Differentiability of Distance Functions in $p$-Normed ...

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Sep 4, 2009 - ABSTRACT. The farthest point mapping in a p-normed space X is studied in virtue of the. Gateaux derivative and the Frechet derivative.
The Australian Journal of Mathematical Analysis and Applications AJMAA Volume 6, Issue 1, Article 10, pp. 1-10, 2009

DIFFERENTIABILITY OF DISTANCE FUNCTIONS IN p-NORMED SPACES M.S. MOSLEHIAN, A. NIKNAM AND S. SHADKAM TORBATI

Special Issue in Honor of the 100th Anniversary of S.M. Ulam Received 20 June, 2008; accepted 20 July, 2008; published 4 September, 2009. D EPARTMENT OF P URE M ATHEMATICS C ENTRE OF E XCELLENCE IN A NALYSIS ON A LGEBRAIC S TRUCTURES (CEAAS) F ERDOWSI U NIVERSITY OF M ASHHAD P. O. B OX 1159, M ASHHAD , I RAN [email protected] [email protected] [email protected]

A BSTRACT. The farthest point mapping in a p-normed space X is studied in virtue of the Gateaux derivative and the Frechet derivative. Let M be a closed bounded subset of X having the uniformly p-Gateaux differentiable norm. Under certain conditions, it is shown that every maximizing sequence is convergent, moreover, if M is a uniquely remotal set then the farthest point mapping is continuous and so M is singleton. In addition, a Hahn–Banach type theorem in p-normed spaces is proved.

Key words and phrases: Frechet derivative, quasi-norm, p-normed space, farthest mapping, Hahn-Banach theorem, remotal set. 2000 Mathematics Subject Classification. Primary 41A52. secondary 41A65, 58C20.

ISSN (electronic): 1449-5910 c 2009 Austral Internet Publishing. All rights reserved.

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M.S. M OSLEHIAN , A. N IKNAM , S. S HADKAM T ORBATI

1. I NTRODUCTION Let X be a real linear space. A quasi-norm is a real-valued function on X satisfying the following conditions: (i) kxk ≥ 0 for all x ∈ X, and kxk = 0 if and only if x = 0. (ii) kλxk = |λ| · kxk for all λ ∈ R and all x ∈ X. (iii) There is a constant K ≥ 1 such that kx + yk ≤ K(kxk + kyk) for all x, y ∈ X. The pair (X, k·k) is called a quasi-normed space if k·k is a quasi-norm on X. The smallest possible K is called the module of concavity of k·k . By a quasi-Banach space we mean a complete quasi-normed space, i.e., a quasi-normed space in which every Cauchy sequence converges in X. This class includes Banach spaces. The most significant class of quasi-Banach spaces, which are not Banach spaces are Lp -spaces for 0 < p < 1 equipped with the Lp -norms k·kp . A quasi-norm k·k is called a p-norm (0 < p < 1) if kx + ykp ≤ kxkp + kykp for all x, y ∈ X. In this case a quasi-normed (quasi Banach) space is called a p-normed (p-Banach) space. By the Aoki–Rolewicz theorem [7] each quasi-norm is equivalent to some p-norm. Since it is much easier to work with p-norms than quasi-norms, henceforth we restrict our attention mainly to p-norms. See [1, 4, 8] and the references therein for more information. Throughout the paper, S(X) and B(X) denote the unit sphere and the unit ball of X, respectivelty. If x∗ is in X ∗ , the dual of X, and x ∈ X we write x∗ (x) as hx∗ , xi. We also consider quasi-norms with K > 1. The case where K = 1 turns out to be the classical normed spaces, so we will not discuss it and refer the interested reader to [2]. Let k > 0. A real valued function f on X is said to be k-Gateaux differentiable at a point x of X if there is an element df (x) of X ∗ such that for each y in X, (1.1)

lim sgn(t)|t|−k (f (x + ty) − f (x)) = hdf (x), yi, t→0

and we call df (x) the k-Gateaux derivative of f at x. We say that f is k-Frechet differentiable at a point x of X if there is an element f 0 (x) of X ∗ such that (1.2)

lim kyk−k (f (x + y) − f (x) − hf 0 (x), yi) = 0,

kyk→0

and we call f 0 (x) the k -Frechet derivative of f at x. See [2]. The norm of X is called uniformly k-Gateaux differentiable if for f (x) = kxkk equality (1.1) holds uniformly for x, y ∈ X, and the operator norm kdf (x)k is less than or equal 1. Then df (x) is denoted by Dk (x). We say that a non-zero element x∗ of X ∗ strongly exposes B(X) at x ∈ S(X), provided hx∗ , xi = kx∗ k, and a sequence {yn } in B(X) converges to x whenever {hx∗ , yn i} converges to hx∗ , xi. The reader is referred to [2, 5, 6] for analogue results concerning normed spaces and to the book of D.H. Hyers, G. Isac and Th.M. Rassias [3] for extensive theory and applications of nonlinear analysis methods. In this paper we use some ideas of [2] to study the farthest point mapping in a p-normed space X by virtue of the Gateaux derivative and the Frechet derivative. Let M be a closed bounded subset of X which has a uniformly p-Gateaux differentiable norm. Under certain conditions, we show that every maximizing sequence is convergent, moreover, if M is a uniquely remotal set then the farthest point mapping is continuous and so M is a singleton. In addition, we prove a Hahn–Banach type theorem for p-normed spaces and give an application.

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D IFFERENTIABILITY OF D ISTANCE F UNCTIONS

2. M AIN R ESULTS We start this section with following definition. Definition 2.1. Let X be a p-normed space and M be a bounded closed subset of X. Setting Ψ(x) := sup{kx − ykp : y ∈ M }

(x ∈ X),

F (x) := {y ∈ M : kx−ykp = Ψ(x)} is called the set of farthest points in M form x (CHECK). We say that M is a uniquely remotal set if F is a singleton for each x ∈ X and then we denote the single element of F (x) by F (x). The map F is called the antiprojection or the farthest point mapping for M . A center of a bounded set M in a p-normed space X is an element c in X such that r(M ) := sup kc − ykp = inf sup kx − yk. x∈X y∈M

y∈M

In fact r(M ), the so called Chebyshev radius of M , is the smallest ball in X containing M . We call a sequence {yn } in M a maximizing sequence for x provided kx − yn kp → Ψ(x) as n → ∞. Lemma 2.1. If M is a nonempty bounded subset of X and x, y ∈ X then |Ψ(x) − Ψ(y)| ≤ kx − ykp .

(2.1) Proof.

Ψ(x) − Ψ(y) = sup{kx − mkp : m ∈ M } − sup{ky − mkp : m ∈ M } ≤ sup{kx − m − y + mkp : m ∈ M } = kx − ykp .

Lemma 2.2. If x ∈ X is a point of k-Gateaux differentiability of Ψ and y ∈ F (x) then ( 0, k < 1; hdΨ(x), kx − yk−1 (x − y)i ≤ pkx − ykp−k , k = 1. Proof. Since y ∈ F (x) we have kx − ykp = Ψ(x). Hence for 0 < t < 1, we get (2.2)

Ψ(x + t(x − y)) − Ψ(x) ≥ kx + t(x − y) − ykp − kx − ykp = (1 + t)p kx − ykp − kx − ykp = ((1 + t)p − 1)kx − ykp .

Let s ∈ [0, ∞). Then (2.3)

hdΨ(x), szi = lim+ t−k (Ψ(x + tsz) − Ψ(x)) t→0 k −k

=s s

lim t−k (Ψ(x + tsz) − Ψ(x))

t→0+ −k −k

= sk lim+ (s t )(Ψ(x + tsz) − Ψ(x)) t→0

k

= s hdΨ(x), zi,

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whence hdΨ(x), kx − yk−1 (x − y)i = lim sgn(t)|t|−k (kx − yk−k )(Ψ(x + t(x − y)) − Ψ(x)) t→0  p  ≥ limt→0+ (|1+t|tk −1) kx − ykp−k , t > 0; =  ≤ lim − (|1+t|p −1) kx − ykp−k , t < 0. t→0 −(−t)k Then ( hdΨ(x), kx − yk−1 (x − y)i =

0,

k < 1;

pkx − ykp−k , k = 1.

Theorem 2.3. Suppose that M is a closed bounded subset of X with a uniformly p-Gateaux differentiable norm, Ψ is p-Gateaux differentiable at a point of x ∈ X\M with y ∈ F (x) and dΨ(x) strongly exposes B(X) at kx − yk−1 (x − y). Then every maximizing sequence for x converges to y. Moreover, if M is a uniquely remotal set then F is continuous at x. Proof. Suppose that {yn } is a maximizing sequence for x. For each t, we have Ψ(x + tz) − Ψ(x) ≥ sup kx + tz − yn kp − lim kx − yn kP n→∞ yn ∈M  ≥ lim sup kx + tz − yn kp − kx − yn kp .

(2.4)

n→∞

Fix z ∈ X. We have (2.5)

hdΨ(x), zi = lim− sgn(t)|t|−p (Ψ(x + tz) − Ψ(x)) t→0

≤ lim− − (−t)−p lim sup kx + tz − yn kp − kx − yn kp t→0



n→∞ −p

 = lim− lim inf − (−t) (kx + tz − yn kp − kx − yn kp ) n→∞

t→0

≤ lim inf hDp (x − yn ), zi n→∞

and hdΨ(x), zi = lim+ sgn(t)|t|−p (Ψ(x + tz) − Ψ(x))

(2.6)

t→0

≥ lim+ t−p lim sup kx + tz − yn kp − kx − yn kp t→0



n→∞

= lim+ lim sup t−p (kx + tz − yn kp − kx − yn kp ) t→0



n→∞

≥ lim suphDp (x − yn ), zi. n→∞

Using (2.5) and (2.6) we get lim hDp (x − yn ), zi = hdΨ(x), zi.

(2.7)

n→∞

We also have (2.8)

hDp (x − yn ), x − yn i = lim+ t−p (kx − yn + t(x − yn )kp − kx − yn kp ) t→0  = lim+ t−p k(x − yn )kp ((1 + t)p − 1) .



t→0

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Using equalities (2.7) for z = x − yn , (2.3) and (2.8) we obtain hdΨ(x), kx − yn k−1 (x − yn )i = lim hDp (x − yn ), kx − yn k−1 (x − yn )i n→∞  = lim lim+ t−p ((1 + t)p − 1) = 0 n→∞ t→0

= hdΨ(x), kx − yk−1 (x − y)i. Since dΨ(x) strongly exposes B(X) at kx − yk−1 (x − y), we get lim kx − yn k−1 (x − yn ) = kx − yk−1 (x − y).

n→∞

Since {yn } is a maximizing sequence for x, we therefore easily get lim yn = y.

(2.9)

n→∞

Next, assume that {xn } converges to x. We shall show that {F (xn )} converges to F (x). Since M is a uniquely remotal set we have kxn −F (xn )kp = sup{kxn −ykp ; y ∈ M }. Hence (2.10)

kxn − F (x)kp ≤ kxn − F (xn )kp .

By (2.10), (2.11)

kx − F (x)kp − kx − F (xn )kp ≤ kx − F (x)kp + kx − xn kp − kxn − F (xn )kp ≤ kx − F (x)kp + kx − xn kp − kxn − F (x)kp ≤ kx − xn kp + kx − F (x) − xn + F (x)kp ≤ 2kx − xn kp .

It follows from (2.11) and the convergence of {xn } to x that lim kx − F (xn )kp = kx − F (x)kp .

n→∞

Hence {F (xn )} is a maximizing sequence for x and so, by the first part of the theorem lim F (xn ) = F (x).

n→∞

Theorem 2.4. Suppose M is a closed bounded subset of a p-normed space X and x ∈ X\M is a point of p-Frechet differentiability of Ψ and if y ∈ F (x) and Ψ0 (x) strongly exposes B(X) at kx − yk−1 (x − y), then every maximizing sequence for x converges to y. Proof. Let {yn } be a maximizing sequence for x and let k = inf n kx−yn k2p . Since M is closed, k > 0. Hence there is N such that for n ≥ N then Ψ(x) − kx − yn kp < k ≤ kx − yn k2p , and so we can choose a sequence {αn } of positive numbers such that αn → 0 and kx − yn k2p > α2n > Ψ(x) − kx − yn kp

(n ∈ N).

Hence (2.12)

(1 + t)p kx − yn kp > (1 + t)p (Ψ(x) − α2n )

(n ∈ N, −1 < t < 1).

Let 0 < t < 1. Since yn ∈ M we have (2.13)

Ψ(x − t(yn − x)) ≥ kx − t(yn − x) − yn kp = (1 + t)p kx − yn kp .

It follows from (2.12) and (2.13) that (2.14)

Ψ(x − t(yn − x) − Ψ(x) ≥ (1 + t)p kx − yn kp − Ψ(x) > Ψ(x)((1 + t)p − 1) − (1 + t)p α2n .

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Fix ε > 0. By the definition of Ψ0 (x), there is δ > 0 such that if kykp < δ then |Ψ(x + y) − Ψ(x) − hΨ0 (x), yi| ≤ εkykp .

(2.15)

Let tpn = αn (kx − yn k)−p < 1 and αn < δ for large n. Replacing y by tn (x − yn ) in (2.15) and noting (2.14), we get εktn (x − yn )kp + hΨ0 (x), tn (x − yn )i ≥ Ψ(x + tn (x − yn )) − Ψ(x) ≥ Ψ(x)((1 + tn )p − 1) − (1 + tn )p α2n , whence 1 hΨ0 (x), tn (x − yn )i αn  1 ≥ Ψ(x)((1 + tn )p − 1) − (1 + tn )p α2n − εktn (yn − x)kp αn 1 ≥ Ψ(x)((1 + tn )p − 1) − (1 + tn )p αn − ε αn 1 ≥ p Ψ(x)((1 + tn )p − 1) − (1 + tn )p αn − ε. p tn k(yn − x)k

(2.16)

Using (2.16), we get (2.17)   kx − yn k−1 0 tn (x − yn ) hΨ (x), kx − yn k (x − yn )i = Ψ (x), tn kx − yn k−p 0 = hΨ (x), tn (x − yn )i tpn 1 = hΨ0 (x), tn (x − yn )i αn 1 ≥ Ψ(x) p ((1 + tn )p − 1) − (1 + tn )p αn − ε. tn k(yn − x)kp 0

−1

Let n → ∞ in (2.17). Then αn → 0, kyn − xk → Ψ(x) and tn → 0+ such that (2.18)

lim hΨ0 (x), kx − yn k−1 (x − yn )i ≥ 0

n→∞

Changing t to −t in (2.13) and (2.14) and utilizing a similar strategy as above we get lim hΨ0 (x), kyn − xk−1 (yn − x)i ≥ 0,

n→∞

whence (2.19)

lim hΨ0 (x), kx − yn k−1 (x − yn )i ≤ 0.

n→∞

By (2.18), (2.19) we have, lim hΨ0 (x), kx − yn k−1 (x − yn )i = 0.

n→∞

By Lemma 2.2, hdΨ(x), kx − yk−1 (x − y)i = 0

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and then 0=

lim

kt(x−y)k→0

 kt(x − y)k−p Ψ(x + t(x − y))) − Ψ(x) − hΨ0 (x), t(x − y)i

 = lim+ kt(x − y)k−p (Ψ(x + (t(x − y))) − Ψ(x)) − kt(x − y)k−p (hΨ0 (x), t(x − y)i) t→0  −p  t (Ψ(x + (t(x − y)) − Ψ(x)) 0 −1 = lim+ − hΨ (x), kx − yk (x − y)i t→0 k(x − y)kp limt→0+ t−p (Ψ(x + (t(x − y)) − Ψ(x)) = − hΨ0 (x), kx − yk−1 (x − y)i k(x − y)kp = kx − yk−p hdΨ(x), (x − y)i − hΨ0 (x), kx − yk−1 (x − y)i = hdΨ(x), kx − yk−1 (x − y)i − hΨ0 (x), kx − yk−1 (x − y)i = hΨ0 (x), kx − yk−1 (x − y)i, therefore lim hΨ0 (x), kx − yn k−1 (x − yn )i = 0 = hΨ0 (x), kx − yk−1 (x − y)i.

n→∞

Since Ψ0 (x) strongly exposes B(X) at kx − yk−1 (x − y), we deduce that kx − yn k−1 (x − yn ) → kx − yk−1 (x − y) which yields yn → y. 3. H AHN –BANACH T HEOREM AND ITS A PPLICATION This section is devoted to one version of the celebrated Hahn–Banach theorem. We follow the strategy and terminology of [9]. We begin with a lemma. Lemma 3.1. Let 0 < p < 1, let M be a linear subspace of a real vector space X and let ρ : X → R be a mapping such that (i) ρ(x + y) ≤ ρ(x) + ρ(y) (x, y ∈ X); (ii) ρ(tx) = tp ρ(x) (x ∈ X, 0 ≤ t ∈ R). 1

If ϕ0 : M → R is a linear mapping such that ϕ0 (x) ≤ ρ p (x) for all x ∈ M , then there is a 1 linear mapping ϕ : X → R such that ϕ|M = φ0 and ϕ(x) ≤ ρ p (x) (x ∈ X). Proof. We use Zorn’s lemma. Consider the partially ordered set P, whose typical member is a pair (Y, ψ), where (i) Y is a linear subspace of X which contains X0 ; and (ii) ψ : Y → R 1 is a linear mapping which is an extension of ϕ0 and satisfies ψ(x) ≤ ρ p (x) ∀x ∈ Y ; the partial order on P is defined by setting (Y1 , ψ 1 ) ≤ (Y2 , ψ 2 ) precisely when (a) Y1 ⊂ Y2 , and (b) ψ 2 |Y1 = ψ 1 . Furthermore, if Γ = {(Yi , ψ i ) : i ∈ I} is any totally ordered set in P, an easy verification shows that an upper bound for the family Γ is given by (Y, ψ), where Y = ∪i∈I Yi and ψ : Y → R is the unique necessarily linear mapping satisfying ψ|Yi = ψ i for all i. Hence, by Zorn’s lemma, the partially ordered set P has a maximal element, call it (Y, ψ). The proof of the lemma will be completed once we have shown that Y = X. Suppose that Y 6= X; fix x0 ∈ X −Y , and let Y1 = Y +Rx0 = {y +tx0 : y ∈ Y, t ∈ R}. The definitions ensure that Y1 is a subspace of X which properly contains Y . Also, notice that any linear mapping ψ 1 : Y1 → R which extends ψ is prescribed uniquely by the number t0 = ψ(x0 ) (and the equation ψ 1 (y + tx0 ) = ψ(x) + tt0 ). We assert that it is possible to find a number t0 ∈ R such that the associated mapping ψ 1 1 would - in addition to extending ψ - also satisfy ψ 1 ≤ ρ p . This would then establish the inequality (Y, ψ) ≤ (Y1 , ψ 1 ), contradicting the maximality of (Y, ψ); which would in turn imply that we must have had Y = X in the first place, and the proof would be complete.

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First, observe that if y1 , y2 ∈ Y are arbitrary, then ψ(y1 ) + ψ(y2 ) = ψ(y1 + y2 ) = ψ(y1 − x0 + y2 + x0 ) = ψ(y1 − x0 ) + ψ(y2 + x0 ) 1

1

≤ ρ p (y1 − x0 ) + ρ p (y2 + x0 ) and consequently, (3.1)

h i h 1 i 1 p p sup ψ(y1 ) − ρ (y1 − x0 ) ≤ inf ρ (y2 + x0 ) − ψ(y2 ) . y2 ∈Y

y1 ∈Y

Let t0 be any real number which lies between the supremum and the infimum appearing in equation (3.1). We now verify that this t0 does the job. Indeed, if t > 0, and if y ∈ Y , then, since the definition of t0 ensures that 1

ψ(y2 ) + t0 ≤ ρ p (y2 + x0 ) ∀y2 ∈ Y, we find that ψ 1 (y + tx0 ) = ψ(y) + tt0 h y  i =t ψ + t0  yt  1 p ≤ tρ + x0 t 1 = ρ p (y + tx0 ). Similarly, if t < 0, then, since the definition of t0 also ensures that 1

ψ(y1 ) − t0 ≤ ρ p (y1 − x0 ) ∀y1 ∈ Y, we find that ψ 1 (y + tx0 ) = ψ(y) + tt0     y = −t ψ − t0 −t   1 y − x0 ≤ −tρ p −t 1

= ρ p (y + tx0 ). 1

Thus, ψ(y + tx0 ) ≤ ρ p (y + tx0 ) ∀y ∈ Y, t ∈ R, and the proof of the lemma is complete. Theorem 3.2. (Hahn–Banach theorem) Let V be a p-normed space and let V0 be a subspace of V . Suppose ϕ0 ∈ V0∗ ; then there exist a ϕ ∈ V ∗ such that (i)ϕ|V0 = ϕ0 (ii)kϕk = kϕ0 k. Proof. Let V0 be a (real) linear subspace of V . Now, apply the preceding lemma with X = 1 V, M = V0 and ρ p (x) = kϕ0 kkxk, to find that the desired conclusion follows immediately. Remark 3.1. One can follow the method of [9] to prove that the Hahn–Banach theorem holds for complex quasi-normed spaces. Corollary 3.3. Let X be a p-normed space and 0 6= x0 ∈ X. Then there exists ϕ ∈ X ∗ such that kϕk = 1 and ϕ(x0 ) = kx0 k.

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Proof. Set X0 = Cx0 = {αx0 : α ∈ C}, consider the linear functional ϕ0 ∈ X ∗ defined by ϕ0 (λx) = λkx0 k, and appeal to the Hahn–Banach theorem. Theorem 3.4. Let X be a p-normed space and let M be a uniquely remotal subset of X ad1 mitting a center. Suppose that for every x in M + r(M ) p B(X) the farthest point mapping F : X → M restricted to the line segment [x, F (x)] is continued at x. Then M is a singleton. Proof. We may assume that − 0kp : y ∈ M }, we  0 is a center of M . Sincer(M ) =1 sup{ky  1 1 have M ⊆ B 0, r(M ) p . For each x ∈ M, 0 ∈ B x, r(M ) p ⊆ M + r(M ) p B(X). Set x0 = F (0) and gn = F ( xn0 ). By the Hahn–Banach theorem for each gn − xn0 there exists a

functional Φn such that kΦn k = 1, and Φn , gn − xn0 = gn − xn0 . Then D D x0 E x0 E Φn , = hΦn , gn i − Φn , gn − n n

x0

≤ kΦn kkgn k − gn − n

x0

= kgn k − gn − , n whence

1 x0

Φn (x0 ) ≤ kgn k − gn − . n n Since kgn kp ≤ sup k0 − gn kp n

≤ sup k0 − ykp y∈M

p

x

o ≤ sup − z n z∈M

x

p  x  p x

0

0

o = −F

= − gn , n n n we get Φn (x0 ) ≤ 0 for all n. The functionF is continuous at Φn gn − xn0 → Φn (x0 ),

x0 , n

so F ( xn0 ) → F (0) hence gn − xn0 → x0 − 0 = x0 . Therefore

x0  x0

lim Φn (x0 ) = lim Φn gn − = lim gn − = kx0 k. n→∞ n→∞ n→∞ n n If kx0 k > 0, then Φn (x0 ) > 0 for some n which is a contradiction. Hence x0 = 0 and so for y ∈ M we have ky − 0kp ≤ kF (0) − 0kp = kx0 kP = 0, whence M = {0}. 

R EFERENCES [1] Y. BENYAMINI and J. LINDENSTRAUSS, Geometric Nonlinear Functional Analysis, Vol. 1, Colloq. Publ. 48, Amer. Math. Soc., Providence, 2000. [2] S. FITZPATRICK, Metric projection and the differentiability of distance functions, Bull. Austral. Math. Soc., 22 (1980), 291–312. [3] D. H. HYERS, G. ISAC and Th. M. RASSIAS, Topics in Nonlinear Analysis and Applications, World Scientific Publishging Company, Singapore, New Jersey, London, 1997. [4] M. S. MOSLEHIAN and Gh. SADEGHI, Stability of linear mappings in quasi-Banach modules, Math. Inequal. Appl., 11 (2008), no. 3, 549–557.

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[5] A. NIKNAM, Continuity of the farthest point map, Indian J. Pure Appl. Math., 18 (1987), No. 7, 630–632. [6] B. B. PANDA and O. P. KAPOOR, On farthest points of set, J. Math. Anal., 62 (1978), 345–353. [7] S. ROLEWICZ, Metric Linear Spaces, PWN-Polish Sci. Publ., Reidel and Dordrecht, 1984. [8] L. SCHWARTZ, Geometry and Probability in Banach Space, Lecture Notes in Mathematics 852, Springer–Verlag, New York, 1981. [9] V. S. SUNDER, Functional Analysis Spectral Theory, Birkhäuser Advanced Texts, Basel, 1997.

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