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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

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.

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 $P(l2p2) (1≤p≤∞)$ [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 $Ls(l2∞3)$

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 = Wj 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 |aii| ≤ 1 for i = 1, 2,3 and $∣aij∣≤12$ 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 $Ls(l2∞3)$.

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 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, $T∈extBL(l2∞3)$.

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 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, $T∈extBL(l2∞3)$.

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(Zj)| = 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 have$extBLs(l2∞3)⊂extBLs(l2∞n)$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, aj, bj ∈ 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 = 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

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

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 bj = 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 cj, 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 |xj| ≤ 1, |yj| ≤ 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 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,

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

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 $T∈extBLs(l2∞k+1)$. 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 ai, bi, cj ∈ ℝ for i = 1, 2, 3, j = 1, …, 4. Note that, for i = 1, 2,

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

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 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

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

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 cj = 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 cj, 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 |xj| ≤ 1, |yj| ≤ 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 cj = 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

and

$T2((x1,…,xn),(y1,…,yn))=(1-a1)x1y1-∑2≤j≤najxjyj-∑1≤i

for some ai, bij ∈ 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 ≤ jn 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 = aj = b1j for every 2 ≤ jn. Let 2 ≤ i < jn be fixed. Since

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

bij = 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 $Ls(l2∞3)$

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

Theorem 3.1

Let$f∈Ls(l2∞3)*$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 |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]) $expBLs(l2∞3)=extBLs(l2∞3)$

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, $-(-12,0,0,12,12,12)t,±(0,-12,0,12,12,12)t,±(0,0,-12,12,12,12)t$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 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 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 $Ls(l2∞3)$

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

Theorem 4.1

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

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

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

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

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

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, αjj = 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 < |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 $f∈Ls(l2∞3)*$ 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 αjj = 1 = −α22 for j = 1,3 and β12 = β23 = 0, β13 = 2. Let n1 ∈ ℕ be such that, for j = 1, 2, 3,

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

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

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

and

$0

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 αjj = 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, ||fk|| = 1 = fk(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, ||ft|| = 1 = ft(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, ||ft|| = 1 = ft(T) for |t| ≤ 2. Hence, $T∉smBLs(l2∞3)$. 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

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

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

Suppose that a11a22 + a33 = 1. Let

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

and

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

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

Suppose that a11a22 + a33 = −1. Let

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

and

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

By Theorem 3.1, ||fk|| = 1 = fk(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 X1X2 ∈ Γ such that |T(Xk)| = 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.

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

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