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Kyungpook Mathematical Journal 2017; 57(2): 223-232

Published online June 23, 2017

Copyright © Kyungpook Mathematical Journal.

Extremal Problems for

Sung Guen Kim

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

Received: March 11, 2015; Accepted: December 4, 2015

We classify the extreme and exposed symmetric bilinear forms of the unit ball of the space of symmetric bilinear forms on ℝ2 with hexagonal norms. We also show that every extreme symmetric bilinear forms of the unit ball of the space of symmetric bilinear forms on ℝ2 with hexagonal norms is exposed.

Keywords:

symmetric bilinear forms, extreme points, exposed ,points, hexagonal norms on $mathbb{R}^2$

We write BEfor the closed unit ball of a real Banach space E and the dual space of E is denoted by E*. xBEis called an extreme point of BEif y, zBEwith x=12(y+z) implies x = y = z. xBEis called an exposed point of BEif there is a fE* so that f(x) = 1 = ||f|| and f(y) < 1 for every yBE {x}. It is easy to see that every exposed point of BEis an extreme point. We denote by extBEand expBEthe sets of extreme and exposed 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)|.s(2E) denotes the subspace of ℒ(2E) of all continuous symmetric bilinear forms on E. ℘(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 multilinear mappings and polynomials on a Banach space, we refer to [7].

In 1998, Choi et al. ([2], [3]) characterized the extreme points of the unit ball of P(l212) and P(l222). In 2007, Kim [11] classified the exposed 2-homogeneous polynomials on P(l2l2)(1p). Kim ([13], [15], [19]) classified the extreme, exposed, smooth points of the unit ball of ℘(2d* (1, w)2), where d* (1, w)2 = ℝ2 with the octagonal norm of weight w. In 2009, Kim [12] classified the extreme, exposed, smooth points of the unit ball of Ls(l22). Kim ([14], [16], [17], [18]) classified the extreme, exposed, smooth points of the unit balls of ℒs(2d*(1, w)2) and ℒ(2d*(1, w)2).

We refer to ([16], [825] 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 ℝ2 with the hexagonal norm of weight w by

h(w)2:={(x,y)2:(x,y)h(w):=max{y,x+(1-w)y}}.

Recently, Kim [20] characterized the extreme points of the unit ball of L(2h(w)2). In this paper, we classify the extreme and exposed symmetric bilinear forms of the unit ball of Ls(2h(w)2). We also show that every extreme symmetric bilinear form of the unit ball of Ls(2h(w)2) is exposed.

Let 0 < w < 1 and T((x1,y1),(x2,y2))=ax1x2+by1y2+c(x1y2+x2y1)Ls(2h(w)2) for some reals a, b, c. For simplicity we will write T((x1, y1), (x2, y2)) = (a, b, c).

Theorem 2.1

Let 0 < w < 1 and T((x1,y1),(x2,y2)):=(a,b,c)Ls(2h(w)2). Then,

T=max{a,aw+c,aw2-b,aw2+b+2wc}.
Proof

By substituting ((x1, y1), (x2, y2)) in T for ((x1,y1), (x2,y2)), we may assume that c ≥ 0. Since {(±1, 0), (w, ± 1), (−w,±1)} is the set of all extreme points of the unit ball of h(w)2 and T is bilinear,

T=max{T((±1,0),(±1,0)),T((±1,0),(w,±1)),T((w,±1),(w,±1))}.

It follows that, by symmetry of T,

T=max{T((1,0),(1,0)),T((1,0),(w,1)),T((1,0),(w,-1)),T((w,1),(w,1)),T((w,-1),(w,-1)),T((w,1),(w,-1))}=max{a,aw+c,aw2-b,aw2+b+2wc}.

Note that if ||T|| = 1, then |a| ≤ 1, |b| ≤ 1 and |c| ≤ 1. Let

Norm(T)={((x1,y1)(x2,y2)){((1,0),(1,0)),((1,0),(w,1)),((1,0),(w,-1)),((w,1),(w,1)),((w,-1),(w,-1)),((w,1),(w,-1))}:T((x1,y1),(x2,y2))=T}.

We call Norm(T) the norming set of T.

Theorem 2.2

Let 0 < w < 1 and T((x1,y1),(x2,y2))=ax1x2+by1y2+c(x1y2+x2y1)Ls(2h(w)2) with ||T|| = 1. Then, T is extreme if and only if Norm(T) has exactly three elements.

Proof

Without loss of generality we may assume that a, c ≥ 0.

(⇐): We have 20 cases as follows:

  • Case 1: Norm(T) = {((1, 0), (1, 0)), ((1, 0), (w, 1)), ((1, 0), (w,−1))}

  • Case 2: Norm(T) = {((1, 0), (1, 0)), ((1, 0), (w, 1)), ((w, 1), (w, 1))}

  • Case 3: Norm(T) = {((1, 0), (1, 0)), ((1, 0), (w, 1)), ((w,−1), (w,−1))}

  • Case 4: Norm(T) = {((1, 0), (1, 0)), ((1, 0), (w, 1)), ((w, 1), (w,−1))}

  • Case 5: Norm(T) = {((1, 0), (1, 0)), ((1, 0), (w,−1)), ((w, 1), (w, 1))}

  • Case 6: Norm(T) = {((1, 0), (1, 0)), ((1, 0), (w,−1)), ((w,−1), (w,−1))}

  • Case 7: Norm(T) = {((1, 0), (1, 0)), ((1, 0), (w,−1)), ((w, 1), (w,−1))}

  • Case 8: Norm(T) = {((1, 0), (1, 0)), ((w, 1), (w, 1)), ((w,−1), (w,−1))}

  • Case 9: Norm(T) = {((1, 0), (1, 0)), ((w, 1), (w, 1)), ((w, 1), (w,−1))}

  • Case 10: Norm(T) = {((1, 0), (1, 0)), ((w,−1), (w,−1)), ((w, 1), (w,−1))}

  • Case 11: Norm(T) = {((1, 0), (w, 1)), ((1, 0), (w,−1)), ((w, 1), (w, 1))}

  • Case 12: Norm(T) = {((1, 0), (w, 1)), ((1, 0), (w,−1)), ((w,−1), (w,−1))}

  • Case 13: Norm(T) = {((1, 0), (w, 1)), ((1, 0), (w,−1)), ((w, 1), (w,−1))}

  • Case 14: Norm(T) = {((1, 0), (w, 1)), ((w, 1), (w, 1)), ((w,−1), (w,−1))}

  • Case 15: Norm(T) = {((1, 0), (w, 1)), ((w, 1), (w, 1)), ((w, 1), (w,−1))}

  • Case 16: Norm(T) = {((1, 0), (w, 1)), ((w,−1), (w,−1)), ((w, 1), (w,−1))}

  • Case 17: Norm(T) = {((1, 0), (w,−1)), ((w, 1), (w, 1)), ((w,−1), (w,−1))}

  • Case 18: Norm(T) = {((1, 0), (w,−1)), ((w, 1), (w, 1)), ((w, 1), (w,−1))}

  • Case 19: Norm(T) = {((1, 0), (w,−1)), ((w,−1), (w,−1)), ((w, 1), (w,−1))}

  • Case 20: Norm(T) = {((w, 1), (w, 1)), ((w,−1), (w,−1)), ((w, 1), (w,−1))}.

We will consider each case.

Case 1: Norm(T) = {((1, 0), (1, 0)), ((1, 0), (w, 1)), ((1, 0), (w,−1))}

Note that T does not exist in case 1.

Case 2: Norm(T) = {((1, 0), (1, 0)), ((1, 0), (w, 1)), ((w, 1), (w, 1))}

Then T = (1, (1−w)2, 1−w) for all 0 < w < 1. Note that T = (1, (1−w)2, 1−w) is extreme for all 0 < w < 1. Indeed, let T1 = (1+ε, (1 − w)2 + δ, 1 − w + γ), T2 = (1 − ε, (1 − w)2δ, 1 − wγ) be such that ||T1|| = 1 = ||T2|| for some ε,δ, γ ∈ ℝ. Since |Ti((1, 0), (1, 0))| ≤ 1, |Ti((1, 0), (w, 1))| ≤ 1, |Ti((w, 1), (w, 1))| ≤ 1, we have 0 = ε = δ = γ.

Case 3: Norm(T) = {((1, 0), (1, 0)), ((1, 0), (w, 1)), ((w,−1), (w,−1))}

Then T = (1,−3w2 + 2w − 1, 1 − w) for all 0<w12. Note that T = (1,−3w2 + 2w − 1, 1 − w) is extreme for all 0<w12. Indeed, let T1 = (1+ ε,−3w2 + 2w − 1 + δ, 1 − w + γ), T2 = (1 − ε,−3w2 + 2w − 1 − δ, 1 − wγ) be such that ||T1|| = 1 = ||T2|| for some ε,δ, γ ∈ ℝ. Since |Ti((1, 0), (1, 0))| ≤ 1, |Ti((1, 0), (w, 1))| ≤ 1, |Ti((w,−1), (w,−1))| ≤ 1, we have

ɛ=0wɛ+γ=0w2ɛ+δ-2wγ=0,

which show that 0 = ε = δ = γ.

Case 4: Norm(T) = {((1, 0), (1, 0)), ((1, 0), (w, 1)), ((w, 1), (w,−1))}

Then T = (1, w2 − 1, 1 − w) for all w12. Note that T = (1, w2 − 1, 1 − w) is extreme for all w12. Indeed, let T1 = (1+ε,w2 − 1 + δ, 1 − w + γ), T2 = (1 − ε,w2 − 1 − δ, 1 − wγ) be such that ||T1|| = 1 = ||T2|| for some ε,δ, γ ∈ ℝ. Since |Ti((1, 0), (1, 0))| ≤ 1, |Ti((1, 0), (w, 1))| ≤ 1, |Ti((w, 1), (w,−1))| ≤ 1, we have

ɛ=0wɛ+γ=0w2ɛ-δ=0,

which show that 0 = ε = δ = γ.

Case 5: Norm(T) = {((1, 0), (1, 0)), ((1, 0), (w,−1)), ((w, 1), (w, 1))}

Note that T does not exist in case 5.

Case 6: Norm(T) = {((1, 0), (1, 0)), ((1, 0), (w,−1)), ((w,−1), (w,−1))}

Note that T does not exist in case 6.

Case 7: Norm(T) = {((1, 0), (1, 0)), ((1, 0), (w,−1)), ((w, 1), (w,−1))}

Note that T does not exist in case 7.

Case 8: Norm(T) = {((1, 0), (1, 0)), ((w, 1), (w, 1)), ((w,−1), (w,−1))}

Then T = (1, 1−w2, 0) for all 0 < w < 1. Note that T = (1, 1−w2, 0) is extreme for all 0 < w < 1. Indeed, let T1 = (1+ε, 1−w2 +δ, γ), T2 = (1−ε, 1−w2δ,γ) be such that ||T1|| = 1 = ||T2|| for some ε,δ, γ ∈ ℝ. Since |Ti((1, 0), (1, 0))| ≤ 1, |Ti((w, 1), (w, 1))| ≤ 1, |Ti((w,−1), (w,−1))| ≤ 1, we have

ɛ=0w2ɛ+δ+2wγ=0w2ɛ+δ-2wγ=0,

which show that 0 = ε = δ = γ.

Case 9: Norm(T) = {((1, 0), (1, 0)), ((w, 1), (w, 1)), ((w, 1), (w,−1))}

Note that T does not exist in case 9.

Case 10: Norm(T) = {((1, 0), (1, 0)), ((w,−1), (w,−1)), ((w, 1), (w,−1))}

Then T = (1, w2 − 1, w) for all 0<w12. Note that T = (1, w2 − 1, w) is extreme for all 0<w12. Indeed, let T1 = (1+ε,w2 − 1 + δ,w + γ), T2 = (1 − ε,w2 − 1 − δ,wγ) be such that ||T1|| = 1 = ||T2|| for some ε,δ, γ ∈ ℝ. Since |Ti((1, 0), (1, 0))| ≤ 1, |Ti((w,−1), (w,−1))| ≤ 1, |Ti((w, 1), (w,−1))| ≤ 1, we have

ɛ=0w2ɛ+δ-2wγ=0w2ɛ-δ=0,

which show that 0 = ε = δ = γ.

Case 11: Norm(T) = {((1, 0), (w, 1)), ((1, 0), (w,−1)), ((w, 1), (w, 1))}

Then T = (0, 1−2w, 1) for all 0<w12. Note that T = (0, 1−2w, 1) is extreme for all 0<w12. Indeed, let T1 = (ε, 1−2w+δ,1+γ), T2 = (−ε, 1−2wδ, 1−γ) be such that ||T1|| = 1 = ||T2|| for some ε,δ, γ ∈ ℝ. Since |Ti((1, 0), (w, 1))| ≤ 1, |Ti((1, 0), (w,−1))| ≤ 1, |Ti((w, 1), (w, 1))| ≤ 1, we have

wɛ+γ=0wɛ-γ=0w2ɛ+δ+2wγ=0,

which show that 0 = ε = δ = γ.

Case 12: Norm(T) = {((1, 0), (w, 1)), ((1, 0), (w,−1)), ((w,−1), (w,−1))}

Note that T does not exist in case 12.

Case 13: Norm(T) = {((1, 0), (w, 1)), ((1, 0), (w,−1)), ((w, 1), (w,−1))}

Note that T does not exist in case 13.

Case 14: Norm(T) = {((1, 0), (w, 1)), ((w, 1), (w, 1)), ((w,−1), (w,−1))}

Then T=(2w-12w2,1-2w2,12w) for all w12. Note that T=(2w-12w2,1-2w2,12w) is extreme for all w12. Indeed, let T1=(2w-12w2+ɛ,1-2w2+δ,12w+γ),T2=(2w-12w2-ɛ,1-2w2-δ,12w-γ) be such that ||T1|| = 1 = ||T2|| for some ε,δ, γ ∈ ℝ. Since |Ti((1, 0), (w, 1))| ≤ 1, |Ti((w, 1), (w, 1))| ≤ 1, |Ti((w,−1), (w,−1))| ≤ 1, we have

ɛ+γ=0w2ɛ+δ+2wγ=0w2ɛ+δ-2wγ=0,

which show that 0 = ε = δ = γ.

Case 15: Norm(T) = {((1, 0), (w, 1)), ((w, 1), (w, 1)), ((w, 1), (w,−1))}

Note that T does not exist in case 15.

Case 16: Norm(T) = {((1, 0), (w, 1)), ((w,−1), (w,−1)), ((w, 1), (w,−1))}

Then T=(12w,w-22,12) for all w12. Note that T=(12w,w-22,12) is extreme for all w12. Indeed, let T1=(12w+ɛ,w-22+δ,12+γ),T2=(12w-ɛ,w-22-δ,12-γ) be such that ||T1|| = 1 = ||T2|| for some ε,δ, γ ∈ ℝ. Since |Ti((1, 0), (w, 1))| ≤ 1, |Ti((w,−1), (w,−1))| ≤ 1, |Ti((w, 1), (w,−1))| ≤ 1, we have

wɛ+γ=0w2ɛ+δ-2wγ=0w2ɛ-δ=0,

which show that 0 = ε = δ = γ.

Case 17: Norm(T) = {((1, 0), (w,−1)), ((w, 1), (w, 1)), ((w,−1), (w,−1))}

Note that T does not exist in case 17.

Case 18: Norm(T) = {((1, 0), (w,−1)), ((w, 1), (w, 1)), ((w, 1), (w,−1))}

Note that T does not exist in case 18.

Case 19: Norm(T) = {((1, 0), (w,−1)), ((w,−1), (w,−1)), ((w, 1), (w,−1))}

Note that T does not exist in case 19.

Case 20: Norm(T) = {((w, 1), (w, 1)), ((w,−1), (w,−1)), ((w, 1), (w,−1))}.

Note that T does not exist in case 20.

(⇒) : By the argument of (⇐), it is enough to show that if Norm(T) has at most two elements, then T is not extreme. For an example, let

Norm(T)={((1,0),(1,0)),((1,0),(w,1))}.

We will show that T is not extreme. Notice that

T((1,0),(1,0))=1=T((1,0),(w,1)),T((w,1),(w,1))<1,T((w,1),(w,-1))<1.

Hence, a = 1, c = 1 − w, |w2b| < 1, |w2 + b| + 2w(1 − w) < 1. Let δ > 0 such that |w2b| +δ < 1, |w2 + b| + 2w(1 − w) +δ < 1. Let T1 = (1, b + δ, 1 − w) and T2 = (1, bδ, 1 − w). By Theorem 2.1, ||Ti|| = 1 for i = 1, 2. Since Ti= T, T=12(T1+T2), T is not extreme. For the other cases, we may show that if Norm(T) has at most two elements, then T is not extreme using Theorem 2.1. Hence, we will omit the proofs. Therefore, we complete the proof.

Now we are in position to describe all the extreme points of the unit ball of Ls(2h(w)2).

Theorem 2.3

  • Let 0<w12. Then,

    extBLs(2h(w)2)={±(0,1,0),±(1,(1-w)2,±(1-w)),±(1,1-w2,0),±(1,w2-1,±w),±(0,1-2w,±1),±(1,-3w2+2w-1,±(1-w))}.

  • Let 12<w<1. Then,

    extBLs(2h(w)2)={±(0,1,0),±(1,(1-w)2,±(1-w)),±(1,1-w2,0),±(1,w2-1,±(1-w)),±(12w,w-22,±12),±(2w-12w2,1-2w2,±12w)}.

Proof

It follows from the proof of Theorem 2.2.

Theorem 3.1

Let 0 < w < 1 and fLs(2h(w)2)* and α = f(x1x2), β = f(y1y2), γ = f(x1y2 + x2y1).

  • Let 0<w12. Then,

    f=max{β,α+(1-w)2β+(1-w)γ,α+(1-w2)β,α-(1-w2)β+wγ,α-(3w2-2w+1)β+(1-w)γ,(1-2w)β+γ}.

  • Let 12<w<1. Then,

    f=max{β,α+(1-w)2β+(1-w)γ,α+(1-w2)β,α-(1-w2)β+(1-w)γ,(12w)α-(2-w2)β+12γ,(2w-12w2)α+(1-2w2)β+12wγ}.

Proof

It follows from Theorem 2.3 and the fact that

f=maxTextBLs(2h(w)2)f(T).

Note that if ||f|| = 1, then |α| ≤ 1, |β| ≤ 1, |γ| ≤ min{1, 2w}.

Theorem 3.2.([17, Theorem 2.3])

Let E be a real 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 yextBEx}. Then, xexpBE.

Now we are in position to describe all the exposed points of the unit ball of Ls(2h(w)2).

Theorem 3.3

For 0 < w < 1, expBLs(2h(w)2)=extBLs(2h(w)2).

Proof

We divide two cases.

Case 1: 0<w12.

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

Let fLs(2h(w)2)* be such that α = 0 = γ, β = 1. Then f(T) = 1, |f(S)| < 1 for every SextBLs(2h(w)2){±T}. By Theorem 3.2, T is exposed.

Claim: T = (1, (1 − w)2, 1 − w) is exposed.

Let fLs(2h(w)2)* be such that α=12-(1-w)2n,β=1n,γ=12(1-w) for a sufficiently large n ∈ ℕ. Then f(T) = 1, |f(S)| < 1 for every SextBLs(2h(w)2){±T}. By Theorem 3.2, T is exposed.

Claim: (1, 1 − w2, 0) is exposed.

Let fLs(2h(w)2)* be such that α=12-(1-w)2n,β=1n,γ=0, γ = 0 for a sufficiently large n ∈ ℕ. Then f(T) = 1, |f(S)| < 1 for every SextBLs(2h(w)2){±T}. By Theorem 3.2, T is exposed.

Claim: (0, 1 − 2w, 1) is exposed.

Let fLs(2h(w)2)* be such that α = 0 = β, γ = 1. Then f(T) = 1, |f(S)| < 1 for every SextBLs(2h(w)2){±T}. By Theorem 3.2, T is exposed.

Claim: T = (1, w2 − 1, w) is exposed.

First suppose that 0<w12. Let fLs(2h(w)2)* be such that α = 0, β = −1, γ = w. Then f(T) = 1, |f(S)| < 1 for every SextBLs(2h(w)2){±T}. By Theorem 3.2, T is exposed.

If w=12, Then T=(1,-34,12). By Theorem 2.3, extBLs(2h(12)2)={±(1,14,±12),±(1,-34,±12),±(1,34,0)(0,0,±1)}.

Let fLs(2h(w)2)* be such that α=14, β = −1, γ = 0. Then f(T) = 1, |f(S)| < 1 for every SextBLs(2h(w)2){±T}. By Theorem 3.2, T is exposed.

Claim: T = (1,−3w2 + 2w − 1, 1 − w) for 0 < w < 1 is exposed.

Let fLs(2h(w)2)* be such that α=12-3w2-2w+1n,β=-1n,γ=12(1-w) for a sufficiently large n ∈ ℕ. Then f(T) = 1, |f(S)| < 1 for every SextBLs(2h(w)2){±T}. By Theorem 3.2, T is exposed.

Case 2: 12<w<1.

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

Let fLs(2h(w)2)* be such that α = 0 = γ, β = 1. Then f(T) = 1, |f(S)| < 1 for every SextBLs(2h(w)2){±T}. By Theorem 3.2, T is exposed.

Claim: T = (1, (1 − w)2, 1 − w) is exposed.

Let fLs(2h(w)2)* be such that α=w-(1-w)2n,β=1n, γ = 1 for a sufficiently large n ∈ ℕ. Then f(T) = 1, |f(S)| < 1 for every SextBLs(2h(w)2){±T}. By Theorem 3.2, T is exposed.

Claim: (1, 1 − w2, 0) is exposed.

Let fLs(2h(w)2)* be such that α=12-1-w2nβ=1n, γ = 0 for a sufficiently large n ∈ ℕ. Then f(T) = 1, |f(S)| < 1 for every SextBLs(2h(w)2){±T}. By Theorem 3.2, T is exposed.

Claim: (2w-12w2,1-2w2,±12w) is exposed.

Let fLs(2h(w)2)* be such that α = 0 = β, γ = 2w. Then f(T) = 1, |f(S)| < 1 for every SextBLs(2h(w)2){±T}. By Theorem 3.2, T is exposed.

Claim: T = (1, w2 − 1, 1 − w) is exposed.

Let fLs(2h(w)2)* be such that α = w, β=-11+w, γ = 0. Then f(T) = 1, |f(S)| < 1 for every SextBLs(2h(w)2){±T}. By Theorem 3.2, T is exposed.

Claim: T=(12w,w-22,12) is exposed.

Let fLs(2h(w)2)* be such that α = 0, β = 1, γ = w. Then f(T) = 1, |f(S)| < 1 for every SextBLs(2h(w)2){±T}. By Theorem 3.2, T is exposed.

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