CONVEX HULL DEVIATION AND CONTRACTIBILITY G.M. IVANOV Abstract. We study the Hausdorff distance between a set and its convex hull. Let X be a Banach space, define the CHD-constant of space X as the supremum of this distance for all subset of the unit ball in X. In the case of finite dimensional 5 Banach spaces we obtain the exact upper bound of the CHD-constant depending 1 on the dimension of the space. We give an upper bound for the CHD-constant in 0 L spaces. We prove that CHD-constant is not greater than the maximum of the p 2 Lipschitz constants of metric projection operator onto hyperplanes. This implies y that for a Hilbert space CHD-constant equals 1. We prove criterion of the Hilbert a space and study the contractibility of proximally smooth sets in uniformly convex M and uniformly smooth Banach spaces. 6 1. Introduction ] A Let X be a Banach space. For a set A ⊂ X by ∂A,intA and coA we denote the F boundary, interior and convex hull of A, respectively. We use hp,xi to denote the value . h of functional p ∈ X∗ at the vector x ∈ X. For R > 0 and c ∈ X we denote by B (c) a R t a closed ball with center c and radius R. m By ρ(x,A) we denote distance between the point x ∈ X and set A. We define the [ deviation from set A to set B as follows 3 (1) h+(A,B) = supρ(x,B). v 6 x∈A 9 In case B ⊂ A, which takes place below, the deviation h+(A,B) coincides with the 5 Hausdorff distance between the sets A and B. 2 0 Given D ⊂ X the deviation h+(coD,D) is called the convex hull deviation (CHD) 1. of D. 0 We define CHD-constant ζ of X as X 5 1 ζ = sup h+(coD,D). X : v D⊂B1(o) Xi Remark 1. Directly from our definition it follows that for any normed linear space X r we have 1 6 ζX 6 2. a We denote by ℓn the n-dimensional real vector space with the p-norm. p This article contains estimates for the CHD-constant for different spaces and some of its geometrical applications. In particular, for finite-dimensional spaces we obtain the exact upper bound of the CHD-constant depending on the dimension of the space: Theorem 1. Let X be a normed linear space, dimX = n > 2, then ζ ≤ 2n−1. If n n Xn n X = ℓn or X = ℓn , then the estimate is reached. n 1 n ∞ Let the sets P and Q be the intersections of the unit ball with two parallel affine hyperplanes of dimension k and P is a central section. In Corollary 1 we obtain the exact upper bound of the homothety coefficient, that provides covering of Q by P. 1 CONVEX HULL DEVIATION AND CONTRACTIBILITY 2 The next theorem gives an estimate for the CHD-constant in the L ,1 6 p 6 +∞ p spaces: Theorem 2. For any p ∈ [1, +∞] (cid:12) (cid:12) (2) ζ 6 2(cid:12)(cid:12)1p−p1′(cid:12)(cid:12), Lp where 1 + 1 = 1. p p′ Theorem 5 shows that CHD-constant is not greater than the maximum of the Lip- schitz constants of metric projection operator onto hyperplanes. This implies that for Hilbert space CHD-constant equals 1. Besides that, we prove the criterion of a Hilbert space in terms of CHD-constant. The idea of the proof is analogous to the idea used by A. L. Garkavi in [1]. Theorem 3. The equation ζ = 1 holds for a Banach space X iff X is an Euclidian X space or dimX = 2. In addition we study the contractibility of a covering of the convex set with balls. Definition 1. A covering of a convex set with balls is called admissible if it consists of a finite number of balls with centers in this set and the same radii. Definition 2. A family of balls is called admissible when it is an admissible covering of the convex hull of its centers. We say that a covering of a set by balls is conractible when the union of these balls is contactible. It is easy to show that in two-dimensional and Hilbert spaces any admissible covering is contractible (see Lemmas 2 and 3). On the other hand, using Theorem 3, we prove the following statement. Theorem 4. In a three dimensional Banach space X every admissible covering is contractible iff X is a Hilbert space. For 3-dimensional spaces we consider an example of an admissible covering of a convex set with four balls that is not contractible. To demonstrate the usefulness of this technics in Theorem 6 we obtain the sufficient condition for the contractibility of the proximally smooth sets in uniformly convex and uniformly smooth Banach space. 2. Proof of Theorem 1 and some other results Lemma 1. Supposethe setB (o)\intB (o ) is nonempty. Thenit is arcwiseconnected. 1 r 1 Proof. We suppose that o 6= o , otherwise the statement is trivial. Let z be the point of 1 intersectionofrayo oandtheboundaryoftheclosedballB (o). Thetriangleinequality 1 1 tells us that B (o)\intB (o ) contains z (because the set is nonempty). We claim that 1 r 1 ∂B (o)\intB (o ) is arcwise connected and thus prove the lemma. It suffices to show 1 r 1 that in the two dimensional case every point of ∂B (o)\ intB (o ) is connected with 1 r 1 z. Suppose, by contradiction, that it is not true. This means that there exist points a ,b ∈ ∂B (o )∩∂B (o) lying on the same side of the line oo such that the arc a b 1 1 r 1 1 1 1 1 of the circle ∂B (o) contains a point c ∈/ B (o ), that is kc−o k > r. 1 r 1 1 Consider two additional rays oa and ob codirectional with o a and o b respectively, 1 1 1 1 where a,b ∈ ∂B (o). Since balls B (o) andB (o ) aresimilar, we have a b k ab. So, the 1 1 r 1 1 1 CONVEX HULL DEVIATION AND CONTRACTIBILITY 3 facts that points a,b,a ,b lie on the same side of oo line, oa∩o a = ∅,ob∩o b = ∅ 1 1 1 1 1 1 1 and that a unit ball is convex, imply that segments ab and a b lie on the same line, 1 1 this contradicts kc−o k > r. 1 Proof of Theorem 1. Denote r = 2n−1. n n Suppose the inequality doesn’t hold. It means that there exists a Banach space X n with dimension n > 2, a set D ⊂ B (o) ⊂ X and a point o ∈ coD, such that 1 n 1 B (o ) ∩ D = ∅. But if o ∈ coD, then o ∈ co(B (o)\intB (o )). According to rn 1 1 1 1 rn 1 Lemma 1 the set B = B (o)\intB (o ) is connected. So, taking into consideration 1 rn 1 the generalized Caratheodory’s theorem ([2], theorem 2.29), we see that the point o 1 is a convex combination of not more than n points from B. These points denoted as a ,··· ,a , k 6 n, may be regarded as vertices of a (k−1)-dimensional simplex A and 1 k point o = α a +···+α a lies in its relative interior (α > 0,α +···+α = 1). 1 1 1 k k i 1 k Let c be the point of intersection of ray a o with the opposite facet of the simplex l l 1 A. So, o = α a +(1−α )c . Then 1 l l l l ko −a k = (1−α )kc −a k. 1 l l l l And [c ,a ] ⊂ A ⊂ B (o) implies that ka − c k 6 2, for all l ∈ 1,k. Therefore l l 1 l l r < ko −a k 6 2(1−α ). Thus α < 1− rn < 1, and finally α +···+α < k 6 1. n 1 l l l 2 n 1 k n Contradiction. Now let us show that the estimate is attained for spaces ℓn, ℓn . 1 ∞ Consider ℓn. Let A = {e }n be a standard basis for ℓn space and 1 i i=1 1 b = 1(e + ... + e ) ∈ co{e ,...,e }. The distance between point b and an arbi- n 1 n 1 n trary point from A is ka −bk = 2n−1. i n Consider ℓn . Let a = (−1)δij, where δ is Kroneker symbol, a = (a ,··· ,a ) ∞ ij ij i i1 in and A = {a }n . Now let b = 1(a +...+a ) = n−2,··· , n−2 ∈ co{a ,...,a }. And i i=1 n 1 n n n 1 n the distance from point b to an arbitrary point from A is ka −bk = 2n−1. (cid:0) i (cid:1) n (cid:4) So, Theorem 1 and inequality ζ ≥ 1 imply the CHD-constant of any 2-dimensional X normed space equals 1. Obviously, CHD-constant of ℓ space equals 2. 1 Remark 2. Let X bea Banach space, dimX = n. Then forevery d < ζ there exists a X set A that consists of not more than n points and meets the condition h+(coA,A) = d. Corollary 1. Let sets P and Q be plane sections of the unit ball with two parallel hyperplanes of dimension k, and let the hyperplane containing P contains 0 as well. Then it is possible to cover Q with the set min{2 k ;ζ }P using parallel translation. k+1 X Proof. Define η = min{2 k ;ζ }. Due Helly theorem it sufficies to prove that we could cover k+1 X any k-simplex ∆ ⊂ Q with the set ηP. Let us consider k-simplex ∆ ⊂ Q with verticex {x ,··· ,x }. Due to the 1 k+1 definition of the ζ and by Theorem 1 for any set of indices I ⊂ 1,(k+1), X we have co{x } ⊂ (B (x ) ∩ ∆). Using KKM theorem [14] we obtain that i η i i∈I i∈I S = (B (x )∩∆S) 6= ∅. Then ∆ ⊂ B (s), where s ∈ S ⊂ ∆. (cid:4) η i η i∈1,(k+1) T CONVEX HULL DEVIATION AND CONTRACTIBILITY 4 Let us show that Hilbert and 2-dimensional Banach spaces meet the requirements of Theorem 4. We consider the area covered with balls to be shaded. Balls’ radii may be taken equal to 1. Lemma 2. Let X be a Banach space, dimX = 2, then any admissible covering is contractible. Proof. Without loss of generality, let we have an admissible covering of a convex set V by balls B (a ), i = 1,n. Let us put S = B (a ). Since the unit ball is a convex 1 i 1 i i∈1,n closed body, the set S is homotopy equivaSlent to its nerve [4], in our case it is finite CW complex. Therefore, S is contractible iff S is connected, simply connected and its homology groups H (S) are trivial for k > 2. Obviously, S is connected set. k Let us show that the set S is simply connected and H (S) = 0 for k > 2. The unit k circle is a continuous closed line without self-intersections, it divides a plane in two parts. Afinite set of circles divides a plane ina finite number of connected components. Let us now shade the unit balls. It is remarkable, that the problem is stable against subtle perturbations of norm. To be more precise: if a norm does not meet the requirements of the theorem, then there exists a polygon norm, which does not meet them too. Let us choose a bounded not-covered area U with shaded boundary. It is possible to put a ball of radius 3ε (ε > 0) inside this area. There exists ε (ε > 0) such 1 1 2 2 that if B (a )∩B (a ) = ∅ for i ,i ∈ 1,n, than B (a )∩B (a ) = ∅. Denote 1 i1 1 iw 1 2 1+ε2 i1 1+ε2 i2 ε = min{ε , ε }. 1 2 Consider the following set Bc(o) = {x : hp,xi 6 1}, 1 p∈C \ where C is a finite set of unit vectors from space X∗, such that C = −C. So, Bc(o) is 1 the unit ball for some norm. According to [5], Corollary 2.6.1, it is possible to pick such a set C, that h+(Bc(o),B (o)) 6 ε. Then the set of balls Bc(a ), i = 1,n is admissible 1 1 1 i covering, contains the boundary of U, because B (o) ⊂ Bc(o), and it does not cover U 1 1 entirely. Furthermore nerve, and consequently homology group, of the sets Bc(a ) 1 i i∈1,n and S are coincide. S Now it suffices to show that the statement of the lemma is true in case of a polygon norm. In this case S is the neighborhood retract in R2 (see [6]), therefore straightfor- ward from Alexander duality (see [7], Chapter 4, §6) we obtain that H (S) = 0 for k k > 2. Now we shall prove that S is simply connected. Assume the contrary, there exist a norm, anadmissible covering of aconvex set V by ballsB (a ), i = 1,nandnon-shaded 1 i bounded set U with a shaded boundary. Note that its boundary appears to be a closed polygonal line without self-intersections. Let us define set A = co{a |i = 1,n}. i Let x be an arbitrary point of the set U. The union of the balls B (a ) is admissible 1 i covering of the set A, thus x 6∈ A. Then there exists a line l that separates x from set a A. This line may serve as a supporting line of set A. Let l k l be a supporting line of a U in a point v, such that sets U and A lie at one side from line ℓ. Line l divides the plane in two semiplanes. Let H be the semiplane that does not contain A, we denote + CONVEX HULL DEVIATION AND CONTRACTIBILITY 5 the other semiplane as H . Let points p,q ∈ l lie on different sides from v. We want − to choose all the edges of polygonal curve ∂U, that contain point v. We will call them vb ,i ∈ 1,k : cos∠pvb > cos∠pvb , i > j. i i j Notethatitisimpossibleforanyoftheedgestolieonlinel.Otherwisel issupporting line for a ball B (a ),p ∈ 1,n, and B (a )∩H 6= ∅, so we come to the contradiction. 1 p 1 p + We may pick such a number ε that the ball B (v) intersects only with particular edges ε of polygonal curve ∂U. From now on we use p, q, b ,i ∈ 1,k for points of intersection i of circle ∂B (v) with corresponding edges. Since v ∈ ∂U, it follows that there exists a ε point z on circle ∂B (v), such that the interior of segment vz lies in U and the ray vz ε lies between vb and vb . Then, since the ball is convex, there is no such ball B (a ), 1 k 1 i that simultaneously covers a point from the interior of vb and a point from vb , i.e. 1 k point v is covered by at least two balls, and the centers of these balls a , a are divided i j by ray vz in semiplane H . Again, since the ball is convex, point x = vz ∩ a a is − i j not covered by balls B (a ), B (a ), thus ka −a k = kx−a k+kx−a k > 2, which 1 i 1 j i j i j contradicts the fact that a and a are contained in ball B (v). (cid:4) i j 1 Lemma 3. Let X be an Euclidean space. Then any admissible covering is contractible. Proof. Let us remind that a closed convex set is contractible and in a Hilbert space the projection onto a closed convex set is unique. Since a projection onto a convex set is a continuous function of the projected point, it is enough to prove that a line segment, which connects a shaded point with its projection onto a convex hull of centers of an admissible covering, is shaded. Suppose that we have an admissible set of balls. The convex hull of its center is a polygon, Let us call it C. If a shaded point a is projected onto the v-vertex of the polygon, then the segment av is shaded as well. Let a shaded point a lying in the ball B (v) from a set of balls be projected onto the point b 6= v. 1 Let L be a hyperplane passing through point b and perpendicular to the line segment [a,b]. It divides the space in two half-spaces. The one with the point a we call H . a C is convex, thus it contains the segment [v,b]. Then it is impossible for point v to lie in H , so ∠abv > π, i.e. kv −ak > kv −bk. Thus, b ∈ B (v) and, consequently, A 2 1 ab ⊂ B (v). (cid:4) 1 3. Upper bound for CHD-constant in a Banach space Let J (x) = {p ∈ X∗ | hp,xi = kpk · kxk = kxk}. Let us introduce the following 1 characteristic of a space: ξ = sup sup kx−hp,xiyk, X kxk=1, p∈J1(y) kyk=1 Note that if y ∈ ∂B (0), p ∈ J (y), then vector (x−hp,xiy) is a metric projection 1 1 p of x onto the hyperplane H = {x ∈ X : hp,xi = 0}. So, ξ = sup sup ξ , p X y∈B1(o) p∈J1(x) X where ξp is half of diameter of a unit ball’s projection onto the hyperplane H . This X p implies the following remark. Let us use ξ for estimation of CHD-constant of X: X Lemma 4. Let y ∈ co[B (o)\intB (y )] and let p ∈ J (y). There is 1 r 1 1 x ∈ B (o)\intB (y ) such that hp,xi = hp,yi. 1 r 1 CONVEX HULL DEVIATION AND CONTRACTIBILITY 6 Then in hyperplane H = {x ∈ X : hp,xi = hp,o i} there exists a point x, such that p 1 x ∈ B (o)\intB (o ). 1 r 1 Proof. Define set B = B (o)\intB (y). Since y ∈ coB, there exist points a ,··· ,a ∈ B and 1 r 1 n a set of positive coefficients λ ,...,λ (λ +...+λ = 1), such that 1 n 1 n (3) y = λ a +...+λ a . 1 1 n n Let H+ = {x ∈ X : hp,xi > hp,yi. According to Lemma 1 set B is connected, thus, p sinceB\H+ isnotempty, ifthestatement weproveisnottrue, wearriveatB∩H+ = ∅. p p Then hp,a i < hp,yi and formula (3) implies i hp,yi = λ hp,a i+...+λ hp,a i < hp,yi. 1 1 n n Contradiction.(cid:4) Lemma 5. (4) ζ 6 sup inf sup kx−yk X kyk=1p∈J1(y) x∈B1(o): hp,x−yi=0 Proof. Let ε be a positive real number. Then, according to the definiton of the CHD-constant, there exists set D ⊂ B (o), such that h+(coD,D) > ζ −ε. It means that there exists 1 X point y ∈ coD : ρ(y,D) > ζ −2ε. Let us put r = ρ(y,D). So, D ⊂ B (o)\intB (y). X 1 r Hence, y ∈ co[B (o)\intB (y)]. Now let p ∈ J (y). 1 r 1 According to Lemma 4 there exists vector x ∈ B (o)\intB (y) : hp,x−yi = 0. And 1 r r 6 kx−yk. Therefore, ζ 6 ρ(y, D)+2ε = r+2ε 6 kx−yk+2ε. Now let ε tend to X zero. The lemma is proved. (cid:4) It becomes obvious that ξ = sup sup kx−yk. X y∈B1(o), x∈B1(o): p∈J1(y) hp,x−yi=0 Then Lemma 5 implies Theorem 5. ζ 6 ξ . X X Using Remark 1 and Theorem 5 we get Corollary 2. If H is a Hilbert space, then ζ = 1. H With the following lemma we can pass to finite subspace limit in CHD-constant calculations. Lemma 6. Let X be a Banach space and {x , x , ···} be a vector system in it, such 1 2 that the subspace Xˇ = Lin{x , x , ···} is dense in X. Then 1 2 (5) ζ = lim ζ , X n→∞ Xn where X = Lin{x ,··· ,x }. n 1 n Proof. Let us set ζ = ζ , and fix a real number ε > 0. Since X ⊂ X ⊂ X, the sequence X n n+1 ζ is monotone and bounded and, consequently, convergent. Let ζ = lim ζ . Since Xn 2 n→∞ Xn CONVEX HULL DEVIATION AND CONTRACTIBILITY 7 X ⊂ X it follows that ζ 6 ζ. According to the CHD-constant definition there exists a n 2 setA ⊂ B (o)andapointd ∈ coA,suchthatρ(d,A) > ζ−ε.Sinced ∈ coA,thereexist 1 2 a natural number N, points a ∈ A, and numbers α > 0, i ∈ 1,N, α +···+α = 1, i i 1 N such that d = α a +···+α a . 1 1 N N Then kd−a k > ζ − ε, i ∈ 1,N Since Xˇ = X, it is possible to pick points i 2 b ∈ B (o) ∩ Xˇ, i ∈ 1,N, so that ka −b k 6 ε. According to the definition of a i 1 i i 4 linear span for some natural n we have: b ∈ X . Let M = maxn ,i ∈ 1,N. Consider i i ni i set B = {b , ··· , b } in the space X . Let d = α b +···+α b ∈ coB, then 1 N M ε 1 1 N N N N ε kd −dk = α (b −a ) 6 α kb −a k 6 , ε j j j j j j 4 (cid:13) (cid:13) (cid:13)Xj=1 (cid:13) Xj=1 (cid:13) (cid:13) (cid:13) (cid:13) so for every i ∈ 1,N we have(cid:13) (cid:13) kd −b k = k(d −d)+(d−a )+(a −b )k > kd−a k−kd −dk−ka −b k > ζ−ε. ε i ε i i i i ε i i Thus ζ −ε 6 h+(coB,B) 6 ζ 6 ζ 6 ζ, and since ε > 0 was chosen arbitrarily, XM 2 ζ = ζ . (cid:4) 2 Let p′ ∈ [1; +∞] be such that 1 + 1 = 1, r = min{p,p′}, r′ = max{p,p′}. p p′ Lemma 7. Given p ∈ [1, +∞]. Let x ⊂ L ,1 6 p 6 ∞,i = 1,··· ,k; i p k k α = 1, α > 0 (i = 1,··· ,k), x = α x . i i 0 i i i=1 i=1 X X Then 1 1 k r k r (6) αikxi −x0krp 6 2−r1′ αiαjkxi −xjkrp , ! ! i=1 i=1,j=1 X X 1 k r (7) αiαjkxi −xjkrp 6 21r max kxikp. ! 16i6k i=1,j=1 X If 1 6 p 6 2, then the latter inequality can be strengthened: 1 k r 2−1 k −1 p (8) αiαjkxi −xjkrp! 6 2r1 k 1m6ia6xkkxikp. i=1,j=1 (cid:18) (cid:19) X Proof. The inequality (7) follow from Schoenberg’s inequalities ([8], Theorem 15.1): 1 1 k r 2−1 k r r αiαjkxi −xjkrp 6 2r1 max{1−αi} αikxikrp . ! 16i6k ! i=1,j=1 (cid:18) (cid:19) i=1 X X The inequality (8) was deduced by S.A. Pichugov and V.I. Ivanov in ([9], Assertion 1). CONVEX HULL DEVIATION AND CONTRACTIBILITY 8 Using the Riesz-Thorin theorem for spaces with a mixed L -norm ([8], §14), p S.A. Pichugov proved the following inequality ([10], Theorem 1): 1 k l r r α β k(x −x )−(y −y )k i j i 0 j 0 p ! i=1 j=1 XX 1 k l r (9) 6 2−r1′ αi1αi2kxi1 −xi2krp + βj1βj2kyj1 −yj2krp , ! i1=X1,i2=1 j1=X1,j2=1 k l k where α = β = 1, α > 0 (i = 1,··· ,k), β > 0 (j = 1,··· ,l), x = α x , i j i j 0 i i i=1 j=1 i=1 lP P P y = β y . 0 j j j=1 SubstPituting y for 0 and β for 1 in (9) we obtain the inequality (6). (cid:4) j j l Proof of Theorem 2. Consider the case of p ∈ (1,+∞). For L spaces and arbitrary set of vectors A = {x ,x ,··· ,x }, such that p 0 1 k x = k α x , k α = 1, α > 0 (i ∈ 1,k), A ⊂ B (0) we have 0 i=1 i i i=1 i i 1 1 P P 1 k r r minkx −x kr 6 α kx −x kr . 0 i p i i 0 p i∈1,k ! (cid:18) (cid:19) i=1 X Using (6) and (7), since set of vectors A was chosen arbitrarily, we get (cid:12) (cid:12) ζLp 6 2(r1−r1′) = 2(cid:12)(cid:12)p1−p1′(cid:12)(cid:12). As it was shown in proof of Theorem 1 that ζ = ζ = 2n−1. Thus, ζ = ζ = 2. ℓn1 ℓn∞ n L1 L∞ (cid:4) Remark 3. If 1 6 p 6 2, then, using in the proof of Theorem 2 inequality (8) instead of (7), we arrive at: (cid:12) (cid:12) (cid:12)1−1(cid:12) n−1 (cid:12)p p′(cid:12) (10) ζ 6 2 . ℓn p n (cid:18) (cid:19) Still without any answer remains the following questions: Question 1. Is the inequality (10) true if p ∈ (2;∞)? Question 2. Is the estimate in the inequality (2) exact in case of p ∈ (1;∞), p 6= 2? 4. Criterion of a Hilbert space In order to prove Theorem 3 we need the following lemma, which follows directly from the KKM theorem [14]. Lemma 8. Let X be a Banach space. Suppose the triangle a a a ⊂ X satisfies the 1 2 3 inequality diama a a 6 2R and is covered by balls B (a ), i = 1,2,3. Then these balls 1 2 3 R i have a common point lying in the plane of the triangle. CONVEX HULL DEVIATION AND CONTRACTIBILITY 9 Taking into account Lemma 8, the proof of Theorem 3 is very similar to the one of Theorem 5 from [1]. Proof of theorem 3. Using Theorem 1 and Corollary 2 it suffices to prove that a Banach space X, with dimX > 3 and ζ = 1 is a Hilbert space. According to the well-known results X obtained by Frechet and Blashke-Kakutani, it is enough to describe only the case when dimX = 3. We need to show that if ζ = 1, then for every 2-dimensional subspace X there exists a unit-norm operator that projects X onto this particular subspace. Let 0 ∈ L be an arbitrary 2-dimensional subspace in X, point c is not contained in L. We denote B2(0) = L∩B (0) (it is a ball of radius n ∈ N in space L). For every n ∈ N n n let us introduce the following notations: E = {x ∈ L : kc−xk ≤ n}, n F = {x ∈ L : kc−xk = n}. n If n is big enough, these sets are nonempty. Let x , x , x be arbitrary points from 1 2 3 E . The CHD-constant of space X equals 1, so the balls B2(x ),i = 1,2,3 cover the n n i triangle x x x . According to Lemma 8, their intersection is not empty. According to 1 2 3 Helly theorem, the set S = B2(x) n n x\∈En is nonempty as well. Let us pick a ∈ S , then by construction we have n n (11) kx−a k 6 kx−ck n for every x ∈ F . Let us show that n kx−a k 6 kx−ck n for every x ∈ E . Suppose that for some x ∈ E n n (12) kx−a k > kx−ck. n According to (11) we may assume that x ∈ E \ F . Set E is bounded and its n n n boundary relatively to subspace L coincides with F , thus there exists point b ∈ F , n n such that x is contained in interval (a , b). Then a −x = λ(a −b), 0 < λ < 1. n n n Note that c − x = (c − a ) + (a − x) = c − a + λ(a − b), then (12) may be n n n n reformulated as kc−a +λ(a −b)k < λka −bk. n n n So, kc−bk = k(c−a )+λ(a −b)+(1−λ)(a −b)k 6 n n n 6 k(c−a )+λ(a −b)k+(1−λ)ka −bk < λka −bk+(1−λ)ka −bk = ka −bk, n n n n n n and it contradicts (11). Consider the sequence {a }. Note that E ⊂ E and ∪∞ E = L. So, starting with n n n+1 i=1 i afixednaturalk,theinclusion0 ∈ E , n > k becomestrue, thuswhenx = 0inequality n (11) implies ka k 6 kck, n > k, i.e. the sequence {a } is bounded. It means that n n sequence {a } has a limit point a. Then every point x ∈ L satisfies kx−ak 6 kx−ck. n Let now represent every element z ∈ X in the form z = tc+x (x ∈ L, t ∈ R). CONVEX HULL DEVIATION AND CONTRACTIBILITY 10 Operator P(z) = P(tc+x) = ta+x projects X onto L. In addition: x x kP(z)k = kta+xk = |t| a+ 6 |t| c+ = ktc+xk = kzk. t t (cid:13) (cid:13) (cid:13) (cid:13) Hence, the kPk = 1 and takin(cid:13)g into(cid:13)consid(cid:13)eration(cid:13)the theorem of Blashke and (cid:13) (cid:13) (cid:13) (cid:13) Kakutani we come to a conclusion that X is a Hilbert space. (cid:4) Proof of Theorem 4. It remains to check that in every Banach space X that is not a Hilbert one, where dimX = 3, there exist a convex set and an admissible and not contractible covering. To make the proof easier we first need to prove a trivial statement from geometry. Let hyperplane H divide space X in two half-spaces H ,H . Let M be a bounded + − set in H. We want to cover set M with balls B = {∪B (a ) | i ∈ 1,n,n ∈ N} and call d i this covering (ε,r,H )-good if h+(B, H ) 6 ε. + − Lemma 9. Let X be a Banach space, 3 6 dimX < +∞. Let hyperplane H divide X in two half-spaces: H and H . Let M be a bounded set in H. Then for every + − ε > 0, d > 0 there exists an admissible set of balls B (a ),i ∈ 1,N,N ∈ N, such d i that set B = B (a ) may be regarded as (ε,d,H )-good covering of set M and d i + i∈1,N S co(M ∪{a }) ⊂ B, i ∈ 1,N. i Proof. Let dimX = n. Without loss of generality we assume that ε < d and H is the sup- porting hyperplane for the ball B (0) and B (0) ⊂ H . For any r > 0 and a ∈ X d d − we use C (a) to denote a (n−1)-dimensional hypercube centered in a that lies in the r hyperplane parallel to H, where r is the length of its edges. Let x ∈ H ∩B (0). Then d h+(B (ε x ),H ) 6 ε. Let D = B (ε x ) ∩ H. Note that x is an inner point of set d kxk − d kxk D relatively to subspace L. In a finite dimensional linear space all norms are equiv- alent, so C (x) ⊂ D for some r > 0. As the ball B (ε x ) is centrally-symmetric, it r d kxk contains affine hypercube co C (x)∪C (ε x ) . Consider next an arbitrary bounded r r kxk set M ⊂ L. Since it is bound(cid:16)ed, M ⊂ C (b), (cid:17)where b ∈ L, R > 0. We suppose that R R = kr, k ∈ N. Lets split hypercube C (b) in hypercubes with edges of length r and R let b ,i ∈ 1,N be the centers of these hypercubes. Hence, from the above, the balls i B (b −(d−ε) x ) give us the necessary covering. (cid:4) d i kxk Let us consider an approach to construct an admissible and not contractible covering of a convex set. Let a Banach space X be a non-Hilbert one and dimX = 3. According to Theorem 3, ζ > 1, by Remark 2, there exists set A = {a , a , a } ⊂ B (0) and point b ∈ coA, X 1 2 3 1 such that ρ(b, A) = 1 + 4ε > 1. According to Theorem 1, o ∈/ H. Consider the balls B (a ), i ∈ 1,3, let B = B (a )∪B (a )∪B (a ). It is obvious that b ∈/ B . 1+ε i 1 1+ε 1 1+ε 2 1+ε 3 1 Since allthe edges oftriangle a a a liein B ,facets 0a a , 0a a , 0a a of tetrahedron 1 2 3 1 1 2 1 3 2 3 0a a a lie in B . Let H be a plane passing through points a , a , a . 1 2 3 1 1 2 3 Let H divide space X in two half-spaces: H and H . Let 0 ∈ H . According + − + to Lemma 9 there exists an (ε,1 + ε,H )-good covering of triangle a a a with an + 1 2 3 admissible set of balls that have centers lying in a set C = {c , i ∈ 1,N}, N ∈ N. Let i