytope of lesser volume. Let be the height of the set K t H (x i, ξ i C i, that is, the smallest number s such that K t H (x i, ξ i C i H + (x i + sξ i, ξ i. Let z i be a point in K t C i where the height is attaine. We may assume that B 2 K cb 2 where c = c c 2. Also we may assume that P n 2cB 2 (3.3 if we allow twice as many faces. This follows from (3.2: There is a polytope Q k such that 2 B 2 Q k B2 an the number of vertices k is smaller than ( 28 7 π( /2. Thus Q k satisfies B 2 Q k 2B 2 an has at most ( 28 7 π( /2 ( -imensional faces. As the new polytope P n we choose the intersection of cq k with the original polytope P n. Since we have by assumption that n is greater than ( 28 7 π( /2 the new polytope has at most 6 et vol (K t \ K. (3.4
220 CARSTEN SCHÜTT z i H(x i, ξ i K t P n K C i 0 Figure 4. ( -imensional faces. We show first that for t with 0 t (5c vol (K an all i, i =,..., n we have (3.5 Assume that there is a face F i with >. Consier the smallest infinite cone D i having z i as vertex an containing K. Since H(x i, ξ i is a supporting hyperplane to K an K c B2 we have K D i H + (x i, ξ i H (x i 2cξ i, ξ i an D i H (x i, ξ = [z i, K] H (x i, ξ We have t = vol ([z i, K] \ K vol ([z i, K] H (x i, ξ i = vol (D i H (x i, ξ i = Thus vol (D i H(x i, ξ i 2 vol (D i H(x i, ξ i vol (D i H(x i, ξ i 2 t (3.6
FLOATING BODY, ILLUMINATION BODY, AND APPROXIMATION 22 Since (3.5 oes not hol we have vol (D i H(x i 2cξ i, ξ i = ( 2c + vol (D i H(x i, ξ i By (3.6 we get Thus (2c + vol (D i H(x i, ξ i. vol (D i H(x i 2cξ i, ξ i (2c + 2 t (3c 2 t. an we conclue that vol (K vol (D i H + (x i, ξ i H (x i 2cξ i, ξ i 2c(3c 2 t (3c + t, t (3c vol (K. This is a contraiction to the assumption on t in the hypothesis of the theorem. Thus we have shown (3.5. We consier now two cases: All those heights that are smaller than 2t/vol (F i an those that are greater. We may assume that, i =,..., k are smaller than 2t/vol (F i an, i = k +,..., n are strictly greater. We have vol ((K t \ P n C i = Since B 2 K P n we get vol ((K t \ P n C i By (3.5 we get For i =,..., k we get Thus i 0 i 0 vol ((K t \ P n C i H(x i + sξ i, ξ i s. vol (F i ( + s s ( + vol (F i. vol ((K t \ P n C i ( + vol (F i. vol ((K t \ P n C i By (3.4 we get vol ((K t \ P n vol ((K t \ P n ( 2t + vol vol (F i (F i 2et. k C i 2ket 2net. i= k C i 8 vol (K t \ K. (3.7 Now we consier the other faces. For i = k +,..., n, we have i= 2t vol (F i. (3.8
222 CARSTEN SCHÜTT We show that, for i = k +,..., n, we have 5c ( 5c vol (F i 2 vol (K. (3.9 Suppose that there is a face F i so that (3.9 oes not hol. Then t = vol ([z i, K] \ K vol ([z i, K] H (x i, ξ i = vol ([z i, K] H(x i, ξ i. Therefore we get, by (3.8, Since K B 2 we have Thus vol ([z i, K] H(x i, ξ i t 2 vol (F i. (3.0 K D i H + (x i, ξ i H (x i 2cξ i, ξ i. vol (K vol (D i H (x i 2cξ i, ξ i. The cone D i H (x i 2cξ i, ξ i has a height equal to 2c +. Therefore vol (K ( 2c + (2c + i i vol (D i H(x i, ξ i. By (3.5 we have. Therefore we get vol (K 3c ( 3c vol (D i H(x i, ξ i = 3c ( 3c vol ([z i, K] H(x i, ξ i. By (3.0 we get vol (K 3c ( 3c vol (F i, 2 which implies (3.9. Let y i be the unique point y i = [0, z i ] H(x i, ξ i. We want to make sure that y i F i [z i, K]. This hols since z i C i H (x i, ξ i an > 0. Since y i F i we have vol (F i = vol (B2 vol 2 ( B2 r i (η µ(η, B 2 where r i (η is the istance of y i to the bounary F i in irection η, η B2, an, since y i F i [z i, K], we have vol (F i [z i, K] = vol (B vol 2 ( B B 2 ρ i (η µ(η,
FLOATING BODY, ILLUMINATION BODY, AND APPROXIMATION 223 where ρ i (η is the istance of y i to the bounary (F i [z i, K]. Consier the set A i = {η ( 4 r i(η ρ i (η }. We show that 4 vol (F i vol (B2 vol 2 ( B r i (η ρ i (η µ(η (3. We have vol (B vol 2 ( B Therefore vol (B vol 2 ( B A c i A i r i (η ρ i (η µ(η A c i vol (B vol 2 ( B 4 vol (B vol 2 ( B r i (η ρ i (η µ(η r i (η ( ( 4 µ(η A i A i r i (η µ(η 4 vol (F i. vol (B2 vol 2 ( B2 r i (η ρ i (η µ(η B 2 vol (B2 vol 2 ( B2 r i (η ρ i (η µ(η A i vol (F i vol (F i [z i, K] 4 vol (F i. By (3.0 this is greater than 4 vol (F i. This implies 4 vol (F i vol (B2 vol 2 ( B r i (η ρ i (η µ(η. Thus we have establishe (3.. We shall show that We have A c i vol ((K t \ P n C i 0 6 e 2 c 2+ vol ((P n \ K C i. (3.2 vol (D c i H + (x i, ξ i C i vol ((P n \ K C i. Compare Figure 5. Therefore, if we want to verify (3.2 it is enough to show that vol ((K t \ P n C i 0 6 e 2 c 2+ vol (D c i H + (x i, ξ i C i. We may assume that y i an z i are orthogonal to H(x i, ξ i. This is accomplishe by a linear, volume preserving map: Any vector orthogonal to ξ i is mappe onto itself an y i is mappe to ξ i, y i ξ i. See Figure 6.
224 CARSTEN SCHÜTT z i α i F i y i y i + r i (ηη δ i β i 0 Figure 5. z i F i y i y i + ρ(ηη α i β i y i + r i (ηη δ i 0 Figure 6.
FLOATING BODY, ILLUMINATION BODY, AND APPROXIMATION 225 Let w i (η D c i H+ (x i, ξ i C i such that w i (η is an element of the 2- imensional subspace containing 0, y i, an y i + η. Let δ i (η be the istance of w i (η to the plane H(x i, ξ i. Then vol (B vol 2 ( B A c i (r i (η ρ i (η δ i (η µ(η Thus, in orer to verify (3.2, it suffices to show that vol ((K t \ P n C i 0 6 e 2 c 2+ vol (B vol 2 ( B A c i vol (D c i H + (x i, ξ i C i. (r i (η ρ i (η δ i (η µ(η. (3.3 In orer to o this we shall show that for all i = k +,..., n an all η A c i there is w i (η such that the istance δ i (η of w i from H(x i, ξ i satisfies { 32c if 0 αi π i 4, δ i if π 4 α i π 2. (3.4 ( 60 c 2 5c vol (F i r i 2 vol (K The angles α i (η an β i (η are given in Figure 6. We have for all η A c i δ i =(r i ρ i sin(α i sin(β i sin(π α i β i, =ρ i tan α i, (3.5 with 0 α i, β i π 2. Thus we get ρ i sin(π α i β i ρ i δ i r i ρ i cos(α i sin(β i (r i ρ i cos(α i sin(β i. By (3. we have ρ i ( 4 r i. Therefore δ i 4 cos(α i sin(β i. Since B2 K P n 2c B2 we have tan β i 4c : Here we have to take into account that we applie a transform to K mapping y i to ξ i, y i ξ i. That leaves the istance of F i to the origin unchange an r i (η is less than 4c. If β i π 4 we have sin β i 2. If β i π 4 then 4c tan β i = sin β i cos β i 2 sin β i. Therefore we get 6 2 c. δ i cos α i Therefore we get, for all 0 α i π 4, δ i 32c.
226 CARSTEN SCHÜTT By (3.9 an (3.5 we get δ i ( sin(π α i β i 5c r i ρ i sin(α i sin(β i 5c vol (F i. 2 vol (K We procee as in the estimate above an obtain 6 2 c 5c δ i sin(α i Thus we get for π 4 α i π 2 δ i r i ( 5c vol (F i 2 vol (K ( 32 c 5c vol (F i 5c. r i 2 vol (K. We verify now (3.3. By the efinition of A i we get vol (B2 vol 2 ( B2 (r i (η ρ i (η δ i (η µ(η A c i ( e vol (B 8 vol 2 ( B2 r i (η δ i µ(η. A c i We get by (3.4 that the last expression is greater than 320c vol (B2 i vol 2 ( B2 ( A c i α i π 4 By (3. we get that either or vol (B vol 2 ( B r i µ + ( 2 vol (K 5c 5c vol (F i Ac i α i > π 4 vol (B2 vol 2 ( B2 A c i α i π 4 A c i α i > π 4 r i µ 8 vol (F i r i µ 8 vol (F i. r i µ In the first case we get for the above estimate vol (B2 vol 2 ( B2 (r i (η ρ i (η δ i (η µ(η A c i 2560c vol (F i 2560ec vol ((K t \ P n C i. The last inequality is obtaine by using (3.5: Since B 2 K we have, for all hyperplanes H that are parallel to F i, vol (K t H C i ( + vol (F i..
FLOATING BODY, ILLUMINATION BODY, AND APPROXIMATION 227 By (3.5 we get vol (K t H C i e vol (F i. In the secon case we have vol (B2 vol 2 ( B2 A c i ( 2 vol (K 5c 5c vol (F i 5c (r i (η ρ i (η δ i (η µ(η ( 2 vol (K 5c vol (F i ( 2 vol (K 5c 5c vol (F i = ( vol (K 5c 20c vol (B2 ( vol (K 5c 20c vol (B2 Since B 2 K we get 320c 320c vol (B2 vol 2 ( B2 A c i α i > π 4 vol (B2 (vol 2 ( B2 ( r i µ A c i α i > π 4 r i µ 320c vol (B ( 8 vol (F i 2560c vol (F i 2560ec vol ((K t \ P n C i. vol (B2 vol 2 ( B2 (r i (η ρ i (η δ i (η µ(η A c i ( 5c 5c vol (B 2 20c vol (B 2560ec vol ((K t \ P n C i ( 20c 2560ec vol ((K t \ P n C i (0 6 ec 2+ vol ((K t \ P n C i. The secon case gives a weaker estimate. Therefore we get for both cases vol ((K t \ P n C i 0 6 ec 2+ vol (B vol 2 ( B2 (r i (η ρ i (η δ i µ(η. A c i Thus we have verifie (3.3 an thereby also (3.2. By (3.2 we get vol ((K t \P n n i=k+ (( n C i 0 6 e 2 c 2+ vol i=k+ C i (P n \K 0 6 e 2 c 2+ vol ((P n \K. (3.6
228 CARSTEN SCHÜTT If the assertion of the theorem oes not hol we have Thus we get vol ((P n \ K vol (K t \ K 0 7 e 2 c 2+ vol ((K t \ P n Together with (3.7 we obtain n i=k+ C i 0 vol (K t \ K.. (3.7 vol (K t \ P n 4 vol (K t \ K 4 {vol (K t \ P n + vol (P n \ K}. (3.8 By (3.7 we have This implies vol (P n \ K vol (K t \ K 0 7 e 2 c 2+ 2 vol (K t \ K 2 vol (K t \ P n + 2 vol (P n \ K. vol (P n \ K vol (K t \ P n. Together with (3.8 we get now the contraiction vol (K t \ P n 2 vol (K t \ P n. References [B] K. Ball, Logarithmically concave functions an sections of convex sets in R n, Stuia Math. 88 (988, 69 84. [BL] I. Bárány an D. G. Larman, Convex boies, economic cap covering, ranom polytopes, Mathematika 35 (988, 274 29. [BI] E. M. Bronshtein an L. D. Ivanov, The approximation of convex sets by polyhera, Siberian Math. J. 6 (975, 0 2. [D ] R. Duley, Metric entropy of some classes of sets with ifferentiable bounaries, J. of Approximation Theory 0 (974, 227 236. [D 2] R. Duley, Correction to Metric entropy of some classes of sets with ifferentiable bounaries, J. of Approximation Theory 26 (979, 92 93. [F T] L. Fejes Toth, Über zwei Maximumsaufgaben bei Polyeern, Tohoku Math. J. 46 (940, 79 83. [GMR ] Y. Goron, M. Meyer, an S. Reisner, Volume approximation of convex boies by polytopes a constructive metho, Stuia Math. (994, 8 95. [GMR 2 ] Y. Goron, M. Meyer an S. Reisner, Constructing a polytope to approximate a convex boy, Geometriae Deicata 57 (995, 27 222. [GRS] Y. Goron, S. Reisner, an C. Schütt, Umbrellas an polytopal approximation of the Eucliean ball, J. of Approximation Theory 90 (997, 9 22. [Gr ] P. M. Gruber, Volume approximation of convex boies by inscribe polytopes, Math. Annalen 28 (988, 292 245.
FLOATING BODY, ILLUMINATION BODY, AND APPROXIMATION 229 [Gr 2] P. M. Gruber, Asymptotic estimates for best an stepwise approximation of convex boies II, Forum Mathematicum (993, 5 52 538. [GK] P. M. Gruber an P. Kenerov, Approximation of convex boies by polytopes, Ren. Circolo Mat. Palermo 3 (982, 95 225. [Grü] B. Grünbaum, Partitions of mass-istributions an of convex boies by hyperplanes, Pacific J. Math. 0 (960, 257 26. [H] D. Hensley, Slicing convex boies-bouns for slice area in terms of the boy s covariance, Proc. Amer. Math. Soc. 79 (980, 69 625. [J] F. John, Extremum problems with inequalities as subsiiary conitions, R. Courant Anniversary Volume, Interscience, New York, 948, 87 204. [Mac] A. M. Macbeath, An extremal property of the hypersphere, Proc. Cambrige Phil. Soc. 47 (95, 245 247. [MP] V. Milman an A. Pajor, Isotropic position an inertia ellipsois an zonois of the unit ball of a norme n-imensional space, Geometric Aspects of Functional Analysis (GAFA 987 88, eite by J. Linenstrauss an V. D. Milman, Springer, 989, 64 04. [Mü] J. S. Müller, Approximation of the ball by ranom polytopes, J. Approximation Theory 63 (990, 98 209. [R] C. A. Rogers, Packing an covering, Cambrige University Press, 964. [S] C. Schütt, The convex floating boy an polyheral approximation, Israel J. Math. 73 (99, 65 77. [SW] C. Schütt an E. Werner, The convex floating boy, Math. Scaninavica 66 (990, 275 290. [W] E. Werner, Illumination boies an the affine surface area, Stuia Math. 0 (994, 257 269. Carsten Schütt Mathematisches Seminar Christian Albrechts Universität D-24098 Kiel Germany CarstenSchuett@compuserve.com