A Note on The Convexity of Chebyshev Sets

Perhaps one of the major unsolved problem in Approximation Theory is : Whether or not every Chebyshev subset of a Hilbert space must be convex. Many partial answers to this problem are available in the literature. R.R. Phelps [Proc. Amer. Math. Soc. 8 (1957), 790-797] showed that a Chebyshev set in an inner product space (or in a strictly convex normed linear space) is convex if the associated metric projection is non-expansive. We extend this result to metric spaces.


Introduction
It is well known (see e.g.[4], p. 35) that a closed convex subset of a Hilbert space is Chebyshev.One of the major unsolved problem in Approximation Theory is its converse i.e. whether every Chebyshev subset of a Hilbert space is convex.Surveys giving various partial answers of this problem were given by Vlasov (1973), Narang (1977), Deutsch (1993) and by Balaganski and Vlasov (1996).Metric projections have been very helpful in giving some partial answers of this problem.Phelps [9] showed that in an inner product space (in a strictly convex normed linear space), a Chebyshev set is convex if the associated metric projection is non-expansive.Here we extend this result to metric spaces.
Before proceeding to our main result, we recall few definitions.

Definitions and Notations
Definition 2.1 Let (X, d) be a metric space and x, y, z ∈ X.We say that z is between x and y if d(x, z) + d(z, y) = d(x, y).For any two points x, y ∈ X, the set Definition 2.2 A metric space (X, d) is said to be convex [10] if for every x, y in X and for every t, 0 ≤ t ≤ 1 there exists at least one point z such that d(x, z) = (1 − t) d(x, y) and d(z, y) = t d(x, y).

Definition 2.3
The space X is said to be strongly convex [10] or an M -space [7] if such a z exists and is unique for each pair x, y of X.
Thus for strongly convex metric spaces each t, 0 ≤ t ≤ 1, determines a unique point of the segment [x, y].Definition 2.4 A metric space (X, d) is called externally convex [6] if for all distinct points x, y such that d(x, y) = λ, and r > λ there exists a unique z of X such that d(x, y) + d(y, z) = d(x, z) = r.Definition 2.5 A strongly convex metric space (X, d) is said to be strictly convex [8] if for every pair x, y of X and r > 0. Therefore, in a strictly convex metric space if x and y are any two points on the boundary of a sphere then ]x, y[ lies strictly inside the sphere.
Remark 2.1: A convex metric space need not be externally convex and an externally convex space need not be convex(see [6]).The unit disc in R 2 with Euclidean distance is a strictly convex M -space and is not a normed linear space.For more examples of convex metric spaces, externally convex metric spaces and M -spaces one may refer to [6] and [11].Definition 2.6 A subset K of a metric space (X, d) is said to be convex (see [8]) if for every x, y ∈ K, any point between x and y is also in K i.e. for each x, y in K, the metric segment [x, y] lies in K. Definition 2.7 Let S be a subset of a metric space (X, d) and x ∈ X.An element s o ∈ S satisfying d(x, s o ) ≡ inf{d(x, y) : y ∈ S} ≡ d(x, S), is called a best approximation to x in S. The set P S (x) = {s o ∈ S : d(x, s o ) = d(x, S)} is called the set of best approximants to x in S. The set S is said to be proximinal if P S (x) = φ for each x ∈ X and is called Chebyshev if P S (x) is exactly singleton for each x ∈ X.The set-valued map f : X → 2 S ≡ collection of subsets of S, taking each x ∈ X to the set P S (x) is called the nearest point map or the metric projection.
It is clear that f exists iff S is proximinal and that f is single-valued iff S is Chebyshev.
Definition 2.8 We say that f shrinks distances or f is non-expansive if d(f (x), f (y)) ≤ d(x, y) for all x, y ∈ X. Definition 2.9 We say that X has Property (P) [9] if the nearest point map f shrinks distances whenever it exists for a closed convex set S ⊆ X.

Convexity of Chebyshev sets
The following lemma, which is easy to prove, will be used in showing that the shrinking property of the metric projection implies the convexity of a Cheybyshev set in externally convex M -spaces : Proof: Suppose S is not convex.Then there exist x 1 , y 1 ∈ S, x 1 = y 1 such that some ( This will be true if we prove that d(x, y) = d(x, f (z)) + d(f (z), y).
{z ∈ X : d(x, z) + d(z, y) = d(x, y)} is called the metric segment and is denoted by [x, y].The set [x, y, −[= {z ∈ X : d(x, y) + d(y, z) = d(x, z)} denotes the half ray starting from x and passing through 2000 Mathematics Subject Classification: 41A65 T.D. Narang and Sangeeta y.Correspondingly, ]−, x, y] is the half ray starting from y and passing through x, ] − x, y, −[ is the line passing through x and y.
d(x, p) ≤ r, d(y, p) ≤ r imply d(z, p) < r unless x = y, where p is arbitrary but fixed point of X and z is any point in the open metric segment ]x, y[.

1 )
r 2 > 0. By the lemma, both K 1 and K 2 are proximinal.Hence, there exists x ∈ K 1 , y ∈ K 2 such that d(u, K 1 ) = d(u, x), d(u, K 2 ) = d(u, y) and the metric segment ]x, y[ lies in X \ S. Let z be the mid point of x and y i.e. d(x, z) = d(z, y) = 1 2 d(x, y).(Firstly, we claim that f (z) = x and f (z) = y.If f (z) = x then d(f (y), f (z)) = d(y, x) > 1 2 d(y, x) = d(y, z), a contradiction to the fact that f is nonexpansive.Same contradiction results if f (z) = y.Therefore f (z) is neither equal to x nor equal to y.Now we claim that d(x, y) < d(x, f (z)) + d(f (z), y).