EUROPEAN JOURNAL OF PURE AND APPLIED MATHEMATICS Vol. 16, No. 4, 2023, 2397-2404 ISSN 1307-5543 – ejpam.com Published by New York Business Global On The Farthest Point Problem In Hilbert Spaces A. Yousef1,2,∗, R. Khalil1, A. Talafha1, B. Mutabagani3 1 Department of Mathematics, School of Science, The University of Jordan, Amman, Jordan 2 Department of Mathematics and Statistics, College of Arts and Sciences, American University of Sharjah, Sharjah, UAE 2 Department of Mathematics, Western Illinois University, Illinois, USA Abstract. In an attempt to solve the farthest point problem, we introduce a new class of uniquely remotal sets. Namely, the class of uniquely distant sets. Then, we prove that in a separable Hilbert space, every uniquely distant set is a singleton. 2020 Mathematics Subject Classifications: 46B20, 41A50, 41A65 Key Words and Phrases: Uniquely remotal sets; Farthest points; Uniquely distant sets. 1. Introduction Let X be a normed space, and E be a closed bounded subset of X. Define the real valued function r(., E) : X → R by r(x,E) = sup{∥x− e∥ : e ∈ E}, which is known as the farthest distance function. The set E is said to be remotal if for every x ∈ X, there exists e ∈ E such that r(x,E) = ∥x − e∥. In this case, we denote the set {e ∈ E : r(x,E) = ∥x − e∥} by F (x,E). It is clear that F (., E) : X → E is a multi-valued function. However, if F (., E) : X → E is a single-valued function, then E is called uniquely remotal, also known as max-Chebyshev set. In this case, we denote F (x,E) by F (x), if no confusion arises. The study of remotal and uniquely remotal sets has attracted many mathematicians in the last decades, due to its connection to the geometry of Banach spaces. We refer the reader to [1], [10], [11], [12], [13], [15], [8], [2] and [9] for samples of these studies. However, uniquely remotal sets are of special interest. ∗Corresponding author. DOI: https://doi.org/10.29020/nybg.ejpam.v16i4.4919 Email addresses: abd.yousef@ju.edu.jo; afyousef@aus.edu (A. Yousef), roshdi@ju.edu.jo (R. Khalil), a.tallafha@ju.edu.jo (A. Talafha), ba-mutabaganial-madani@wiu.edu (B. Mutabagani) https://www.ejpam.com 2397 © 2023 EJPAM All rights reserved. A. Yousef et al. / Eur. J. Pure Appl. Math, 16 (4) (2023), 2397-2404 2398 One of the most interesting and challenging problems in this field, is known as the farthest point problem, which is stated as: Q) If E is a uniquely remotal set in a normed space X, does it follow that E is a singleton? This problem became more important when Klee [14] proved that singletoness of closed uniquely remotal sets is equivalent to convexity of Chebyshev sets in Hilbert spaces (which is an open problem too, in the theory of nearest points). Since then, a considerable work has been done to answer this question, and many partial results have been obtained in the positive direction. An element c in a normed space X is called a Chebyshev center of E ⊂ X if r(c, E) = inf x∈X r(x,E). Whether a set has a Chebyshev center or not is still an open question. However, in inner product spaces, any closed bounded set does have a Chebyshev center [7]. Chebyshev Centers of sets have played a major role in the study of uniquely remotal sets. We refer the reader to [4], [5], [3] and[6]. For x, y ∈ X, [x, y] is the line segment joining x and y. In [16] it was proved that if E is a uniquely remotal subset of a normed space, admitting a Chebyshev center c, and if F , restricted to the line segment [c, F (c)] is continuous at c, then E is a singleton. Partially continuity was introduced in [17] as follows: Let G : A ⊂ X → X be a function, and let a ∈ A. Then G is partially continuous at a if there exists a nonconstant sequence (an) ⊂ A, such that an → a and G(an) → G(a). In [17], the following result was proved: Theorem 1. Let E be a closed subset of a normed space X, admitting a Chebyshev center c. If E is uniquely remotal, and if F : [c, F (c)] → E is partially continuous at c, then E is a singleton. One of the important notions that appeared in the literature while studying remotal and uniquely remotal sets is the so-called storngly remotal set. It was first introduced by Kalil et al in [8] for general Banach spaces as follows: Let X be a Banach space and E be a nonempty closed convex bounded subset of X. Let M = {ϕ : ϕ : [0,∞) → [0,∞) be strictly increasing function , ϕ(0) = 0}, and N = {ψ : ψ : X → (0, 1) such that ψ(x) ≤ ψ(y) whenever ||x|| < ||y||}. E is called stongly remotal in X if there exist ϕ ∈ M and ψ ∈ N , with inf y∈X ψ(y) > 0, such that for A. Yousef et al. / Eur. J. Pure Appl. Math, 16 (4) (2023), 2397-2404 2399 each x ∈ X there exists y ∈ E such that for all z ∈ E, the following inequality holds ϕ(||x− y||) ≥ ϕ(||x− z||) + ψ(y)ϕ(||y − z||) (1) It was also proved in [8] that strong remotality of E, with the associated functions ϕ and ψ, is equivalent to saying that: For every x ∈ X, there exists y ∈ E such that inf z∈E\{y} { ϕ(||x− y||)− ϕ(||x− z||) ϕ(||y − z||) } > 0. (2) clearly, strong remotal sets in Hilbert space can be defined similarly. We provide the definition below with ϕ(t) = t2. Definition 1. Let E be a non-empty closed convex bounded set in a Hilbert space H. Then E is called strongly remotal in H if for every x ∈ H there exists y ∈ E such that inf z∈E\{y} { ∥x− y∥2 − ∥x− z∥2 ∥y − z∥2 } > 0 (3) It is easy to see that inequality 1, with ϕ(t) = t2, can be written as ||x− y||2 ≥ ||x− z||2 + ψ(y)||y − z||2. (4) The main purpose of this article is to give an answer for the farthest point problem in separable Hilbert spaces, considering a special class of uniquely remotal sets that satisfy a certain condition. 2. Main Result In this section, we introduce the so-called uniquely distant sets, and prove that every uniquely distant set in a separable Hilbert space is a singleton. First we prove the follow- ing result that plays an important role in the proof of our main result. Theorem 2. Let E be a non-empty strongly remotal subset of a separable Hilbert space H. Then the mapping F : H → E, defined by F (x) = F (x,E) is continuous on H. Proof. Let x ∈ H be such that r(x,E) = r0. Let (xn) be a sequence in H such that xn → x. We claim that F (xn) → F (x) . E is strongly remotal, so ∥x− F (x) ∥2 ≥ ∥x− z∥2 + ψ (z) ∥F (x)− z∥2, (5) for all z ∈ E. A. Yousef et al. / Eur. J. Pure Appl. Math, 16 (4) (2023), 2397-2404 2400 Let zn = F (xn). Then, ∥x− F (x) ∥2 ≥ ∥x− F (xn) ∥2 + ψ (y) ∥F (x)− F (xn) ∥2 for all n. (6) This implies ∥F (x)− F (xn) ∥2 ≤ 1 ψ (y) [ ∥x− F (x) ∥2 − ∥x− F (xn) ∥2] (7) Since F (xn) is the farthest point in E from xn we have ∥xn − F (xn)∥ ≥ ∥xn − F (x)∥ , it follows that ∥xn − F (x)∥ ≤ ∥xn − F (xn)∥ ≤ ∥xn − x∥+ ∥x− F (xn)∥ , for all n ∈ N (8) Taking the limits on both sides of 8 as n→ ∞, we obtain ∥x− F (x)∥ ≤ lim n→∞ ∥x− F (xn)∥ (9) Inequality 7 is true for all n, which implies lim n→∞ ∥F (x)− F (xn) ∥2 ≤ 1 ψ (y) [∥x− F (x) ∥2 − lim n→∞ ∥x− F (xn) ∥2] (10) Substitute 9 in inequality 10 to get lim n→∞ ∥F (x)− F (xn) ∥2 = 0 So F (xn) → F (x), which completes the proof. Theorems 2 and 1 imply the following important result: Theorem 3. Every strongly remotal set in a Hilbert space is a singleton. Remark 1. Theorems 1 and 2 imply Garkavi’s result in [7], which states that a Chebyshev center in a Hilbert space exists and is unique. Next, we introduce the so-called uniquely distant sets. Definition 2. Let H be a Hilbert space and E ⊂ H be a closed bounded subset. Then E is said to be uniquely distant set in H if the following two conditions are satisfied: (i) E is uniquely remotal A. Yousef et al. / Eur. J. Pure Appl. Math, 16 (4) (2023), 2397-2404 2401 (ii) If x ∈ H, and y is the farthest point from x in E, then for every ε > 0, there exists δ > 0 such that r(x,E\B(y, δ)) ≤ r(x,E)− ε. Now, we are ready to prove the main result of this paper. Theorem 4. Every uniquely distant subset of a separable Hilbert space is a singleton. Proof. Let E be a nonsingleton uniquely distant subset in a Hilbert space H. Using Theorem 2.6 in [10], we can assume, without loss of generality, that E is convex. Let x ∈ H be any element, and y is a farthest element of x in E. Now, assume with no loss of generality that ∥x− y∥ = r(x,E) = 1. We claim that inf{∥x− y∥2 − ∥x− z∥2 ∥y − z∥2 : z ∈ E\{y}} > 0. Assume on the contrary that inf{∥x− y∥2 − ∥x− z∥2 ∥y − z∥2 : z ∈ E\{y}} = 0 Then, there exists a sequence zn ∈ E such that lim n→∞ ∥x− y∥2 − ∥x− zn∥2 ∥y − zn∥2 = 0 (11) Since E is bounded, this implies that lim n→∞ ∥x− zn∥ = ∥x− y∥ (12) We claim that zn → y. If not, then there exists an open ball B(y, δ) such that zn /∈ B(y, δ) ∩ E for all n. The set E\B(y, δ) ⊊ E is a closed set. Since E is uniquely distant then for any ε > 0, we haveB(y, δ) such that r(x,E\B(y, δ)) ≤ 1− ε. In this case equation 12 is not true. Thus, we must have limn→∞ zn = y. Now, let [p, q] be the line segment joining p and q. Since H is a Hilbert space, it follows that for each zn there exists wn in [x, y] such that [zn, wn] is orthogonal to [x, y]. Assume that ∥zn − wn∥ = dn, and ∥y − wn∥ = an and ∥x− wn∥ = bn. From basic geometry in Hilbert spaces we have: A. Yousef et al. / Eur. J. Pure Appl. Math, 16 (4) (2023), 2397-2404 2402 ∥x− y∥2 = 1 = (an + bn) 2, ∥x− zn∥2 = d2n + b2n, and ∥y − zn∥2 = d2n + a2n. Thus ∥x− y∥2 − ∥x− zn∥2 ∥y − zn∥2 = (an + bn) 2 − (d2n + b2n) d2n + a2n . (13) But (an + bn) 2 − (d2n + b2n) = a2n + b2n + 2anbn − (d2n + b2n) = 2a2n + 2anbn − (d2n + a2n). Since E is uniquely remotal and r(x,E) = ∥x− y∥ , so ∥x− y∥ > ∥x− zn∥ for all n, It follows that 2a2n + 2anbn − (d2n + a2n) > 0 But 2a2n + 2anbn = 2an(an + bn) = 2an. Thus 2an − (d2n + a2n) > 0. (14) Equations 13 and 14 gives ∥x− y∥2 − ∥x− zn∥2 ∥y − zn∥2 = 2an − (d2n + a2n) d2n + a2n = 2an d2n + a2n − 1. (15) From 14 we have 2an d2n + a2n − 1 > 0, which is equivalent to 2an d2n + a2n > 1. (16) Now, one should remark that as zn approaches y, then (an, dn) → (0, 0). Equation 16 implies that an does not converge to zero before dn, otherwise we obtain 2an d2n + a2n → 0, which contradicts 16. Therefore, we can assume that dn converges to 0 as a function of an, since ∥y − zn∥2 = d2n + a2n. Also, by sequential criterion, assume with no loss of generality that dn is a dif- ferentiable function of an. Now, we can think of the following limit: lim z→y ∥x− y∥2 − ∥x− zn∥2 ∥y − zn∥2 = lim (an,dn)→(0,0) ( 2an d2n + a2n − 1), (17) as lim w→0 ( 2w P 2(w) + w2 − 1 ) . Using L’Hopital’s Rule we get: REFERENCES 2403 lim (an,dn)→(0,0) ( 2an d2n + a2n − 1) = lim (an,dn)→(0,0) ( 2 2dn.d′n + 2an − 1). (18) Clearly, the limit in 18 is certainly bounded away from zero noting that by unique remotality, d′n can not equal to infinity. Otherwise, ∥x− zn∥ will reach ∥x− y∥ before zn reaches y. Hence E is strongly remotal, and by Theorem 3, E is a singleton. This completes the proof. Clearly, one can see that every uniquely distant set is uniquely remotal. However, we could not find a uniquely remotal set that is not uniquely distant. So, we believe that they are equivalent, and we conjecture the following: Conjecture: Every uniquely remotal set in Hilbert space is uniquely distant. 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