• Title, Summary, Keyword: bilinear forms

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Extremal Problems for 𝓛s(22h(w))

  • Kim, Sung Guen
    • Kyungpook Mathematical Journal
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    • v.57 no.2
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    • pp.223-232
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    • 2017
  • We classify the extreme and exposed symmetric bilinear forms of the unit ball of the space of symmetric bilinear forms on ${\mathbb{R}}^2$ with hexagonal norms. We also show that every extreme symmetric bilinear forms of the unit ball of the space of symmetric bilinear forms on ${\mathbb{R}}^2$ with hexagonal norms is exposed.

THE UNIT BALL OF THE SPACE OF BILINEAR FORMS ON ℝ3 WITH THE SUPREMUM NORM

  • Kim, Sung Guen
    • Communications of the Korean Mathematical Society
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    • v.34 no.2
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    • pp.487-494
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    • 2019
  • We classify all the extreme and exposed bilinear forms of the unit ball of ${\mathcal{L}}(^2l^3_{\infty})$ which leads to a complete formula of ${\parallel}f{\parallel}$ for every $f{\in}{\mathcal{L}}(^2l^3_{\infty})^*$. It follows from this formula that every extreme bilinear form of the unit ball of ${\mathcal{L}}(^2l^3_{\infty})$ is exposed.

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

  • Kim, Sung Guen
    • Kyungpook Mathematical Journal
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    • v.56 no.3
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    • pp.781-791
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    • 2016
  • Let $d_*(1,w)^2 ={\mathbb{R}}^2$ 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: $$S_{{\mathcal{L}}_s(^2d_*(1,w)^2)}=smB_{{\mathcal{L}}_s(^2d_*(1,w)^2)}{\bigcup}extB_{{\mathcal{L}}_s(^2d_*(1,w)^2)}{\bigcup}A$$, where the set A consists of $ax_1x_2+by_1y_2+c(x_1y_2+x_2y_1)$ with (a = b = 0, $c={\pm}{\frac{1}{1+w^2}}$), ($a{\neq}b$, $ab{\geq}0$, c = 0), (a = b, 0 < ac, 0 < ${\mid}c{\mid}$ < ${\mid}a{\mid}$), ($a{\neq}{\mid}c{\mid}$, a = -b, 0 < ac, 0 < ${\mid}c{\mid}$), ($a={\frac{1-w}{1+w}}$, b = 0, $c={\frac{1}{1+w}}$), ($a={\frac{1+w+w(w^2-3)c}{1+w^2}}$, $b={\frac{w-1+(1-3w^2)c}{w(1+w^2)}}$, ${\frac{1}{2+2w}}$ < c < ${\frac{1}{(1+w)^2(1-w)}}$, $c{\neq}{\frac{1}{1+2w-w^2}}$), ($a={\frac{1+w(1+w)c}{1+w}}$, $b={\frac{-1+(1+w)c}{w(1+w)}}$, 0 < c < $\frac{1}{2+2w}$) or ($a={\frac{1=w(1+w)c}{1+w}}$, $b={\frac{1-(1+w)c}{1+w}}$, $\frac{1}{1+w}$ < c < $\frac{1}{(1+w)^2(1-w)}$).