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

Published online September 30, 2020

Copyright © Kyungpook Mathematical Journal.

Extreme Points, Exposed Points and Smooth Points of the Space LS(2l3)

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

We present a complete description of all the extreme points of the unit ball of Ls(l23) which leads to a complete formula for ||f|| for every fLs(l23)*. We also show that extBLs(l23)extBLs(l2n) for every n ≥ 4. Using the formula for ||f|| for every fLs(l23)*, we show that every extreme point of the unit ball of Ls(l23) is exposed. We also characterize all the smooth points of the unit ball of Ls(l23).

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

We denote by BE the closed unit ball of a real Banach space E and by E* the dual space of E. A point xBE is called an extreme point of BE if the equation x=12(y+z) for some y, zBE implies x = y = z. A point xBE is called an exposed point of BE if there is fE* so that f(x) = 1 = ||f|| and f(y) < 1 for every yBE \ {x}. A point xBE is called a smooth point of BE if there is a unique fE* so that f(x) = 1 = ||f||. It is easy to see that every exposed point of BE is an extreme point. We denote by extBE, expBE and smBE the set of extreme points, the set of exposed points and the set of smooth points of BE, 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 xE. 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 ℒs(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)(1p) [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(2h(12)2), where h(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(l23). 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(l32),L(l32),Ls(d2*(1,w)2),L(d2*(1,w)2),Ls(2h(w)2)and L(2h(w)2),

where 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 Ls(l2n) and L(l2n) for n ≥ 2.

Let n ≥ 2 and ln:=n with the supremum norm. Given {aij}i,j=1n, let TL(l2n) be defined by the rule

T((x1,x2,,xn),(y1,y2,,yn))=1i,jnaijxiyj.

If n = 2, for simplicity, we will write

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

as a 4 × 1 column vector. If TLs(l22), 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 TLs(l23), we will write

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

as a 6 × 1 column vector.

In [12] it was shown:

  • (a) extBLs(l22)={±(1,0,0)t,±(0,1,0)t,±(12,-12,1)t,±(12,-12,-1)t};

  • (b) extBLs(l22)=extBLs(l22).

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(l23) which leads to a complete formula of ||f|| for every fLs(l23)*. We also show that extBLs(l22)extBLs(l2n) for every n ≥ 4. Using the formula of ||f|| for every fLs(l23)*, we show that every extreme point of the unit ball of Ls(l23) 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(l23).

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)tLs(l23) with ||T|| = 1. Then, TextBLs(l23) 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 SLs(l23)and T(Zj)=1for j=1,,6.

    Let A be the invertible 6 × 6 matrix such that the j-row vector of A as [Row(A)]j = Wj for j = 1, …, 6. Then, AS = (S(Z1), · · ·, S(Z6))t for all SLs(l23).

  • (b) expBLs(l22)=extBLs(l23).

Theorem 2.1.([22])

Let T=(a11,a22,a33,2a12,2a13,2a23)tLs(l23). Then,

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

Note that if ||T|| = 1, then |aii| ≤ 1 for i = 1, 2,3 and aij12 for 1 ≤ ij ≤ 3. With the aid of Wolfram Mathematica 8, we present a complete description of the 42 extreme points of the unit ball of Ls(l23).

Theorem 2.2

extBLs(l23)={±(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)textBLs(l23). 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 εij ∈ 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 εij = 0 for i, j = 1, 2, 3. Hence, TextBL(l23).

Claim: T=(12,0,0,-12,12,12)textBLs(l23). 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 εij ∈ 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 εij = 0 for i, j = 1, 2, 3. Hence, TextBL(l23).

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)tLs(l23) with ||T|| = 1. By the comments made right before Theorem 2.1, TextBLs(l23) if and only if T = A−1(T(Z1), · · ·, T(Z6))t for some Z1, …, Z6 ∈ Γ with |T(Zj)| = 1 (j = 1, …, 6). Therefore, we may classify all the extreme points of BLs(l23) 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(l23). 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(l23).

Theorem 2.3

We haveextBLs(l23)extBLs(l2n)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, TextBLs(l2n). Use induction on n.

Case 1: n = 4.

Suppose that T=12(S1+S2) for some S1,S2Ls(l24) 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, aj, bj ∈ ℝ for j = 1, …, 4.

Note that, for i = 1, 2,

Si((x1,x2,x3,0),(y1,y2,y3,0))Ls(l23),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 TextBLs(l23), T((x1, x2, x3), (y1, y2, y3)) = S1((x1, x2, x3, 0), (y1, y2, y3, 0)), which shows that ε1 = ε2 = aj = 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

1max{Si((1,1,1,x4),(1,-1,1,y4)),Si((1,1,1,x4),(1,1,-1,y4)):x41,y41,i=1,2}=max{1+b1(y4+x4)+b2(y4-x4)+b3(y4+x4)+b4x4y4,+1b1(y4+x4)+b2(y4+x4)+b3(y4-x4)+b4x4y4:x41,y41},

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 bj = 0 for j = 1, …, 4. Therefore, TextBLs(l24).

Suppose that for n = k, TextBLs(l2k). We will show that TextBLs(l2k+1). Suppose that T=12(W1+W2) for some W1,W2Ls(l2k+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))+1jkcj(xjyk+1+xk+1yj)+dxk+1yk+1

and

W2((x1,,xk+1),(y1,,yk+1))=T((x1,x2,x3),(y1,y2,y3))-1jkcj(xjyk+1+xk+1yj)-dxk+1yk+1

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

1max{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)):xj1,yj1,j=4,,k+1,i=1,2}=max{1+c1(yk+1+xk+1)+c2(yk+1-xk+1)+c3(yk+1+xk+1)+4jkcj(xjyk+1+xk+1yj)+dxk+1yk+1,1+c1(yk+1+xk+1)+c2(yk+1+xk+1)+c3(yk+1-xk+1)+4jkcj(xjyk+1+xk+1yj)+dxk+1yk+1,1+c1(yk+1+xk+1)+c2(yk+1-xk+1)+c3(yk+1-xk+1)+4jkcj(xjyk+1+xk+1yj)+dxk+1yk+1:xj1,yj1,j=4,,k+1},

which shows that, for |xj| ≤ 1, |yj| ≤ 1, j = 4, …, k + 1,

0=c1(yk+1+xk+1)+c2(yk+1-xk+1)+c3(yk+1+xk+1)+4jkcj(xjyk+1+xk+1yj)+dxk+1yk+1,0=c1(yk+1+xk+1)+c2(yk+1+xk+1)+c3(yk+1-xk+1)+4jkcj(xjyk+1+xk+1yj)+dxk+1yk+1,0=c1(yk+1+xk+1)+c2(yk+1-xk+1)+c3(yk+1-xk+1)+4jkcj(xjyk+1+xk+1yj)+dxk+1yk+1.

If xj = yj = 0 for j = 4, …, k, then, for |xk+1| ≤ 1, |yk+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 cj = 0 = d for j = 1, 2, 3. Hence,

(*)0=4jkcj(xjyk+1+xk+1yj)forxj1,yj1,j=4,,k+1.

We claim that cj = 0 for all j = 4, …, k. Let 4 ≤ j0k be fixed. Let xj = yj = 1 for 4 ≤ jj0k and yk+1 = −yj0 = 1 = −xk+1 = xj0. By (*), we have cj0 = 0. Hence, W1 = T = W2, so TextBLs(l2k+1). Therefore, we have shown that TextBLs(l2n) 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 SextBLs(l2n). Use induction on n.

Case 1: n = 4

Suppose that S=12(T1+T2) for some T1,T2Ls(l24) 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 ai, bi, cj ∈ 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(l23),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 SextBLs(l23), S((x1, x2, x3), (y1, y2, y3)) = T1((x1, x2, x3, 0), (y1, y2, y3, 0)), which shows that ai = bi = 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

1max{Ti((1,1,1,x4),(1,-1,1,y4)),Ti((1,1,1,x4),(1,1,-1,y4)):x41,y41,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:x41,y41},

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 cj = 0 for j = 1, …, 4. Therefore, SextBLs(l24).

Suppose that for n = k, SextBLs(l2k). We will show that SextBLs(l2k+1). Suppose that S=12(W1+W2) for some W1,W2Ls(l2k+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))+1jkcj(xjyk+1+xk+1yj)+dxk+1yk+1

and

W2((x1,,xk+1),(y1,,yk+1))=S((x1,x2,x3),(y1,y2,y3))-1jkcj(xjyk+1+xk+1yj)-dxk+1yk+1

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

1max{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)):xj1,yj1,j=4,,k+1,i=1,2}=max{1+c1(yk+1+xk+1)+c2(yk+1-xk+1)+c3(yk+1+xk+1)+4jkcj(xjyk+1+xk+1yj)+dxk+1yk+1,1+c1(yk+1+xk+1)+c2(yk+1+xk+1)+c3(yk+1-xk+1)+4jkcj(xjyk+1+xk+1yj)+dxk+1yk+1,1+c1(yk+1+xk+1)+c2(yk+1-xk+1)+c3(yk+1-xk+1)+4jkcj(xjyk+1+xk+1yj)+dxk+1yk+1:xj1,yj1,j=4,,k+1},

which shows that, for |xj| ≤ 1, |yj| ≤ 1, j = 4, …, k + 1,

0=c1(yk+1+xk+1)+c2(yk+1-xk+1)+c3(yk+1+xk+1)+4jkcj(xjyk+1+xk+1yj)+dxk+1yk+1,0=c1(yk+1+xk+1)+c2(yk+1+xk+1)+c3(yk+1-xk+1)+4jkcj(xjyk+1+xk+1yj)+dxk+1yk+1,0=c1(yk+1+xk+1)+c2(yk+1-xk+1)+c3(yk+1-xk+1)+4jkcj(xjyk+1+xk+1yj)+dxk+1yk+1.

By Claim 1, we conclude that cj = 0 = d for j = 1, …, k. Hence, W1 = S = W2, so SextBLs(l2k+1). Therefore, we have shown that SextBLs(l2n) for n ≥ 4.

Let

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

Claim 3: for n ≥ 4, SextBLs(l2n).

Suppose that S=12(T1+T2) for some T1,T2Ls(l2n) with ||T1|| = 1 = ||T2||.

Write

T1((x1,,xn),(y1,,yn))=(1+a1)x1y1+2jnajxjyj+1i<jnbij(xiyj+xjyi)

and

T2((x1,,xn),(y1,,yn))=(1-a1)x1y1-2jnajxjyj-1i<jnbij(xiyj+xjyi)

for some ai, bij ∈ ℝ for i, j = 1, …, n. It follows that

1max{Tl(e1,e1):l=1,2}=1+a1,

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

1max{Tl(e1+tej,e1+tej):l=1,2,t1}=1+ajt2+2bijt

and

0=ajt2+2b1jt

for every |t| ≤ 1. Hence, 0 = aj = b1j for every 2 ≤ jn. Let 2 ≤ i < jn be fixed. Since

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

bij = 0. Hence, S = T1, so SextBLs(l2n) 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(l2n). We complete the proof.

Theorem 2.2 leads to a complete formula of ||f|| for every fLs(l23)*.

Theorem 3.1

LetfLs(l23)*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,12α11-α22+12β12,12α22-α33+12β23,12α11-α33+12β13,14(2αjj+β12+β13-β23),14(2αjj-β12+β13+β23)}.
Proof

By the Krein-Milman Theorem, f=supTextBLs(l23)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,12α11-α22+12β12,12α22-α33+12β23,12α11-α33+12β13,14(2αjj+β12+β13-β23),14(2αjj-β12+β13+β23)}.

Note that if ||f|| = 1, then |αjj| ≤ 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 extBEis finite. Suppose that xextBEsatisfies that there exists an fE*with f(x) = 1 = ||f|| and |f(y)| < 1 for every yextBE\{±x}. Then, xexpBE.

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

Theorem 3.3.([22]) expBLs(l23)=extBLs(l23)

Proof

It suffices to show that if TextBLs(l23), 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(l23)*. Then, by Theorems 3.1 and 3.2, ||f|| = 1 = f(T) and |f(S)| < 1 for SextBLs(l23)\{±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(l23)*. Then, by Theorems 3.1 and 3.2, ||f|| = 1 = f(T) and |f(S)| < 1 for SextBLs(l23)\{±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(l23)* . Then, by Theorems 3.1 and 3.2, ||f|| = 1 = f(T) and |f(S)| < 1 for SextBLs(l23)\{±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(l23)* . Then, by Theorems 3.1 and 3.2, ||f|| = 1 = f(T) and |f(S)| < 1 for SextBLs(l23)\{±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(l23)* . Then, by Theorems 3.1 and 3.2, ||f|| = 1 = f(T) and |f(S)| < 1 for SextBLs(l23)\{±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(l23)*. Then, by Theorems 3.1 and 3.2, ||f|| = 1 = f(T) and |f(S)| < 1 for SextBLs(l23)\{±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(l23)* . Then, by Theorems 3.1 and 3.2, ||f|| = 1 = f(T) and |f(S)| < 1 for SextBLs(l23)\{±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(l23)*. Then, by Theorems 3.1 and 3.2, ||f|| = 1 = f(T) and |f(S)| < 1 for SextBLs(l23)\{±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(l23)*. Then, by Theorems 3.1 and 3.2, ||f|| = 1 = f(T) and |f(S)| < 1 for SextBLs(l23)\{±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(l23)*. Then, by Theorems 3.1 and 3.2, ||f|| = 1 = f(T) and |f(S)| < 1 for SextBLs(l23)\{±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(l23)*. Then, by Theorems 3.1 and Theorem 3.2, ||f|| = 1 = f(T) and |f(S)| < 1 for SextBLs(l23)\{±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.

In this section we will characterize all the smooth points of the unit ball of Ls(l23).

Theorem 4.1

LetT=(a11,a22,a33,2a12,2a13,2a23)Ls(l23). Then, TsmBLs(l23)if and only if

(T((1,1,1),(1,1,1))=1,0<a11+a22+a33+2a23<1,T(Y)<1forallYΓ\{[(1,1,1),(1,1,1)]})or(T((-1,1,1),(-1,1,1))=1,0<a11+a22+a33+2a23<1,T(Y)<1forallYΓ\{[(-1,1,1),(-1,1,1)]})or(T((1,-1,1),(1,-1,1))=1,0<a11+a22+a33+2a13<1,T(Y)<1forallYΓ\{(1,-1,1),(1,-1,1)})or(T((1,1,-1),(1,1,-1))=1,0<a11+a22+a33+2a12<1,T(Y)<1forallYΓ\{[(1,1,-1),(1,1,-1)]})or(T((1,1,1),(1,-1,1))=1,0<a11-a22+a33<1,T(Y)<1forallYΓ\{[(1,1,1),(1,-1,1)]})or(T((1,1,1),(1,1,-1))=1,0<a11+a22-a33<1,T(Y)<1forallYΓ\{[(1,1,1),(1,1,-1)]})or(T((1,1,1),(-1,1,1))=1,0<-a11+a22+a33<1,T(Y)<1forallYΓ\{[(1,1,1),(-1,1,1)]})or(T((1,-1,1),(1,1,-1))=1,0<-a11+a22+a33<1,T(Y)<1forallYΓ\{[(1,-1,1),(1,1,-1)]})or(T((1,-1,1),(-1,1,1))=1,0<a11+a22-a33<1,T(Y)<1forallYΓ\{[(1,-1,1),(-1,1,1)]})or(T((1,1,-1),(-1,1,1))=1,0<a11-a22+a33<1,T(Y)<1forallYΓ\{[(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, |ajj| < 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)>0and a11+a22+a33+2a23>0.

Let fLs(l23)* 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 αjj = 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<1for all YΓ\{[(1,1,1),(1,1,1)]}.

By Theorem 2.1, for n > n1,

1=(a11±1n,2a12,2a13,a221n,2a23,a33)t,1=(a11±1n,2a12,2a13,a22,2a23,a331n)t,1=(a11,2a12±1n,2a131n,a22,2a23,a33)t,1=(a11±1n,2a12,2a13,a22,2a231n,a33)t.

It follows that for n > n1,

  • (1) 1f((a11±1n,2a12,2a13,a221n,2a23,a33)t)=1±1n(α11-α22),

  • (2) 1f((a11±1n,2a12,2a13,a22,2a23,a331n)t)=1±1n(α11-α33),

  • (3) 1f((a11,2a12±1n,2a131n,a22,2a23,a33)t)=1±12n(β12-β13),

  • (4) 1f((a11±1n,2a12,2a13,a22,2a231n,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,2a1212n,2a1312n,a22±12n,2a231n,a33±12n)t.

Since

1f(a11±1n,2a1212n,2a1312n,a22±12n,2a231n,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, αjj = 1 for j = 1, 2,3 and β12 = β13 = β23 = 2. Hence, T would be smooth in Ls(l23).

Case 2: |T((1, 1, 1), (1,−1, 1))| = 1, 0 < |a11a22+a33| < 1, |T(Y )| < 1 for all Y ∈ Γ\{[(1, 1, 1), (1,−1, 1)>

By Theorem 2.1, ||T|| = 1, |ajj| < 1 for j = 1, 2, 3, a12<12,a23<12. Let l := T((1,−1, 1), (1, 1, 1)) = a11a22 + 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 fLs(l23)* 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 αjj = 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<1for all YΓ\{[(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,a331n)t,1=(a11,2a12±1n,2a13,a22,2a231n,a33)t,1=(a11,2a12±1n,2a13,a22,2a23±1n,a33)t,1=(a11±1n,2a12,2a131n,a22,2a23,a33)t

It follows that for n > n2,

  • (1′) 1f((a11±1n,2a12,2a13,a22±1n,2a23,a33)t)=1±1n(α11+α22),

  • (2′) 1f((a11±1n,2a12,2a13,a22,2a23,a331n)t)=1±1n(α11-α33),

  • (3′) 1f((a11,2a12±1n,2a13,a22,2a231n,a33)t)=1±12n(β12-β23),

  • (4′) 1f((a11,2a12±1n,2a13,a22,2a23±1n,a33)t)=1±12n(β12+β23),

  • (5′) 1f((a11±1n,2a12,2a131n,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 αjj = 1 = −α22 for j = 1, 3. Hence, T would be smooth in Ls(l23). 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):={XΓ:S(X)=S}

for SLs(l23).

Notice that

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

We have ten subcases as follows:

(T((1,1,1),(1,1,1))=1,a11+a22+a33+2a23=0or ±1,T(Y)<1for all YΓ\{[(1,1,1),(1,1,1)]})or(T((-1,1,1),(-1,1,1))=1,a11+a22+a33+2a23=0,or ±1,T(Y)<1for all YΓ\{[(-1,1,1),(-1,1,1)]})or(T((1,-1,1),(1,-1,1))=1,a11+a22+a33+2a13=0,or ±1,T(Y)<1for all YΓ\{(1,-1,1),(1,-1,1)})or(T((1,1,-1),(1,1,-1))=1,a11+a22+a33+2a12=0,or ±1,T(Y)<1for all YΓ\{[(1,1,-1),(1,1,-1)]})or(T((1,1,1),(1,-1,1))=1,a11-a22+a33=0or ±1,T(Y)<1for all YΓ\{[(1,1,1),(1,-1,1)]})or(T((1,1,1),(1,1,-1))=1,a11+a22-a33=0or ±1,T(Y)<1for all YΓ\{[(1,1,1),(1,1,-1)]})or(T((1,1,1),(-1,1,1))=1,-a11+a22+a33=0or ±1,T(Y)<1for all YΓ\{[(1,1,1),(-1,1,1)]})or(T((1,-1,1),(1,1,-1))=1,-a11+a22+a33=0or ±1,T(Y)<1for all YΓ\{[(1,-1,1),(1,1,-1)]})or(T((1,-1,1),(-1,1,1))=1,a11+a22-a33=0or ±1,T(Y)<1for all YΓ\{[(1,-1,1),(-1,1,1)]})or(T((1,1,-1),(-1,1,1))=1,a11-a22+a33=0or ±1,T(Y)<1for all YΓ\{[(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(l23)*.

By Theorem 3.1, ||fk|| = 1 = fk(T) for k = 1, 2. Hence, TsmBLs(l23). This is a contradiction.

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

ft(1,1,1,t,t,0)Ls(l23)*.

By Theorem 3.1, ||ft|| = 1 = ft(T) for |t| ≤ 2. Hence, TsmBLs(l23). This is a contradiction.

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

ft(-1,-1,-1,t,t,0)Ls(l23)*.

By Theorem 3.1, ||ft|| = 1 = ft(T) for |t| ≤ 2. Hence, TsmBLs(l23). This is a contradiction.

Subcase 2: |T((1, 1,−1), (−1, 1, 1))| = 1, a11a22 + a33 = 0 or ± 1, |T(Y )| < 1 for all Y ∈ Γ\{[(1, 1,−1), (−1, 1, 1)>.

Suppose that a11a22 + 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(l23)*.

By Theorem 3.1, ||fk|| = 1 = fk(T) for k = 1, 2. Hence, TsmBLs(l23).

Suppose that a11a22 + 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(l23)*.

By Theorem 3.1, ||fk|| = 1 = fk(T) for k = 1, 2. Hence, TsmBLs(l23).

Suppose that a11a22 + 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(l23)*.

By Theorem 3.1, ||fk|| = 1 = fk(T) for k = 1, 2. Hence, TsmBLs(l23).

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

Case 2: |Norm(T)| ≥ 2.

There exist X1X2 ∈ Γ such that |T(Xk)| = 1 for k = 1, 2. Note that sign(T(Xk))δXkLs(l23)* 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.

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

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