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 $L_{S} (^{2} l_{\infty}^{3})$

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 $L_{s} ({{}^{2}l}_{\infty}^{3})$
3. The Exposed Points of the Unit Ball of $L_{s} ({{}^{2}l}_{\infty}^{3})$
4. The Smooth Points of the Unit Ball of $L_{s} ({{}^{2}l}_{\infty}^{3})$
Acknowledgements
References

We present a complete description of all the extreme points of the unit ball of $L_{s} ({{}^{2}l}_{\infty}^{3})$ which leads to a complete formula for ||f|| for every $f \in L_{s} {({{}^{2}l}_{\infty}^{3})}^{*}$ . We also show that $e x t B_{L_{s} ({{}^{2}l}_{\infty}^{3})} \subset e x t B_{L_{s} ({{}^{2}l}_{\infty}^{n})}$ for every n ≥ 4. Using the formula for ||f|| for every $f \in L_{s} {({{}^{2}l}_{\infty}^{3})}^{*}$ , we show that every extreme point of the unit ball of $L_{s} ({{}^{2}l}_{\infty}^{3})$ is exposed. We also characterize all the smooth points of the unit ball of $L_{s} ({{}^{2}l}_{\infty}^{3})$ .

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 $L_{s} ({{}^{2}l}_{\infty}^{3})$
3. The Exposed Points of the Unit Ball of $L_{s} ({{}^{2}l}_{\infty}^{3})$
4. The Smooth Points of the Unit Ball of $L_{s} ({{}^{2}l}_{\infty}^{3})$
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 $x = \frac{1}{2} (y + z)$ 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 $P ({{}^{2}l}_{1}^{2})$ and $P ({{}^{2}l}_{2}^{2})$ . Kim classified the exposed 2-homogeneous polynomials in $P ({{}^{2}l}_{p}^{2}) (1 \leq p \leq \infty)$ [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 $P ({{}^{2}ℝ}_{h (\frac{1}{2})}^{2})$ , where $ℝ_{h (\frac{1}{2})}^{2}$ is the plane ℝ² with the hexagonal norm of weight $\frac{1}{2}$ .

In 2009, Kim [12] initiated the classification of the extreme points, exposed points, and smooth points of the unit ball of $L_{s} ({{}^{2}l}_{\infty}^{3})$ . 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

L_{s} ({{}^{3}l}_{\infty}^{2}), L ({{}^{3}l}_{\infty}^{2}), L_{s} ({{}^{2}d}_{*} {(1, w)}^{2}), L ({{}^{2}d}_{*} {(1, w)}^{2}), L_{s} ({{}^{2}ℝ}_{h (w)}^{2}) and L ({{}^{2}ℝ}_{h (w)}^{2}),

where $ℝ_{h (w)}^{2}$ 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 $L_{s} ({{}^{2}l}_{\infty}^{n})$ and $L ({{}^{2}l}_{\infty}^{n})$ for n ≥ 2.

Let n ≥ 2 and $l_{\infty}^{n} : = ℝ^{n}$ with the supremum norm. Given ${a_{i j}}_{i, j = 1}^{n} \subset ℝ$ , let $T \in L ({{}^{2}l}_{\infty}^{n})$ be defined by the rule

T ((x_{1}, x_{2}, \dots, x_{n}), (y_{1}, y_{2}, \dots, y_{n})) = \sum_{1 \leq i, j \leq n} a_{i j} x_{i} y_{j} .

If n = 2, for simplicity, we will write

T = {(a_{11}, a_{22}, a_{12}, a_{21})}^{t}

as a 4 × 1 column vector. If $T \in L_{s} ({{}^{2}l}_{\infty}^{2})$ , we will write

T = {(a_{11}, a_{22}, 2 a_{12})}^{t}

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

T = {(a_{11}, a_{22}, a_{33}, a_{12}, a_{21}, a_{13}, a_{31}, a_{23}, a_{32})}^{t}

as a 9 × 1 column vector. If $T \in L_{s} ({{}^{2}l}_{\infty}^{3})$ , we will write

T = {(a_{11}, a_{22}, a_{33}, 2 a_{12}, 2 a_{13}, 2 a_{23})}^{t}

as a 6 × 1 column vector.

In [12] it was shown:

(a) $e x t B_{L_{s} ({{}^{2}l}_{\infty}^{2})} = {\pm {(1, 0, 0)}^{t}, \pm {(0, 1, 0)}^{t}, \pm {(\frac{1}{2}, - \frac{1}{2}, 1)}^{t}, \pm {(\frac{1}{2}, - \frac{1}{2}, - 1)}^{t}}$ ;
(b) $e x t B_{L_{s} ({{}^{2}l}_{\infty}^{2})} = e x t B_{L_{s} ({{}^{2}l}_{\infty}^{2})}$ .

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 $L_{s} ({{}^{2}l}_{\infty}^{3})$ which leads to a complete formula of ||f|| for every $f \in L_{s} {({{}^{2}l}_{\infty}^{3})}^{*}$ . We also show that $e x t B_{L_{s} ({{}^{2}l}_{\infty}^{2})} \subset e x t B_{L_{s} ({{}^{2}l}_{\infty}^{n})}$ for every n ≥ 4. Using the formula of ||f|| for every $f \in L_{s} {({{}^{2}l}_{\infty}^{3})}^{*}$ , we show that every extreme point of the unit ball of $L_{s} ({{}^{2}l}_{\infty}^{3})$ 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 $L_{s} ({{}^{2}l}_{\infty}^{3})$ .

2. The Extreme Points of the Unit Ball of $L_{s} ({{}^{2}l}_{\infty}^{3})$

Go to

Abstract
1. Introduction
2. The Extreme Points of the Unit Ball of $L_{s} ({{}^{2}l}_{\infty}^{3})$
3. The Exposed Points of the Unit Ball of $L_{s} ({{}^{2}l}_{\infty}^{3})$
4. The Smooth Points of the Unit Ball of $L_{s} ({{}^{2}l}_{\infty}^{3})$
Acknowledgements
References

Recently, Kim [22] showed the following: Let

\begin{array}{l} 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)} \end{array}

and

\begin{array}{l} 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)} . \end{array}

The following statements hold true.

(a) Let $T = {(a_{11}, a_{22}, a_{33}, 2 a_{12}, 2 a_{13} 2 a_{23})}^{t} \in L_{s} ({{}^{2}l}_{\infty}^{3})$ with ||T|| = 1. Then, $T \in e x t B_{L_{s} ({{}^{2}l}_{\infty}^{3})}$ if and only if there exist at least 6 linearly independent vectors W₁, …, W₆ ∈ Ω and Z₁, …, Z₆ ∈ Γ such that

$W_{j} \cdot S = S (Z_{j}) for all S \in L_{s} ({{}^{2}l}_{\infty}^{3}) and ∣ T (Z_{j}) ∣ = 1 for j = 1, \dots, 6.$

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 $S \in L_{s} ({{}^{2}l}_{\infty}^{3})$ .
(b) $e x p B_{L_{s} ({{}^{2}l}_{\infty}^{2})} = e x t B_{L_{s} ({{}^{2}l}_{\infty}^{3})}$ .

Theorem 2.1.([22])

Let $T = {(a_{11}, a_{22}, a_{33}, 2 a_{12}, 2 a_{13}, 2 a_{23})}^{t} \in L_{s} ({{}^{2}l}_{\infty}^{3})$ . Then,

\begin{array}{l} ‖ T ‖ = max {2 ∣ a_{12} ∣ + ∣ a_{11} + a_{22} - a_{33} ∣, 2 ∣ a_{13} ∣ + ∣ a_{11} - a_{22} + a_{33} ∣, \\ 2 ∣ a_{23} ∣ + ∣ - a_{11} + a_{22} + a_{33} ∣, 2 ∣ a_{12} + a_{13} ∣ + ∣ a_{11} + a_{22} + a_{33} + 2 a_{23} ∣, \\ 2 ∣ a_{12} - a_{13} ∣ + ∣ a_{11} + a_{22} + a_{33} - 2 a_{23} ∣} . \end{array}

Note that if ||T|| = 1, then |a_ii| ≤ 1 for i = 1, 2,3 and $∣ a_{i j} ∣ \leq \frac{1}{2}$ 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 $L_{s} ({{}^{2}l}_{\infty}^{3})$ .

Theorem 2.2

$\begin{array}{l} e x t B_{L_{s} ({{}^{2}l}_{\infty}^{3})} = {\pm {(1, 0, 0, 0, 0, 0)}^{t}, \pm {(0, 1, 0, 0, 0, 0)}^{t}, \pm {(0, 0, 1, 0, 0, 0)}^{t}, \\ \pm {(\frac{1}{2}, - \frac{1}{2}, 0, 1, 0, 0)}^{t}, \pm {(\frac{1}{2}, - \frac{1}{2}, 0, - 1, 0, 0)}^{t}, \pm {(\frac{1}{2}, 0, - \frac{1}{2}, 0, 1, 0)}^{t}, \\ \pm {(\frac{1}{2}, 0, - \frac{1}{2}, 0, - 1, 0)}^{t}, \pm {(0, \frac{1}{2}, - \frac{1}{2}, 0, 0, 1)}^{t}, \pm {(0, \frac{1}{2}, - \frac{1}{2}, 0, 0, - 1)}^{t} \\ \pm {(\frac{1}{2}, 0, 0, - \frac{1}{2}, \frac{1}{2}, \frac{1}{2})}^{t}, \pm {(\frac{1}{2}, 0, 0, \frac{1}{2}, - \frac{1}{2}, \frac{1}{2})}^{t}, \pm {(\frac{1}{2}, 0, 0, \frac{1}{2}, \frac{1}{2}, - \frac{1}{2})}^{t}, \\ \pm {(0, \frac{1}{2}, 0, - \frac{1}{2}, \frac{1}{2}, \frac{1}{2})}^{t}, \pm {(0, \frac{1}{2}, 0, \frac{1}{2}, - \frac{1}{2}, \frac{1}{2})}^{t}, \pm {(0, \frac{1}{2}, 0, \frac{1}{2}, \frac{1}{2}, - \frac{1}{2})}^{t}, \\ \pm {(0, 0, \frac{1}{2}, - \frac{1}{2}, \frac{1}{2}, \frac{1}{2})}^{t}, \pm {(0, 0, \frac{1}{2}, \frac{1}{2}, - \frac{1}{2}, \frac{1}{2})}^{t}, \pm {(0, 0, \frac{1}{2}, \frac{1}{2}, \frac{1}{2}, - \frac{1}{2})}^{t}, \\ \pm {(- \frac{1}{2}, 0, 0, \frac{1}{2}, \frac{1}{2}, \frac{1}{2})}^{t}, \pm {(0, - \frac{1}{2}, 0, \frac{1}{2}, \frac{1}{2}, \frac{1}{2})}^{t}, \pm {(0, 0, - \frac{1}{2}, \frac{1}{2}, \frac{1}{2}, \frac{1}{2})}^{t}} . \end{array}$

Proof

Claim: $T = {(\frac{1}{2}, - \frac{1}{2}, 0, 1, 0, 0)}^{t} \in e x t B_{L_{s} ({{}^{2}l}_{\infty}^{3})}$ . Let

T_{1} = {(\frac{1}{2} + ɛ_{11}, - \frac{1}{2} + ɛ_{22}, ɛ_{33}, 1 + ɛ_{12}, ɛ_{13}, ɛ_{23})}^{t}

and

T_{2} = {(\frac{1}{2} - ɛ_{11}, - \frac{1}{2} - ɛ_{22}, - ɛ_{33}, 1 - ɛ_{12}, - ɛ_{13}, - ɛ_{23})}^{t}

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

\begin{array}{l} 0 = ɛ_{12} = ɛ_{13} = ɛ_{23} \\ 0 = ɛ_{11} + ɛ_{22} + ɛ_{33}, \\ 0 = - ɛ_{11} + ɛ_{22} + ɛ_{33}, \\ 0 = ɛ_{11} - ɛ_{22} + ɛ_{33}, \end{array}

which show that ε_ij = 0 for i, j = 1, 2, 3. Hence, $T \in e x t B_{L ({{}^{2}l}_{\infty}^{3})}$ .

Claim: $T = {(\frac{1}{2}, 0, 0, - \frac{1}{2}, \frac{1}{2}, \frac{1}{2})}^{t} \in e x t B_{L_{s} ({{}^{2}l}_{\infty}^{3})}$ . Let

T_{1} = {(\frac{1}{2} + ɛ_{11}, ɛ_{22}, ɛ_{33}, - \frac{1}{2} + ɛ_{12}, \frac{1}{2} + ɛ_{13}, \frac{1}{2} + ɛ_{23})}^{t}

and

T_{2} = {(\frac{1}{2} = ɛ_{11}, - ɛ_{22}, - ɛ_{33}, - \frac{1}{2} - ɛ_{12}, \frac{1}{2} - ɛ_{13}, \frac{1}{2} - ɛ_{23})}^{t}

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

\begin{array}{l} 0 = ɛ_{12} - ɛ_{13} = ɛ_{12} + ɛ_{13} \\ 0 = ɛ_{11} + ɛ_{22} + ɛ_{33} - ɛ_{23}, \\ 0 = ɛ_{11} + ɛ_{22} + ɛ_{33} + ɛ_{23}, \\ 0 = - ɛ_{11} + ɛ_{22} + ɛ_{33}, \\ 0 = ɛ_{11} - ɛ_{22} + ɛ_{33}, \end{array}

which show that ε_ij = 0 for i, j = 1, 2, 3. Hence, $T \in e x t B_{L ({{}^{2}l}_{\infty}^{3})}$ .

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 $T = {(a_{11}, a_{22}, a_{33}, 2 a_{12}, 2 a_{13}, 2 a_{23})}^{t} \in L_{s} ({{}^{2}l}_{\infty}^{3})$ with ||T|| = 1. By the comments made right before Theorem 2.1, $T \in e x t B_{L_{s} ({{}^{2}l}_{\infty}^{3})}$ 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 $B_{L_{s} ({{}^{2}l}_{\infty}^{3})}$ 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} {(b_{1}, \dots, b_{6})}^{t} ‖ = 1

for some b₁, …, b₆ = ±1.

Those T = A⁻¹(b₁, · · ·, b₆)^t are all the extreme points of $B_{L_{s} ({{}^{2}l}_{\infty}^{3})}$ . 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 $L_{s} ({{}^{2}l}_{\infty}^{3})$ .

Theorem 2.3

We have $e x t B_{L_{s} ({{}^{2}l}_{\infty}^{3})} \subset e x t B_{L_{s} ({{}^{2}l}_{\infty}^{n})}$ for every n ≥ 4.

Proof

Let

\begin{array}{l} T ((x_{1}, x_{2}, x_{3}), (y_{1}, y_{2}, y_{3})) = \frac{1}{2} x_{3} y_{3} - \frac{1}{4} (x_{1} y_{2} + x_{2} y_{1}) + \frac{1}{4} (x_{1} y_{3} + x_{3} y_{1}) + \frac{1}{4} (x_{2} y_{3} + x_{3} y_{2}) \\ : = {(0, 0, \frac{1}{2}, - \frac{1}{2}, \frac{1}{2}, \frac{1}{2})}^{t} . \end{array}

Claim 1: for n ≥ 4, $T \in e x t B_{L_{s} ({{}^{2}l}_{\infty}^{n})}$ . Use induction on n.

Case 1: n = 4.

Suppose that $T = \frac{1}{2} (S_{1} + S_{2})$ for some $S_{1}, S_{2} \in L_{s} ({{}^{2}l}_{\infty}^{4})$ with ||S₁|| = 1 = ||S₂||.

Write

\begin{array}{l} S_{1} ((x_{1}, x_{2}, x_{3}, x_{4}), (y_{1}, y_{2}, y_{3}, y_{4})) \\ = ɛ_{1} x_{1} y_{1} + ɛ_{2} x_{2} y_{2} + (\frac{1}{2} + a_{1}) x_{3} y_{3} + (- \frac{1}{4} + a_{2}) (x_{1} y_{2} + x_{2} y_{1}) \\ + (\frac{1}{4} + a_{3}) (x_{1} y_{3} + x_{3} y_{1}) + (\frac{1}{4} + a_{4}) (x_{2} y_{3} + x_{3} y_{2}) \\ + b_{1} (x_{4} y_{1} + x_{1} y_{4}) + b_{2} (x_{2} y_{4} + x_{4} y_{2}) + b_{3} (x_{3} y_{4} + x_{4} y_{3}) + b_{4} x_{4} y_{4} \end{array}

and

\begin{array}{l} S_{2} ((x_{1}, x_{2}, x_{3}, x_{4}), (y_{1}, y_{2}, y_{3}, y_{4})) \\ = ɛ_{1} x_{1} y_{1} - ɛ_{2} x_{2} y_{2} + (\frac{1}{2} - a_{1}) x_{3} y_{3} + (- \frac{1}{4} - a_{2}) (x_{1} y_{2} + x_{2} y_{1}) \\ + (\frac{1}{4} - a_{3}) (x_{1} y_{3} + x_{3} y_{1}) + (\frac{1}{4} - a_{4}) (x_{2} y_{3} + x_{3} y_{2}) \\ - b_{1} (x_{4} y_{1} + x_{1} y_{4}) - b_{2} (x_{2} y_{4} + x_{4} y_{2}) - b_{3} (x_{3} y_{4} + x_{4} y_{3}) - b_{4} x_{4} y_{4} \end{array}

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

Note that, for i = 1, 2,

S_{i} ((x_{1}, x_{2}, x_{3}, 0), (y_{1}, y_{2}, y_{3}, 0)) \in L_{s} ({{}^{2}l}_{\infty}^{3}), ‖ S_{i} ((x_{1}, x_{2}, x_{3}, 0), (y_{1}, y_{2}, y_{3}, 0)) ‖ \leq 1,

and

T = \frac{1}{2} (S_{1} ((x_{1}, x_{2}, x_{3}, 0), (y_{1}, y_{2}, y_{3}, 0)) + S_{2} ((x_{1}, x_{2}, x_{3}, 0), (y_{1}, y_{2}, y_{3}, 0)) .

Since $T \in e x t B_{L_{s} ({{}^{2}l}_{\infty}^{3})}$ , 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,

\begin{array}{l} S_{1} ((x_{1}, x_{2}, x_{3}, x_{4}), (y_{1}, y_{2}, y_{3}, y_{4})) = T ((x_{1}, x_{2}, x_{3}), (y_{1}, y_{2}, y_{3})) + b_{1} (x_{4} y_{1} + x_{1} y_{4}) \\ + b_{2} (x_{2} y_{4} + x_{4} y_{2}) + b_{3} (x_{3} y_{4} + x_{4} y_{3}) + b_{4} x_{4} y_{4} \end{array}

and

\begin{array}{l} S_{2} ((x_{1}, x_{2}, x_{3}, x_{4}), (y_{1}, y_{2}, y_{3}, y_{4})) = T ((x_{1}, x_{2}, x_{3}), (y_{1}, y_{2}, y_{3})) - (b_{1} (x_{4} y_{1} + x_{1} y_{4}) \\ + b_{2} (x_{2} y_{4} + x_{4} y_{2}) + b_{3} (x_{3} y_{4} + x_{4} y_{3}) + b_{4} x_{4} y_{4}) . \end{array}

It follows that

\begin{array}{l} 1 \geq max {∣ S_{i} ((1, 1, 1, x_{4}), (1, - 1, 1, y_{4})) ∣, ∣ S_{i} ((1, 1, 1, x_{4}), (1, 1, - 1, y_{4})) ∣ : ∣ x_{4} ∣ \leq 1, ∣ y_{4} ∣ \leq 1, i = 1, 2} \\ = max {1 + ∣ b_{1} (y_{4} + x_{4}) + ∣ b_{2} (y_{4} - x_{4}) + b_{3} (y_{4} + x_{4}) + b_{4} x_{4} y_{4} ∣, \\ + 1 ∣ b_{1} (y_{4} + x_{4}) + b_{2} (y_{4} + x_{4}) + b_{3} (y_{4} - x_{4}) + b_{4} x_{4} y_{4} ∣ : ∣ x_{4} ∣ \leq 1, ∣ y_{4} ∣ \leq 1}, \end{array}

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

\begin{array}{l} 0 = b_{1} (y_{4} + x_{4}) + b_{2} (y_{4} - x_{4}) + b_{3} (y_{4} + x_{4}) + b_{4} x_{4} y_{4} \\ 0 = b_{1} (y_{4} + x_{4}) + b_{2} (y_{4} + x_{4}) + b_{3} (y_{4} - x_{4}) + b_{4} x_{4} y_{4}, \end{array}

which shows that b_j = 0 for j = 1, …, 4. Therefore, $T \in e x t B_{L_{s} ({{}^{2}l}_{\infty}^{4})}$ .

Suppose that for n = k, $T \in e x t B_{L_{s} ({{}^{2}l}_{\infty}^{k})}$ . We will show that $T \in e x t B_{L_{s} ({{}^{2}l}_{\infty}^{k + 1})}$ . Suppose that $T = \frac{1}{2} (W_{1} + W_{2})$ for some W₁, $W_{2} \in L_{s} ({{}^{2}l}_{\infty}^{k + 1})$ with ||W₁|| = 1 = ||W₂||. By the above argument, we may assume that

W_{1} ((x_{1}, \dots, x_{k + 1}), (y_{1}, \dots, y_{k + 1})) = T ((x_{1}, x_{2}, x_{3}), (y_{1}, y_{2}, y_{3})) + \sum_{1 \leq j \leq k} c_{j} (x_{j} y_{k + 1} + x_{k + 1} y_{j}) + d x_{k + 1} y_{k + 1}

and

W_{2} ((x_{1}, \dots, x_{k + 1}), (y_{1}, \dots, y_{k + 1})) = T ((x_{1}, x_{2}, x_{3}), (y_{1}, y_{2}, y_{3})) - \sum_{1 \leq j \leq k} c_{j} (x_{j} y_{k + 1} + x_{k + 1} y_{j}) - d x_{k + 1} y_{k + 1}

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

\begin{array}{l} 1 \geq max {∣ W_{i} ((1, 1, 1, x_{4}, \dots, x_{k + 1}), (1, - 1, 1, y_{4}, \dots, y_{k + 1})) ∣, \\ ∣ W_{i} ((1, 1, 1, x_{4}, \dots, x_{k + 1}), (1, 1, - 1, y_{4}, \dots, y_{k + 1})) ∣, \\ ∣ W_{i} ((1, 1, 1, x_{4}, \dots, x_{k + 1}), (1, - 1, - 1, y_{4}, \dots, y_{k + 1})) ∣ : ∣ x_{j} ∣ \leq 1, ∣ y_{j} ∣ \leq 1, \\ j = 4, \dots, k + 1, i = 1, 2} \\ = max {1 + ∣ c_{1} (y_{k + 1} + x_{k + 1}) + c_{2} (y_{k + 1} - x_{k + 1}) + c_{3} (y_{k + 1} + x_{k + 1}) \\ + \sum_{4 \leq j \leq k} c_{j} (x_{j} y_{k + 1} + x_{k + 1} y_{j}) + d x_{k + 1} y_{k + 1} ∣, \\ 1 + ∣ c_{1} (y_{k + 1} + x_{k + 1}) + c_{2} (y_{k + 1} + x_{k + 1}) + c_{3} (y_{k + 1} - x_{k + 1}) \\ + \sum_{4 \leq j \leq k} c_{j} (x_{j} y_{k + 1} + x_{k + 1} y_{j}) + d x_{k + 1} y_{k + 1} ∣, \\ 1 + ∣ c_{1} (y_{k + 1} + x_{k + 1}) + c_{2} (y_{k + 1} - x_{k + 1}) + c_{3} (y_{k + 1} - x_{k + 1}) \\ + \sum_{4 \leq j \leq k} c_{j} (x_{j} y_{k + 1} + x_{k + 1} y_{j}) + d x_{k + 1} y_{k + 1} ∣ \\ : ∣ x_{j} ∣ \leq 1, ∣ y_{j} ∣ \leq 1, j = 4, \dots, k + 1}, \end{array}

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

\begin{array}{l} 0 = c_{1} (y_{k + 1} + x_{k + 1}) + c_{2} (y_{k + 1} - x_{k + 1}) + c_{3} (y_{k + 1} + x_{k + 1}) \\ + \sum_{4 \leq j \leq k} c_{j} (x_{j} y_{k + 1} + x_{k + 1} y_{j}) + d x_{k + 1} y_{k + 1}, \\ 0 = c_{1} (y_{k + 1} + x_{k + 1}) + c_{2} (y_{k + 1} + x_{k + 1}) + c_{3} (y_{k + 1} - x_{k + 1}) \\ + \sum_{4 \leq j \leq k} c_{j} (x_{j} y_{k + 1} + x_{k + 1} y_{j}) + d x_{k + 1} y_{k + 1}, \\ 0 = c_{1} (y_{k + 1} + x_{k + 1}) + c_{2} (y_{k + 1} - x_{k + 1}) + c_{3} (y_{k + 1} - x_{k + 1}) \\ + \sum_{4 \leq j \leq k} c_{j} (x_{j} y_{k + 1} + x_{k + 1} y_{j}) + d x_{k + 1} y_{k + 1} . \end{array}

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

\begin{array}{l} 0 = c_{1} (y_{k + 1} + x_{k + 1}) + c_{2} (y_{k + 1} - x_{k + 1}) + c_{3} (y_{k + 1} + x_{k + 1}) + d x_{k + 1} y_{k + 1}, \\ 0 = c_{1} (y_{k + 1} + x_{k + 1}) + c_{2} (y_{k + 1} + x_{k + 1}) + c_{3} (y_{k + 1} - x_{k + 1}) + d x_{k + 1} y_{k + 1}, \\ 0 = c_{1} (y_{k + 1} + x_{k + 1}) + c_{2} (y_{k + 1} - x_{k + 1}) + c_{3} (y_{k + 1} - x_{k + 1}) + d x_{k + 1} y_{k + 1} \end{array}

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

(*) 0 = \sum_{4 \leq j \leq k} c_{j} (x_{j} y_{k + 1} + x_{k + 1} y_{j}) f o r ∣ x_{j} ∣ \leq 1, ∣ y_{j} ∣ \leq 1, j = 4, \dots, 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 $T \in e x t B_{L_{s} ({{}^{2}l}_{\infty}^{k + 1})}$ . Therefore, we have shown that $T \in e x t B_{L_{s} ({{}^{2}l}_{\infty}^{n})}$ for n ≥ 4.

Let

\begin{array}{l} S ((x_{1}, x_{2}, x_{3}), (y_{1}, y_{2}, y_{3})) = \frac{1}{2} x_{1} y_{1} - \frac{1}{2} x_{2} y_{2} + \frac{1}{2} (x_{1} y_{2} + x_{2} y_{1} \\ : = {(\frac{1}{2}, - \frac{1}{2}, 0, 1, 0, 0)}^{t} . \end{array}

Claim 2: for $S \in e x t B_{L_{s} ({{}^{2}l}_{\infty}^{n})}$ . Use induction on n.

Case 1: n = 4

Suppose that $S = \frac{1}{2} (T_{1} + T_{2})$ for some $T_{1}, T_{2} \in L_{s} ({{}^{2}l}_{\infty}^{4})$ with ||T₁|| = 1 = ||T₂||.

Write

\begin{array}{l} T_{1} ((x_{1}, x_{2}, x_{3}, x_{4}), (y_{1}, y_{2}, y_{3}, y_{4})) = (\frac{1}{2} + a_{1}) x_{1} y_{1} + (- \frac{1}{2} + a_{2}) x_{2} y_{2} + a_{3} x_{3} y_{3} \\ + (\frac{1}{2} + b_{1}) (x_{1} y_{2} + x_{2} y_{1}) + b_{2} (x_{2} y_{3} + x_{3} y_{2}) \\ + b_{3} (x_{3} y_{1} + x_{1} y_{3}) + c_{1} (x_{1} y_{4} + x_{4} y_{1}) \\ + c_{2} (x_{2} y_{4} + x_{4} y_{2}) + c_{3} (x_{3} y_{4} + x_{4} y_{3}) + c_{4} x_{4} y_{4} \end{array}

and

\begin{array}{l} T_{2} ((x_{1}, x_{2}, x_{3}, x_{4}), (y_{1}, y_{2}, y_{3}, y_{4})) = (\frac{1}{2} - a_{1}) x_{1} y_{1} + (- \frac{1}{2} - a_{2}) x_{2} y_{2} - a_{3} x_{3} y_{3} \\ + (\frac{1}{2} - b_{1}) (x_{1} y_{2} + x_{2} y_{1}) - b_{2} (x_{2} y_{3} + x_{3} y_{2}) \\ - b_{3} (x_{3} y_{1} + x_{1} y_{3}) - c_{1} (x_{1} y_{4} + x_{4} y_{1}) \\ - c_{2} (x_{2} y_{4} + x_{4} y_{2}) - c_{3} (x_{3} y_{4} + x_{4} y_{3}) - c_{4} x_{4} y_{4} \end{array}

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

T_{i} ((x_{1}, x_{2}, x_{3}, 0), (y_{1}, y_{2}, y_{3}, 0)) \in L_{s} ({{}^{2}l}_{\infty}^{3}), ‖ T_{i} ((x_{1}, x_{2}, x_{3}, 0), (y_{1}, y_{2}, y_{3}, 0)) ‖ \leq 1

and

S = \frac{1}{2} (T_{1} ((x_{1}, x_{2}, x_{3}, 0), (y_{1}, y_{2}, y_{3}, 0)) + T_{2} ((x_{1}, x_{2}, x_{3}, 0), (y_{1}, y_{2}, y_{3}, 0)) .

Since $S \in e x t B_{L_{s} ({{}^{2}l}_{\infty}^{3})}$ , 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,

\begin{array}{l} T_{1} ((x_{1}, x_{2}, x_{3}, x_{4}), (y_{1}, y_{2}, y_{3}, y_{4})) = S ((x_{1}, x_{2}, x_{3}), (y_{1}, y_{2}, y_{3})) + c_{1} (x_{1} y_{4} + x_{4} y_{1}) \\ + c_{2} (x_{2} y_{4} + x_{4} y_{2}) + c_{3} (x_{3} y_{4} + x_{4} y_{3}) + c_{4} x_{4} y_{4} \end{array}

and

\begin{array}{l} S_{2} ((x_{1}, x_{2}, x_{3}, x_{4}), (y_{1}, y_{2}, y_{3}, y_{4})) = T ((x_{1}, x_{2}, x_{3}), (y_{1}, y_{2}, y_{3})) - (c_{1} (x_{1} y_{4} + x_{4} y_{1}) \\ + c_{2} (x_{2} y_{4} + x_{4} y_{2}) + c_{3} (x_{3} y_{4} + x_{4} y_{3}) + c_{4} x_{4} y_{4}) . \end{array}

It follows that

\begin{array}{l} 1 \geq max {∣ T_{i} ((1, 1, 1, x_{4}), (1, - 1, 1, y_{4})) ∣, ∣ T_{i} ((1, 1, 1, x_{4}), (1, 1, - 1, y_{4})) ∣ : ∣ x_{4} ∣ \leq 1, ∣ y_{4} ∣ \leq 1, i = 1, 2} \\ = max {1 + ∣ c_{1} (y_{4} + x_{4}) + c_{2} (y_{4} - x_{4}) + c_{3} (y_{4} + x_{4}) + c_{4} x_{4} y_{4} ∣, \\ 1 + ∣ c_{1} (y_{4} + x_{4}) + c_{2} (y_{4} + x_{4}) + c_{3} (y_{4} - x_{4}) + c_{4} x_{4} y_{4} ∣ : ∣ x_{4} ∣ \leq 1, ∣ y_{4} ∣ \leq 1}, \end{array}

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

\begin{array}{l} 0 = c_{1} (y_{4} + x_{4}) + c_{2} (y_{4} - x_{4}) + c_{3} (y_{4} + x_{4}) + c_{4} x_{4} y_{4} \\ 0 = c_{1} (y_{4} + x_{4}) + c_{2} (y_{4} + x_{4}) + c_{3} (y_{4} - x_{4}) + c_{4} x_{4} y_{4}, \end{array}

which shows that c_j = 0 for j = 1, …, 4. Therefore, $S \in e x t B_{L_{s} ({{}^{2}l}_{\infty}^{4})}$ .

Suppose that for n = k, $S \in e x t B_{L_{s} ({{}^{2}l}_{\infty}^{k})}$ . We will show that $S \in e x t B_{L_{s} ({{}^{2}l}_{\infty}^{k + 1})}$ . Suppose that $S = \frac{1}{2} (W_{1} + W_{2})$ for some W₁, $W_{2} \in L_{s} ({{}^{2}l}_{\infty}^{k + 1})$ with ||W₁|| = 1 = ||W₂||. By the above argument, we may assume that

W_{1} ((x_{1}, \dots, x_{k + 1}), (y_{1}, \dots, y_{k + 1})) = S ((x_{1}, x_{2}, x_{3}), (y_{1}, y_{2}, y_{3})) + \sum_{1 \leq j \leq k} c_{j} (x_{j} y_{k + 1} + x_{k + 1} y_{j}) + d x_{k + 1} y_{k + 1}

and

W_{2} ((x_{1}, \dots, x_{k + 1}), (y_{1}, \dots, y_{k + 1})) = S ((x_{1}, x_{2}, x_{3}), (y_{1}, y_{2}, y_{3})) - \sum_{1 \leq j \leq k} c_{j} (x_{j} y_{k + 1} + x_{k + 1} y_{j}) - d x_{k + 1} y_{k + 1}

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

\begin{array}{l} 1 \geq max {∣ W_{i} ((1, 1, 1, x_{4}, \dots, x_{k + 1}), (1, - 1, 1, y_{4}, \dots, y_{k + 1})) ∣, \\ ∣ W_{i} ((1, 1, 1, x_{4}, \dots, x_{k + 1}), (1, 1, - 1, y_{4}, \dots, y_{k + 1})) ∣, \\ ∣ W_{i} ((1, 1, 1, x_{4}, \dots, x_{k + 1}), (1, - 1, - 1, y_{4}, \dots, y_{k + 1})) ∣ : \\ ∣ x_{j} ∣ \leq 1, ∣ y_{j} ∣ \leq 1, j = 4, \dots, k + 1, i = 1, 2} \\ = max {1 + ∣ c_{1} (y_{k + 1} + x_{k + 1}) + c_{2} (y_{k + 1} - x_{k + 1}) + c_{3} (y_{k + 1} + x_{k + 1}) \\ + \sum_{4 \leq j \leq k} c_{j} (x_{j} y_{k + 1} + x_{k + 1} y_{j}) + d x_{k + 1} y_{k + 1} ∣, \\ 1 + ∣ c_{1} (y_{k + 1} + x_{k + 1}) + c_{2} (y_{k + 1} + x_{k + 1}) + c_{3} (y_{k + 1} - x_{k + 1}) \\ + \sum_{4 \leq j \leq k} c_{j} (x_{j} y_{k + 1} + x_{k + 1} y_{j}) + d x_{k + 1} y_{k + 1} ∣, \\ 1 + ∣ c_{1} (y_{k + 1} + x_{k + 1}) + c_{2} (y_{k + 1} - x_{k + 1}) + c_{3} (y_{k + 1} - x_{k + 1}) \\ + \sum_{4 \leq j \leq k} c_{j} (x_{j} y_{k + 1} + x_{k + 1} y_{j}) + d x_{k + 1} y_{k + 1} ∣ : \\ ∣ x_{j} ∣ \leq 1, ∣ y_{j} ∣ \leq 1, j = 4, \dots, k + 1}, \end{array}

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

\begin{array}{l} 0 = c_{1} (y_{k + 1} + x_{k + 1}) + c_{2} (y_{k + 1} - x_{k + 1}) + c_{3} (y_{k + 1} + x_{k + 1}) \\ + \sum_{4 \leq j \leq k} c_{j} (x_{j} y_{k + 1} + x_{k + 1} y_{j}) + d x_{k + 1} y_{k + 1}, \\ 0 = c_{1} (y_{k + 1} + x_{k + 1}) + c_{2} (y_{k + 1} + x_{k + 1}) + c_{3} (y_{k + 1} - x_{k + 1}) \\ + \sum_{4 \leq j \leq k} c_{j} (x_{j} y_{k + 1} + x_{k + 1} y_{j}) + d x_{k + 1} y_{k + 1}, \\ 0 = c_{1} (y_{k + 1} + x_{k + 1}) + c_{2} (y_{k + 1} - x_{k + 1}) + c_{3} (y_{k + 1} - x_{k + 1}) \\ + \sum_{4 \leq j \leq k} c_{j} (x_{j} y_{k + 1} + x_{k + 1} y_{j}) + d x_{k + 1} y_{k + 1} . \end{array}

By Claim 1, we conclude that c_j = 0 = d for j = 1, …, k. Hence, W₁ = S = W₂, so $S \in e x t B_{L_{s} ({{}^{2}l}_{\infty}^{k + 1})}$ . Therefore, we have shown that $S \in e x t B_{L_{s} ({{}^{2}l}_{\infty}^{n})}$ for n ≥ 4.

Let

S ((x_{1}, x_{2}, x_{3}), (y_{1}, y_{2}, y_{3})) = x_{1} y_{1} : = {(1, 0, 0, 0, 0, 0)}^{t} .

Claim 3: for n ≥ 4, $S \in e x t B_{L_{s} ({{}^{2}l}_{\infty}^{n})}$ .

Suppose that $S = \frac{1}{2} (T_{1} + T_{2})$ for some T₁, $T_{2} \in L_{s} ({{}^{2}l}_{\infty}^{n})$ with ||T₁|| = 1 = ||T₂||.

Write

T_{1} ((x_{1}, \dots, x_{n}), (y_{1}, \dots, y_{n})) = (1 + a_{1}) x_{1} y_{1} + \sum_{2 \leq j \leq n} a_{j} x_{j} y_{j} + \sum_{1 \leq i < j \leq n} b_{i j} (x_{i} y_{j} + x_{j} y_{i})

and

T_{2} ((x_{1}, \dots, x_{n}), (y_{1}, \dots, y_{n})) = (1 - a_{1}) x_{1} y_{1} - \sum_{2 \leq j \leq n} a_{j} x_{j} y_{j} - \sum_{1 \leq i < j \leq n} b_{i j} (x_{i} y_{j} + x_{j} y_{i})

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

1 \geq max {∣ T_{l} (e_{1}, e_{1}) ∣ : l = 1, 2} = 1 + ∣ a_{1} ∣,

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

1 \geq max {∣ T_{l} (e_{1} + t e_{j}, e_{1} + t e_{j}) ∣ : l = 1, 2, ∣ t ∣ \leq 1} = 1 + ∣ a_{j} t^{2} + 2 ∣ b_{i j} t ∣

and

0 = a_{j} t^{2} + 2 b_{1 j} t

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

1 \geq max {∣ T_{l} (e_{1} + e_{i} + e_{j}, e_{1} + e_{i} + e_{j}) ∣ : l = 1, 2, = 1 + 2 b_{i j} ∣,

b_ij = 0. Hence, S = T₁, so $S \in e x t B_{L_{s} ({{}^{2}l}_{\infty}^{n})}$ 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 $\frac{1}{2}$ ’s and those having four $\frac{1}{2}$ ’s. Therefore, the other 39 extreme points in the list of Theorem 2.2 are extreme in the unit ball of $L_{s} ({{}^{2}l}_{\infty}^{n})$ . We complete the proof.

3. The Exposed Points of the Unit Ball of $L_{s} ({{}^{2}l}_{\infty}^{3})$

Go to

Abstract
1. Introduction
2. The Extreme Points of the Unit Ball of $L_{s} ({{}^{2}l}_{\infty}^{3})$
3. The Exposed Points of the Unit Ball of $L_{s} ({{}^{2}l}_{\infty}^{3})$
4. The Smooth Points of the Unit Ball of $L_{s} ({{}^{2}l}_{\infty}^{3})$
Acknowledgements
References

Theorem 2.2 leads to a complete formula of ||f|| for every $f \in L_{s} {({{}^{2}l}_{\infty}^{3})}^{*}$ .

Theorem 3.1

Let $f \in L_{s} {({{}^{2}l}_{\infty}^{3})}^{*}$ with α₁₁ := f(x₁x₂), α₂₂ := f(y₁y₂), α₃₃ := f(z₁z₂), β₁₂ := f(x₁y₂ + x₂y₁), β₁₃ := f(x₁z₂ + x₂z₁), β₂₃ := f(y₁z₂ + y₂z₁). Then,

\begin{array}{l} ‖ f ‖ = max_{j = 1, 2, 3} {∣ α_{j j} ∣, \frac{1}{2} ∣ α_{11} - α_{22} ∣ + \frac{1}{2} ∣ β_{12} ∣, \frac{1}{2} ∣ α_{22} - α_{33} ∣ + \frac{1}{2} ∣ β_{23} ∣, \\ \frac{1}{2} ∣ α_{11} - α_{33} ∣ + \frac{1}{2} ∣ β_{13} ∣, \frac{1}{4} (∣ 2 α_{j j} + β_{12} ∣ + ∣ β_{13} - β_{23} ∣), \\ \frac{1}{4} (∣ 2 α_{j j} - β_{12} ∣ + ∣ β_{13} + β_{23} ∣)} . \end{array}

Proof

By the Krein-Milman Theorem, $‖ f ‖ = {sup}_{T \in e x t B_{L_{s} ({{}^{2}l}_{\infty}^{3})}} ∣ f (T) ∣$ . By Theorem 2.2, it follows that

\begin{array}{l} ‖ 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 ({(\frac{1}{2}, - \frac{1}{2}, 0, 1, 0, 0)}^{t}) ∣, ∣ f ({(\frac{1}{2}, - \frac{1}{2}, 0, - 1, 0, 0)}^{t}) ∣, ∣ f ({(\frac{1}{2}, 0, - \frac{1}{2}, 0, 1, 0)}^{t}) ∣, \\ ∣ f ({(\frac{1}{2}, 0, - \frac{1}{2}, 0, - 1, 0)}^{t}) ∣, ∣ f ({(0, \frac{1}{2}, - \frac{1}{2}, 0, 0, 1)}^{t}) ∣, ∣ f ({(0, \frac{1}{2}, - \frac{1}{2}, 0, 0, - 1)}^{t}) ∣, \\ ∣ f ({(\frac{1}{2}, 0, 0, - \frac{1}{2}, \frac{1}{2}, \frac{1}{2})}^{t}) ∣, ∣ f ({(\frac{1}{2}, 0, 0, \frac{1}{2}, - \frac{1}{2}, \frac{1}{2})}^{t}) ∣, ∣ f ({(\frac{1}{2}, 0, 0, \frac{1}{2}, \frac{1}{2}, - \frac{1}{2})}^{t}) ∣, \\ ∣ f ({(0, \frac{1}{2}, 0, - \frac{1}{2}, \frac{1}{2}, \frac{1}{2})}^{t}) ∣, ∣ f ({(0, \frac{1}{2}, 0, \frac{1}{2}, - \frac{1}{2}, \frac{1}{2})}^{t}) ∣, ∣ f ({(0, \frac{1}{2}, 0, \frac{1}{2}, \frac{1}{2}, - \frac{1}{2})}^{t}) ∣, \\ ∣ f ({(0, 0, \frac{1}{2}, - \frac{1}{2}, \frac{1}{2}, \frac{1}{2})}^{t}) ∣, ∣ f ({(0, 0, \frac{1}{2}, \frac{1}{2}, - \frac{1}{2}, \frac{1}{2})}^{t}) ∣, ∣ f ({(0, 0, \frac{1}{2}, \frac{1}{2}, \frac{1}{2} - \frac{1}{2})}^{t}) ∣, \\ ∣ f ({(- \frac{1}{2}, 0, 0, \frac{1}{2}, \frac{1}{2}, \frac{1}{2})}^{t}) ∣, ∣ f ({(0, - \frac{1}{2}, 0, \frac{1}{2}, \frac{1}{2}, \frac{1}{2})}^{t}) ∣, ∣ f ({(0, 0, - \frac{1}{2}, \frac{1}{2}, \frac{1}{2}, \frac{1}{2})}^{t}) ∣} \\ = max_{j = 1, 2, 3} {∣ α_{j j} ∣, \frac{1}{2} ∣ α_{11} - α_{22} ∣ + \frac{1}{2} ∣ β_{12} ∣, \frac{1}{2} ∣ α_{22} - α_{33} ∣ + \frac{1}{2} ∣ β_{23} ∣, \\ \frac{1}{2} ∣ α_{11} - α_{33} ∣ + \frac{1}{2} ∣ β_{13} ∣, \frac{1}{4} (∣ 2 α_{j j} + β_{12} ∣ + ∣ β_{13} - β_{23} ∣), \\ \frac{1}{4} (∣ 2 α_{j j} - β_{12} ∣ + ∣ β_{13} + β_{23} ∣)} . \end{array}

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]) $e x p B_{L_{s} ({{}^{2}l}_{\infty}^{3})} = e x t B_{L_{s} ({{}^{2}l}_{\infty}^{3})}$

Proof

It suffices to show that if $T \in e x t B_{L_{s} ({{}^{2}l}_{\infty}^{3})}$ , then it is exposed.

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

Let $f = (α_{11}, α_{22}, α_{33}, β_{12}, β_{13}, β_{23}) = (1, 0, 0, 0, 0, 0) \in L_{s} {({{}^{2}l}_{\infty}^{3})}^{*}$ . Then, by Theorems 3.1 and 3.2, ||f|| = 1 = f(T) and |f(S)| < 1 for $S \in e x t B_{L_{s} ({{}^{2}l}_{\infty}^{3})} \ {\pm T}$ . 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: $T = {(\frac{1}{2}, - \frac{1}{2}, 0, 1, 0, 0)}^{t}$ is exposed.

Let $f = (α_{11}, α_{22}, α_{33}, β_{12}, β_{13}, β_{23}) = (\frac{1}{4}, 0, 0, \frac{7}{4}, 0, 0) \in L_{s} {({{}^{2}l}_{\infty}^{3})}^{*}$ . Then, by Theorems 3.1 and 3.2, ||f|| = 1 = f(T) and |f(S)| < 1 for $S \in e x t B_{L_{s} ({{}^{2}l}_{\infty}^{3})} \ {\pm T}$ . Similarly, $- {(\frac{1}{2}, - \frac{1}{2}, 0, 1, 0, 0)}^{t}$ is exposed.

Claim: $T = {(\frac{1}{2}, - \frac{1}{2}, 0, - 1, 0, 0)}^{t}$ is exposed.

Let $f = (α_{11}, α_{22}, α_{33}, β_{12}, β_{13}, β_{23}) = (\frac{1}{4}, 0, 0, - \frac{7}{4}, 0, 0) \in L_{s} {({{}^{2}l}_{\infty}^{3})}^{*}$ . Then, by Theorems 3.1 and 3.2, ||f|| = 1 = f(T) and |f(S)| < 1 for $S \in e x t B_{L_{s} ({{}^{2}l}_{\infty}^{3})} \ {\pm T}$ . Similarly, $- {(\frac{1}{2}, - \frac{1}{2}, 0, - 1, 0, 0)}^{t}$ is exposed.

Claim: $T = {(\frac{1}{2}, 0, - \frac{1}{2}, 0, 1, 0)}^{t}$ is exposed.

Let $f = (α_{11}, α_{22}, α_{33}, β_{12}, β_{13}, β_{23}) = (\frac{1}{4}, 0, 0, 0, \frac{7}{4}, 0) \in L_{s} {({{}^{2}l}_{\infty}^{3})}^{*}$ . Then, by Theorems 3.1 and 3.2, ||f|| = 1 = f(T) and |f(S)| < 1 for $S \in e x t B_{L_{s} ({{}^{2}l}_{\infty}^{3})} \ {\pm T}$ . Similarly, $- {(\frac{1}{2}, 0, - \frac{1}{2}, 0, 1, 0)}^{t}$ is exposed.

Claim: $T = {(\frac{1}{2}, 0, - \frac{1}{2}, 0, - 1, 0)}^{t}$ is exposed.

Let $f = (α_{11}, α_{22}, α_{33}, β_{12}, β_{13}, β_{23}) = (\frac{1}{4}, 0, 0, 0, - \frac{7}{4}, 0) \in L_{s} {({{}^{2}l}_{\infty}^{3})}^{*}$ . Then, by Theorems 3.1 and 3.2, ||f|| = 1 = f(T) and |f(S)| < 1 for $S \in e x t B_{L_{s} ({{}^{2}l}_{\infty}^{3})} \ {\pm T}$ . Similarly, $- {(\frac{1}{2}, 0, - \frac{1}{2}, 0, - 1, 0)}^{t}$ is exposed.

Claim: $T = {(0, \frac{1}{2}, - \frac{1}{2}, 0, 0, 1)}^{t}$ is exposed.

Let $f = (α_{11}, α_{22}, α_{33}, β_{12}, β_{13}, β_{23}) = (0, \frac{1}{4}, 0, 0, 0, \frac{7}{4}) \in L_{s} {({{}^{2}l}_{\infty}^{3})}^{*}$ . Then, by Theorems 3.1 and 3.2, ||f|| = 1 = f(T) and |f(S)| < 1 for $S \in e x t B_{L_{s} ({{}^{2}l}_{\infty}^{3})} \ {\pm T}$ . Similarly, $- {(0, \frac{1}{2}, - \frac{1}{2}, 0, 0, 1)}^{t}$ is exposed.

Claim: $T = {(0, \frac{1}{2}, - \frac{1}{2}, 0, 0, - 1)}^{t}$ is exposed.

Let $f = (α_{11}, α_{22}, α_{33}, β_{12}, β_{13}, β_{23}) = (0, \frac{1}{4}, 0, 0, 0, - \frac{7}{4}) \in L_{s} {({{}^{2}l}_{\infty}^{3})}^{*}$ . Then, by Theorems 3.1 and 3.2, ||f|| = 1 = f(T) and |f(S)| < 1 for $S \in e x t B_{L_{s} ({{}^{2}l}_{\infty}^{3})} \ {\pm T}$ . Similarly, $- {(0, \frac{1}{2}, - \frac{1}{2}, 0, 0, - 1)}^{t}$ is exposed.

Claim: $T = {(- \frac{1}{2}, 0, 0, \frac{1}{2}, \frac{1}{2}, \frac{1}{2})}^{t}$ is exposed.

Let $f = (α_{11}, α_{22}, α_{33}, β_{12}, β_{13}, β_{23}) = (- \frac{1}{4}, 0, 0, \frac{5}{4}, 1, \frac{5}{4}) \in L_{s} {({{}^{2}l}_{\infty}^{3})}^{*}$ . Then, by Theorems 3.1 and 3.2, ||f|| = 1 = f(T) and |f(S)| < 1 for $S \in e x t B_{L_{s} ({{}^{2}l}_{\infty}^{3})} \ {\pm T}$ . Similarly, $- {(- \frac{1}{2}, 0, 0, \frac{1}{2}, \frac{1}{2}, \frac{1}{2})}^{t}, \pm {(0, - \frac{1}{2}, 0, \frac{1}{2}, \frac{1}{2}, \frac{1}{2})}^{t}, \pm {(0, 0, - \frac{1}{2}, \frac{1}{2}, \frac{1}{2}, \frac{1}{2})}^{t}$ are exposed.

Claim: $T = {(\frac{1}{2}, 0, 0, - \frac{1}{2}, \frac{1}{2}, \frac{1}{2})}^{t}$ is exposed.

Let $f = (α_{11}, α_{22}, α_{33}, β_{12}, β_{13}, β_{23}) = (\frac{1}{4}, 0, 0, - \frac{5}{4}, 1, \frac{5}{4}) \in L_{s} {({{}^{2}l}_{\infty}^{3})}^{*}$ . Then, by Theorems 3.1 and 3.2, ||f|| = 1 = f(T) and |f(S)| < 1 for $S \in e x t B_{L_{s} ({{}^{2}l}_{\infty}^{3})} \ {\pm T}$ . Similarly, $- {(\frac{1}{2}, 0, 0, - \frac{1}{2}, \frac{1}{2}, \frac{1}{2})}^{t}, \pm {(0, \frac{1}{2}, 0, - \frac{1}{2}, \frac{1}{2}, \frac{1}{2})}^{t}, \pm {(0, 0, \frac{1}{2}, - \frac{1}{2}, \frac{1}{2}, \frac{1}{2})}^{t}$ are exposed.

Claim: $T = {(\frac{1}{2}, 0, 0, \frac{1}{2}, - \frac{1}{2}, \frac{1}{2})}^{t}$ is exposed.

Let $f = (α_{11}, α_{22}, α_{33}, β_{12}, β_{13}, β_{23}) = (\frac{1}{4}, 0, 0, \frac{5}{4}, - 1, \frac{5}{4}) \in L_{s} {({{}^{2}l}_{\infty}^{3})}^{*}$ . Then, by Theorems 3.1 and 3.2, ||f|| = 1 = f(T) and |f(S)| < 1 for $S \in e x t B_{L_{s} ({{}^{2}l}_{\infty}^{3})} \ {\pm T}$ . Similarly, $- {(\frac{1}{2}, 0, 0, \frac{1}{2}, - \frac{1}{2}, \frac{1}{2})}^{t}, \pm {(0, \frac{1}{2}, 0, \frac{1}{2}, - \frac{1}{2}, \frac{1}{2})}^{t}, \pm {(0, 0, \frac{1}{2}, \frac{1}{2}, - \frac{1}{2}, \frac{1}{2})}^{t}$ are exposed.

Claim: $T = {(\frac{1}{2}, 0, 0, \frac{1}{2}, \frac{1}{2}, - \frac{1}{2})}^{t}$ is exposed.

Let $f = (α_{11}, α_{22}, α_{33}, β_{12}, β_{13}, β_{23}) = (\frac{1}{4}, 0, 0, \frac{5}{4}, 1, - \frac{5}{4}) \in L_{s} {({{}^{2}l}_{\infty}^{3})}^{*}$ . Then, by Theorems 3.1 and Theorem 3.2, ||f|| = 1 = f(T) and |f(S)| < 1 for $S \in e x t B_{L_{s} ({{}^{2}l}_{\infty}^{3})} \ {\pm T}$ . Similarly, $- {(\frac{1}{2}, 0, 0, \frac{1}{2}, \frac{1}{2}, - \frac{1}{2})}^{t}, \pm {(0, \frac{1}{2}, 0, \frac{1}{2}, \frac{1}{2}, - \frac{1}{2})}^{t}, \pm {(0, 0, \frac{1}{2}, \frac{1}{2}, \frac{1}{2}, - \frac{1}{2})}^{t}$ are exposed.

4. The Smooth Points of the Unit Ball of $L_{s} ({{}^{2}l}_{\infty}^{3})$

Go to

Abstract
1. Introduction
2. The Extreme Points of the Unit Ball of $L_{s} ({{}^{2}l}_{\infty}^{3})$
3. The Exposed Points of the Unit Ball of $L_{s} ({{}^{2}l}_{\infty}^{3})$
4. The Smooth Points of the Unit Ball of $L_{s} ({{}^{2}l}_{\infty}^{3})$
Acknowledgements
References

In this section we will characterize all the smooth points of the unit ball of $L_{s} ({{}^{2}l}_{\infty}^{3})$ .

Theorem 4.1

Let $T = (a_{11}, a_{22}, a_{33}, 2 a_{12}, 2 a_{13}, 2 a_{23}) \in L_{s} ({{}^{2}l}_{\infty}^{3})$ . Then, $T \in s m B_{L_{s} ({{}^{2}l}_{\infty}^{3})}$ if and only if

\begin{array}{l} (∣ T ((1, 1, 1), (1, 1, 1)) ∣ = 1, 0 < ∣ a_{11} + a_{22} + a_{33} + 2 a_{23} ∣ < 1, \\ ∣ T (Y) ∣ < 1 f o r a l l Y \in Γ \ {[(1, 1, 1), (1, 1, 1)]}) \\ o r & (∣ T ((- 1, 1, 1), (- 1, 1, 1)) ∣ = 1, 0 < ∣ a_{11} + a_{22} + a_{33} + 2 a_{23} ∣ < 1, \\ ∣ T (Y) ∣ < 1 f o r a l l Y \in Γ \ {[(- 1, 1, 1), (- 1, 1, 1)]}) \\ o r & (∣ T ((1, - 1, 1), (1, - 1, 1)) ∣ = 1, 0 < ∣ a_{11} + a_{22} + a_{33} + 2 a_{13} ∣ < 1, \\ ∣ T (Y) ∣ < 1 f o r a l l Y \in Γ \ {(1, - 1, 1), (1, - 1, 1)}) \\ o r & (∣ T ((1, 1, - 1), (1, 1, - 1)) ∣ = 1, 0 < ∣ a_{11} + a_{22} + a_{33} + 2 a_{12} ∣ < 1, \\ ∣ T (Y) ∣ < 1 f o r a l l Y \in Γ \ {[(1, 1, - 1), (1, 1, - 1)]}) \\ o r & (∣ T ((1, 1, 1), (1, - 1, 1)) ∣ = 1, 0 < ∣ a_{11} - a_{22} + a_{33} ∣ < 1, \\ ∣ T (Y) ∣ < 1 f o r a l l Y \in Γ \ {[(1, 1, 1), (1, - 1, 1)]}) \\ o r & (∣ T ((1, 1, 1), (1, 1, - 1)) ∣ = 1, 0 < ∣ a_{11} + a_{22} - a_{33} ∣ < 1, \\ ∣ T (Y) ∣ < 1 f o r a l l Y \in Γ \ {[(1, 1, 1), (1, 1, - 1)]}) \\ o r & (∣ T ((1, 1, 1), (- 1, 1, 1)) ∣ = 1, 0 < ∣ - a_{11} + a_{22} + a_{33} ∣ < 1, \\ ∣ T (Y) ∣ < 1 f o r a l l Y \in Γ \ {[(1, 1, 1), (- 1, 1, 1)]}) \\ o r & (∣ T ((1, - 1, 1), (1, 1, - 1)) ∣ = 1, 0 < ∣ - a_{11} + a_{22} + a_{33} ∣ < 1, \\ ∣ T (Y) ∣ < 1 f o r a l l Y \in Γ \ {[(1, - 1, 1), (1, 1, - 1)]}) \\ o r & (∣ T ((1, - 1, 1), (- 1, 1, 1)) ∣ = 1, 0 < ∣ a_{11} + a_{22} - a_{33} ∣ < 1, \\ ∣ T (Y) ∣ < 1 f o r a l l Y \in Γ \ {[(1, - 1, 1), (- 1, 1, 1)]}) \\ o r & (∣ T ((1, 1, - 1), (- 1, 1, 1)) ∣ = 1, 0 < ∣ a_{11} - a_{22} + a_{33} ∣ < 1, \\ ∣ T (Y) ∣ < 1 f o r a l l Y \in Γ \ {[(1, 1, - 1), (- 1, 1, 1)]}) . \end{array}

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, $∣ a_{12} ∣ < \frac{1}{2}, ∣ a_{13} ∣ < \frac{1}{2}, ∣ a_{23} ∣ < \frac{1}{2}$ . 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 (a_{12} + a_{13}) > 0 and a_{11} + a_{22} + a_{33} + 2 a_{23} > 0.

Let $f \in L_{s} {({{}^{2}l}_{\infty}^{3})}^{*}$ 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,

\begin{array}{l} ∣ a_{j j} ∣ + \frac{1}{n_{1}} < 1, ∣ a_{12} ∣ + \frac{1}{n_{1}} < \frac{1}{2}, ∣ a_{13} ∣ + \frac{1}{n_{1}} < \frac{1}{2}, ∣ a_{23} ∣ + \frac{1}{n_{1}} < \frac{1}{2}, \\ ∣ T (Y) ∣ + \frac{10}{n_{1}} < 1 for all Y \in Γ \ {[(1, 1, 1), (1, 1, 1)]} . \end{array}

By Theorem 2.1, for n > n₁,

\begin{array}{l} 1 = ‖ {(a_{11} \pm \frac{1}{n}, 2 a_{12}, 2 a_{13}, a_{22} \mp \frac{1}{n}, 2 a_{23}, a_{33})}^{t} ‖, \\ 1 = ‖ {(a_{11} \pm \frac{1}{n}, 2 a_{12}, 2 a_{13}, a_{22}, 2 a_{23}, a_{33} \mp \frac{1}{n})}^{t} ‖, \\ 1 = ‖ {(a_{11}, 2 a_{12} \pm \frac{1}{n}, 2 a_{13} \mp \frac{1}{n}, a_{22}, 2 a_{23}, a_{33})}^{t} ‖, \\ 1 = ‖ {(a_{11} \pm \frac{1}{n}, 2 a_{12}, 2 a_{13}, a_{22}, 2 a_{23} \mp \frac{1}{n}, a_{33})}^{t} ‖ . \end{array}

It follows that for n > n₁,

(1) $1 \geq ∣ f ({(a_{11} \pm \frac{1}{n}, 2 a_{12}, 2 a_{13}, a_{22} \mp \frac{1}{n}, 2 a_{23}, a_{33})}^{t}) ∣ = ∣ 1 \pm \frac{1}{n} (α_{11} - α_{22}) ∣,$
(2) $1 \geq ∣ f ({(a_{11} \pm \frac{1}{n}, 2 a_{12}, 2 a_{13}, a_{22}, 2 a_{23}, a_{33} \mp \frac{1}{n})}^{t}) ∣ = ∣ 1 \pm \frac{1}{n} (α_{11} - α_{33}) ∣,$
(3) $1 \geq ∣ f ({(a_{11}, 2 a_{12} \pm \frac{1}{n}, 2 a_{13} \mp \frac{1}{n}, a_{22}, 2 a_{23}, a_{33})}^{t}) ∣ = ∣ 1 \pm \frac{1}{2 n} (β_{12} - β_{13}) ∣,$
(4) $1 \geq ∣ f ({(a_{11} \pm \frac{1}{n}, 2 a_{12}, 2 a_{13}, a_{22}, 2 a_{23} \mp \frac{1}{n}, a_{33})}^{t}) ∣ = ∣ 1 \pm \frac{1}{n} (α_{11} - \frac{1}{2} β_{23}) ∣ .$

By (1) − (4), α₁₁ = α₂₂ = α₃₃, β₁₂ = β₁₃, $α_{11} = \frac{1}{2} β_{23}$ . Let n₂ ∈ ℕ be such that n₂ > n₁ and

\begin{array}{l} 0 < 2 (a_{12} + a_{13}) - \frac{1}{n_{2}} < 2 (a_{12} + a_{13}) + \frac{1}{n_{2}} < 1, \\ 0 < a_{11} + a_{22} + a_{33} + 2 a_{23} - \frac{1}{n_{2}} < a_{11} + a_{22} + a_{33} + 2 a_{23} + \frac{1}{n_{2}} < 1. \end{array}

By Theorem 2.1, for n > n₂,

1 = ‖ {(a_{11} \pm \frac{1}{n}, 2 a_{12} \mp \frac{1}{2 n}, 2 a_{13} \mp \frac{1}{2 n}, a_{22} \pm \frac{1}{2 n}, 2 a_{23} \mp \frac{1}{n}, a_{33} \pm \frac{1}{2 n})}^{t} ‖ .

Since

1 \geq ∣ f {(a_{11} \pm \frac{1}{n}, 2 a_{12} \mp \frac{1}{2 n}, 2 a_{13} \mp \frac{1}{2 n}, a_{22} \pm \frac{1}{2 n}, 2 a_{23} \mp \frac{1}{n}, a_{33} \pm \frac{1}{2 n})}^{t} ∣ = ∣ 1 \pm \frac{1}{n} (α_{11} - \frac{1}{2} β_{12}) ∣,

so, $α_{11} = \frac{1}{2} β_{12}$ , hence, β₁₂ = β₁₃ = β₂₃. Therefore,

\begin{array}{l} 1 = f (T) = \sum_{j = 1}^{3} a_{j j} α_{j j} + a_{12} β_{12} + a_{13} β_{13} + a_{23} β_{23} \\ = (a_{11} + a_{22} + a_{33} + 2 a_{12} + 2 a_{13} + 2 a_{23}) α_{11} \\ = α_{11}, \end{array}

hence, α_jj = 1 for j = 1, 2,3 and β₁₂ = β₁₃ = β₂₃ = 2. Hence, T would be smooth in $L_{s} ({{}^{2}l}_{\infty}^{3})$ .

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, $∣ a_{12} ∣ < \frac{1}{2}, ∣ a_{23} ∣ < \frac{1}{2}$ . 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,

2 a_{13} > 0, a_{11} - a_{22} + a_{33} > 0.

Let $f \in L_{s} {({{}^{2}l}_{\infty}^{3})}^{*}$ 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,

\begin{array}{l} ∣ a_{j j} ∣ + \frac{1}{n_{1}} < 1, ∣ a_{12} ∣ + \frac{1}{n_{1}} < \frac{1}{2}, ∣ a_{23} ∣ + \frac{1}{n_{1}} < \frac{1}{2}, \\ ∣ T (Y) ∣ + \frac{10}{n_{1}} < 1 for all Y \in Γ \ {[(1, - 1, 1), (1, 1, 1)]} . \end{array}

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

0 < 2 a_{13} - \frac{1}{n_{2}} < 2 a_{13} + \frac{1}{n_{2}} < 1

and

0 < a_{11} - a_{22} + a_{33} - \frac{1}{n_{2}} < a_{11} - a_{22} + a_{33} + \frac{1}{n_{2}} < 1.

By Theorem 2.1, for n > n₂,

\begin{array}{l} 1 = ‖ {(a_{11} \pm \frac{1}{n}, 2 a_{12}, 2 a_{13}, a_{22} \pm \frac{1}{n}, 2 a_{23}, a_{33})}^{t} ‖, \\ 1 = ‖ {(a_{11} \pm \frac{1}{n}, 2 a_{12}, 2 a_{13}, a_{22}, 2 a_{23}, a_{33} \mp \frac{1}{n})}^{t} ‖, \\ 1 = ‖ {(a_{11}, 2 a_{12} \pm \frac{1}{n}, 2 a_{13}, a_{22}, 2 a_{23} \mp \frac{1}{n}, a_{33})}^{t} ‖, \\ 1 = ‖ {(a_{11}, 2 a_{12} \pm \frac{1}{n}, 2 a_{13}, a_{22}, 2 a_{23} \pm \frac{1}{n}, a_{33})}^{t} ‖, \\ 1 = ‖ {(a_{11} \pm \frac{1}{n}, 2 a_{12}, 2 a_{13} \mp \frac{1}{n}, a_{22}, 2 a_{23}, a_{33})}^{t} \end{array}

It follows that for n > n₂,

(1′) $1 \geq ∣ f ({(a_{11} \pm \frac{1}{n}, 2 a_{12}, 2 a_{13}, a_{22} \pm \frac{1}{n}, 2 a_{23}, a_{33})}^{t}) ∣ = ∣ 1 \pm \frac{1}{n} (α_{11} + α_{22}) ∣,$
(2′) $1 \geq ∣ f ({(a_{11} \pm \frac{1}{n}, 2 a_{12}, 2 a_{13}, a_{22}, 2 a_{23}, a_{33} \mp \frac{1}{n})}^{t}) ∣ = ∣ 1 \pm \frac{1}{n} (α_{11} - α_{33}) ∣,$
(3′) $1 \geq ∣ f ({(a_{11}, 2 a_{12} \pm \frac{1}{n}, 2 a_{13}, a_{22}, 2 a_{23} \mp \frac{1}{n}, a_{33})}^{t}) ∣ = ∣ 1 \pm \frac{1}{2 n} (β_{12} - β_{23}) ∣,$
(4′) $1 \geq ∣ f ({(a_{11}, 2 a_{12} \pm \frac{1}{n}, 2 a_{13}, a_{22}, 2 a_{23} \pm \frac{1}{n}, a_{33})}^{t}) ∣ = ∣ 1 \pm \frac{1}{2 n} (β_{12} + β_{23}) ∣,$
(5′) $1 \geq ∣ f ({(a_{11} \pm \frac{1}{n}, 2 a_{12}, 2 a_{13} \mp \frac{1}{n}, a_{22}, 2 a_{23}, a_{33})}^{t}) ∣ = ∣ 1 \pm \frac{1}{n} (α_{11} + \frac{1}{2} β_{13}) ∣ .$

By (1′) − (5′), α₁₁ = −α₂₂ = α₃₃, β₁₂ = β₂₃ = 0, $α_{11} = \frac{1}{2} β_{13}$ . It follows that

\begin{array}{l} 1 = f (T) = \sum_{j = 1}^{3} a_{j j} α_{j j} + a_{12} β_{12} + a_{13} β_{13} + a_{23} β_{23} \\ = (a_{11} - a_{22} + a_{33}) α_{11} + 2 a_{13} α_{11} \\ = (1 - 2 a_{13}) α_{11} + 2 a_{13} α_{11} \\ = α_{11}, \end{array}

hence, β₁₃ = 2 and α_jj = 1 = −α₂₂ for j = 1, 3. Hence, T would be smooth in $L_{s} ({{}^{2}l}_{\infty}^{3})$ . 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

N o r m (S) : = {X \in Γ : ∣ S (X) ∣ = ‖ S ‖}

for $S \in L_{s} ({{}^{2}l}_{\infty}^{3})$ .

Notice that

\begin{array}{l} 1 \geq ∣ a_{11} + a_{22} + a_{33} + 2 a_{12} ∣, \\ 1 \geq ∣ a_{11} + a_{22} + a_{33} + 2 a_{13} ∣, \\ 1 \geq ∣ a_{11} + a_{22} + a_{33} + 2 a_{23} ∣, \\ 1 \geq ∣ - a_{11} + a_{22} + a_{33} ∣, \\ 1 \geq ∣ a_{11} - a_{22} + a_{33} ∣, \\ 1 \geq ∣ a_{11} + a_{22} - a_{33} ∣ . \end{array}

We have ten subcases as follows:

\begin{array}{l} (∣ T ((1, 1, 1), (1, 1, 1)) ∣ = 1, a_{11} + a_{22} + a_{33} + 2 a_{23} = 0 or \pm 1, \\ ∣ T (Y) ∣ < 1 for all Y \in Γ \ {[(1, 1, 1), (1, 1, 1)]}) \\ or & (∣ T ((- 1, 1, 1), (- 1, 1, 1)) ∣ = 1, a_{11} + a_{22} + a_{33} + 2 a_{23} = 0, or \pm 1, \\ ∣ T (Y) ∣ < 1 for all Y \in Γ \ {[(- 1, 1, 1), (- 1, 1, 1)]}) \\ or & (∣ T ((1, - 1, 1), (1, - 1, 1)) ∣ = 1, a_{11} + a_{22} + a_{33} + 2 a_{13} = 0, or \pm 1, \\ ∣ T (Y) ∣ < 1 for all Y \in Γ \ {(1, - 1, 1), (1, - 1, 1)}) \\ or & (∣ T ((1, 1, - 1), (1, 1, - 1)) ∣ = 1, a_{11} + a_{22} + a_{33} + 2 a_{12} = 0, or \pm 1, \\ ∣ T (Y) ∣ < 1 for all Y \in Γ \ {[(1, 1, - 1), (1, 1, - 1)]}) \\ or & (∣ T ((1, 1, 1), (1, - 1, 1)) ∣ = 1, a_{11} - a_{22} + a_{33} = 0 or \pm 1, \\ ∣ T (Y) ∣ < 1 for all Y \in Γ \ {[(1, 1, 1), (1, - 1, 1)]}) \\ or & (∣ T ((1, 1, 1), (1, 1, - 1)) ∣ = 1, a_{11} + a_{22} - a_{33} = 0 or \pm 1, \\ ∣ T (Y) ∣ < 1 for all Y \in Γ \ {[(1, 1, 1), (1, 1, - 1)]}) \\ or & (∣ T ((1, 1, 1), (- 1, 1, 1)) ∣ = 1, - a_{11} + a_{22} + a_{33} = 0 or \pm 1, \\ ∣ T (Y) ∣ < 1 for all Y \in Γ \ {[(1, 1, 1), (- 1, 1, 1)]}) \\ or & (∣ T ((1, - 1, 1), (1, 1, - 1)) ∣ = 1, - a_{11} + a_{22} + a_{33} = 0 or \pm 1, \\ ∣ T (Y) ∣ < 1 for all Y \in Γ \ {[(1, - 1, 1), (1, 1, - 1)]}) \\ or & (∣ T ((1, - 1, 1), (- 1, 1, 1)) ∣ = 1, a_{11} + a_{22} - a_{33} = 0 or \pm 1, \\ ∣ T (Y) ∣ < 1 for all Y \in Γ \ {[(1, - 1, 1), (- 1, 1, 1)]}) \\ or & (∣ T ((1, 1, - 1), (- 1, 1, 1)) ∣ = 1, a_{11} - a_{22} + a_{33} = 0 or \pm 1, \\ ∣ T (Y) ∣ < 1 for all Y \in Γ \ {[(1, 1, - 1), (- 1, 1, 1)]}) . \end{array}

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

f_{1} = (α_{11}, α_{22}, α_{33}, β_{12}, β_{13}, β_{23}) = (0, 0, 0, 2, 2, 0)

and

f_{2} = (α_{11}, α_{22}, α_{33}, β_{12}, β_{13}, β_{23}) = (1, 1, 1, 2, 2, 2) \in L_{s} {({{}^{2}l}_{\infty}^{3})}^{*} .

By Theorem 3.1, ||f_k|| = 1 = f_k(T) for k = 1, 2. Hence, $T \notin s m B_{L_{s} ({{}^{2}l}_{\infty}^{3})}$ . This is a contradiction.

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

f_{t} (1, 1, 1, t, t, 0) \in L_{s} {({{}^{2}l}_{\infty}^{3})}^{*} .

By Theorem 3.1, ||f_t|| = 1 = f_t(T) for |t| ≤ 2. Hence, $T \notin s m B_{L_{s} ({{}^{2}l}_{\infty}^{3})}$ . This is a contradiction.

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

f_{t} (- 1, - 1, - 1, t, t, 0) \in L_{s} {({{}^{2}l}_{\infty}^{3})}^{*} .

By Theorem 3.1, ||f_t|| = 1 = f_t(T) for |t| ≤ 2. Hence, $T \notin s m B_{L_{s} ({{}^{2}l}_{\infty}^{3})}$ . 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 $a_{13} = \frac{1}{2}$ . Let

f_{1} = (α_{11}, α_{22}, α_{33}, β_{12}, β_{13}, β_{23}) = (1, - 1, 1, 0, 2, 0)

and

f_{2} = (α_{11}, α_{22}, α_{33}, β_{12}, β_{13}, β_{23}) = (0, 0, 0, 0, 2, 0) \in L_{s} {({{}^{2}l}_{\infty}^{3})}^{*} .

By Theorem 3.1, ||f_k|| = 1 = f_k(T) for k = 1, 2. Hence, $T \notin s m B_{L_{s} ({{}^{2}l}_{\infty}^{3})}$ .

Suppose that a₁₁ − a₂₂ + a₃₃ = 1. Let

f_{1} = (α_{11}, α_{22}, α_{33}, β_{12}, β_{13}, β_{23}) = (1, - 1, 1, 0, 0, 0)

and

f_{2} = (α_{11}, α_{22}, α_{33}, β_{12}, β_{13}, β_{23}) = (1, - 1, 1, 0, 1, 0) \in L_{s} {({{}^{2}l}_{\infty}^{3})}^{*} .

By Theorem 3.1, ||f_k|| = 1 = f_k(T) for k = 1, 2. Hence, $T \notin s m B_{L_{s} ({{}^{2}l}_{\infty}^{3})}$ .

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

f_{1} = (α_{11}, α_{22}, α_{33}, β_{12}, β_{13}, β_{23}) = - (1, - 1, 1, 0, 0, 0)

and

f_{2} = (α_{11}, α_{22}, α_{33}, β_{12}, β_{13}, β_{23}) = - (1, - 1, 1, 0, 1, 0) \in L_{s} {({{}^{2}l}_{\infty}^{3})}^{*} .

By Theorem 3.1, ||f_k|| = 1 = f_k(T) for k = 1, 2. Hence, $T \notin s m B_{L_{s} ({{}^{2}l}_{\infty}^{3})}$ .

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 $s i g n (T (X_{k})) δ_{X_{k}} \in L_{s} {({{}^{2}l}_{\infty}^{3})}^{*}$ and

‖ s i g n (T (X_{k})) δ_{X_{k}} ‖ = 1 = s i g n (T (X_{k})) δ_{X_{k}} (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 $L_{s} ({{}^{2}l}_{\infty}^{3})$
3. The Exposed Points of the Unit Ball of $L_{s} ({{}^{2}l}_{\infty}^{3})$
4. The Smooth Points of the Unit Ball of $L_{s} ({{}^{2}l}_{\infty}^{3})$
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 $L_{s} ({{}^{2}l}_{\infty}^{3})$
3. The Exposed Points of the Unit Ball of $L_{s} ({{}^{2}l}_{\infty}^{3})$
4. The Smooth Points of the Unit Ball of $L_{s} ({{}^{2}l}_{\infty}^{3})$
Acknowledgements
References

RM. Aron, YS. Choi, SG. Kim, and M. Maestre. Local properties of polynomials on a Banach space. Illinois J Math., 45(2001), 25-39.
YS. Choi, H. Ki, and SG. Kim. Extreme polynomials and multilinear forms on l₁. J Math Anal Appl., 228(1998), 467-482.
YS. Choi, and SG. Kim. The unit ball of . Arch Math (Basel)., 71(1998), 472-480.
YS. Choi, and SG. Kim. Extreme polynomials on c₀. Indian J Pure Appl Math., 29(1998), 983-989.
YS. Choi, and SG. Kim. Smooth points of the unit ball of the space ℘(²l₁). Results Math., 36(1999), 26-33.
YS. Choi, and SG. Kim. Exposed points of the unit balls of the spaces . Indian J Pure Appl Math., 35(2004), 37-41.
S. Dineen. Complex analysis on infinite dimensional spaces, , Springer-Verlag, London, 1999.
S. Dineen. Extreme integral polynomials on a complex Banach space. Math Scand., 92(2003), 129-140.
BC. Grecu. Geometry of 2-homogeneous polynomials on l_p spaces, 1 < p < ∞. J Math Anal Appl., 273(2002), 262-282.
BC. Grecu, GA. Munoz-Fernandez, and JB. Seoane-Sepulveda. Unconditional constants and polynomial inequalities. J Approx Theory., 161(2009), 706-722.
SG. Kim. Exposed 2-homogeneous polynomials on . Math Proc R Ir Acad., 107A(2007), 123-129.
SG. Kim. The unit ball of . Extracta Math., 24(2009), 17-29.
SG. Kim. The unit ball of ℘(²d_*(1, w)²). Math Proc R Ir Acad., 111A(2011), 79-94.
SG. Kim. The unit ball of ℒ_s(²d_*(1, w)²). Kyungpook Math J., 53(2013), 295-306.
SG. Kim. Smooth polynomials of ℘(²d_*(1, w)²). Math Proc R Ir Acad., 113A(2013), 45-58.
SG. Kim. Extreme bilinear forms of ℒ(²d_*(1, w)²). Kyungpook Math J., 53(2013), 625-638.
SG. Kim. Exposed symmetric bilinear forms of ℒ_s(²d_*(1, w)²). Kyungpook Math J., 54(2014), 341-347.
SG. Kim. Exposed bilinear forms of ℒ(²d_*(1, w)²). Kyungpook Math J., 55(2015), 119-126.
SG. Kim. Exposed 2-homogeneous polynomials on the two-dimensional real predual of Lorentz sequence space. Mediterr J Math., 13(2016), 2827-2839.
SG. Kim. The unit ball of . Bull Korean Math Soc., 54(2017), 417-428.
SG. Kim. Extremal problems for . Kyungpook Math J., 57(2017), 223-232.
SG. Kim. The unit ball of . Comment Math Prace Mat., 57(2017), 1-7.
SG. Kim. The geometry of . Commun Korean Math Soc., 32(2017), 991-997.
SG. Kim. The geometry of and optimal constants in the Bohnenblust-Hille inequality for multilinear forms and polynomials. Extracta Math., 33(1)(2018), 51-66.
SG. Kim. Extreme bilinear form on ℝⁿ with the supremum norm. Period Math Hungar., 77(2018), 274-290.
SG. Kim. Exposed polynomials of . Extracta Math., 33(2)(2018), 127-143.
SG. Kim. The unit ball of the space of bilinear forms on ℝ³ with the supremum norm. Commun Korean Math Soc., 34(2019), 487-494.
SG. Kim. Smooth polynomials of the space ., Preprint.
SG. Kim, and SH. Lee. Exposed 2-homogeneous polynomials on Hilbert spaces. Proc Amer Math Soc., 131(2003), 449-453.
GA. Munoz-Fernandez, S. Revesz, and JB. Seoane-Sepulveda. Geometry of homogeneous polynomials on non symmetric convex bodies. Math Scand., 105(2009), 147-160.
GA. Munoz-Fernandez, and JB. Seoane-Sepulveda. Geometry of Banach spaces of trinomials. J Math Anal Appl., 340(2008), 1069-1087.
RA. Ryan, and B. Turett. Geometry of spaces of polynomials. J Math Anal Appl., 221(1998), 698-711.

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