ON PROJECTION CONSTANT PROBLEMS AND THE EXISTENCE OF METRIC PROJECTIONS IN NORMED SPACES

We give the sufficient conditions for the existence of a metric projection onto convex closed subsets of normed linear spaces which are reduced conditions than that in the case of reflexive Banach spaces and we find a general formula for the projections onto the maximal proper subspaces of the classical Banach spaces lp, 1 ≤ p < ∞ and c0. We also give the sufficient and necessary conditions for an infinite matrix to represent a projection operator from lp, 1 ≤ p < ∞ or c0 onto anyone of their maximal proper subspaces.


Introduction
A subset C of a normed linear space X is called an existence subset of X if and only if for every element x ∈ X there is an element y ∈ C, where y is called the best approximation of x denoted by b(x, C) [2,3] such that x − y = dist(x, C) := inf x − y : y ∈ C . (1.1) Needless to say, best approximations play a major role in many applications, including approximation theory, optimization and applications in mathematical economics and engineering.Thus, the mathematical analysis of the properties of the best approximation elements has drawn much attention in research.
In [11,13], the authors there used the terms Čebyšev subsets and proximal subsets instead of the existence subsets and studied the characterizations of the existence subsets of Banach spaces.
It is shown that if X is a reflexive Banach space and C is a closed convex subset, then for every x ∈ X the best approximation element b(x, C) exists and is unique.
In this paper, reduced assumptions on a normed linear space for a closed convex subset to exist are given, instead of the reflexivity and the completeness assumptions of the given normed space.
On the other hand, it is known that the existence of a projection P from a Banach space X onto its closed subspace Y is equivalent to the existence of an extension T of any operator T from Y into W to an operator from X into W such that T ≤ P T .The two equivalent problems, that is (1) how small can the norm of the extended operator be made?and (2) what is the projection of smallest norm?, are challenging to the study of the relative projection constant λ(Y, X) of Y in X, where and the absolute projection constant of Y , λ(Y ), where In [10], the upper estimate for the absolute projection constant λ(Y ) of a finitedimensional space Y with dim Y = n is found in the form , in the complex field. (1.4) The precise values for l n 1 , l n 2 , and l p n , p = 1, p = 2, have been calculated by Grünbaum [8], Gordon [7], Garling and Gordon [6], Rutovitz [12], and König et al. [9].
In [5], interesting results have been given for the injective and projective tensor products.
For a finite codimensional subspaces, Garling and Gordon [6] showed that if Y is a finite codimensional subspace of the space X with co-dimension n, then for every > 0 there exists a projection P from X onto Y with norm

Notation and basic definitions
We will use the same notation that is given in [1,4,14].

E. M. El-Shobaky et al. 403
Let C be a nonempty closed convex subset of a normed space X.If for every x ∈ X there is a unique b(x, C) in C, then the mapping b(x, C) is said to be a metric projection onto C, in this case we have Clearly, if X is a Hilbert space and C is a nonempty closed convex subset of X, then there is a metric projection from X onto C, see [14].
As a direct consequence of the separation theorem, if X is a locally convex linear topological space, then a nonempty convex subset C of X is closed in the strong topology of X if and only if C is closed in the weak topology of X, see [4].
A relative projection constant of the closed subspace Y in the space X is said to be exact if and only if there is a projection P from X onto Y at which the infimum of (1.2) is attained.
A subspace Y of the space X is said to be a hyperplane (maximal proper subspace) of the space X if and only if X contains Y as a subspace of deficiency 1.
It is known that a subspace Y is a hyperplane of the space X if and only if there is a functional f ∈ X * such that Y = f −1 ({0}); see [1].
Let P be an operator on the space X.Then the point x ∈ X is said to be a maximal point of the operator P if and only if P = P (x) .
If X is either the Banach space l ∞ , the Banach space of all bounded scalarvalued functions {x n } ∞ n=1 on a countably infinite set N , or l p the Banach space of all scalar-valued functions x = {x n } ∞ n=1 on a countably infinite set N , such that ∞ n=1 |x n | p < ∞ or c 0 the closed subspace of the Banach space l ∞ , then the norms on X are defined as follows: Our first result is the following theorem.
Theorem 2.1.Let X be a normed space in which every Cauchy sequence has a weakly convergent subsequence and the parallelogram law holds.Let C be a nonempty closed convex subset of X.Then the metric projection b(•, C) from X onto C exists.
Proof.Let x ∈ X and consider the distance function dist(x, C), there exists a sequence Taking the limit as n → ∞, we get lim n→∞ x −y n = dist(x, C).The sequence {y n } ∞ n=1 is a Cauchy sequence in X.In fact, using the convexity of C, we have (1/2)(y i + y j ) ∈ C, also using the parallelogram law, we have (2.4) Therefore, using the assumption, every Cauchy sequence has a weakly convergent subsequence, the sequence {y n } ∞ n=1 has a subsequence {y n i } ∞ i=1 converging weakly to some point y 0 in X, y n i i → ∞ − −− → weakly y 0 .Since C is convex and is closed in the strong topology, C is closed in the weak topology, then C contains as well all its weak limits y 0 ∈ C. Now, consider the proper convex lower semi-continuous real-valued function g on C defined by we have and the mapping g attains its minimum at y 0 .In fact, let > 0. Then the set is a convex closed subset of X.Using the convexity of G g(y 0 )− , for every > 0 the set G g(y 0 )− is a weakly closed subset of C and hence for every > 0 the set g(y 0 )− , there is a neighborhood V of y 0 (w.r.t. the weak topology) such that V ⊂ G c(C) g(y 0 )− .Using the weak limit point definition of the sequence (2.10) Since > 0 is arbitrary and y 0 ∈ C, we have (2.11) Thus for every x ∈ X there is y 0 ∈ C such that (2.12) To show that such a point y 0 is unique, let g(y 0 ) = 0. Since x − y 0 = 0, x = y 0 , y 0 is unique.Let g(y 0 ) > 0 and let z 0 be an element in C with and z 0 = y 0 ( z 0 − y 0 > 0), since C is convex, (1/2)(y 0 + z 0 ) ∈ C, using the parallelogram law, we have = g y 0 . (2.14) This is a contradiction.Therefore no such z 0 exists, and y 0 is unique.Now, define the mapping b(x, C) from X onto C by b(x, C) = y 0 , the mapping b(x, C) is the required metric projection.
Our result concerning the maximal proper subspaces of the classical Banach spaces c 0 or l p for 1 ≤ p < ∞ is the following result.
Then the general formula of any projection from X onto Y is given by (2.15) x i e i , any operator P on X is completely determined by P (e i ), suppose that P (e i ) = {e ik } ∞ k=1 , we have x i e ik ∞ k=1 . (2.16) Define the element z = {α k } ∞ k=1 ∈ X as follows (2.17) Since f is a nonzero element of the space l 1 or l q , where 1/p +1/q = 1, f = 0, there is at least one index i for which f i = 0, for this index multiplying (2.17) by f k and summing with respect to k, we get (2.18) Since P (e i ) ∈ Y and f (P (e i )) = 0, that is, k≥1 f k e ik = 0. Therefore, f i k≥1 f k α k = f i , this proves that k≥1 f k α k = 1, that is, f (z) = 1.On the other hand, we have e ik = δ ik − α k f i .Thus the representation of P is as follows: (

2.19)
The converse direction is clearly true.To calculate the norm of the given projection P , we have two distinct cases, the first is when X = c 0 in which the norm is as follows: (2.20) E. M. El-Shobaky et al. 407 The second case is when X = l p in which the norm of P is as follows: i=1 is an element of the space (l 1 ) * = l ∞ and Y is a subspace of the space l 1 , Y = f −1 ({0}), then some of the maximal points of any projection P from l 1 onto Y lie in the set {e i } ∞ i=1 , where {e i } ∞ i=1 is the canonical basis of l 1 .
Proof.According to Theorem 2.2, the norm of any projection in this case is given by (2.22) Therefore the norm of any such projection is attained at some points in the set {e i } ∞ i=1 .
Remark 2.4.If one of the coordinates of f , say f i = 0, then e i ∈ Y (for f (e i ) = f i = 0).And to get a norm one projection the effect of this projection on the basis elements must not exceed one.
Corollary 2.5.Let X denote one of the spaces c 0 or l p , 1 < p < ∞, and let f ∈ X * .Then an infinite matrix P = {p in } i,n∈N represents a projection operator from X onto its hyperplane f −1 ({0}) if and only if the matrix α = {δ in − p in } i,n∈N satisfies the following conditions: (1) trace α = 1, (2) each row of the matrix α is a scalar multiple of f .
Proof.Let P = {p in } i,n∈N be an infinite matrix satisfying conditions (1) and ( 2).Then P = I − α, since each row of the matrix α is a scalar multiple of f , α(y) = 0 for all y ∈ f −1 ({0}).And so, P (y) = y for all y ∈ f −1 ({0}).If f n denotes the nth row of the matrix α, using condition (2) we obtain for each n ∈ N a scalar α n such that f n = α n f , using condition (1) we have where α in = z i f n .This proves that the operator P has the matrix representation P = {δ in − α in } i,n∈N .Clearly the matrix α = {α in } i,n∈N satisfies conditions (1) and ( 2).
E. M. El-Shobaky et al. 409 Corollary 2.6.Let f = δ{ k } n k=1 , n > 2, be a sequence of the space l 1 n , where δ is a nonzero scalar and i = ±1.Then the relative projection constant of the (n − 1)-dimensional subspace f −1 ({0}) in the space l ∞ n is given by (2.25) Moreover, the minimal norm projection is given by Proof.As given in Theorem 2.2 the norm of any projection corresponding to the element z = {α k } n k=1 is given by (2.27) Assume that the minimal projection is a norm one projection.Then there is z ∈ l 1 n and 1 For n > 2, this is true only if α k = 0 for every k = 1, 2, . . ., n.This is an obvious contradiction, thus there is no norm one projection from l 1 n onto f −1 ({0}).Now, let x = {x k } n k=1 ∈ l ∞ n be an arbitrary point.To project this point to the point x 0 = {x 0 k } n k=1 in the space f −1 ({0}) with a minimal available distance between the points x = {x k } n k=1 and x 0 = {x 0 k } n k=1 , the sequence x 0 − x = {x 0 k − x k } n k=1 must be parallel to the line passing through f = {f k } n k=1 and perpendicular to the plane f −1 ({0}).Thus there is a scalar λ such that x 0 −x = λf .On the other hand, since Now we are going to show that this projection is a minimal norm projection.Assume the contrary, that is, there exists an element z ∈ l 1 n such that f (z) = 1 and the corresponding projection P satisfies P < (2 − 2/n), according to (2.27), we have for every k ∈ {1, 2, . . ., n}, from which we get and so for such z we have Proof.As given in Theorem 2.2, the norm of any projection corresponding to the element z is given by