On nearest and furthest points

Balaganskii, V. S.

doi:10.1007/bf02308765

Cited by 4 publications

(6 citation statements)

References 6 publications

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“…Thus Balaganski [3] proved that there are no antiproximinal sets with smooth complement in any reflexive Banach spaces. Other similar results can be found, for example, in [4], [6], [9], [13], [21], [29], [33].…”

Section: Theorem 5 a Bounded Set A Is Antiremotal If And Only If The ...supporting

confidence: 80%

Relationships Between Farthest Point Problem and Best Approximation Problem

Precupanu¹

2011

Annals of the Alexandru Ioan Cuza University - Mathematics

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Section: Theorem 5 a Bounded Set A Is Antiremotal If And Only If The ...supporting

confidence: 80%

Relationships Between Farthest Point Problem and Best Approximation Problem

Precupanu¹

2011

Annals of the Alexandru Ioan Cuza University - Mathematics

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“…It is proved in [5] that A is antiproximinal in L 1 [0, 1] and M is a set without farthest points in L 1 [0, 1]. Nevertheless, we have the following fact.…”

Section: Proof Of Statement (B) Assume the Contrary: There Existsmentioning

confidence: 75%

“…Consequently, the set N (α) is not antiproximinal and, since α is arbitrary, statement (a) is proved. Statement (b) was proved in [5].…”

Section: Definitionmentioning

confidence: 85%

“…In [3], it was proved that there are no antiproximinal sets with convex complement in a reflexive Banach space. Edelstein [6] constructed in l 1 an example of a bounded centrally symmetric convex set without farthest points; in [5], such sets were constructed in a rather wide class of Banach spaces.…”

Section: Doi: 101134/s0081543812050069mentioning

confidence: 99%

“…Convex closed bounded antiproximinal sets were considered by many authors (see [1][2][3][4][5][6] and references therein). It is known [6] that an antiproximinal set with bounded convex complement can be constructed in the space l 1 with some new equivalent norm.…”

Section: Doi: 101134/s0081543812050069mentioning

confidence: 99%

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On convex closed bounded bodies without farthest points such that the closure of their complement is antiproximinal

Balaganskii

2012

Proc. Steklov Inst. Math.

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A bounded closed convex Chebyshev approximatively compact body M ⊂ X = L 1 [0, 1] without farthest points is constructed such that X\M is antiproximinal.We consider the connection between the antiproximinality of a convex bounded closed body M , on the one hand, and the antiproximinality of the closure of its complement and the absence of farthest points in M , on the other. Let us introduce the notation: X is a real Banach space (X ∈ (B)),is the unit sphere in X * centered at 0, M and ∂M are the closure and the boundary of the set M , and int M is the set of interior points of M . A nonempty subset M = X of the Banach space X is called antiproximinal if, for any point x ∈ X\M , the set M contains no nearest points, i.e., if P M (x) = ∅.A nonempty subset N = X of the Banach space X is called a set without farthest points if, for any point x ∈ X, there are no farthest points in the set N , i.e., if F N (x) = ∅.A set M ⊂ X of the Banach space X is called an existence set if, for any point x ∈ X, the set P M (x) is nonempty.A set M ⊂ X of the Banach space X is called a Chebyshev set if, for any point x ∈ X, the set P M (x) is a singleton.A set M ⊂ X of the Banach space X is called approximatively compact if, for any point x ∈ X, each minimizing sequence {y n } ∈ M has a subsequence convergent to an element from M .We will say that F satisfies the Lipschitz condition (with Lipschitz constant l) on Ω if, for any x 1 , x 2 ∈ Ω, F (x 1 ) − F (x 2 ) ≤ l x 1 − x 2 .

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