# Proximinality and co-proximinality in metric linear spaces

Proximinality and co-proximinality in metric linear spaces spaces. The present paper is also a step in this direction. In this paper, we discuss the existence and uniqueness results on best approximation and best coapproximation in metric linear spaces thereby generalizing the various known results. We start with a few definitions. The research work of the author has been supported by U.G.C., India under Emeritus Fellowship. 2 The research work of the author has been supported by U.G.C., India under Senior Research Fellowship. 2010 Mathematics Subject Classification. 41A50, 41A52, 41A65, 35E10. Key words and phrases. Best approximation, best coapproximation, proximinal set, co-proximinal set, Chebyshev set, co-Chebyshev set. Let G be a non-empty subset of a metric space (X, d). An element g0 G is called a best approximation (best coapproximation) to x X if d(x, g0 ) d(x, g) (d(g0 , g) d(x, g)) for all g G. The set of all such g0 G is denoted by PG (x)(RG (x)). The set G is called proximinal (co-proximinal) if PG (x) (RG (x)) contains at least one element for every x X. If for each x X, PG (x)(RG (x)) has exactly one element in G, then the set G is called Chebyshev (coChebyshev). If x, y and z are any three points in a metric space (X, d) then z is said to be a between point of x and y if d(x, z) + d(z, y) = d(x, y). A metric d defined on X is said to be a convex (strongly convex) metric if for each pair x, y X, d determine at least one (exactly one) between point. The space X together with a convex metric d is called a convex metric space. A linear space X with a translation invariant metric d (i.e. d(x+z, y+z) = d(x, y) for all x, y, z X) such that addition and scalar multiplication are continuous on (X, d) is called a metric linear space. The space X of all entire functions, i.e. X= f : f (z) = n=0 an z n , |an | n 0 as n with the metric d defined by d(f, g) = max{|a0 - b0 |, |an - bn | n , n 1}, n n where f (z) = n=0 an z , g(z) = n=0 bn z , is a non-normable metric linear space (see [10], p. 238). Remarks. (i) A proximinal (co-proximinal) subset of a metric space is closed. (ii) Every singleton subset of a metric space is Chebyshev (co-Chebyshev). (iii) Every closed interval in R is proximinal (co-proximinal). (iv) PG (PG (x)) = PG (x) and RG (RG (x)) = RG (x). (v) PG (x) = {g0 G : d(x, g0 ) d(x, g) for every g G} = G B(x, d(x, G)). (vi) RG (x) = {g0 G : d(g0 , g) d(x, g) for every g G} = G [ {B(g, d(x, g)} : g G}]. -1 (vii) If G is a linear subspace of a metric linear space (X, d) then PG (0) -1 -1 G = {0} and RG (0) G = {0}, where PG (0) = {x X : 0 PG (x)} and -1 RG (0) = {x X : 0 RG (x)}. -1 (viii) If G is a linear subspace of a metric linear space (X, d), then d(g, RG (0)) = d(g, 0) for every g G. (ix) If G is a linear subspace of a metric linear space (X, d), then g0 -1 -1 PG (x) (RG (x)) if and only if x - g0 PG (0) (RG (0)). Let (X, d) be a metric linear space and x, y X, then we say that x is orthogonal to y, x y if d(x, 0) d(x, y) for every scalar . We say that G x if g x for every g G. Concerning orthogonality and best coapproximation, we have Theorem 1. Let G be a linear subspace of a metric linear space (X, d) such that G(x - g0 ), then g0 RG (x). Proof. Since G(x - g0 ), we have g(x - g0 ) for every g G i.e. d(g, 0) d(g, (x - g0 )) for every scalar . Take = 1, we get d(g, 0) d(g, x - g0 ) for every g G. This gives d(g0 , g + g0 ) d(x, g + g0 ) for every g G. Hence g0 RG (x). Note. The converse of the above theorem is also true in normed linear spaces (see [2]). We do not know whether this is true in metric linear spaces. However, in metric linear spaces, we have Theorem 2. Let G be a linear subspace of a metric linear space (X, d) and g0 G. Then g0 RG (x) for every scalar if and only if G(x - g0 ). Proof. Let g0 RG (x), then d(g0 , g) d(x, g) for every g G i.e. d(0, g - g0 ) d(x - g0 , g - g0 ) for every g G. This gives d(0, g ) d(x - g0 , g ) for every g G i.e. d(g , 0) d(g , x - g0 ) for every scalar . Hence G(x - g0 ). Conversely, assume G(x - g0 ) i.e. g (x - g0 ) for every g G. This implies that d(g , 0) d(g , (x - g0 )) for every scalar and for every g G. This gives d(g0 , g + g0 ) d(x, g + g0 ) for every g G i.e. d(g0 , g) d(x, g) for every g G. Hence g0 RG (x). Before proving the next theorem, we prove the following lemmas. Lemma 1. Let G be a closed linear subspace of a metric linear space (X, d). If x G is such that x has a best coapproximation in G for every scalar / , then every element of the subspace {x, G} (the subspace generated by {x} G) has a best coapproximation in G. Proof. Let x + g {x, G} and g0 RG (x) i.e. d(g0 , g) d(x, g) f or every g G. This implies d(g0 + g , g + g ) d(x + g , g + g )) for every g G i.e. d(g0 + g , g ) d(x + g , g ) for every g G and so g0 + g RG (x + g ). Hence every element of the subspace {x, G} has a best coapproximation in G. Lemma 2. If G and H are two subspaces of a metric linear space (X, d) such that G H. If x H has a best coapproximation in H and if every / element of H has a best coapproximation in G, then x has a best coapproximation in G. Proof. Let x H be such that h0 RH (x), i.e. d(h0 , h) d(x, h) for every / h H. Now, for h0 H, let g0 G be such that d(g0 , g) d(h0 , g) for every g G. Then d(g0 , g) d(x, g) for every g G, as G H. Hence g0 RG (x). Theorem 3. If for every subspace G of a metric linear space (X, d), there exists at least one element x X\G such that x has a best coapproximation in G for every scalar , then for any subspace G of X every element of X has a best coapproximation in G. Proof. Using Lemmas 1 and 2 and proceeding as in Theorem 4.1 [5], we obtain the result. For normed linear spaces, the following result was proved in [3]. Theorem 4. Let G be a proximinal subspace of a metric linear space (X, d). -1 If PG (0) is a convex set, then G is Chebyshev. Proof. Suppose x X and g1 , g2 PG (x). Since g1 , g2 PG (x), we -1 have x - g1 , x - g2 PG (0). Put x - g1 = g and x - g2 = g , where -1 -1 g , g PG (0). We first claim that g1 - x PG (0). Since 0 PG (x - g1 ), we have d(x - g1 , 0) d(x - g1 , g) f or every g G. This implies d(g1 - x, 0) d(-g, g1 - x) for every g G i.e. d(g1 - x, 0) d(g1 - x, g ) for -1 every g G. Therefore, g1 - x PG (0). This proves our claim. Now, -1 -1 x-g2 , g1 -x PG (0) and PG (0) is convex, we have 1 [(x-g2 )+(g1 -x)] 2 -1 -1 PG (0) i.e. 1 [g - g ] PG (0). Also 1 [g - g ] = 1 [g1 - g2 ] G and 2 2 2 -1 so 1 [g1 - g2 ] PG (0) G = {0}. This implies g1 = g2 . Hence G is 2 Chebyshev. Remark. If we take G to be a proximinal subset of a metric linear space -1 (X, d), then the convexity of PG (0) need not imply the Chebyshevity of G. Example 1. Let X = R and G = {0, 1, 2, 3, ..., 10} then G, being a compact -1 set is proximinal (see [1]) and PG (0) = (-, 0.5] is a convex set but G is not Chebyshev as PG (0.5) = {0, 1}. Analogously, concerning the co-Chebyshevity of G, we have the following result: Theorem 5. Let G be a co-proximinal subspace of a metric linear space -1 (X, d). If RG (0) is a convex set, then G is co-Chebyshev. Proof. The proof runs on similar lines as that of Theorem 4. Remark. If we take G to be a co-proximinal subset of a metric linear space -1 (X, d), then the convexity of RG (0) need not imply the co-Chebyshevity of G. -1 Example 2. Let X = R and G = [0, ), then RG (0) = (-, 0] and -1 RG (-1) = [0, 1] i.e. RG (0) is a convex set but G is not co-Chebyshev. Before proceeding further, we recall the following results on coapproximation proved in [8]. Lemma 3 ([8] Theorems 5 and 6). Let G be a closed linear subspace of a metric linear space (X, d), then the following statements are equivalent: (i) G is co-proximinal. -1 (ii) X = G + RG (0). (iii) For the canonical mapping wG : X X/G defined by wG (x) = x + G, we have -1 wG (RG (0)) = X/G. Lemma 4 ([8], Theorem 7). For a closed linear subspace G of a metric linear space (X, d), the following statements are equivalent: (i) G is co-Chebyshev. -1 (ii) X = G RG (0), where means that the sum decomposition of each x X is unique. -1 -1 (iii) G is co-proximinal and [RG (0) - RG (0)] G = {0}. -1 (iv) G is co-proximinal and the restriction map wG (RG (0)) is one to one. Remark. For best approximation, Lemmas 3 and 4 were proved in [7]. -1 Concerning the proximinality of RG (0), we have Theorem 6. Let G be a co-proximinal subspace of a metric linear space -1 -1 (X, d). If RG (0) is a subspace of X then RG (0) is proximinal in X. -1 Proof. Since RG (0) is linear subspace of the metric linear space (X, d), G is -1 co-Chebyshev in X, by Theorem 5. Therefore, X = GRG (0) by Lemma 4. -1 -1 Let x X\RG (0) be arbitrary then x = g1 + g2 , g1 G, g2 RG (0). -1 Consider d(x, g2 ) = d(x - g2 , 0) = d(g1 , 0) = d(g1 , RG (0)). This gives -1 -1 -1 d(x, g2 ) = d(x - g2 , RG (0)) = d(x, RG (0)). Hence RG (0) is proximinal in X. Before proving the next theorem, we prove the following lemma. Lemma 5. Let H be a co-proximinal linear subspace of a metric linear space (X, d), then there exists an element z X\{0} such that 0 RH (z). Proof. Let x X\H. Since H is co-proximinal, there exists y0 RH (x) -1 and so x - y0 RH (0). Hence 0 RH (x - y0 ), x - y0 = 0. Theorem 7. Let G be a co-proximinal linear subspace of a metric linear space (X, d), then G is closed and in every linear subspace Fx X(x X\G) of form Fx = G [x] there exists an element z Fx \{0} such that 0 RG (z). Proof. Since G is co-proximinal, G is closed and G is co-proximinal in every subspace Fx X(x X\G) of form Fx = G [x]. Then by using the above lemma, there exists z Fx \{0} such that 0 RG (z). Note. A similar result for proximinality was proved in [11] for normed linear spaces. For the metric coprojection RG : X 2G , the graph of RG is denoted by G(RG ) i.e. G(RG ) = {(x, RG (x)) : x X}. Concerning the graph of RG , we have the following theorem (a similar result for the metric projection PG was proved in [6]). Theorem 8. If G is a co-Chebyshev subset of a metric space (X, d) then the graph of the metric coprojection RG is closed. Proof. Let (y, z) be a limit point of G(RG ) = {(x, RG (x)) : x X}. Then there exists a sequence (yn , RG (yn )) in G(RG ) such that (yn , RG (yn )) (y, z) i.e. yn y, RG (yn ) z. Since d(RG (yn ), g) d(yn , g) for every g G, we get d(z, g) d(y, g) for every g G and so z RG (y). Since G is co-Chebyshev, {z} = RG (y). Therefore (y, z) G(RG ) and hence G(RG ) is closed. Remarks. (1) A proximinal (co-proximinal) subset of a metric space is closed but the converse is not true. Example 3. Let X = R - {p}, then M = (p, p + 1] is a closed subset of the metric space X with usual metric but it is not proximinal. Example 4. The set of natural numbers is a closed subset of the real line R but it is not co-proximinal. (2) Whereas a compact subset of a metric space is proximinal (see [1]), it need not be co-proximinal. Example 5. Let X = R2 and M = {(x, y) R2 : x2 + y 2 = 1}, then M is a compact subset of R2 and hence proximinal. However, M is not co-proximinal as (0, 0) R2 does not have any best coapproximation in M . This example also shows that a proximinal subset of a metric space need not be co-proximinal. (3) A co-proximinal subset of a metric space need not be proximinal. Example 6. Let X = R - {1} and M = (1, 2], then M a is co-proximinal subset of X but it is not proximinal. (4) A Chebyshev subset of a metric space need not be co-Chebyshev. Example 7. Let X = R and G = [1, 2], then G is Chebyshev in the real line R but it is not co-Chebyshev. (5) A co-Chebyshev subset of a metric space need not be Chebyshev. Example 8. Let X = R2 with the metric d((x1 , y1 ), (x2 , y2 )) = |x1 - x2 | + |y1 - y2 | and G = {(x, y) R2 : x = y}. Then, X is a real Banach space and G is a proximinal subspace of X. We have PG (x, y) = {(x, x) + (1 - )(y, y) : 0 1} and RG (x, y) = {( x+y , x+y )}. Hence G is 2 2 co-Chebyshev but not Chebyshev. (6) In a Hilbert space H, it is known (see [2]) that if M is a subspace of H, then the set of best approximations and set of best coapproximations are the same. But if we take M to be a subset of H, then this need not be so. Example 9. Let X = R2 and M = {(x, y) R2 : x2 + y 2 = 1}, then every element of M is best approximation to (0, 0), but (0, 0) does not have any best coapproximation in M . (7) It is known (see [6]) that if G is a subset of a convex metric space (X, d) -1 -1 and x PG (g0 ) = {x0 X : d(x0 , g0 ) = d(x0 , G)}, then z PG (g0 ) for every z between x and g0 . But such a result is not true in case of best coapproximation. -1 / Example 10. Let X = R and G = (-1, 1) then -2 RG (0), but -1 -1 RG (0). (8) It is known (see [6]) that if G is a Chebyshev subset of a convex metric space (X, d), then PG (z) = PG (x), where z X is any element between x and PG (x). Does such a result hold for best coapproximation? Acknowledgement. The authors are thankful to the learned referee for valuable comments and suggestions leading to an improvement of the paper. http://www.deepdyve.com/assets/images/DeepDyve-Logo-lg.png Annales UMCS, Mathematica de Gruyter

# Proximinality and co-proximinality in metric linear spaces

, Volume 69 (1) – Jun 1, 2015
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### Abstract

spaces. The present paper is also a step in this direction. In this paper, we discuss the existence and uniqueness results on best approximation and best coapproximation in metric linear spaces thereby generalizing the various known results. We start with a few definitions. The research work of the author has been supported by U.G.C., India under Emeritus Fellowship. 2 The research work of the author has been supported by U.G.C., India under Senior Research Fellowship. 2010 Mathematics Subject Classification. 41A50, 41A52, 41A65, 35E10. Key words and phrases. Best approximation, best coapproximation, proximinal set, co-proximinal set, Chebyshev set, co-Chebyshev set. Let G be a non-empty subset of a metric space (X, d). An element g0 G is called a best approximation (best coapproximation) to x X if d(x, g0 ) d(x, g) (d(g0 , g) d(x, g)) for all g G. The set of all such g0 G is denoted by PG (x)(RG (x)). The set G is called proximinal (co-proximinal) if PG (x) (RG (x)) contains at least one element for every x X. If for each x X, PG (x)(RG (x)) has exactly one element in G, then the set G is called Chebyshev (coChebyshev). If x, y and z are any three points in a metric space (X, d) then z is said to be a between point of x and y if d(x, z) + d(z, y) = d(x, y). A metric d defined on X is said to be a convex (strongly convex) metric if for each pair x, y X, d determine at least one (exactly one) between point. The space X together with a convex metric d is called a convex metric space. A linear space X with a translation invariant metric d (i.e. d(x+z, y+z) = d(x, y) for all x, y, z X) such that addition and scalar multiplication are continuous on (X, d) is called a metric linear space. The space X of all entire functions, i.e. X= f : f (z) = n=0 an z n , |an | n 0 as n with the metric d defined by d(f, g) = max{|a0 - b0 |, |an - bn | n , n 1}, n n where f (z) = n=0 an z , g(z) = n=0 bn z , is a non-normable metric linear space (see [10], p. 238). Remarks. (i) A proximinal (co-proximinal) subset of a metric space is closed. (ii) Every singleton subset of a metric space is Chebyshev (co-Chebyshev). (iii) Every closed interval in R is proximinal (co-proximinal). (iv) PG (PG (x)) = PG (x) and RG (RG (x)) = RG (x). (v) PG (x) = {g0 G : d(x, g0 ) d(x, g) for every g G} = G B(x, d(x, G)). (vi) RG (x) = {g0 G : d(g0 , g) d(x, g) for every g G} = G [ {B(g, d(x, g)} : g G}]. -1 (vii) If G is a linear subspace of a metric linear space (X, d) then PG (0) -1 -1 G = {0} and RG (0) G = {0}, where PG (0) = {x X : 0 PG (x)} and -1 RG (0) = {x X : 0 RG (x)}. -1 (viii) If G is a linear subspace of a metric linear space (X, d), then d(g, RG (0)) = d(g, 0) for every g G. (ix) If G is a linear subspace of a metric linear space (X, d), then g0 -1 -1 PG (x) (RG (x)) if and only if x - g0 PG (0) (RG (0)). Let (X, d) be a metric linear space and x, y X, then we say that x is orthogonal to y, x y if d(x, 0) d(x, y) for every scalar . We say that G x if g x for every g G. Concerning orthogonality and best coapproximation, we have Theorem 1. Let G be a linear subspace of a metric linear space (X, d) such that G(x - g0 ), then g0 RG (x). Proof. Since G(x - g0 ), we have g(x - g0 ) for every g G i.e. d(g, 0) d(g, (x - g0 )) for every scalar . Take = 1, we get d(g, 0) d(g, x - g0 ) for every g G. This gives d(g0 , g + g0 ) d(x, g + g0 ) for every g G. Hence g0 RG (x). Note. The converse of the above theorem is also true in normed linear spaces (see [2]). We do not know whether this is true in metric linear spaces. However, in metric linear spaces, we have Theorem 2. Let G be a linear subspace of a metric linear space (X, d) and g0 G. Then g0 RG (x) for every scalar if and only if G(x - g0 ). Proof. Let g0 RG (x), then d(g0 , g) d(x, g) for every g G i.e. d(0, g - g0 ) d(x - g0 , g - g0 ) for every g G. This gives d(0, g ) d(x - g0 , g ) for every g G i.e. d(g , 0) d(g , x - g0 ) for every scalar . Hence G(x - g0 ). Conversely, assume G(x - g0 ) i.e. g (x - g0 ) for every g G. This implies that d(g , 0) d(g , (x - g0 )) for every scalar and for every g G. This gives d(g0 , g + g0 ) d(x, g + g0 ) for every g G i.e. d(g0 , g) d(x, g) for every g G. Hence g0 RG (x). Before proving the next theorem, we prove the following lemmas. Lemma 1. Let G be a closed linear subspace of a metric linear space (X, d). If x G is such that x has a best coapproximation in G for every scalar / , then every element of the subspace {x, G} (the subspace generated by {x} G) has a best coapproximation in G. Proof. Let x + g {x, G} and g0 RG (x) i.e. d(g0 , g) d(x, g) f or every g G. This implies d(g0 + g , g + g ) d(x + g , g + g )) for every g G i.e. d(g0 + g , g ) d(x + g , g ) for every g G and so g0 + g RG (x + g ). Hence every element of the subspace {x, G} has a best coapproximation in G. Lemma 2. If G and H are two subspaces of a metric linear space (X, d) such that G H. If x H has a best coapproximation in H and if every / element of H has a best coapproximation in G, then x has a best coapproximation in G. Proof. Let x H be such that h0 RH (x), i.e. d(h0 , h) d(x, h) for every / h H. Now, for h0 H, let g0 G be such that d(g0 , g) d(h0 , g) for every g G. Then d(g0 , g) d(x, g) for every g G, as G H. Hence g0 RG (x). Theorem 3. If for every subspace G of a metric linear space (X, d), there exists at least one element x X\G such that x has a best coapproximation in G for every scalar , then for any subspace G of X every element of X has a best coapproximation in G. Proof. Using Lemmas 1 and 2 and proceeding as in Theorem 4.1 [5], we obtain the result. For normed linear spaces, the following result was proved in [3]. Theorem 4. Let G be a proximinal subspace of a metric linear space (X, d). -1 If PG (0) is a convex set, then G is Chebyshev. Proof. Suppose x X and g1 , g2 PG (x). Since g1 , g2 PG (x), we -1 have x - g1 , x - g2 PG (0). Put x - g1 = g and x - g2 = g , where -1 -1 g , g PG (0). We first claim that g1 - x PG (0). Since 0 PG (x - g1 ), we have d(x - g1 , 0) d(x - g1 , g) f or every g G. This implies d(g1 - x, 0) d(-g, g1 - x) for every g G i.e. d(g1 - x, 0) d(g1 - x, g ) for -1 every g G. Therefore, g1 - x PG (0). This proves our claim. Now, -1 -1 x-g2 , g1 -x PG (0) and PG (0) is convex, we have 1 [(x-g2 )+(g1 -x)] 2 -1 -1 PG (0) i.e. 1 [g - g ] PG (0). Also 1 [g - g ] = 1 [g1 - g2 ] G and 2 2 2 -1 so 1 [g1 - g2 ] PG (0) G = {0}. This implies g1 = g2 . Hence G is 2 Chebyshev. Remark. If we take G to be a proximinal subset of a metric linear space -1 (X, d), then the convexity of PG (0) need not imply the Chebyshevity of G. Example 1. Let X = R and G = {0, 1, 2, 3, ..., 10} then G, being a compact -1 set is proximinal (see [1]) and PG (0) = (-, 0.5] is a convex set but G is not Chebyshev as PG (0.5) = {0, 1}. Analogously, concerning the co-Chebyshevity of G, we have the following result: Theorem 5. Let G be a co-proximinal subspace of a metric linear space -1 (X, d). If RG (0) is a convex set, then G is co-Chebyshev. Proof. The proof runs on similar lines as that of Theorem 4. Remark. If we take G to be a co-proximinal subset of a metric linear space -1 (X, d), then the convexity of RG (0) need not imply the co-Chebyshevity of G. -1 Example 2. Let X = R and G = [0, ), then RG (0) = (-, 0] and -1 RG (-1) = [0, 1] i.e. RG (0) is a convex set but G is not co-Chebyshev. Before proceeding further, we recall the following results on coapproximation proved in [8]. Lemma 3 ([8] Theorems 5 and 6). Let G be a closed linear subspace of a metric linear space (X, d), then the following statements are equivalent: (i) G is co-proximinal. -1 (ii) X = G + RG (0). (iii) For the canonical mapping wG : X X/G defined by wG (x) = x + G, we have -1 wG (RG (0)) = X/G. Lemma 4 ([8], Theorem 7). For a closed linear subspace G of a metric linear space (X, d), the following statements are equivalent: (i) G is co-Chebyshev. -1 (ii) X = G RG (0), where means that the sum decomposition of each x X is unique. -1 -1 (iii) G is co-proximinal and [RG (0) - RG (0)] G = {0}. -1 (iv) G is co-proximinal and the restriction map wG (RG (0)) is one to one. Remark. For best approximation, Lemmas 3 and 4 were proved in [7]. -1 Concerning the proximinality of RG (0), we have Theorem 6. Let G be a co-proximinal subspace of a metric linear space -1 -1 (X, d). If RG (0) is a subspace of X then RG (0) is proximinal in X. -1 Proof. Since RG (0) is linear subspace of the metric linear space (X, d), G is -1 co-Chebyshev in X, by Theorem 5. Therefore, X = GRG (0) by Lemma 4. -1 -1 Let x X\RG (0) be arbitrary then x = g1 + g2 , g1 G, g2 RG (0). -1 Consider d(x, g2 ) = d(x - g2 , 0) = d(g1 , 0) = d(g1 , RG (0)). This gives -1 -1 -1 d(x, g2 ) = d(x - g2 , RG (0)) = d(x, RG (0)). Hence RG (0) is proximinal in X. Before proving the next theorem, we prove the following lemma. Lemma 5. Let H be a co-proximinal linear subspace of a metric linear space (X, d), then there exists an element z X\{0} such that 0 RH (z). Proof. Let x X\H. Since H is co-proximinal, there exists y0 RH (x) -1 and so x - y0 RH (0). Hence 0 RH (x - y0 ), x - y0 = 0. Theorem 7. Let G be a co-proximinal linear subspace of a metric linear space (X, d), then G is closed and in every linear subspace Fx X(x X\G) of form Fx = G [x] there exists an element z Fx \{0} such that 0 RG (z). Proof. Since G is co-proximinal, G is closed and G is co-proximinal in every subspace Fx X(x X\G) of form Fx = G [x]. Then by using the above lemma, there exists z Fx \{0} such that 0 RG (z). Note. A similar result for proximinality was proved in [11] for normed linear spaces. For the metric coprojection RG : X 2G , the graph of RG is denoted by G(RG ) i.e. G(RG ) = {(x, RG (x)) : x X}. Concerning the graph of RG , we have the following theorem (a similar result for the metric projection PG was proved in [6]). Theorem 8. If G is a co-Chebyshev subset of a metric space (X, d) then the graph of the metric coprojection RG is closed. Proof. Let (y, z) be a limit point of G(RG ) = {(x, RG (x)) : x X}. Then there exists a sequence (yn , RG (yn )) in G(RG ) such that (yn , RG (yn )) (y, z) i.e. yn y, RG (yn ) z. Since d(RG (yn ), g) d(yn , g) for every g G, we get d(z, g) d(y, g) for every g G and so z RG (y). Since G is co-Chebyshev, {z} = RG (y). Therefore (y, z) G(RG ) and hence G(RG ) is closed. Remarks. (1) A proximinal (co-proximinal) subset of a metric space is closed but the converse is not true. Example 3. Let X = R - {p}, then M = (p, p + 1] is a closed subset of the metric space X with usual metric but it is not proximinal. Example 4. The set of natural numbers is a closed subset of the real line R but it is not co-proximinal. (2) Whereas a compact subset of a metric space is proximinal (see [1]), it need not be co-proximinal. Example 5. Let X = R2 and M = {(x, y) R2 : x2 + y 2 = 1}, then M is a compact subset of R2 and hence proximinal. However, M is not co-proximinal as (0, 0) R2 does not have any best coapproximation in M . This example also shows that a proximinal subset of a metric space need not be co-proximinal. (3) A co-proximinal subset of a metric space need not be proximinal. Example 6. Let X = R - {1} and M = (1, 2], then M a is co-proximinal subset of X but it is not proximinal. (4) A Chebyshev subset of a metric space need not be co-Chebyshev. Example 7. Let X = R and G = [1, 2], then G is Chebyshev in the real line R but it is not co-Chebyshev. (5) A co-Chebyshev subset of a metric space need not be Chebyshev. Example 8. Let X = R2 with the metric d((x1 , y1 ), (x2 , y2 )) = |x1 - x2 | + |y1 - y2 | and G = {(x, y) R2 : x = y}. Then, X is a real Banach space and G is a proximinal subspace of X. We have PG (x, y) = {(x, x) + (1 - )(y, y) : 0 1} and RG (x, y) = {( x+y , x+y )}. Hence G is 2 2 co-Chebyshev but not Chebyshev. (6) In a Hilbert space H, it is known (see [2]) that if M is a subspace of H, then the set of best approximations and set of best coapproximations are the same. But if we take M to be a subset of H, then this need not be so. Example 9. Let X = R2 and M = {(x, y) R2 : x2 + y 2 = 1}, then every element of M is best approximation to (0, 0), but (0, 0) does not have any best coapproximation in M . (7) It is known (see [6]) that if G is a subset of a convex metric space (X, d) -1 -1 and x PG (g0 ) = {x0 X : d(x0 , g0 ) = d(x0 , G)}, then z PG (g0 ) for every z between x and g0 . But such a result is not true in case of best coapproximation. -1 / Example 10. Let X = R and G = (-1, 1) then -2 RG (0), but -1 -1 RG (0). (8) It is known (see [6]) that if G is a Chebyshev subset of a convex metric space (X, d), then PG (z) = PG (x), where z X is any element between x and PG (x). Does such a result hold for best coapproximation? Acknowledgement. The authors are thankful to the learned referee for valuable comments and suggestions leading to an improvement of the paper.

### Journal

Annales UMCS, Mathematicade Gruyter

Published: Jun 1, 2015

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