Extreme Points, Exposed Points and Smooth Points of the Space LS(2l∞3)

Sung Guen Kim

doi:10.5666/KMJ.2020.60.3.485

Article

Kyungpook Mathematical Journal 2020; 60(3): 485-505

Published online September 30, 2020

Extreme Points, Exposed Points and Smooth Points of the Space LS(2l∞3)

Sung Guen Kim

Department of Mathematics, Kyungpook National University, Daegu 41566, Republic of Korea
e-mail : sgk317@knu.ac.kr

Received: September 5, 2019; Revised: March 10, 2020; Accepted: April 21, 2020

Abstract

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Abstract
1. Introduction
2. The Extreme Points of the Unit Ball of Ls(l2∞3)
3. The Exposed Points of the Unit Ball of Ls(l2∞3)
4. The Smooth Points of the Unit Ball of Ls(l2∞3)
Acknowledgements
References

We present a complete description of all the extreme points of the unit ball of Ls(l2∞3) which leads to a complete formula for ||f|| for every f∈Ls(l2∞3)*. We also show that extBLs(l2∞3)⊂extBLs(l2∞n) for every n ≥ 4. Using the formula for ||f|| for every f∈Ls(l2∞3)*, we show that every extreme point of the unit ball of Ls(l2∞3) is exposed. We also characterize all the smooth points of the unit ball of Ls(l2∞3).

Keywords: symmetric bilinear forms on ℝ,³ with the supremum norm, extreme points, exposed points, smooth points.

1. Introduction

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Abstract
1. Introduction
2. The Extreme Points of the Unit Ball of Ls(l2∞3)
3. The Exposed Points of the Unit Ball of Ls(l2∞3)
4. The Smooth Points of the Unit Ball of Ls(l2∞3)
Acknowledgements
References

We denote by B_E the closed unit ball of a real Banach space E and by E^* the dual space of E. A point x ∈ B_E is called an extreme point of B_E if the equation x=12(y+z) for some y, z ∈ B_E implies x = y = z. A point x ∈ B_E is called an exposed point of B_E if there is f ∈ E^* so that f(x) = 1 = ||f|| and f(y) < 1 for every y ∈ B_E \ {x}. A point x ∈ B_E is called a smooth point of B_E if there is a unique f ∈ E^* so that f(x) = 1 = ||f||. It is easy to see that every exposed point of B_E is an extreme point. We denote by extB_E, expB_E and smB_E the set of extreme points, the set of exposed points and the set of smooth points of B_E, respectively. A mapping P: E → ℝ is a continuous 2-homogeneous polynomial if there exists a continuous bilinear form L on the product E × E such that P(x) = L(x, x) for every x ∈ E. We denote by ℒ(²E) the Banach space of all continuous bilinear forms on E endowed with the norm ||L|| = sup_{||x||=||y||=1} |L(x, y)|. The subspace of all continuous symmetric bilinear forms on E is denoted by ℒ_s(²E). We denote by ℘(²E) the Banach space of all continuous 2-homogeneous polynomials from E into ℝ endowed with the norm ||P|| = sup_||x||=1 |P(x)|. For more details about the theory of multilinear mappings and polynomials on a Banach space, we refer to [7].

In 1998, Choi et al. [2, 3] initiated the classification of the extreme points of the unit ball of P(l212) and P(l222). Kim classified the exposed 2-homogeneous polynomials in P(l2p2) (1≤p≤∞) [11] and the extreme points, exposed points, and smooth points of the unit ball of ℘(²d_*(1, w)²) [13, 15, 19], where d_*(1, w)² = ℝ² with the octagonal norm of weight w. Recently, Kim [25, 26, 28] classified all the extreme points, exposed points, and smooth points of the unit ball of P(ℝ2h(12)2), where ℝh(12)2 is the plane ℝ² with the hexagonal norm of weight 12.

In 2009, Kim [12] initiated the classification of the extreme points, exposed points, and smooth points of the unit ball of Ls(l2∞3). Kim [14, 16, 17, 18, 20, 21, 22, 23, 24] classified the extreme points, exposed points, and smooth points of the unit balls of the spaces

Ls(l3∞2),L(l3∞2),Ls(d2*(1,w)2),L(d2*(1,w)2),Ls(ℝ2h(w)2) and L(ℝ2h(w)2),

where ℝh(w)2 is the plane ℝ² with the hexagonal norm of weight w, ||(x, y)||_h(w) = max{|y|, |x| + (1 − w)|y|}. Recently, Kim [25] characterized the extreme points of the spaces Ls(l2∞n) and L(l2∞n) for n ≥ 2.

Let n ≥ 2 and l∞n:=ℝn with the supremum norm. Given {aij}i,j=1n⊂ℝ, let T∈L(l2∞n) be defined by the rule

T((x1,x2,…,xn),(y1,y2,…,yn))=∑1≤i,j≤naijxiyj.

If n = 2, for simplicity, we will write

T=(a11,a22,a12,a21)t

as a 4 × 1 column vector. If T∈Ls(l2∞2), we will write

T=(a11,a22,2a12)t

as a 3 × 1 column vector. If n = 3, for simplicity, we will write

T=(a11,a22,a33,a12,a21,a13,a31,a23,a32)t

as a 9 × 1 column vector. If T∈Ls(l2∞3), we will write

T=(a11,a22,a33,2a12,2a13,2a23)t

as a 6 × 1 column vector.

In [12] it was shown:

(a) extBLs(l2∞2)={±(1,0,0)t,±(0,1,0)t,±(12,-12,1)t,±(12,-12,-1)t};
(b) extBLs(l2∞2)=extBLs(l2∞2).

We refer to ([1, 2, 3, 4, 5, 6, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 20, 21, 22, 23, 24, 25, 26, 27, 28, 29, 30, 31, 32] for some recent works about extremal properties of multilinear mappings and homogeneous polynomials on some classical Banach spaces.

In this paper we present a complete description of the 42 extreme points of the unit ball of Ls(l2∞3) which leads to a complete formula of ||f|| for every f∈Ls(l2∞3)*. We also show that extBLs(l2∞2)⊂extBLs(l2∞n) for every n ≥ 4. Using the formula of ||f|| for every f∈Ls(l2∞3)*, we show that every extreme point of the unit ball of Ls(l2∞3) is exposed. The main result about smooth points is known to “the Mazur density theorem.” Recall that the Mazur density theorem says that the set of all the smooth points of a solid closed convex subset of a separable Banach space is a residual subset of its boundary. Motivated by the Mazur density theorem we characterize all the smooth points of the unit ball of Ls(l2∞3).

2. The Extreme Points of the Unit Ball of Ls(l2∞3)

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Abstract
1. Introduction
2. The Extreme Points of the Unit Ball of Ls(l2∞3)
3. The Exposed Points of the Unit Ball of Ls(l2∞3)
4. The Smooth Points of the Unit Ball of Ls(l2∞3)
Acknowledgements
References

Recently, Kim [22] showed the following: Let

Ω={(1,1,1,1,1,1),(1,-1,1,0,1,0),(1,1,-1,1,0,0),(-1,1,1,0,0,1),(1,1,1,-1,1,-1),(1,-1,-1,0,0,1),(-1,-1,1,1,1,0,0),(1,1,1,1,-1,-1)(-1,1,-1,0,1,0),(1,1,1,-1,-1,1)}

and

Γ={[(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),(1,1,-1)],[(1,1,-1),(-1,1,1)],[(-1,1,1),(-1,1,1)}.

The following statements hold true.

(a) Let T=(a11,a22,a33,2a12,2a132a23)t∈Ls(l2∞3) with ||T|| = 1. Then, T∈extBLs(l2∞3) if and only if there exist at least 6 linearly independent vectors W₁, …, W₆ ∈ Ω and Z₁, …, Z₆ ∈ Γ such that

Wj·S=S(Zj) for all S∈Ls(l2∞3) and ∣T(Zj)∣ =1 for j=1,…,6.

Let A be the invertible 6 × 6 matrix such that the j-row vector of A as [Row(A)]_j = W_j for j = 1, …, 6. Then, AS = (S(Z₁), · · ·, S(Z₆))^t for all S∈Ls(l2∞3).
(b) expBLs(l2∞2)=extBLs(l2∞3).

Theorem 2.1.([22])

Let T=(a11,a22,a33,2a12,2a13,2a23)t∈Ls(l2∞3). Then,

‖T‖=max{2∣a12∣+∣a11+a22-a33∣,2∣a13∣+∣a11-a22+a33∣,2∣a23∣+∣-a11+a22+a33∣,2∣a12+a13∣+∣a11+a22+a33+2a23∣,2∣a12-a13∣+∣a11+a22+a33-2a23∣}.

Note that if ||T|| = 1, then |a_ii| ≤ 1 for i = 1, 2,3 and ∣aij∣≤12 for 1 ≤ i ≠ j ≤ 3. With the aid of Wolfram Mathematica 8, we present a complete description of the 42 extreme points of the unit ball of Ls(l2∞3).

Theorem 2.2

extBLs(l2∞3)={±(1,0,0,0,0,0)t,±(0,1,0,0,0,0)t,±(0,0,1,0,0,0)t,±(12,-12,0,1,0,0)t,±(12,-12,0,-1,0,0)t,±(12,0,-12,0,1,0)t,±(12,0,-12,0,-1,0)t,±(0,12,-12,0,0,1)t,±(0,12,-12,0,0,-1)t±(12,0,0,-12,12,12)t,±(12,0,0,12,-12,12)t,±(12,0,0,12,12,-12)t,±(0,12,0,-12,12,12)t,±(0,12,0,12,-12,12)t,±(0,12,0,12,12,-12)t,±(0,0,12,-12,12,12)t,±(0,0,12,12,-12,12)t,±(0,0,12,12,12,-12)t,±(-12,0,0,12,12,12)t,±(0,-12,0,12,12,12)t,±(0,0,-12,12,12,12)t}.

Proof

Claim: T=(12,-12,0,1,0,0)t∈extBLs(l2∞3). Let

T1=(12+ɛ11,-12+ɛ22,ɛ33,1+ɛ12,ɛ13,ɛ23)t

and

T2=(12-ɛ11,-12-ɛ22,-ɛ33,1-ɛ12,-ɛ13,-ɛ23)t

with ||T₁|| = ||T₂|| = 1 for some ε_ij ∈ ℝ for i, j = 1, 2, 3. By Theorem 2.1, it follows that

0=ɛ12=ɛ13=ɛ230=ɛ11+ɛ22+ɛ33,0=-ɛ11+ɛ22+ɛ33,0=ɛ11-ɛ22+ɛ33,

which show that ε_ij = 0 for i, j = 1, 2, 3. Hence, T∈extBL(l2∞3).

Claim: T=(12,0,0,-12,12,12)t∈extBLs(l2∞3). Let

T1=(12+ɛ11,ɛ22,ɛ33,-12+ɛ12,12+ɛ13,12+ɛ23)t

and

T2=(12=ɛ11,-ɛ22,-ɛ33,-12-ɛ12,12-ɛ13,12-ɛ23)t

with ||T₁|| = ||T₂|| = 1 for some ε_ij ∈ ℝ for i, j = 1, 2, 3. By Theorem 2.1, it follows that

0=ɛ12-ɛ13=ɛ12+ɛ130=ɛ11+ɛ22+ɛ33-ɛ23,0=ɛ11+ɛ22+ɛ33+ɛ23,0=-ɛ11+ɛ22+ɛ33,0=ɛ11-ɛ22+ɛ33,

which show that ε_ij = 0 for i, j = 1, 2, 3. Hence, T∈extBL(l2∞3).

The other 40 bilinear forms in the list of Theorem 2.2 can be proved to be extreme in a similar way. We leave this to the reader.

Let T=(a11,a22,a33,2a12,2a13,2a23)t∈Ls(l2∞3) with ||T|| = 1. By the comments made right before Theorem 2.1, T∈extBLs(l2∞3) if and only if T = A⁻¹(T(Z₁), · · ·, T(Z₆))^t for some Z₁, …, Z₆ ∈ Γ with |T(Z_j)| = 1 (j = 1, …, 6). Therefore, we may classify all the extreme points of BLs(l2∞3) by the following steps: There are 210 choices of 6 vectors among the 10 elements of Ω, that is ¹⁰C₆ = 210. First, among 210 cases, find W₁, …, W₆ ∈ Ω such that the corresponding matrix A with rows W₁, …, W₆ is invertible, and next solve A⁻¹ and using Theorem 2.1, obtain T = A⁻¹(b₁, · · ·, b₆)^t satisfying

‖A-1(b1,⋯,b6)t‖=1

for some b₁, …, b₆ = ±1.

Those T = A⁻¹(b₁, · · ·, b₆)^t are all the extreme points of BLs(l2∞3). With the aid of Wolfram Mathematica 8, we may check the 42 bilinear forms in the list of Theorem 2.2 are all the extreme points of the unit ball of Ls(l2∞3).

Theorem 2.3

We haveextBLs(l2∞3)⊂extBLs(l2∞n)for every n ≥ 4.

Proof

Let

T((x1,x2,x3),(y1,y2,y3))=12x3y3-14(x1y2+x2y1)+14(x1y3+x3y1)+14(x2y3+x3y2):=(0,0,12,-12,12,12)t.

Claim 1: for n ≥ 4, T∈extBLs(l2∞n). Use induction on n.

Case 1: n = 4.

Suppose that T=12(S1+S2) for some S1,S2∈Ls(l2∞4) with ||S₁|| = 1 = ||S₂||.

Write

S1((x1,x2,x3,x4),(y1,y2,y3,y4))=ɛ1x1y1+ɛ2x2y2+(12+a1)x3y3+(-14+a2) (x1y2+x2y1)+(14+a3) (x1y3+x3y1)+(14+a4) (x2y3+x3y2)+b1(x4y1+x1y4)+b2(x2y4+x4y2)+b3(x3y4+x4y3)+b4x4y4

and

S2((x1,x2,x3,x4),(y1,y2,y3,y4))=ɛ1x1y1-ɛ2x2y2+(12-a1)x3y3+(-14-a2) (x1y2+x2y1)+(14-a3) (x1y3+x3y1)+(14-a4) (x2y3+x3y2)-b1(x4y1+x1y4)-b2(x2y4+x4y2)-b3(x3y4+x4y3)-b4x4y4

for some ε₁, ε₂, a_j, b_j ∈ ℝ for j = 1, …, 4.

Note that, for i = 1, 2,

Si((x1,x2,x3,0),(y1,y2,y3,0))∈Ls(l2∞3),‖Si((x1,x2,x3,0),(y1,y2,y3,0))‖≤1,

and

T=12(S1((x1,x2,x3,0),(y1,y2,y3,0))+S2((x1,x2,x3,0),(y1,y2,y3,0)).

Since T∈extBLs(l2∞3), T((x₁, x₂, x₃), (y₁, y₂, y₃)) = S₁((x₁, x₂, x₃, 0), (y₁, y₂, y₃, 0)), which shows that ε₁ = ε₂ = a_j = 0 for j = 1, …, 4. Hence,

S1((x1,x2,x3,x4),(y1,y2,y3,y4))=T((x1,x2,x3),(y1,y2,y3))+b1(x4y1+x1y4)+b2(x2y4+x4y2)+b3(x3y4+x4y3)+b4x4y4

and

S2((x1,x2,x3,x4),(y1,y2,y3,y4))=T((x1,x2,x3),(y1,y2,y3))-(b1(x4y1+x1y4)+b2(x2y4+x4y2)+b3(x3y4+x4y3)+b4x4y4).

It follows that

1≥max{∣Si((1,1,1,x4),(1,-1,1,y4))∣,∣Si((1,1,1,x4),(1,1,-1,y4))∣:∣x4∣≤1,∣y4∣≤1,i=1,2}=max{1+∣b1(y4+x4)+∣b2(y4-x4)+b3(y4+x4)+b4x4y4∣,+1∣b1(y4+x4)+b2(y4+x4)+b3(y4-x4)+b4x4y4∣:∣x4∣≤1,∣y4∣≤1},

which imply that, for all |x₄| ≤ 1, |y₄| ≤ 1,

0=b1(y4+x4)+b2(y4-x4)+b3(y4+x4)+b4x4y40=b1(y4+x4)+b2(y4+x4)+b3(y4-x4)+b4x4y4,

which shows that b_j = 0 for j = 1, …, 4. Therefore, T∈extBLs(l2∞4).

Suppose that for n = k, T∈extBLs(l2∞k). We will show that T∈extBLs(l2∞k+1). Suppose that T=12(W1+W2) for some W₁,W2∈Ls(l2∞k+1) with ||W₁|| = 1 = ||W₂||. By the above argument, we may assume that

W1((x1,…,xk+1),(y1,…,yk+1))=T((x1,x2,x3),(y1,y2,y3))+∑1≤j≤kcj(xjyk+1+xk+1yj)+dxk+1yk+1

and

W2((x1,…,xk+1),(y1,…,yk+1))=T((x1,x2,x3),(y1,y2,y3))-∑1≤j≤kcj(xjyk+1+xk+1yj)-dxk+1yk+1

for some c_j, d ∈ ℝ (j = 1, …, k). It follows that

1≥max{∣Wi((1,1,1,x4,…,xk+1),(1,-1,1,y4,…,yk+1))∣,∣Wi((1,1,1,x4,…,xk+1),(1,1,-1,y4,…,yk+1))∣,∣Wi((1,1,1,x4,…,xk+1),(1,-1,-1,y4,…,yk+1))∣:∣xj∣≤1,∣yj∣≤1,j=4,…,k+1,i=1,2}=max{1+∣c1(yk+1+xk+1)+c2(yk+1-xk+1)+c3(yk+1+xk+1)+∑4≤j≤kcj(xjyk+1+xk+1yj)+dxk+1yk+1∣,1+∣c1(yk+1+xk+1)+c2(yk+1+xk+1)+c3(yk+1-xk+1)+∑4≤j≤kcj(xjyk+1+xk+1yj)+dxk+1yk+1∣,1+∣c1(yk+1+xk+1)+c2(yk+1-xk+1)+c3(yk+1-xk+1)+∑4≤j≤kcj(xjyk+1+xk+1yj)+dxk+1yk+1∣:∣xj∣≤1,∣yj∣≤1,j=4,…,k+1},

which shows that, for |x_j| ≤ 1, |y_j| ≤ 1, j = 4, …, k + 1,

0=c1(yk+1+xk+1)+c2(yk+1-xk+1)+c3(yk+1+xk+1)+∑4≤j≤kcj(xjyk+1+xk+1yj)+dxk+1yk+1,0=c1(yk+1+xk+1)+c2(yk+1+xk+1)+c3(yk+1-xk+1)+∑4≤j≤kcj(xjyk+1+xk+1yj)+dxk+1yk+1,0=c1(yk+1+xk+1)+c2(yk+1-xk+1)+c3(yk+1-xk+1)+∑4≤j≤kcj(xjyk+1+xk+1yj)+dxk+1yk+1.

If x_j = y_j = 0 for j = 4, …, k, then, for |x_k+1| ≤ 1, |y_k+1| ≤ 1,

0=c1(yk+1+xk+1)+c2(yk+1-xk+1)+c3(yk+1+xk+1)+dxk+1yk+1,0=c1(yk+1+xk+1)+c2(yk+1+xk+1)+c3(yk+1-xk+1)+dxk+1yk+1,0=c1(yk+1+xk+1)+c2(yk+1-xk+1)+c3(yk+1-xk+1)+dxk+1yk+1

which shows that c_j = 0 = d for j = 1, 2, 3. Hence,

(*) 0=∑4≤j≤kcj(xjyk+1+xk+1yj) for ∣xj∣≤1,∣yj∣≤1,j=4,…,k+1.

We claim that c_j = 0 for all j = 4, …, k. Let 4 ≤ j₀ ≤ k be fixed. Let x_j = y_j = 1 for 4 ≤ j ≠ j₀ ≤ k and y_k+1 = −y_j0 = 1 = −x_k+1 = x_j0. By (*), we have c_j0 = 0. Hence, W₁ = T = W₂, so T∈extBLs(l2∞k+1). Therefore, we have shown that T∈extBLs(l2∞n) for n ≥ 4.

Let

S((x1,x2,x3),(y1,y2,y3))=12x1y1-12x2y2+12(x1y2+x2y1:=(12,-12,0,1,0,0)t.

Claim 2: for S∈extBLs(l2∞n). Use induction on n.

Case 1: n = 4

Suppose that S=12(T1+T2) for some T1,T2∈Ls(l2∞4) with ||T₁|| = 1 = ||T₂||.

Write

T1((x1,x2,x3,x4),(y1,y2,y3,y4))=(12+a1)x1y1+(-12+a2)x2y2+a3x3y3+(12+b1) (x1y2+x2y1)+b2(x2y3+x3y2)+b3(x3y1+x1y3)+c1(x1y4+x4y1)+c2(x2y4+x4y2)+c3(x3y4+x4y3)+c4x4y4

and

T2((x1,x2,x3,x4),(y1,y2,y3,y4))=(12-a1)x1y1+(-12-a2)x2y2-a3x3y3+(12-b1) (x1y2+x2y1)-b2(x2y3+x3y2)-b3(x3y1+x1y3)-c1(x1y4+x4y1)-c2(x2y4+x4y2)-c3(x3y4+x4y3)-c4x4y4

for some a_i, b_i, c_j ∈ ℝ for i = 1, 2, 3, j = 1, …, 4. Note that, for i = 1, 2,

Ti((x1,x2,x3,0),(y1,y2,y3,0))∈Ls(l2∞3),‖Ti((x1,x2,x3,0),(y1,y2,y3,0))‖≤1

and

S=12(T1((x1,x2,x3,0),(y1,y2,y3,0))+T2((x1,x2,x3,0),(y1,y2,y3,0)).

Since S∈extBLs(l2∞3), S((x₁, x₂, x₃), (y₁, y₂, y₃)) = T₁((x₁, x₂, x₃, 0), (y₁, y₂, y₃, 0)), which shows that a_i = b_i = 0 for i = 1, 2, 3. Hence,

T1((x1,x2,x3,x4),(y1,y2,y3,y4))=S((x1,x2,x3),(y1,y2,y3))+c1(x1y4+x4y1)+c2(x2y4+x4y2)+c3(x3y4+x4y3)+c4x4y4

and

S2((x1,x2,x3,x4),(y1,y2,y3,y4))=T((x1,x2,x3),(y1,y2,y3))-(c1(x1y4+x4y1)+c2(x2y4+x4y2)+c3(x3y4+x4y3)+c4x4y4).

It follows that

1≥max{∣Ti((1,1,1,x4),(1,-1,1,y4))∣,∣Ti((1,1,1,x4),(1,1,-1,y4))∣:∣x4∣≤1,∣y4∣≤1,i=1,2}=max{1+∣c1(y4+x4)+c2(y4-x4)+c3(y4+x4)+c4x4y4∣,1+∣c1(y4+x4)+c2(y4+x4)+c3(y4-x4)+c4x4y4∣:∣x4∣≤1,∣y4∣ ≤1},

which imply that, for all |x₄| ≤ 1, |y₄| ≤ 1,

0=c1(y4+x4)+c2(y4-x4)+c3(y4+x4)+c4x4y40=c1(y4+x4)+c2(y4+x4)+c3(y4-x4)+c4x4y4,

which shows that c_j = 0 for j = 1, …, 4. Therefore, S∈extBLs(l2∞4).

Suppose that for n = k, S∈extBLs(l2∞k). We will show that S∈extBLs(l2∞k+1). Suppose that S=12(W1+W2) for some W₁,W2∈Ls(l2∞k+1) with ||W₁|| = 1 = ||W₂||. By the above argument, we may assume that

W1((x1,…,xk+1),(y1,…,yk+1))=S((x1,x2,x3),(y1,y2,y3))+∑1≤j≤kcj(xjyk+1+xk+1yj)+dxk+1yk+1

and

W2((x1,…,xk+1),(y1,…,yk+1))=S((x1,x2,x3),(y1,y2,y3))-∑1≤j≤kcj(xjyk+1+xk+1yj)-dxk+1yk+1

for some c_j, d ∈ ℝ (j = 1, …, k). It follows that

1≥max{∣Wi((1,1,1,x4,…,xk+1),(1,-1,1,y4,…,yk+1))∣,∣Wi((1,1,1,x4,…,xk+1),(1,1,-1,y4,…,yk+1))∣,∣Wi((1,1,1,x4,…,xk+1),(1,-1,-1,y4,…,yk+1))∣:∣xj∣≤1,∣yj∣≤1,j=4,…,k+1,i=1,2}=max{1+∣c1(yk+1+xk+1)+c2(yk+1-xk+1)+c3(yk+1+xk+1)+∑4≤j≤kcj(xjyk+1+xk+1yj)+dxk+1yk+1∣,1+∣c1(yk+1+xk+1)+c2(yk+1+xk+1)+c3(yk+1-xk+1)+∑4≤j≤kcj(xjyk+1+xk+1yj)+dxk+1yk+1∣,1+∣c1(yk+1+xk+1)+c2(yk+1-xk+1)+c3(yk+1-xk+1)+∑4≤j≤kcj(xjyk+1+xk+1yj)+dxk+1yk+1∣:∣xj∣≤1,∣yj∣≤1,j=4,…,k+1},

which shows that, for |x_j| ≤ 1, |y_j| ≤ 1, j = 4, …, k + 1,

0=c1(yk+1+xk+1)+c2(yk+1-xk+1)+c3(yk+1+xk+1)+∑4≤j≤kcj(xjyk+1+xk+1yj)+dxk+1yk+1,0=c1(yk+1+xk+1)+c2(yk+1+xk+1)+c3(yk+1-xk+1)+∑4≤j≤kcj(xjyk+1+xk+1yj)+dxk+1yk+1,0=c1(yk+1+xk+1)+c2(yk+1-xk+1)+c3(yk+1-xk+1)+∑4≤j≤kcj(xjyk+1+xk+1yj)+dxk+1yk+1.

By Claim 1, we conclude that c_j = 0 = d for j = 1, …, k. Hence, W₁ = S = W₂, so S∈extBLs(l2∞k+1). Therefore, we have shown that S∈extBLs(l2∞n) for n ≥ 4.

Let

S((x1,x2,x3),(y1,y2,y3))=x1y1:=(1,0,0,0,0,0)t.

Claim 3: for n ≥ 4, S∈extBLs(l2∞n).

Suppose that S=12(T1+T2) for some T₁,T2∈Ls(l2∞n) with ||T₁|| = 1 = ||T₂||.

Write

T1((x1,…,xn),(y1,…,yn))=(1+a1)x1y1+∑2≤j≤najxjyj+∑1≤i<j≤nbij(xiyj+xjyi)

and

T2((x1,…,xn),(y1,…,yn))=(1-a1)x1y1-∑2≤j≤najxjyj-∑1≤i<j≤nbij(xiyj+xjyi)

for some a_i, b_ij ∈ ℝ for i, j = 1, …, n. It follows that

1≥max{∣Tl(e1,e1)∣:l=1,2}=1+∣a1∣,

which imply that a₁ = 0. Let 2 ≤ j ≤ n be fixed. Since

1≥max{∣Tl(e1+tej,e1+tej)∣:l=1,2,∣t∣≤1}=1+∣ajt2+2∣bijt∣

and

0=ajt2+2b1jt

for every |t| ≤ 1. Hence, 0 = a_j = b_1j for every 2 ≤ j ≤ n. Let 2 ≤ i < j ≤ n be fixed. Since

1≥max{∣Tl(e1+ei+ej,e1+ei+ej)∣:l=1,2,=1+2bij∣,

b_ij = 0. Hence, S = T₁, so S∈extBLs(l2∞n) for n ≥ 4.

Notice that the other 39 cases are similar since, essentially there are three groups of extreme points, those having an 1, those having two 12’s and those having four 12’s. Therefore, the other 39 extreme points in the list of Theorem 2.2 are extreme in the unit ball of Ls(l2∞n). We complete the proof.

3. The Exposed Points of the Unit Ball of Ls(l2∞3)

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Abstract
1. Introduction
2. The Extreme Points of the Unit Ball of Ls(l2∞3)
3. The Exposed Points of the Unit Ball of Ls(l2∞3)
4. The Smooth Points of the Unit Ball of Ls(l2∞3)
Acknowledgements
References

Theorem 2.2 leads to a complete formula of ||f|| for every f∈Ls(l2∞3)*.

Theorem 3.1

Letf∈Ls(l2∞3)*with α₁₁ := f(x₁x₂), α₂₂ := f(y₁y₂), α₃₃ := f(z₁z₂), β₁₂ := f(x₁y₂ + x₂y₁), β₁₃ := f(x₁z₂ + x₂z₁), β₂₃ := f(y₁z₂ + y₂z₁). Then,

‖f‖=maxj=1,2,3{∣αjj∣,12∣α11-α22∣+12∣β12∣,12∣α22-α33∣+12∣β23∣,12∣α11-α33∣+12∣β13∣,14(∣2αjj+β12∣+∣β13-β23∣),14(∣2αjj-β12∣+∣β13+β23∣)}.

Proof

By the Krein-Milman Theorem, ‖f‖=supT∈extBLs(l2∞3)∣f(T)∣. By Theorem 2.2, it follows that

‖f‖=max{∣f(1,0,0,0,0,0)t)∣,∣f(0,1,0,0,0,0)t)∣,∣f((0,0,1,0,0,0)t)∣,∣f((12,-12,0,1,0,0)t)∣,∣f((12,-12,0,-1,0,0)t)∣,∣f((12,0,-12,0,1,0)t)∣,∣f((12,0,-12,0,-1,0)t)∣,∣f((0,12,-12,0,0,1)t)∣,∣f((0,12,-12,0,0,-1)t)∣,∣f((12,0,0,-12,12,12)t)∣,∣f((12,0,0,12,-12,12)t)∣,∣f((12,0,0,12,12,-12)t)∣,∣f((0,12,0,-12,12,12)t)∣,∣f((0,12,0,12,-12,12)t)∣,∣f((0,12,0,12,12,-12)t)∣,∣f((0,0,12,-12,12,12)t)∣,∣f((0,0,12,12,-12,12)t)∣,∣f((0,0,12,12,12-12)t)∣,∣f((-12,0,0,12,12,12)t)∣,∣f((0,-12,0,12,12,12)t)∣,∣f((0,0,-12,12,12,12)t)∣}=maxj=1,2,3{∣αjj∣,12∣α11-α22∣+12∣β12∣,12∣α22-α33∣+12∣β23∣,12∣α11-α33∣+12∣β13∣,14(∣2αjj+β12∣+∣β13-β23∣),14(∣2αjj-β12∣+∣β13+β23∣)}.

Note that if ||f|| = 1, then |α_jj| ≤ 1 for j = 1, 2,3 and |β₁₂| ≤ 2, |β₁₃| ≤ 2 and |β₂₃| ≤ 2.

Theorem 3.2.([17])

Let E be a real finite dimensional Banach space such that extB_Eis finite. Suppose that x ∈ extB_Esatisfies that there exists an f ∈ E^*with f(x) = 1 = ||f|| and |f(y)| < 1 for every y ∈ extB_E\{±x}. Then, x ∈ expB_E.

We give another proof of the following theorem which was shown in [22].

Theorem 3.3.([22]) expBLs(l2∞3)=extBLs(l2∞3)

Proof

It suffices to show that if T∈extBLs(l2∞3), then it is exposed.

Claim: T = (1, 0, 0, 0, 0, 0)^t is exposed.

Let f=(α11,α22,α33,β12,β13,β23)=(1,0,0,0,0,0)∈Ls(l2∞3)*. Then, by Theorems 3.1 and 3.2, ||f|| = 1 = f(T) and |f(S)| < 1 for S∈extBLs(l2∞3)\{±T}. Similarly, −(1, 0, 0, 0, 0, 0)^t,±(0, 1, 0, 0, 0, 0)^t,±(0, 0, 1, 0, 0, 0)^t are exposed.

Claim: T=(12,-12,0,1,0,0)t is exposed.

Let f=(α11,α22,α33,β12,β13,β23)=(14,0,0,74,0,0)∈Ls(l2∞3)*. Then, by Theorems 3.1 and 3.2, ||f|| = 1 = f(T) and |f(S)| < 1 for S∈extBLs(l2∞3)\{±T}. Similarly, -(12,-12,0,1,0,0)t is exposed.

Claim: T=(12,-12,0,-1,0,0)t is exposed.

Let f=(α11,α22,α33,β12,β13,β23)=(14,0,0,-74,0,0)∈Ls(l2∞3)* . Then, by Theorems 3.1 and 3.2, ||f|| = 1 = f(T) and |f(S)| < 1 for S∈extBLs(l2∞3)\{±T}. Similarly, -(12,-12,0,-1,0,0)t is exposed.

Claim: T=(12,0,-12,0,1,0)t is exposed.

Let f=(α11,α22,α33,β12,β13,β23)=(14,0,0,0,74,0)∈Ls(l2∞3)* . Then, by Theorems 3.1 and 3.2, ||f|| = 1 = f(T) and |f(S)| < 1 for S∈extBLs(l2∞3)\{±T}. Similarly, -(12,0,-12,0,1,0)t is exposed.

Claim: T=(12,0,-12,0,-1,0)t is exposed.

Let f=(α11,α22,α33,β12,β13,β23)=(14,0,0,0,-74,0)∈Ls(l2∞3)* . Then, by Theorems 3.1 and 3.2, ||f|| = 1 = f(T) and |f(S)| < 1 for S∈extBLs(l2∞3)\{±T}. Similarly, -(12,0,-12,0,-1,0)t is exposed.

Claim: T=(0,12,-12,0,0,1)t is exposed.

Let f=(α11,α22,α33,β12,β13,β23)=(0,14,0,0,0,74)∈Ls(l2∞3)*. Then, by Theorems 3.1 and 3.2, ||f|| = 1 = f(T) and |f(S)| < 1 for S∈extBLs(l2∞3)\{±T}. Similarly, -(0,12,-12,0,0,1)t is exposed.

Claim: T=(0,12,-12,0,0,-1)t is exposed.

Let f=(α11,α22,α33,β12,β13,β23)=(0,14,0,0,0,-74)∈Ls(l2∞3)* . Then, by Theorems 3.1 and 3.2, ||f|| = 1 = f(T) and |f(S)| < 1 for S∈extBLs(l2∞3)\{±T}. Similarly, -(0,12,-12,0,0,-1)t is exposed.

Claim: T=(-12,0,0,12,12,12)t is exposed.

Let f=(α11,α22,α33,β12,β13,β23)=(-14,0,0,54,1,54)∈Ls(l2∞3)*. Then, by Theorems 3.1 and 3.2, ||f|| = 1 = f(T) and |f(S)| < 1 for S∈extBLs(l2∞3)\{±T}. Similarly, -(-12,0,0,12,12,12)t,±(0,-12,0,12,12,12)t,±(0,0,-12,12,12,12)tare exposed.

Claim: T=(12,0,0,-12,12,12)t is exposed.

Let f=(α11,α22,α33,β12,β13,β23)=(14,0,0,-54,1,54)∈Ls(l2∞3)*. Then, by Theorems 3.1 and 3.2, ||f|| = 1 = f(T) and |f(S)| < 1 for S∈extBLs(l2∞3)\{±T}. Similarly, -(12,0,0,-12,12,12)t,±(0,12,0,-12,12,12)t,±(0,0,12,-12,12,12)tare exposed.

Claim: T=(12,0,0,12,-12,12)tis exposed.

Let f=(α11,α22,α33,β12,β13,β23)=(14,0,0,54,-1,54)∈Ls(l2∞3)*. Then, by Theorems 3.1 and 3.2, ||f|| = 1 = f(T) and |f(S)| < 1 for S∈extBLs(l2∞3)\{±T}. Similarly, -(12,0,0,12,-12,12)t,±(0,12,0,12,-12,12)t,±(0,0,12,12,-12,12)t are exposed.

Claim: T=(12,0,0,12,12,-12)t is exposed.

Let f=(α11,α22,α33,β12,β13,β23)=(14,0,0,54,1,-54)∈Ls(l2∞3)*. Then, by Theorems 3.1 and Theorem 3.2, ||f|| = 1 = f(T) and |f(S)| < 1 for S∈extBLs(l2∞3)\{±T}. Similarly, -(12,0,0,12,12,-12)t,±(0,12,0,12,12,-12)t,±(0,0,12,12,12,-12)t are exposed.

4. The Smooth Points of the Unit Ball of Ls(l2∞3)

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Abstract
1. Introduction
2. The Extreme Points of the Unit Ball of Ls(l2∞3)
3. The Exposed Points of the Unit Ball of Ls(l2∞3)
4. The Smooth Points of the Unit Ball of Ls(l2∞3)
Acknowledgements
References

In this section we will characterize all the smooth points of the unit ball of Ls(l2∞3).

Theorem 4.1

LetT=(a11,a22,a33,2a12,2a13,2a23)∈Ls(l2∞3). Then, T∈smBLs(l2∞3)if and only if

(∣T((1,1,1),(1,1,1))∣ =1,0<∣a11+a22+a33+2a23∣<1, ∣T(Y)∣<1 for all Y∈Γ\{[(1,1,1),(1,1,1)]})or (∣T((-1,1,1),(-1,1,1))∣ =1,0<∣a11+a22+a33+2a23∣<1, ∣T(Y)∣<1 for all Y∈Γ\{[(-1,1,1),(-1,1,1)]})or (∣T((1,-1,1),(1,-1,1))∣ =1,0<∣a11+a22+a33+2a13∣<1, ∣T(Y)∣<1 for all Y∈Γ\{(1,-1,1),(1,-1,1)})or (∣T((1,1,-1),(1,1,-1))∣ =1,0<∣a11+a22+a33+2a12∣<1, ∣T(Y)∣<1 for all Y∈Γ\{[(1,1,-1),(1,1,-1)]})or (∣T((1,1,1),(1,-1,1))∣ =1,0<∣a11-a22+a33∣<1, ∣T(Y)∣<1 for all Y∈Γ\{[(1,1,1),(1,-1,1)]})or (∣T((1,1,1),(1,1,-1))∣ =1,0<∣a11+a22-a33∣<1, ∣T(Y)∣<1 for all Y∈Γ\{[(1,1,1),(1,1,-1)]})or (∣T((1,1,1),(-1,1,1))∣ =1,0<∣-a11+a22+a33∣<1, ∣T(Y)∣<1 for all Y∈Γ\{[(1,1,1),(-1,1,1)]})or (∣T((1,-1,1),(1,1,-1))∣ =1,0<∣-a11+a22+a33∣<1, ∣T(Y)∣<1 for all Y∈Γ\{[(1,-1,1),(1,1,-1)]})or (∣T((1,-1,1),(-1,1,1))∣ =1,0<∣a11+a22-a33∣<1, ∣T(Y)∣<1 for all Y∈Γ\{[(1,-1,1),(-1,1,1)]})or (∣T((1,1,-1),(-1,1,1))∣ =1,0<∣a11-a22+a33∣<1, ∣T(Y)∣<1 for all Y∈Γ\{[(1,1,-1),(-1,1,1)]}).

Proof

(⇐): Case 1: |T((1, 1, 1), (1, 1, 1))| = 1, 0 < |a₁₁ + a₂₂ + a₃₃ + 2a₂₃| < 1, |T(Y )| < 1 for all Y ∈ Γ\{[(1, 1, 1), (1, 1, 1)>.

By Theorem 2.1, ||T|| = 1, |a_jj| < 1 for j = 1, 2, 3, ∣a12∣<12,∣a13∣<12,∣a23∣<12. Let l := T((1, 1, 1), (1, 1, 1)) = a₁₁+a₂₂+a₃₃+2a₁₂+2a₁₃+2a₂₃ for some l ∈ {1,−1}.

Without loss of generality, we may assume that l = 1. Obviously,

2(a12+a13)>0 and a11+a22+a33+2a23>0.

Let f∈Ls(l2∞3)* be such that f(T) = 1 = ||f|| with α₁₁ := f(x₁x₂), α₂₂ := f(y₁y₂), α₃₃ := f(z₁z₂), β₁₂ := f(x₁y₂+x₂y₁), β₁₃ := f(x₁z₂+x₂z₁), β₂₃ := f(y₁z₂+ y₂z₁).

We claim that α_jj = 1 (j = 1, 2, 3) and β₁₂ = β₁₃ = β₂₃ = 2. Let n₁ ∈ ℕ be such that, for j = 1, 2, 3,

∣ajj∣+1n1<1,∣a12∣+1n1<12,∣a13∣+1n1<12,∣a23∣+1n1<12,∣T(Y)∣+10n1<1 for all Y∈Γ\{[(1,1,1),(1,1,1)]}.

By Theorem 2.1, for n > n₁,

1=‖(a11±1n,2a12,2a13,a22∓1n,2a23,a33)t‖,1=‖(a11±1n,2a12,2a13,a22,2a23,a33∓1n)t‖,1=‖(a11,2a12±1n,2a13∓1n,a22,2a23,a33)t‖,1=‖(a11±1n,2a12,2a13,a22,2a23∓1n,a33)t‖.

It follows that for n > n₁,

(1) 1≥∣f((a11±1n,2a12,2a13,a22∓1n,2a23,a33)t)∣ =∣1±1n(α11-α22)∣,
(2) 1≥∣f((a11±1n,2a12,2a13,a22,2a23,a33∓1n)t)∣ =∣1±1n(α11-α33)∣,
(3) 1≥∣f((a11,2a12±1n,2a13∓1n,a22,2a23,a33)t)∣ =∣1±12n(β12-β13)∣,
(4) 1≥∣f((a11±1n,2a12,2a13,a22,2a23∓1n,a33)t)∣ =∣1±1n(α11-12β23)∣.

By (1) − (4), α₁₁ = α₂₂ = α₃₃, β₁₂ = β₁₃, α11=12β23. Let n₂ ∈ ℕ be such that n₂ > n₁ and

0<2(a12+a13)-1n2<2(a12+a13)+1n2<1,0<a11+a22+a33+2a23-1n2<a11+a22+a33+2a23+1n2<1.

By Theorem 2.1, for n > n₂,

1=‖(a11±1n,2a12∓12n,2a13∓12n,a22±12n,2a23∓1n,a33±12n)t‖.

Since

1≥∣f(a11±1n,2a12∓12n,2a13∓12n,a22±12n,2a23∓1n,a33±12n)t∣ =∣1±1n(α11-12β12)∣,

so, α11=12β12, hence, β₁₂ = β₁₃ = β₂₃. Therefore,

1=f(T)=∑j=13ajjαjj+a12β12+a13β13+a23β23=(a11+a22+a33+2a12+2a13+2a23)α11=α11,

hence, α_jj = 1 for j = 1, 2,3 and β₁₂ = β₁₃ = β₂₃ = 2. Hence, T would be smooth in Ls(l2∞3).

Case 2: |T((1, 1, 1), (1,−1, 1))| = 1, 0 < |a₁₁−a₂₂+a₃₃| < 1, |T(Y )| < 1 for all Y ∈ Γ\{[(1, 1, 1), (1,−1, 1)>

By Theorem 2.1, ||T|| = 1, |a_jj| < 1 for j = 1, 2, 3, ∣a12∣<12,∣a23∣<12. Let l := T((1,−1, 1), (1, 1, 1)) = a₁₁ − a₂₂ + a₃₃ + 2a₁₃ for some l ∈ {1,−1}. Without loss of generality, we may assume that l = 1. Obviously,

2a13>0,a11-a22+a33>0.

Let f∈Ls(l2∞3)* be such that f(T) = 1 = ||f|| with α₁₁ := f(x₁x₂), α₂₂ := f(y₁y₂), α₃₃ := f(z₁z₂), β₁₂ := f(x₁y₂+x₂y₁), β₁₃ := f(x₁z₂+x₂z₁), β₂₃ := f(y₁z₂+ y₂z₁). We claim that α_jj = 1 = −α₂₂ for j = 1,3 and β₁₂ = β₂₃ = 0, β₁₃ = 2. Let n₁ ∈ ℕ be such that, for j = 1, 2, 3,

∣ajj∣+1n1<1,∣a12∣+1n1<12,∣a23∣+1n1<12,∣T(Y)∣+10n1<1 for all Y∈Γ\{[(1,-1,1),(1,1,1)]}.

Let n₂ ∈ ℕ be such that, n₂ > n₁ and for j = 1, 2, 3,

0<2a13-1n2<2a13+1n2<1

and

0<a11-a22+a33-1n2<a11-a22+a33+1n2<1.

By Theorem 2.1, for n > n₂,

1=‖(a11±1n,2a12,2a13,a22±1n,2a23,a33)t‖,1=‖(a11±1n,2a12,2a13,a22,2a23,a33∓1n)t‖,1=‖(a11,2a12±1n,2a13,a22,2a23∓1n,a33)t‖,1=‖(a11,2a12±1n,2a13,a22,2a23±1n,a33)t‖,1=‖(a11±1n,2a12,2a13∓1n,a22,2a23,a33)t

It follows that for n > n₂,

(1′) 1≥∣f((a11±1n,2a12,2a13,a22±1n,2a23,a33)t)∣ =∣1±1n(α11+α22)∣,
(2′) 1≥∣f((a11±1n,2a12,2a13,a22,2a23,a33∓1n)t)∣ =∣1±1n(α11-α33)∣,
(3′) 1≥∣f((a11,2a12±1n,2a13,a22,2a23∓1n,a33)t)∣ =∣1±12n(β12-β23)∣,
(4′) 1≥∣f((a11,2a12±1n,2a13,a22,2a23±1n,a33)t)∣ =∣1±12n(β12+β23)∣,
(5′) 1≥∣f((a11±1n,2a12,2a13∓1n,a22,2a23,a33)t)∣ =∣1±1n(α11+12β13)∣.

By (1′) − (5′), α₁₁ = −α₂₂ = α₃₃, β₁₂ = β₂₃ = 0, α11=12β13. It follows that

1=f(T)=∑j=13ajjαjj+a12β12+a13β13+a23β23=(a11-a22+a33)α11+2a13α11=(1-2a13)α11+2a13α11=α11,

hence, β₁₃ = 2 and α_jj = 1 = −α₂₂ for j = 1, 3. Hence, T would be smooth in Ls(l2∞3). Since the proofs of other cases are similar as those in the cases 1 and 2, we omit the proofs.

(⇒): If not, then we have two cases.

Case 1: Norm(T) is a singleton, where

Norm(S):={X∈Γ:∣S(X)∣ =‖S‖}

for S∈Ls(l2∞3).

Notice that

1≥∣a11+a22+a33+2a12∣,1≥∣a11+a22+a33+2a13∣,1≥∣a11+a22+a33+2a23∣,1≥∣-a11+a22+a33∣,1≥∣a11-a22+a33∣,1≥∣a11+a22-a33∣.

We have ten subcases as follows:

(∣T((1,1,1),(1,1,1))∣ =1,a11+a22+a33+2a23=0 or ±1, ∣T(Y)∣<1 for all Y∈Γ\{[(1,1,1),(1,1,1)]})or (∣T((-1,1,1),(-1,1,1))∣ =1,a11+a22+a33+2a23=0, or ±1, ∣T(Y)∣<1 for all Y∈Γ\{[(-1,1,1),(-1,1,1)]})or (∣T((1,-1,1),(1,-1,1))∣ =1,a11+a22+a33+2a13=0, or ±1, ∣T(Y)∣<1 for all Y∈Γ\{(1,-1,1),(1,-1,1)})or (∣T((1,1,-1),(1,1,-1))∣ =1,a11+a22+a33+2a12=0, or ±1, ∣T(Y)∣<1 for all Y∈Γ\{[(1,1,-1),(1,1,-1)]})or (∣T((1,1,1),(1,-1,1))∣ =1,a11-a22+a33=0 or ±1, ∣T(Y)∣<1 for all Y∈Γ\{[(1,1,1),(1,-1,1)]})or (∣T((1,1,1),(1,1,-1))∣ =1,a11+a22-a33=0 or ±1, ∣T(Y)∣<1 for all Y∈Γ\{[(1,1,1),(1,1,-1)]})or (∣T((1,1,1),(-1,1,1))∣ =1,-a11+a22+a33=0 or ±1, ∣T(Y)∣<1 for all Y∈Γ\{[(1,1,1),(-1,1,1)]})or (∣T((1,-1,1),(1,1,-1))∣ =1,-a11+a22+a33=0 or ±1, ∣T(Y)∣<1 for all Y∈Γ\{[(1,-1,1),(1,1,-1)]})or (∣T((1,-1,1),(-1,1,1))∣ =1,a11+a22-a33=0 or ±1, ∣T(Y)∣<1 for all Y∈Γ\{[(1,-1,1),(-1,1,1)]})or (∣T((1,1,-1),(-1,1,1))∣ =1,a11-a22+a33=0 or ±1, ∣T(Y)∣<1 for all Y∈Γ\{[(1,1,-1),(-1,1,1)]}).

Subcase 1: |T((1, 1, 1), (1, 1, 1))| = 1, a₁₁ + a₂₂ + a₃₃ + 2a₂₃ = 0 or ± 1, |T(Y )| < 1 for all Y ∈ Γ\{[(1, 1, 1), (1, 1, 1)>.

Suppose that a₁₁ + a₂₂ + a₃₃ + 2a₂₃ = 0. Let

f1=(α11,α22,α33,β12,β13,β23)=(0,0,0,2,2,0)

and

f2=(α11,α22,α33,β12,β13,β23)=(1,1,1,2,2,2)∈Ls(l2∞3)*.

By Theorem 3.1, ||f_k|| = 1 = f_k(T) for k = 1, 2. Hence, T∉smBLs(l2∞3). This is a contradiction.

Suppose that a₁₁ + a₂₂ + a₃₃ + 2a₂₃ = 1. For |t| ≤ 2, let

ft(1,1,1,t,t,0)∈Ls(l2∞3)*.

By Theorem 3.1, ||f_t|| = 1 = f_t(T) for |t| ≤ 2. Hence, T∉smBLs(l2∞3). This is a contradiction.

Suppose that a₁₁ + a₂₂ + a₃₃ + 2a₂₃ = −1. For |t| ≤ 2, let

ft(-1,-1,-1,t,t,0)∈Ls(l2∞3)*.

By Theorem 3.1, ||f_t|| = 1 = f_t(T) for |t| ≤ 2. Hence, T∉smBLs(l2∞3). This is a contradiction.

Subcase 2: |T((1, 1,−1), (−1, 1, 1))| = 1, a₁₁ − a₂₂ + a₃₃ = 0 or ± 1, |T(Y )| < 1 for all Y ∈ Γ\{[(1, 1,−1), (−1, 1, 1)>.

Suppose that a₁₁ − a₂₂ + a₃₃ = 0. Then a13=12. Let

f1=(α11,α22,α33,β12,β13,β23)=(1,-1,1,0,2,0)

and

f2=(α11,α22,α33,β12,β13,β23)=(0,0,0,0,2,0)∈Ls(l2∞3)*.

By Theorem 3.1, ||f_k|| = 1 = f_k(T) for k = 1, 2. Hence, T∉smBLs(l2∞3).

Suppose that a₁₁ − a₂₂ + a₃₃ = 1. Let

f1=(α11,α22,α33,β12,β13,β23)=(1,-1,1,0,0,0)

and

f2=(α11,α22,α33,β12,β13,β23)=(1,-1,1,0,1,0)∈Ls(l2∞3)*.

By Theorem 3.1, ||f_k|| = 1 = f_k(T) for k = 1, 2. Hence, T∉smBLs(l2∞3).

Suppose that a₁₁ − a₂₂ + a₃₃ = −1. Let

f1=(α11,α22,α33,β12,β13,β23)=-(1,-1,1,0,0,0)

and

f2=(α11,α22,α33,β12,β13,β23)=-(1,-1,1,0,1,0)∈Ls(l2∞3)*.

By Theorem 3.1, ||f_k|| = 1 = f_k(T) for k = 1, 2. Hence, T∉smBLs(l2∞3).

Since the proofs of other subcases are similar as those of the above, we omit the proofs.

Case 2: |Norm(T)| ≥ 2.

There exist X₁ ≠ X₂ ∈ Γ such that |T(X_k)| = 1 for k = 1, 2. Note that sign(T(Xk))δXk∈Ls(l2∞3)* and

‖sign(T(Xk))δXk‖=1=sign(T(Xk))δXk(T)

for k = 1, 2, which shows that T is not smooth. This is a contradiction. Therefore, we complete the proof.

Acknowledgements

Go to

Abstract
1. Introduction
2. The Extreme Points of the Unit Ball of Ls(l2∞3)
3. The Exposed Points of the Unit Ball of Ls(l2∞3)
4. The Smooth Points of the Unit Ball of Ls(l2∞3)
Acknowledgements
References

The author is thankful to the referee for the careful reading and considered suggestions leading to a better presented paper.

References

Go to

Abstract
1. Introduction
2. The Extreme Points of the Unit Ball of Ls(l2∞3)
3. The Exposed Points of the Unit Ball of Ls(l2∞3)
4. The Smooth Points of the Unit Ball of Ls(l2∞3)
Acknowledgements
References

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