� EXTRACTA MATHEMATICAE Volumen 33, Número 2, 2018 instituto de investigación de matemáticas de la universidad de extremadura EXTRACTA MATHEMATICAE Vol. 35, Num. 2 (2020), 127 – 135 doi:10.17398/2605-5686.35.2.127 Available online May 7, 2020 Extreme and exposed points of L(nl2∞) and Ls( nl2∞) Sung Guen Kim Department of Mathematics, Kyungpook National University Daegu 702-701, South Korea sgk317@knu.ac.kr Received November 26, 2019 Presented by Jesús M.F. Castillo Accepted April 10, 2020 Abstract: For every n ≥ 2 this paper is devoted to the description of the sets of extreme and exposed points of the closed unit balls of L(nl2∞) and Ls(nl2∞), where L(nl2∞) is the space of n-linear forms on R2 with the supremum norm, and Ls(nl2∞) is the subspace of L(nl2∞) consisting of symmetric n-linear forms. First we classify the extreme points of the closed unit balls of L(nl2∞) and Ls(nl2∞), correspondingly. As corollaries we obtain |ext BL(nl2∞)| = 2 (2n) and |ext BLs(nl2∞)| = 2 n+1. We also show that exp BL(nl2∞) = ext BL(nl2∞) and exp BLs(nl2∞) = ext BLs(nl2∞) . Key words: n-linear forms, symmetric n-linear forms, extreme points, exposed points. AMS Subject Class. (2010): 46A22. 1. Introduction Let n ∈ N,n ≥ 2. We write BE for the unit ball of a real Banach space E and the dual space of E is denoted by E∗. An element x ∈ BE is called an extreme point of BE if y,z ∈ BE with x = 12 (y + z) implies x = y = z. We denote by ext BE the set of all the extreme points of BE. An element x ∈ BE is called an exposed point of BE if there is a f ∈ E∗ so that f(x) = 1 = ‖f‖ and f(y) < 1 for every y ∈ BE\{x}. It is easy to see that every exposed point of BE is an extreme point. We denote by exp BE the set of exposed points of BE. We denote by L(nE) the Banach space of all continuous n-linear forms on E endowed with the norm ‖T‖ = sup‖xk‖=1 |T(x1, · · · ,xn)|. Ls( nE) denote the closed subspace of all continuous symmetric n-linear forms on E. Let us say about the history of the classifications of extreme and exposed points of the unit ball of continuous (symmetric) multilinear forms on a Banach space. Kim [1] initiated and classified ext BLs(2l2∞) and exp BLs(2l2∞), where ln∞ = R n with the supremum norm. It was shown that ext BLs(2l2∞) = exp BLs(2l2∞). Kim [2, 3, 4, 5] classified ext BLs(2d∗(1,w)2), ext BL(2d∗(1,w)2), exp BLs(2d∗(1,w)2), and exp BL(2d∗(1,w)2), where d∗(1,w) 2 = R2 with the octago- nal norm ‖(x,y)‖w = max { |x|, |y|, |x|+|y| 1+w } . Kim [6, 7] classified ext BLs(2R2h(w)) ISSN: 0213-8743 (print), 2605-5686 (online) c©The author(s) - Released under a Creative Commons Attribution License (CC BY-NC 3.0) https://doi.org/10.17398/2605-5686.35.2.127 mailto:sgk317@knu.ac.kr https://www.eweb.unex.es/eweb/extracta/ https://creativecommons.org/licenses/by-nc/3.0/ 128 s. g. kim and ext BL(2R2 h(w) ), where where R 2 h(w) = R2 with the hexagonal norm ‖(x,y)‖h(w) = max{|y|, |x|+ (1−w)|y|}. Kim [8, 9, 10] classified ext BLs(2l3∞), ext BLs(3l2∞) and ext BL(3l2∞). It was shown that every extreme point is exposed in each space. Kim [11] characterized ext BL(2ln∞) and ext BLs(2ln∞). Recently, Kim [12] classified ext BL(2l3∞) and showed exp BL(2l3∞) = ext BL(2l3∞). 2. The extreme and exposed points of the unit ball of L(nl2∞) Let l2∞ = {(x,y) ∈ R2 : ‖(x,y)‖∞ = max(|x|, |y|)}. For n ≥ 2, we denote Wn := {[(1,w1), . . . , (1,wn)] : wj = ±1 for j = 1, . . . ,n}. Note that Wn has 2n elements in Sl2∞ ×···×Sl2∞. Recall that the Krein-Milman Theorem [13] say that every nonempty com- pact convex subset of a Housdorff locally convex space is the closed convex hull of its set of extreme points. Hence, the unit ball of l2∞ is the closed convex hull of {(1, 1), (−1, 1), (1,−1), (−1,−1)}. Theorem 2.1. Let n ≥ 2 and T ∈L(nl2∞). Then, ‖T‖ = sup W∈Wn |T(W)|. Proof. It follows that from the Krein-Milman theorem and multi- linearity of T. Let Z1, . . . ,Z2n be an ordering of the monomials xl1 · · ·xljyk1 · · ·ykn−j with {l1, · · · , lj,k1, · · · ,kn−j} = {1, · · · ,n}. Note that {Z1, . . . ,Z2n} is a basis for L(nl2∞). Hence, dim(L(nl2∞)) = 2n. If T ∈L(nl2∞), then, T = 2n∑ k=1 akZk for some a1, . . . ,a2n ∈ R. By simplicity, we denote T = (a1, · · · ,a2n )t. Let W1, . . . ,W2n be an ordering of the elements of Wn. Let M(Z1, . . . ,Z2n ; W1, . . . ,W2n ) = [Zi(Wj)] be the 2n × 2n matrix. Note that, for every T ∈L(nl2∞), M(Z1, . . . ,Z2n ; W1, . . . ,W2n )T = (T(W1), . . . ,T(W2n )) t. Here, (�1, . . . ,�2n ) t denote the transpose of (�1, . . . ,�2n ). extreme and exposed points of L(nl2∞) and Ls(nl2∞) 129 Theorem 2.2. Let n ≥ 2. Then, ext BL(nl2∞) = { M(Z1, . . . ,Z2n ; W1, . . . ,W2n ) −1(�1, . . . ,�2n ) t : �j = ±1, j = 1, . . . , 2n } . Proof. Claim 1: M(Z1, . . . ,Z2n ; W1, . . . ,W2n ) is invertible. Consider the equation M(Z1, . . . ,Z2n ; W1, . . . ,W2n )(t1, . . . , t2n ) t = (0, . . . , 0)t. (*) Let a1, · · · ,a2n be a solution of (*) and let T = ∑2n k=1 akZk ∈L( nl2∞). Then, T(Wj) = 0 j = 1, . . . , 2 n. By Theorem 2.1, ‖T‖ = 0, hence T = 0. Since Z1, . . . ,Z2n are linearly independent in L(nl2∞), we have aj = 0 for all j = 1, . . . , 2n. Hence, the equation (*) has only zero solution. Therefore, M(Z1, . . . ,Z2n ; W1, . . . ,W2n ) is invertible. Claim 2: M(Z1, . . . ,Z2n ; W1, . . . ,W2n ) −1(�1, . . . ,�2n ) t is an extreme point for �j = ±1, (j = 1, . . . , 2n). Let T := M(Z1, . . . ,Z2n ; W1, . . . ,W2n ) −1(�1, . . . ,�2n ) t. Since M(Z1, . . . ,Z2n ; W1, . . . ,W2n )T = (�1, . . . ,�2n ) t, T(Wj) = �j for j = 1, . . . , 2 n. By Theorem 2.1, ‖T‖ = sup 1≤j≤2n |T(Wj)| = sup 1≤j≤2n |�j| = 1. Suppose that T = 1 2 (T1 + T2) for some Tk ∈ BL(nl2∞) (k = 1, 2). We may write T1 = M(Z1, . . . ,Z2n ; W1, . . . ,W2n ) −1(�1, . . . ,�2n ) t + (δ1, . . . ,δ2n ) t and T2 = M(Z1, . . . ,Z2n ; W1, . . . ,W2n ) −1(�1, . . . ,�2n ) t − (δ1, . . . ,δ2n )t for some δj ∈ R (j = 1, . . . , 2n). Note that (Tk(W1), . . . ,Tk(W2n )) t = M(Z1, . . . ,Z2n ; W1, . . . ,W2n )Tk for k = 1, 2. 130 s. g. kim Therefore, (T1(W1), . . . ,T1(W2n )) t = (�1, . . . ,�2n ) t + M(Z1, . . . ,Z2n ; W1, . . . ,W2n )(δ1, . . . ,δ2n ) t and (T2(W1), . . . ,T2(W2n )) t = (�1, . . . ,�2n ) t −M(Z1, . . . ,Z2n ; W1, . . . ,W2n )(δ1, . . . ,δ2n )t. Hence, for j = 1, . . . , 2n, T1(Wj) = �j + (Z1(Wj), . . . ,Z2n (Wj))(δ1, . . . ,δ2n ) t, and T2(Wj) = �j − (Z1(Wj), . . . ,Z2n (Wj))(δ1, . . . ,δ2n )t. It follows that, for j = 1, . . . , 2n, 1 ≥ max{|T1(Wj)|, |T2(Wj)|} = |�j| + |(Z1(Wj), . . . ,Z2n (Wj))(δ1, . . . ,δ2n )t| = 1 + |(Z1(Wj), . . . ,Z2n (Wj))(δ1, . . . ,δ2n )t|, which shows that (Z1(Wj), . . . ,Z2n (Wj))(δ1, . . . ,δ2n ) t = 0 for j = 1, . . . , 2n. Hence, M(Z1, . . . ,Z2n ; W1, . . . ,W2n )(δ1, . . . ,δ2n ) t = 0 . Therefore, (δ1, . . . ,δ2n ) t = M(Z1, . . . ,Z2n ; W1, . . . ,W2n ) −1(0, . . . , 0)t = (0, . . . , 0)t. Hence, Tk = T for k = 1, 2. Therefore, T is extreme. Suppose that T ∈ ext BL(nl2∞). Note that (T(W1), . . . ,T(W2n )) t = M(Z1, . . . ,Z2n ; W1, . . . ,W2n )T. extreme and exposed points of L(nl2∞) and Ls(nl2∞) 131 Claim 3: |T(Wj)| = 1 for all j = 1, . . . , 2n. If not. There exists 1 ≤ j0 ≤ 2n such that |T(Wj0 )| < 1. Let δ0 > 0 such that |T(Wj0 )| + δ0 < 1. Let T1 = M(Z1, . . . ,Z2n ; W1, . . . ,W2n ) −1 × (T(W1), . . . ,T(Wj0−1),T(Wj0 ) + δ0,T(Wj0+1), . . . ,T(W2n )) t and T2 = M(Z1, . . . ,Z2n ; W1, . . . ,W2n ) −1 × (T(W1), . . . ,T(Wj0−1),T(Wj0 ) −δ0,T(Wj0+1), . . . ,T(W2n )) t. Hence, T1(Wj0 ) = T(Wj0 ) + δ0,T2(Wj0 ) = T(Wj0 ) −δ0,T1(Wj) = T2(Wj) = T(Wj) (j 6= j0). Obviously, T 6= Tk for k = 1, 2. By Theorem 2.1, ‖Tk‖ = 1 for k = 1, 2 and T = 1 2 (T1 + T2), which is a contradiction. Therefore, T = M(Z1, . . . ,Z2n ; W1, . . . ,W2n ) −1(T(W1), . . . ,T(W2n )) t with |T(Wj)| = 1 for all j = 1, . . . , 2n. Kim [10] characterized ext BL(3l2∞). Notice that using Wolfram Math- ematica 8 and Theorem 2.2, we can exclusively describe ext BL(nl2∞) for a given n ≥ 2. For every T ∈L(nl2∞), we let Norm(T) := { [(1,w1), . . . , (1,wn)] ∈Wn : |T((1,w1), . . . , (1,wn))| = ‖T‖ } . We call Norm(T) the set of the norming points of T. Corollary 2.3. (a) Let n ≥ 2. ext BL(nl2∞) has exactly 2 (2n) elements. (b) Let n ≥ 2 and T ∈ L(nl2∞) with ‖T‖ = 1. Then T ∈ ext BL(nl2∞) if and only if Norm(T) = Wn. Theorem 2.4. ([4]) Let E be a real Banach space such that ext BE is finite. Suppose that x ∈ ext BE satisfies that there exists an f ∈ E∗ with f(x) = 1 = ‖f‖ and |f(y)| < 1 for every y ∈ ext BE\{±x}. Then x ∈ exp BE. 132 s. g. kim Theorem 2.5. Let n ≥ 2. Then, exp BL(nl2∞) = ext BL(nl2∞). Proof. Let T ∈ ext BL(nl2∞) and let f := 1 2n ∑ 1≤j≤2n sign(T(Wj))δWj ∈L( nl2∞) ∗. Note that 1 = ‖f‖ = f(T). Let S ∈ ext BL(nl2∞) be such that |f(S)| = 1. We will show that S = T or S = −T. It follows that 1 = |f(S)| = | 1 2n ∑ 1≤j≤2n sign(T(Wj))S(Wj)| ≤ 1 2n ∑ 1≤j≤2n |S(Wj)| ≤ 1, which shows that S(Wj) = sign(T(Wj)) (1 ≤ j ≤ 2n) or S(Wj) = −sign(T(Wj)) (1 ≤ j ≤ 2n). Suppose that S(Wj) = −sign(T(Wj)) (1 ≤ j ≤ 2n). It follows that S = M(Z1, . . . ,Z2n ; W1, . . . ,W2n ) −1(S(W1), . . . ,S(W2n )) t = M(Z1, . . . ,Z2n ; W1, . . . ,W2n ) −1(−sign(T(W1)), . . . ,−sign(T(W2n )))t = M(Z1, . . . ,Z2n ; W1, . . . ,W2n ) −1(−T(W1), . . . ,−T(W2n ))t = −T. Note that if S(Wj) = sign(T(Wj)) (1 ≤ j ≤ 2n), then S = T. By Theorem 2.4, T is exposed. extreme and exposed points of L(nl2∞) and Ls(nl2∞) 133 3. The extreme and exposed points of the unit ball of Ls(nl2∞) Let n ≥ 2 and Un := { [(1, 1), (1, 1), . . . , (1, 1)], [(1,−1), (1, 1), . . . , (1, 1)], [(1,−1), (1,−1), (1, 1), . . . , (1, 1)], [(1,−1), (1,−1), (1 − 1), (1, 1), . . . , (1, 1)], . . . , [(1,−1), (1,−1), . . . , (1,−1), (1, 1)], [(1,−1), (1,−1), . . . , (1,−1), (1,−1)] } . Note that Un has n + 1 elements in Sl2∞ ×···×Sl2∞. Theorem 3.1. Let n ≥ 2 and T ∈Ls(nl2∞). Then, ‖T‖ = sup U∈Un |T(U)| . Proof. It follows that from Theorem 2.1 and symmetry of T. For j = 0, . . . ,n, we let Fj = ∑ {l1,··· ,lj,k1,··· ,kn−j}={1,··· ,n} xl1 · · ·xljyk1 · · ·ykn−j. Then, {F0, . . . ,Fn} is a basis for Ls(nl2∞). Hence, dim(Ls(nl2∞)) = n + 1. If T ∈Ls(nl2∞), then, T = n∑ j=0 bjFj for some b0, . . . ,bn ∈ R. By simplicity, we denote T = (b0, · · · ,bn)t. For j = 0, . . . ,n, we let Uj = [(1,u1), . . . , (1,un)] ∈Un, where uk = −1 for 1 ≤ k ≤ j and uk = 1 for j + 1 ≤ k ≤ n. Let M(F0, . . . ,Fn; U0, . . . ,Un) = [Fi(Uj)] be the (n + 1) × (n + 1) matrix. Note that, for every T ∈Ls(nl2∞), M(F0, . . . ,Fn; U0, . . . ,Un)T = (T(U0), . . . ,T(Un)) t. By analogous arguments in the claim 1 of Theorem 2.2, M(F0, . . . ,Fn; U0, . . . ,Un) is invertible. 134 s. g. kim Theorem 3.2. Let n ≥ 2. Then, ext BLs(nl2∞) = { M(F0, . . . ,Fn; U0, . . . ,Un) −1(�0, . . . ,�n) t : �j = ±1,j = 0, . . . ,n } . Proof. It follows by Theorem 3.1 and analogous arguments in the claims 2 and 3 of Theorem 2.2. Notice that using Wolfram Mathematica 8 and Theorem 3.2, we can ex- clusively describe ext BLs(nl2∞) for a given n ≥ 2. For every T ∈Ls(nl2∞), we let Norm(T) := { [(1,u1), . . . , (1,un)] ∈Un : |T((1,u1), . . . , (1,un))| = ‖T‖ } . We call Norm(T) the set of the norming points of T. Corollary 3.3. (a) Let n ≥ 2. ext BLs(nl2∞) has exactly 2 n+1 elements. (b) Let n ≥ 2 and T ∈ Ls(nl2∞) with ‖T‖ = 1. Then T ∈ ext BLs(nl2∞) if and only if Norm(T) = Un. Theorem 3.4. Let n ≥ 2. Then, exp BLs(nl2∞) = ext BLs(nl2∞). Proof. Let T ∈ ext BLs(nl2∞) and let f := 1 n + 1 ∑ 0≤j≤n sign(T(Uj))δUj ∈Ls( nl2∞) ∗. Note that 1 = ‖f‖ = f(T). By analogous arguments in the proof of Theorem 2.5, f exposes T. Therefore, T is exposed. Questions. (a) Let n ≥ 2 and �1, . . . ,�2n be fixed with �j = ±1, (j = 1, . . . , 2n). Is it true that ext BL(nl2∞) = { M(Z1, . . . ,Z2n ; W1, . . . ,W2n ) −1(�1, . . . ,�2n ) t : Z1, . . . ,Z2n,W1, . . . ,W2n are any ordering } ? (b) By Theorem 2.2, M(Z1, . . . ,Z2n ; W1, . . . ,W2n ) −1(�1, . . . ,�2n ) t is ex- treme if Z1, . . . ,Z2n , W1, . . . ,W2n are any ordering. Similarly, we may ask the following: Let n ≥ 2 and δ0, . . . ,δn be fixed with δk = ±1, (k = 0, . . . ,n). Is it true that ext BLs(nl2∞) = { M(F0, . . . ,Fn; U0, . . . ,Un) −1(δ0, . . . ,δn) t : F0, . . . ,Fn,U0, . . . ,Un are any ordering } ? extreme and exposed points of L(nl2∞) and Ls(nl2∞) 135 Acknowledgements The author is thankful to the referee for the careful reading and considered suggestions leading to a better presented paper. References [1] S.G. Kim, The unit ball of Ls(2l2∞), Extracta Math. 24 (2009), 17 – 29. [2] S.G. Kim, The unit ball of Ls(2d∗(1,w)2), Kyungpook Math. J. 53 (2013), 295 – 306. [3] S.G. Kim, Extreme bilinear forms of L(2d∗(1,w)2), Kyungpook Math. J. 53 (2013), 625 – 638. [4] S.G. Kim, Exposed symmetric bilinear forms of Ls(2d∗(1,w)2), Kyungpook Math. J. 54 (2014), 341 – 347. [5] S.G. Kim, Exposed bilinear forms of L(2d∗(1,w)2), Kyungpook Math. J. 55 (2015), 119 – 126. [6] S.G. Kim, The unit ball of L(2R2 h(w) ), Bull. Korean Math. Soc. 54 (2017), 417 – 428. [7] S.G. Kim, Extremal problems for Ls(2R2h(w)), Kyungpook Math. J. 57 (2017), 223 – 232. [8] S.G. Kim, The unit ball of Ls(2l3∞), Comment. Math. 57 (2017), 1 – 7. [9] S.G. Kim, The geometry of Ls(3l2∞), Commun. Korean Math. Soc. 32 (2017), 991 – 997. [10] S.G. Kim, The geometry of L(3l2∞) and optimal constants in the Bohnenblust- Hill inequality for multilinear forms and polynomials, Extracta Math. 33 (1) (2018), 51 – 66. [11] S.G. Kim, Extreme bilinear forms on Rn with the supremum norm, Period. Math. Hungar. 77 (2018), 274 – 290. [12] S.G. Kim, The unit ball of the space of bilinear forms on R3 with the supremum norm, Commun. Korean Math. Soc. 34 (2) (2019), 487 – 494. [13] M.G. Krein, D.P. Milman, On extreme points of regular convex sets, Studia Math. 9 (1940), 133 – 137. Introduction The extreme and exposed points of the unit ball of L(nl2) The extreme and exposed points of the unit ball of Ls(nl2)