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Kyungpook Mathematical Journal -0001; 56(3): 781-791

Published online November 30, -0001

Copyright © Kyungpook Mathematical Journal.

The Geometry of the Space of Symmetric Bilinear Forms on ℝ2 with Octagonal Norm

Sung Guen Kim

Department of Mathematics, Kyungpook National University, Daegu 702-701, Korea

Received: July 11, 2014; Accepted: December 1, 2014

Let d*(1, w)2 = ℝ2 with the octagonal norm of weight w. It is the two dimensional real predual of Lorentz sequence space. In this paper we classify the smooth points of the unit ball of the space of symmetric bilinear forms on d*(1, w)2. We also show that the unit sphere of the space of symmetric bilinear forms on d*(1, w)2 is the disjoint union of the sets of smooth points, extreme points and the set A as follows: SLs(d2*(1,w)2)=smBLs(d2*(1,w)2)extBLs(d2*(1,w)2)A,

where the set A consists of ax1x2 + by1y2 + c(x1y2 + x2y1) with (a = b = 0, c=±11+w2), (ab, ab ≥ 0, c = 0), (a = b, 0 < ac, 0 < |c| < |a|), (a ≠ |c|, a = −b, 0 < ac, 0 < |c|), (a=1-w1+w,b=0,c=11+w),(a=1+w+w(w2-3)c1+w2,b=w-1+(1-3w2)cw(1+w2),12+2w<c<1(1+w)2(1-w),c11+2w-w2),(a=1+w(1+w)c1+w,b=-1+(1+w)cw(1+w),0<c<12+2w) or ( a=1-w(1+w)c1+w,b=1-(1+w)c1+w,11+w<c<1(1+ω)2(1-ω)).

Keywords:

Symmetric bilinear forms, extreme points, ,smooth points, the octagonal norm on $mathbb{R}^2$ with weight ,$w$, the two dimensional ,real predual of the Lorentz sequence space.

We write BE for the closed unit ball of a real Banach space E and the dual space of E is denoted by E*. xBE is called an extreme point of BE if y, zBE with x=12(y+z) implies x = y = z. xBE is called a smooth point of BE if there is a unique fE* so that f(x) = 1 = ||f||. We denote by extBE and smBE the sets of extreme and smooth points of BE, respectively. A mapping P : E → ℝ is a continuous 2-homogeneous polynomial if there exists a continuous symmetric bilinear form L on the product E×E such that P(x) = L(x, x) for every xE. We denote by ℒs(2E) the Banach space of all continuous symmetric bilinear forms on E endowed with the norm ||L|| = sup||x||=||y||=1 |L(x, y)|. ℘(2E) denotes 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 polynomials on a Banach space, we refer to [7].

In 1998, Choi and the author [3] characterized the smooth points of the unit ball of P(l222) and in 1999, Choi and the author [5] characterized the smooth points of the unit ball of P(l212) and studied smooth polynomials of ℘(2l1). In 2009, the author [10] classified the smooth symmetric bilinear forms of Ls(l22). We refer to ([1], [36], [819] and references therein) for some recent work about extremal properties of multilinear mappings and homogeneous polynomials on some classical Banach spaces. Let 0 < w < 1 be fixed. We denote the two dimensional real predual of Lorentz sequence space by

d*(1,w)2:={(x,y)2:(x,y)d*:=max{x,y,x+y1+w}}.

In fact, d*(1, w)2 = ℝ2 with the octagonal norm of weight w. We will denote by T((x1, x2), (y1, y2)) = ax1x2 +by1y2 +c(x1y2 +x2y1) a symmetric bilinear form on d*(1, w)2. Recently, the author [12] computed the norm of T ∈ ℒs(2d*(1, w)2) in terms of their real coefficients and determined all the extreme symmetric bilinear forms of the unit ball of ℒs(2d*(1, w)2). In this paper, using results of the previous work [12], we classify the smooth symmetric bilinear forms of the unit ball of the space ℒs(2d*(1, w)2). We also show that the unit sphere Ss(2d*(1,w)2) is the disjoint union of the sets of smooth points, extreme points and the set A as follows:

SLs(d2*(1,w)2)=smBLs(d2*(1,w)2)extBLs(d2*(1,w)2)A,

where A consists of ax1x2 + by1y2 + c(x1y2 + x2y1) with (a = b = 0, c=±11+w2), (ab, ab ≥ 0, c = 0), (a = b, 0 < ac, 0 < |c| < |a|), (a ≠ |c|, a = −b, 0 < ac, 0 < |c|), (a=1-w1+w,b=0,c=11+w),(a=1+w+w(w2-3)c1+w2,b=w-1+(1-3w2)cw(1+w2),12+2w<c<1(1+w)2(1-w),c11+2w-w2),(a=1+w(1+w)c1+w,b=-1+(1+w)cw(1+w),0<c<12+2w) or ( a=1-w(1+w)c1+w,b=1-(1+w)c1+w,11+w<c<1(1+ω)2(1-ω)).

Let T((x1, y1), (x2, y2)) = ax1x2 + by1y2 + c(x1y2 + x2y1) ∈ ℒs(2d*(1, w)2) for some reals a, b, c. By substituting ((x1, y1), (x2, y2)) in T for ((x1, y1), (−x2, −y2)) or ((x1, −y1), (x2, −y2)) or ((y1, x1), (y2, x2)), we may assume that |b| ≤ a, c ≥ 0.

Theorem 2.1.([12, Theorem 2.1])

Let T((x1, y1), (x2, y2)) = ax1x2 + by1y2 + c(x1y2 + x2y1) ∈ ℒs(2d*(1, w)2) with |b| ≤ a, c ≥ 0. Then

T=max{bw2+2cw+a,a-bw2,(a+b)w+c(1+w2),(a-b)w+c(1-w2)}.

In fact, we have the following:

Case 1: b ≥ 0

Subcase 1: c > a

  • If wc-ac-b, then ||T|| = (a + b)w + c(1 + w2).

  • If w>c-ac-b, then ||T|| = bw2 + 2cw + a.

Subcase 2: If ca, ||T|| = bw2 + 2cw + a.

Case 2: b < 0

Subcase 1: c < |b|

  • If wcb, then ||T|| = max{bw2 + 2cw + a, (ab)w + c(1 − w2)}.

  • If w>cb, then ||T|| = max{abw2, (ab)w + c(1 − w2)}.

Subcase 2: c ≥ |b|

  • If wbc, then ||T|| = max{bw2 + 2cw + a, (ab)w + c(1 − w2)}.

  • If w>bc, then ||T|| = max{bw2 + 2cw + a, (a + b)w + c(1 + w2)}.

By Theorem 2.1, if ||T|| = 1, then |a| ≤ 1, |b| ≤ 1, c11+w2.

Theorem 2.2

Let T((x1, y1), (x2, y2)) = ax1x2 + by1y2 + c(x1y2 + x2y1) ∈ ℒs(2d*(1, w)2). Then the following are equivalent:

  • ax1x2 + by1y2 + c(x1y2 + x2y1) is a smooth point ofs(2d*(1, w)2);

  • − (ax1x2 + by1y2 + c(x1y2 + x2y1)) is a smooth point ofs(2d*(1, w)2);

  • ax1x2 + by1y2c(x1y2 + x2y1) is a smooth point ofs(2d*(1, w)2);

  • bx1x2 + ay1y2 + c(x1y2 + x2y1) is a smooth point ofs(2d*(1, w)2).

Proof

Let S((x1, y1), (x2, y2)) := T((u1, v1), (u2, v2)) for some ((u1, v1), (u2, v2)) = ((x1, y1), (−x2,−y2)) or ((x1,−y1), (x2,−y2)) or ((y1, x1), (y2, x2)). Then S ∈ ℒs(2d*(1, w)2), and T is smooth if and only if S is smooth.

Theorem 2.3.([12, Theorem 2.3])

Let T((x1, y1), (x2, y2)) = ax1x2 + by1y2 + c(x1y2 + x2y1) ∈ ℒs(2d*(1, w)2). Then

  • Let w<2-1. Then T is an extreme point ofs(2d*(1, w)2) if and only ifT{±x1x2,±y1y2,±11+w2(x1x2+y1y2),±1(1+w)2[x1x2+y1y2±(x1y2+x2y1)],±11+2w-w2[x1x2-y1y2±(x1y2+x2y1)],±11+w2[x1x2-y1y2±w(x1y2+x2y1),±11+w2[wx1x2-wy1y2±(x1y2+x2y1)],±1(1+w)2(1-w)[(1-w-w2)x1x2-wy1y2±(x1y2+x2y1)],±1(1+w)2(1-w)[wx1x2-(1-w-w2)y1y2±(x1y2+x2y1)]}.

  • Let w<2-1. Then T is an extreme point ofs(2d*(1, w)2) if and only ifT{±x1x2,±y1y2,±2+24(x1x2+y1y2),±12[x1x2+y1y2±(x1y2+x2y1)],±24[x1x2+y1y2±(2+1)(x1y2+x2y1)],±24[(2+1)(x1y2-x2y1)±(x1y2+x2y1)]}.

  • Let w<2-1. Then T is an extreme point ofs(2d*(1, w)2) if and only ifT{±x1x2,±y1y2,±11+w2(x1x2+y1y2),±1(1+w)2[x1x2+y1y2±(x1y2+x2y1)],±11+2w-w2[x1x2-y1y2±(x1y2+x2y1)],±11+w2[x1x2-y1y2±1-w1+w(x1y2+x2y1)],±11+w2[1-w1+w(x1x2-y1y2)±(x1y2+x2y1)],±12+2w[(2+w)x1x2-1wy1y2±(x1y2+x2y1)],±12+2w[1wx1x2-(2+w)y1y2±(x1y2+x2y1)]}.

Theorem 2.4

Let f ∈ ℒs(2d*(1, w)2)*and α = f(x1x2), β = f(y1y2), γ = f(x1y2 + x2y1).

  • Let w<2-1. Thenf=max{α,β,11+w2α+β,1(1+w)2(α+β+γ),11+2w-w2(α-β+γ),11+w2(α-β+wγ),11+w2(wα-β+γ),1(1+w)2(1-w)((1-w-w2)α-wβ+γ),1(1+w)2(1-w)(wα-(1-w-w2)β+γ)}.

  • Let w<2-1. Thenf=max{α,β,2+24α+β,12(α+β+γ),24(α-β+(2+1)γ),24((2+1)α-β+γ)}.

  • Let 2-1<w. Thenf=max{α,β,11+w2α+β,1(1+w)2(α+β+γ),11+2w-w2(α-β+γ,11+w2(α-β+1-w1+wγ),11+w2(1-w1+wα-β+γ),12+2w((2+w)α-1wβ+γ),12+2w(1wα-(2+w)β+γ)}.

Proof

It follows from Theorem 2.3 since

f=sup{f(T):TextBLs(d2*(1,w)2)}.

Theorem 2.5

Let T((x1, y1), (x2, y2)) = ax1x2 + by1y2 + c(x1y2 + x2y1) ∈ ℒs(2d*(1, w)2) with |b| < a, c > 0. Let S = {bw2 + 2cw + a, abw2, (a + b)w + c(1 + w2), (ab)w + c(1 − w2)}. Then TsmBs(2d*(1,w)2)if and only if there exists a unique lS such that l = 1.

Proof
(⇒):

For (u1, v1), (u2, v2) ∈ Sd*(1,w)2, let δ(u1,v1),(u2,v2) ∈ ℒs(2d*(1, w)2)* such that δ(u1,v1),(u2,v2)(L) = L((u1, v1), (u2, v2)) for L ∈ ℒs(2d*(1, w)2). Then ||δ(u1,v1),(u2,v2) || ≤ 1. Note that, by Theorem 2.4, 1 = ||δ(1,w),(1,w)|| = ||δ(1,−w),(1,w) || = ||δ(1,w),(w,1) || = || δ(1,−w),(w,1) ||. Obviously,

δ(1,w),(1,w)(T)=bw2+2cw+a,δ(1,-w),(1,w)(T)=a-bw2,δ(1,w),(w,1)(T)=(a+b)w+c(1+w2),δ(1,-w),(w,1)(T)=(a-b)w+c(1-w2).

Hence, if TsmBs(2d*(1,w)2), then, by Theorem 2.1, there exists a unique lS such that l = 1.

(⇐):

Let f ∈ ℒs(2d*(1, w)2)* such that 1 = ||f|| = f(T) with α = f(x1x2), β = f(y1y2), γ = f(x1y2 + x2y1).

Case 1: l = bw2 + 2cw + a = 1

Then

bw2+2cw+a=1=aα+bβ+cγ.

By Theorem 2.1, it follows that, for a sufficiently large n ∈ ℕ,

1=(a±1n)x1x2+by1y2+(c12wn)(x1y2+x2y1)=ax1x2(b±1n)y1y2+(c12n)(x1y2+x2y1).

From (**), 1f((a±1n)x1x2+by1y2+(c12wn)(x1y2+x2y1))=1+1nα-12wnγ, hence α=12wγ and 1f(ax1x2+(b±1n)y1y2+(cw2n)(x1y2+x2y1))=1+1nβ-w2γ, hence β=w2γ. It follows that, by (*),

1=aα+bβ+cγ=γ2w(bw2+2cw+a)=γ2w.

Therefore, α = 1, β = w2, γ = 2w, hence f = δ(1,w),(1,w) is uniquely determined.

Case 2: l = abw2 = 1

Then

a-bw2=1=aα+bβ+cγ.

By Theorem 2.1, it follows that, for a sufficiently large n ∈ ℕ,

1=(a±1n)x1x2+(b±1nw2)y1y2+c(x1y2+x2y1)=ax1x2+by1y2+(c±1n)(x1y2+x2y1).

From (**), 1f((a±1n)x1x2+(b±1nw2)y1y2+c(x1y2+x2y1))=1+1nα+1w2β, hence α=-1w2β and 1f(ax1x2+by1y2+(c±1n)(x1y2+x2y1))=1+1nγ, hence γ = 0. It follows that, by (*),

w2=w2(aα+bβ+cγ)=β(bw2-a)=-β.

Therefore, α = 1, β = −w2, γ = 0, hence f = δ(1,−w),(1,w) is uniquely determined.

Case 3: l = (a + b)w + c(1 + w2) = 1

Then

(a+b)w+c(1+w2)=1=aα+bβ+cγ.

By Theorem 2.1, it follows that, for a sufficiently large n ∈ ℕ,

1=(a±1n)x1x2+by1y2+(cwn(1+w2))(x1y2+x2y1)=ax1x2+(b±1n)y1y2+(cwn(1+w2))(x1y2+x2y1).

From (**), 1f((a±1n)x1x2+by1y2+(cwn(1+w2))(x1y2+x2y1))=1+1nα-w1+w2γ, hence α=w1+w2γ and 1f(ax1x2+(b±1n)y1y2+(cwn(1+w2))(x1y2+x2y1))=1+1nβ-w1+w2γ, hence β=w1+w2γ. It follows that, by (*),

1=aα+bβ+cγ=γ1+w2((a+b)w+c(1+w2))=γ1+w2.

Therefore, α = β = w, γ = 1+w2, hence f = δ(1,w),(w,1) is uniquely determined.

Case 4: l = (ab)w + c(1 − w2) = 1

Then

(a-b)w+c(1-w2)=1=aα+bβ+cγ.

By Theorem 2.1, it follows that, for a sufficiently large n ∈ ℕ,

1=(a±1n)x1x2+by1y2+(cwn(1-w2))(x1y2+x2y1)=ax1x2+(b±1n)y1y2+(c±wn(1-w2))(x1y2+x2y1).

From (**), 1f((a±1n)x1x2+by1y2+(cwn(1+w2))(x1y2+x2y1))=1+1nα-w1-w2γ.

Hence α=w1+w2γ and 1f(ax1x2+(b±1n)y1y2+(c±wn(1-w2))(x1y2+x2y1))=1+1nβ+w1-w2γ, hence β=-w1-w2γ. It follows that, by (*),

1=aα+bβ+cγ=γ1-w2((a-b)w+c(1-w2))=γ1-w2.

Therefore, α = w, β = −w, γ = 1 − w2, hence f = δ(1,−w),(w,1) is uniquely determined.

We are in position to classify the smooth symmetric bilinear forms of the unit ball of ℒs(2d*(1, w)2).

Let T = ax1x2 +by1y2 +c(x1y2 +x2y1) ∈ Ss(2d*(1,w)2). Then TsmBs(2d*(1,w)2)if and only if (|a| = 1, b = 0 = c), (a = b = 0, c=±11+w2), (ab ≥ 0, c = 0), (a = b, 0 < ac, 0 < |c| ≤ |a|), (a = −b, 0 < ac, 0 < |c|), (a=1-w1+w,b=0,c=11+w),(a=1+w+w(w2-3)c1+w2,b=w-1+(1-3w2)cw(1+w2),12+2wc1(1+w)2(1-w)),(a=1+w(1+w)c1+w,b=-1+(1+w)cw(1+w),0<c<12+2w)or ( a=1-w(1+w)c1+w,b=1-(1+w)c1+w,11+w<c<1(1+ω)2(1-ω)).

Proof

Without loss of generality, we may assume that |b| ≤ a, c ≥ 0. Let T = ax1x2 + by1y2 + c(x1y2 + x2y1) ∈ Ss(2d*(1,w)2) and let f ∈ ℒs(2d*(1, w)2)* such that 1 = ||f|| = f(T) with α = f(x1x2), β = f(y1y2), γ = f(x1y2 + x2y1). If a = 1, then T = x1x2. We claim that T is not smooth. Indeed, let g, h ∈ ℒs(2d*(1, w)2)* such that g(x1x2) = 1, g(y1y2) = 0, g(x1y2 + x2y1) = 0, h(x1x2) = 1, h(y1y2) = 0, h(x1y2 + x2y1) = w2. Theorem 2.4 shows that ||g|| = 1 = ||h|| = g(T) = h(T), which implies that T is not smooth. If a = 0, then T=11+w2(x1y2+x2y1)

We claim that T is not smooth. Indeed, let g, h ∈ ℒs(2d*(1, w)2)* such that g(x1x2) = 0 = g(y1y2), g(x1y2 + x2y1) = 1+w2, h(x1x2) = w = h(y1y2), h(x1y2 + x2y1) = 1+w2. Theorem 2.4 shows that ||g|| = 1 = ||h|| = g(T) = h(T), which implies that T is not smooth. Suppose that 0 < a < 1. We will consider the three cases (c = 0) or (a = |b|, c > 0) or (|b| < a,c > 0).

Case 1: c=0

We claim that b ≠ 0 since if not, then a = 1, which is impossible. If b > 0, let g, h ∈ ℒs(2d*(1, w)2)* such that g(x1x2) = 1, g(y1y2) = w2, g(x1y2 + x2y1) = 2w, h(x1x2) = 1, h(y1y2) = w2, h(x1y2 + x2y1) = 0. Theorem 2.4 shows that ||g|| = 1 = ||h|| = g(T) = h(T), which implies that T is not smooth. In particular, extreme ±11+w2(x1x2+y1y2) is not smooth. If b < 0, then T = ax1x2 −|b|y1y2 and 1 = ||T|| = a + bw2. We will show that T is smooth. By Theorem 2.1,

a+bw2=1=aα+bβ.

By Theorem 2.1, it follows that, for a sufficiently large n ∈ ℕ,

1=(a±1n)x1x2+(b±1nw2)y1y2=ax1x2+by1y2±1n(x1y2+x2y1).

From (**), 1f((a±1n)x1x2+(b±1nw2)y1y2)=1+1nα+1w2β, hence α=-1w2β and 1f(ax1x2+by1y2±1n(x1y2+x2y1))=1+1nγ , hence γ = 0. It follows that, by (*),

1=aα+bβ=-βw2(a-bw2)=-βw2.

Therefore, α = 1, β = −w2, γ = 0, hence f = δ(1,w),(1,w) is uniquely determined.

Case 2: a = |b|, c > 0

Then (a = b, c > 0) or (a = −b, c > 0). First suppose that a = b, c > 0. If c > a, then we claim that T is smooth. By Theorem 2.1,

2aw+c(1+w2)=1=aα+aβ+cγ.

By Theorem 2.1, it follows that, for a sufficiently large n ∈ ℕ,

1=(a±1n)x1x2+(a1n)y1y2+c(x1y2+x2y1)=ax1x2+(a±1n)y1y2+(cwn(1+w2))(x1y2+x2y1).

From (**), 1f((a±1n)x1x2+(a1n)y1y2+c(x1y2+x2y1))=1+1nα-β, hence α = β and 1f(ax1x2+(a±1n)y1y2+(cwn(1+w2))(x1y2+x2y1))=1+1nβ-w1+w2γ, hence β=w1+w2γ. It follows that, by (*),

1=aα+aβ+cγ=γ1+w2(2aw+c(1-w2))=γ1+w2.

Therefore, α = w = β, γ = 1+w2, hence f = δ(1,w),(1,w) is uniquely determined.

If ca, let g, h ∈ ℒs(2d*(1, w)2)* such that g(x1x2) = 1, g(y1y2) = w2, g(x1y2+ x2y1) = 2w, h(x1x2)=1+w22=h(y1y2), h(x1y2 + x2y1) = 2w. Theorem 2.4 shows that ||g|| = 1 = ||h|| = g(T) = h(T), which implies that T is not smooth. In particular, extreme 1(1+w)2=(x1x2+y1y2+(x1y2+x2y1)) is not smooth. Suppose that a = −b, c > 0. We claim that T is not smooth.

If c < |b|, wcb and w2-1, let g, h ∈ ℒs(2d*(1, w)2)* such that g(x1x2) = 1, g(y1y2) = w2, g(x1y2+x2y1) = 2w, h(x1x2)=1-w22,h(y1y2)=-(1-w2)2, h(x1y2+ x2y1) = 2w. Theorem 2.4 shows that ||g|| = 1 = ||h|| = g(T) = h(T), which implies that T is not smooth.

If c < |b|, wcb and w2-1, let g, h ∈ ℒs(2d*(1, w)2)* such that g(x1x2) = w, g(y1y2) = −w, g(x1y2 + x2y1) = 1 − w2, h(x1x2) = wε, h(y1y2) = −(wε), h(x1y2 + x2y1) = 1 − w2 for a sufficiently small ε > 0. Theorem 2.4 shows that ||g|| = 1 = ||h|| = g(T) = h(T), which implies that T is not smooth. In particular, extreme 1(1+w)2(x1x2-y1y2+w(x1y2+x2y1)) is not smooth.

If c < |b|, w>cb, then w>2-1. Let g, h ∈ ℒs(2d*(1, w)2)* such that g(x1x2) = w, g(y1y2) = −w, g(x1y2 + x2y1) = 1 − w2, h(x1x2) = wε, h(y1y2) = −(wε, ), h(x1y2+x2y1) = 1−w2. Theorem 2.4 shows that ||g|| = 1 = ||h|| = g(T) = h(T), which implies that T is not smooth. In particular, extreme 11+w2(x1x2-y1y2+1-w1+w(x1y2+x2y1)) is not smooth.

If c > |b|, wbc and w<2-1, let g, h ∈ ℒs(2d*(1, w)2)* such that g(x1x2) = w, g(y1y2) = −w, g(x1y2 + x2y1) = 1 − w2, h(x1x2) = wε, h(y1y2) = −(wε, ), h(x1y2 + x2y1) = 1 − w2 for a sufficiently small ε > 0. Theorem 2.4 shows that ||g|| = 1 = ||h|| = g(T) = h(T), which implies that T is not smooth.

If c > |b|, wbc and w<2-1, let g, h ∈ ℒs(2d*(1, w)2)* such that g(x1x2) = 1, g(y1y2) = w2, g(x1y2+x2y1) = 2w, h(x1x2)=1-w22,h(y1y2)=-(1-w2)2, h(x1y2+ x2y1) = 2w. Theorem 2.4 shows that ||g|| = 1 = ||h|| = g(T) = h(T), which implies that T is not smooth.

If c > |b|, w>bc and bc>1-w1+w, let g, h ∈ ℒs(2d*(1, w)2)* such that g(x1x2) = 1, g(y1y2) = w2, g(x1y2+x2y1) = 2w, h(x1x2)=1-w22,h(y1y2)=-(1-w2)2, h(x1y2+ x2y1) = 2w. Theorem 2.4 shows that ||g|| = 1 = ||h|| = g(T) = h(T), which implies that T is not smooth.

If c > |b|, w>bc and bc<1-w1+w, let g, h ∈ ℒs(2d*(1, w)2)* such that g(x1x2) = w, g(y1y2) = w, g(x1y2+x2y1) = 1+w2, h(x1x2) = wε = h(y1y2), h(x1y2+x2y1) = 1 + w2 for a sufficiently small ε > 0. Theorem 2.4 shows that ||g|| = 1 = ||h|| = g(T) = h(T), which implies that T is not smooth.

If c = |b|, then T=11+2w-w2[x1x2-y1y2+(x1y2+x2y1)] is an extreme point, so it is not smooth. Indeed, let 2w < gε (x1y2 + x2y1) < 1 − w2 and gɛ(x1x2)=1+2w-w2-γ2, gε (y1y2) = −α. Theorem 2.4 shows that ||gε|| = 1 = gε(T), which implies that T is not smooth.

Case 3: |b| < a and c > 0

Suppose that T is not smooth. If b ≥ 0, then, by Theorem 2.5, w=c-ac-b and a < c. Then

bw2+2cw+a=1=(a+b)w+c(1+w2).

Simple computation shows that T=(1-w1+w)x1x2+11+w(x1y2+x2y1). Suppose that b < 0. If wcb, then, by Theorem 2.5,

bw2+2cw+a=1=(a-b)w+c(1-w2).

Simple computation shows that T=(1+w+w(w2-3)c1+w2)x1x2+(w-1+(1-3w2)cw(1+w2))y1y2+c(x1y2+x2y1) for 12+2wc<11+2w-w2. In particular, if c=12+2w, then extreme 12+2w[(2+w)x1x2-1wy1y2+(x1y2+x2y1)] is not smooth.

If w>cb, then, by Theorem 2.5,

a-bw2=1=(a-b)w+c(1-w2).

Simple computation shows that T=(1+w(1+w)c1+w)x1x2+(-1+(1+w)cw(1+w))y1y2+c(x1y2+x2y1) for 0<c<12+2w.

If wbc, then, by Theorem 2.5,

bw2+2cw+a=1=(a-b)w+c(1-w2).

Simple computation shows that T=(1+w+w(w2-3)c1+w2)x1x2+(w-1+(1-3w2)cw(1+w2))y1y2+c(x1y2+x2y1) for 12+2w-w2c1(1+w)2(1-w). In particular, if c=1(1+w)2(1-w), then extreme 1(1+w)2(1-w)((1-w-w2)x1x2-wy1y2+(x1y2+x2y1)) is not smooth is not smooth.

If w>bc, then, by Theorem 2.5,

bw2+2cw+a=1=(a+b)w+c(1+w2).

Simple computation shows that T=(1-w(1+w)c1+w)x1x2+(1-(1+w)c1+w)y1y2+c(x1y2+x2y1) for 11+w<c<1(1+w)2(1-w). Therefore, it completes the proof.

We show that the unit sphere Ss(2d*(1,w)2) is the disjoint union of three nonempty subsets as follows:

SLs(d2*(1,w)2)=smBLs(d2*(1,w)2)extBLs(d2*(1,w)2)A,

where A consists of ax1x2 + by1y2 + c(x1y2 + x2y1) with (a = b = 0, c=±11+w2), (ab, ab ≥ 0, c = 0), (a = b, 0 < ac, 0 < |c| < |a|), (a ≠ |c|, a = −b, 0 < ac, 0 < |c|), (a=1-w1+w,b=0,c=11+w),(a=1+w+w(w2-3)c1+w2,b=w-1+(1-3w2)cw(1+w2),12+2w<c<1(1+w)2(1-w),c11+2w-w2,(a=1+w(1+w)c1+w,b=-1+(1+w)cw(1+w),0<c<12+2w)or ( a=1-w(1+w)c1+w,b=1-(1+w)c1+w,11+w<c<1(1+ω)2(1-ω)).

Proof

Note that in the proof of Theorem 2.6 it has shown that every extreme symmetric bilinear form is not smooth. It follows from Theorems 2.3 and 2.6.

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