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

Abstract

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 .