On Continuity of Functions Between Vector Metric Spaces

Hindawi Publishing Corporation Journal of Function Spaces Volume 2014, Article ID 753969, 6 pages http://dx.doi.org/10.1155/2014/753969

Research Article On Continuity of Functions between Vector Metric Spaces

Cüneyt Çevik

Department of Mathematics, Faculty of Science, Gazi University, Teknikokullar, 06500 Ankara, Turkey

Correspondence should be addressed to Cuneyt¨ C¸ evik; [email protected]

Received 17 August 2014; Accepted 8 October 2014; Published 12 November 2014

Academic Editor: Han Ju Lee

Copyright © 2014 Cuneyt¨ C¸ evik. This is an open access article distributed under the Creative Commons Attribution License, which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited.

We introduce vectorial and topological continuities for functions defined on vector metric spaces and illustrate spaces of such functions. Also, we describe some fundamental classes of vector valued functions and extension theorems.

1. Introduction and Preliminaries vector space has started by Zabrejko in [8]. The distance map in the sense of Zabrejko takes values from an ordered vector Let 𝐸 be a Riesz space. If every nonempty bounded above space. We use the structure of lattice with the vector metrics (countable) subset of 𝐸 has a supremum, then 𝐸 is called having values in Riesz spaces; then we have new results as Dedekind (𝜎-) complete.Asubsetin𝐸 is called order bounded mentioned above. This paper is on some concepts and related if it is bounded both above and below. We write 𝑎𝑛 ↓𝑎if (𝑎𝑛) results about continuity in vector metric spaces. is a decreasing sequence in 𝐸 such that inf 𝑎𝑛 =𝑎.TheRiesz −1 space 𝐸 is said to be Archimedean if 𝑛 𝑎↓0holds for every Definition 1. Let 𝑋 be a nonempty set and let 𝐸 be a Riesz 𝑑:𝑋×𝑋→ 𝐸 𝑎∈𝐸+.Asequence(𝑏𝑛) is said to be 𝑜-convergent (or order space. The function is said to be a vector 𝐸 convergent)to𝑏 if there is a sequence (𝑎𝑛) in 𝐸 such that 𝑎𝑛 ↓0 metric (or -metric) if it satisfies the following properties: and |𝑏𝑛−𝑏| ≤𝑛 𝑎 for all 𝑛,where|𝑎| = 𝑎∨(−𝑎) for any 𝑎∈𝐸.We 𝑜 (vm1) 𝑑(𝑥, 𝑦) =0 if and only if 𝑥=𝑦, will denote this order convergence by 𝑏𝑛 →𝑏󳨀 .Furthermore,a sequence (𝑏𝑛) is said to be 𝑜-Cauchy if there exists a sequence (vm2) 𝑑(𝑥, 𝑦) ≤ 𝑑(𝑥, 𝑧) +𝑑(𝑦,𝑧) for all 𝑥, 𝑦,. 𝑧∈𝑋 (𝑎𝑛) in 𝐸 such that 𝑎𝑛 ↓0and |𝑏𝑛 −𝑏𝑛+𝑝|≤𝑎𝑛 for all 𝑛 and 𝑝. The Riesz space 𝐸 is said to be 𝑜-Cauchy complete if every Also the triple (𝑋, 𝑑, 𝐸) is said to be vector metric space. 𝑜-Cauchy sequence is 𝑜-convergent. (𝑋, 𝑑, 𝐸) An operator 𝑇:𝐸between →𝐹 two Riesz spaces is Definition 2. Let be a vector metric space. positive if 𝑇(𝑥) ≥0 for all 𝑥≥0.Theoperator𝑇 is said (𝑥 ) 𝑋 to be order bounded if it maps order bounded subsets of 𝐸 (a) A sequence 𝑛 in is vectorially convergent (or is 𝐸 𝑥∈𝑋 to order bounded subsets of 𝐹.Theoperator𝑇 is called 𝜎- -convergent) to some ,ifthereisasequence 𝑜 𝑜 (𝑎𝑛) in 𝐸 such that 𝑎𝑛 ↓0and 𝑑(𝑥𝑛,𝑥)≤𝑛 𝑎 for all 𝑛. order continuous if 𝑥𝑛 →0󳨀 in 𝐸 implies 𝑇(𝑥𝑛) →0󳨀 in 𝑑,𝐸 𝐹.Every𝜎-order continuous operator is order bounded. If We will denote this vectorial convergence by 𝑥𝑛 󳨀󳨀→ 𝑇(𝑥∨𝑦)= 𝑇(𝑥)∨𝑇(𝑦)for all 𝑥, 𝑦,theoperator ∈𝐸 𝑇 is 𝑥. a lattice homomorphism. For further information about Riesz (b) A sequence (𝑥𝑛) in 𝑋 is called 𝐸-Cauchy sequence spaces and the operators on Riesz spaces, we refer to [1, 2]. whenever there exists a sequence (𝑎𝑛) in 𝐸 such that In [3], a vector metric space is defined with a distance map 𝑎𝑛 ↓0and 𝑑(𝑥𝑛,𝑥𝑛+𝑝)≤𝑎𝑛 for all 𝑛 and 𝑝. having values in a Riesz space, and some results in metric space theory are generalized to vector metric space theory. (c) The vector metric space 𝑋 is called 𝐸-complete if each Some fixed point theorems in vector metric spaces are given 𝐸-Cauchy sequence in 𝑋 is 𝐸-convergenttoalimitin in [3–7].Actually,thestudyofmetricspaceshavingvalueona 𝑋. 2 Journal of Function Spaces

One of the main goals of this paper is to demonstrate the Corollary 5. For a function 𝑓:𝑋between →𝑌 two vector properties of functions on vector metric spaces in a context metric spaces (𝑋, 𝑑, 𝐸) and (𝑌,𝜌,𝐹) the following statements more general than the continuity in metric analysis. Hence, hold. the properties of Riesz spaces will be the tools for the study of (a) If 𝐹 is Dedekind 𝜎-complete and 𝑓 is vectorially contin- continuity of vector valued functions. The result we get from 𝜌(𝑓(𝑥 ), 𝑓(𝑥)) ↓0 𝑑(𝑥 ,𝑥)↓0 here is continuity in general sense, an order property rather uous, then 𝑛 whenever 𝑛 . than a topological feature. (b) If 𝐸 is Dedekind 𝜎-complete and 𝜌(𝑓(𝑥𝑛), 𝑓(𝑥)) ↓ In Section 2, we consider two types of continuity on vec- 0 whenever 𝑑(𝑥𝑛,𝑥) ↓, 0 then the function 𝑓 is tor metric spaces. This approach distinguishes continuities vectorially continuous. vectorially and topologically. Moreover, vectorial continuity (c) Suppose that 𝐸 and 𝐹 are Dedekind 𝜎-complete. Then, examples are given and the relationship between vectorial the function 𝑓 is vectorially continuous if and only if continuity of a function and its graph is demonstrated. In 𝜌(𝑓(𝑥𝑛), 𝑓(𝑥)) ↓0 whenever 𝑑(𝑥𝑛,𝑥)↓0. Section 3, equivalent vector metrics, vectorial isometry, vec- 𝑑,𝐸 torial homeomorphism definitions, and examples related to 𝑑(𝑥 ,𝑥) ↓ 0 𝑥 󳨀󳨀→𝑥 these concepts are given. In Section 4,uniformcontinuityis Proof. (a) If 𝑛 ,then 𝑛 .Bythevectorial continuity of the function 𝑓,thereisasequence(𝑏𝑛) in 𝐹 discussed and some extension theorems for functions defined 𝑏 ↓0 𝜌(𝑓(𝑥 ), 𝑓(𝑥)) ≤𝑏 𝑛 𝑓 on vector metric spaces are given. Finally, in Section 5, such that 𝑛 and 𝑛 𝑛 for all .Since is Dedekind 𝜎-complete, 𝜌(𝑓(𝑥𝑛), 𝑓(𝑥)) ↓0 holds. a uniform limit theorem on a vector metric space is given, 𝑑,𝐸 and the structure of vectorial continuous function spaces is (b) Let 𝑥𝑛 󳨀󳨀→𝑥in 𝑋. Then there is a sequence (𝑎𝑛) in demonstrated. 𝐸 such that 𝑎𝑛 ↓0and 𝑑(𝑥𝑛,𝑥) ≤𝑛 𝑎 for all 𝑛.Since𝐸 is Dedekind 𝜎-complete, 𝑑(𝑥𝑛,𝑥)↓ 0holds. By the hypothesis, 𝜌,𝐹 2. Topological and Vectorial Continuity 𝜌(𝑓(𝑥𝑛), 𝑓(𝑥)),andso ↓0 𝑓(𝑥𝑛) 󳨀󳨀→ 𝑓(𝑥) in 𝑌. (c) Proof is a consequence of (a) and (b). We now introduce two types of continuity in vector metric 𝑎 𝑏 𝑑,𝐸 spaces. For two elements and in a Riesz space, we write (𝑋,𝑑,𝐸) 𝑥 󳨀󳨀→𝑥 𝑎<𝑏 𝑎≤𝑏 𝑎 =𝑏̸ Example 6. Let be a vector metric space. If 𝑛 to indicate that but . 𝑑,𝐸 𝑜 and 𝑦𝑛 󳨀󳨀→𝑦,then𝑑(𝑥𝑛,𝑦𝑛) →󳨀 𝑑(𝑥,;thatis,thevector 𝑦) (𝑋, 𝑑, 𝐸) (𝑌,𝜌, 𝐹) 2 Definition 3. Let and be vector metric metric map 𝑑 from 𝑋 to 𝐸 is vectorially continuous. Here, 𝑥∈𝑋 2 ̃ spaces, and let . 𝑋 is equipped with the 𝐸-valued vector metric 𝑑 defined 𝑑(𝑧,̃ 𝑤) = 𝑑(𝑥 ,𝑥 )+𝑑(𝑦,𝑦 ) 𝑧=(𝑥,𝑦 ), 𝑤 = (a) A function 𝑓:𝑋is →𝑌 said to be topologically as 1 2 1 2 for all 1 1 (𝑥 ,𝑦 )∈𝑋2 𝐸 continuous at 𝑥 if for every 𝑏>0in 𝐹 there exists 2 2 ,and is equipped with the absolute valued |⋅| some 𝑎 in 𝐸 such that 𝜌(𝑓(𝑥), 𝑓(𝑦)) <𝑏 whenever vector metric . 𝑦∈𝑋 𝑑(𝑥, 𝑦) <𝑎 𝑓 and .Thefunction is said 𝑈 (𝑋, 𝑑, 𝐸) to be topologically continuous if it is topologically We recall that a subset of a vector metric space 𝑋 is called 𝐸-closed (or vector closed) whenever (𝑥𝑛)⊆𝑈and continuous at each point of . 𝑑,𝐸 𝑥𝑛 󳨀󳨀→𝑥imply 𝑥∈𝑈. (b) A function 𝑓:𝑋is →𝑌 said to be vectorially 𝑑,𝐸 𝜌,𝐹 continuous at 𝑥 if 𝑥𝑛 󳨀󳨀→𝑥in 𝑋 implies 𝑓(𝑥𝑛) 󳨀󳨀→ Theorem 7. Let (𝑋, 𝑑, 𝐸) and (𝑌,𝜌,𝐹)be vector metric spaces. 𝑓(𝑥) in 𝑌.Thefunction𝑓 is said to be vectorially If a function 𝑓:𝑋is →𝑌 vectorially continuous, then for −1 continuous if it is vectorially continuous at each point every 𝐹-closed subset 𝐵 of 𝑌 the set 𝑓 (𝐵) is 𝐸-closed in 𝑋. of 𝑋. −1 Proof. For any 𝑥∈𝑓 (𝐵), there exists a sequence (𝑥𝑛) in Theorem 4. (𝑋, 𝑑, 𝐸) (𝑌,𝜌,𝐹) 𝑑,𝐸 Let and be vector metric spaces 𝑓−1(𝐵) 𝑥 󳨀󳨀→𝑥 𝑓 𝐹 𝑓:𝑋 →𝑌 such that 𝑛 . Since the function is vectorially where is Archimedean. If a function is 𝜌,𝐹 topologically continuous, then 𝑓 is vectorially continuous. continuous, 𝑓(𝑥𝑛) 󳨀󳨀→ 𝑓(𝑥).Buttheset𝐵 is 𝐹-closed, so 𝑥∈ −1 −1 𝑓 (𝐵).Then,theset𝑓 (𝐵) is 𝐸-closed. 𝑑,𝐸 Proof. Suppose that 𝑥𝑛 󳨀󳨀→𝑥. Then there exists a sequence If 𝐸 and 𝐹 are two Riesz spaces, then 𝐸×𝐹is also a Riesz (𝑎𝑛) in 𝐸 such that 𝑎𝑛 ↓0and 𝑑(𝑥𝑛,𝑥)≤𝑎𝑛 for all 𝑛.Let𝑏 be space with coordinatewise ordering defined as any nonzero positive element in 𝐹. By topological continuity 𝑓 𝑥 𝑏 =𝑏((1/𝑛)𝑏; 𝑥) 𝐸 of at , there exist elements 𝑛 𝑛 in such that (𝑒1,𝑓1)≤(𝑒2,𝑓2)⇐⇒𝑒1 ≤𝑒2,𝑓1 ≤𝑓2 (1) 𝑑(𝑥,𝑛 𝑥 )<𝑏𝑛 implies 𝜌(𝑓(𝑥),𝑛 𝑓(𝑥 )) < (1/𝑛)𝑏 for all 𝑛.Then 𝑐 =𝑎 ∨𝑏 𝐸 𝑑(𝑥 ,𝑥) ≤ there exist elements 𝑛 𝑛 𝑛 in such that 𝑛 for all (𝑒1,𝑓1), (𝑒2,𝑓2)∈𝐸×𝐹. The Riesz space 𝐸×𝐹is 𝑎𝑛 <𝑐𝑛 implies 𝜌(𝑓(𝑥𝑛), 𝑓(𝑥)) < (1/𝑛)𝑏 for all 𝑛.Since𝐹 is a vector metric space, equipped with the biabsolute valued 𝜌,𝐹 vector metric |⋅|defined as Archimedean, (1/𝑛)𝑏.Hence, ↓0 𝑓(𝑥𝑛) 󳨀󳨀→ 𝑓(𝑥). 󵄨 󵄨 󵄨 󵄨 |𝑎−𝑏| = (󵄨𝑒1 −𝑒2󵄨 , 󵄨𝑓1 −𝑓2󵄨) (2) Vectorially continuous functions have a number of nice characterizations. for all 𝑎=(𝑒1,𝑓1), 𝑏=(𝑒2,𝑓2)∈𝐸×𝐹. Journal of Function Spaces 3

Let 𝑑 and 𝜌 be two vector metrics on 𝑋 which are 𝐸- function ℎ from 𝑋×𝑌to 𝐺 defined by ℎ(𝑥, 𝑦) = |𝑓(𝑥) −𝑔(𝑦)| valued and 𝐹-valued, respectively. Then the map 𝛿 defined for all 𝑥∈𝑋, 𝑦∈𝑌is vectorially continuous with the 𝐸×𝐹- as valued product vector metric 𝜋 and the absolute valued vector metric |⋅|. 𝛿 (𝑥,) 𝑦 = (𝑑 (𝑥,) 𝑦 ,𝜌(𝑥,)) 𝑦 (3) (c) If 𝑓 : (𝑋, 𝑑, 𝐸) → (𝑍, 𝜂,𝐺) and 𝑔 : (𝑌,𝜌,𝐹) → 𝑥, 𝑦 ∈𝑋 𝐸×𝐹 𝑋 (𝑊,𝜉, 𝐻) are vectorially continuous functions, then the func- for all is an -valued vector metric on .Wewill ℎ 𝑋×𝑌 𝑍×𝑊 ℎ(𝑥, 𝑦) = (𝑓(𝑥), 𝑔(𝑦)) call 𝛿 double vector metric. tion from to defined by for all 𝑥∈𝑋, 𝑦∈𝑌is vectorially continuous with the 𝐸×𝐹- 𝐺×𝐻 Example 8. Let (𝑋,𝑑,𝐸) and (𝑋, 𝜌, 𝐹) be vector metric valued and -valued product vector metrics. spaces. We have the next proposition for any product vector (i) Suppose that 𝑓:𝑋and →𝐸 𝑔:𝑋are →𝐹 metric. vectorially continuous functions. Then the function ℎ from 𝑋 to 𝐸×𝐹defined by ℎ(𝑥) = (𝑓(𝑥), 𝑔(𝑥)) for all Proposition 12. Let (𝑧𝑛)=(𝑥𝑛,𝑦𝑛) be a sequence in (𝑋 × 𝑥∈𝑋 𝜋,𝐸×𝐹 is vectorially continuous with the double vector 𝑌, 𝜋, 𝐸 ×𝐹) and let 𝑧 = (𝑥,𝑦)∈.Then, 𝑋×𝑌 𝑧𝑛 󳨀󳨀󳨀󳨀→𝑧 metric 𝛿 and the biabsolute valued vector metric |⋅|. 𝑑,𝐸 𝜌,𝐹 ifandonlyif𝑥𝑛 󳨀󳨀→𝑥and 𝑦𝑛 󳨀󳨀→𝑦. (ii) Let 𝑝𝐸 :𝐸×𝐹→and 𝐸 𝑝𝐹 :𝐸×𝐹→be 𝐹 the ℎ:𝑋 → 𝐸×𝐹 projection maps. Any function can be Now let us give relevance between vectorial continuity of ℎ(𝑥) = (𝑓(𝑥), 𝑔(𝑥)) 𝑥∈𝑋 𝑓= written as for all where a function and closeness of its graph. 𝑝𝐸ℎ and 𝑔=𝑝𝐹ℎ.Ifℎ is vectorially continuous, so are 𝑓 𝑔 𝑝 𝑝 and since 𝐸 and 𝐹 are vectorially continuous. Corollary 13. Let (𝑋,𝑑,𝐸) and (𝑌,𝜌,𝐹) be vector metric spaces and let 𝑓 be a function from 𝑋 to 𝑌.Forthegraph𝐺𝑓 (𝑋,𝑑,𝐸) (𝑌,𝜌,𝐹) Let and be vector metric spaces. Then of 𝑓, the following statements hold. 𝑋×𝑌is a vector metric space, equipped with the 𝐸×𝐹-valued 𝐺 𝐸×𝐹 (𝑋 × 𝑌,𝜋, 𝐸 ×𝐹) product vector metric 𝜋 defined as (a) The graph 𝑓 is -closed in if 𝑑,𝐸 𝜋 (𝑧,) 𝑤 = (𝑑 (𝑥 ,𝑥 ) ,𝜌(𝑦 ,𝑦 )) and only if for every sequence (𝑥𝑛) with 𝑥𝑛 󳨀󳨀→𝑥and 1 2 1 2 (4) 𝜌,𝐹 𝑓(𝑥𝑛) 󳨀󳨀→𝑦we have 𝑦=𝑓(𝑥). for all 𝑧=(𝑥1,𝑦1), 𝑤=(𝑥2,𝑦2)∈𝑋×𝑌. 𝑓 Consider the vector metric space (𝑋×𝑌,𝜋,𝐸×𝐹).The (b) If the function is vectorially continuous, then the 𝐺 𝐸×𝐹 projection maps 𝑝𝑋 and 𝑝𝑌 defined on 𝑋×𝑌which are 𝑋- graph 𝑓 is -closed. valued and 𝑌-valued, respectively, are vectorially continuous. (c) If the function 𝑓 is vectorially continuous at 𝑥0 ∈𝑋, For any function 𝑓 from a vector metric space (𝑍, 𝛿, 𝐺) to then the induced function ℎ:𝑋 →𝑓 𝐺 defined by (𝑋 × 𝑌,𝜋, 𝐸,thefunction ×𝐹) 𝑓 is vectorially continuous if ℎ(𝑥) = (𝑥, 𝑓(𝑥)) is vectorially continuous at 𝑥0 ∈𝑋. and only if both 𝑝𝑋𝑓 and 𝑝𝑌𝑓 are vectorially continuous. Proof. For the proof of (a), suppose that the graph 𝐺𝑓 is Example 9. Let (𝑋,𝑑,𝐸)and (𝑌,𝜌,𝐸)be vector metric spaces. 𝑑,𝐸 𝜌,𝐹 𝑓:𝑋 →𝐸 𝑔:𝑌 →𝐸 𝐸×𝐹-closed. If 𝑥𝑛 󳨀󳨀→𝑥and 𝑓(𝑥𝑛) 󳨀󳨀→𝑦,thenwehave Suppose that and are vectorially 𝜋,𝐸×𝐹 continuous functions. Then the function ℎ from 𝑋×𝑌to 𝐸 (𝑥𝑛,𝑓(𝑥𝑛)) 󳨀󳨀󳨀󳨀→(𝑥,𝑦)by Proposition 12.Hence(𝑥, 𝑦) ∈ defined by ℎ(𝑥, 𝑦) = |𝑓(𝑥) −𝑔(𝑦)| for all 𝑥∈𝑋, 𝑦∈𝑌is 𝐺𝑓,andso𝑦=𝑓(𝑥).Conversely,supposethat(𝑧𝑛)= vectorially continuous with the product vector metric 𝜋 and 𝜋,𝐸×𝐹 (𝑥𝑛,𝑓(𝑥𝑛)) is a sequence in 𝐺𝑓 such that 𝑧𝑛 󳨀󳨀󳨀󳨀→𝑧= theabsolutevaluedvectormetric|⋅|. 𝑑,𝐸 𝜌,𝐹 (𝑥, 𝑦) ∈ 𝑋× 𝑌.ByProposition 12, 𝑥𝑛 󳨀󳨀→𝑥and 𝑓(𝑥𝑛) 󳨀󳨀→𝑦. (𝑋,𝑑,𝐸) (𝑌,𝜌,𝐹) Example 10. Let and be vector metric Then 𝑦=𝑓(𝑥),andso𝑧∈𝐺𝑓. 𝑓:𝑋 →𝐸 𝑔:𝑌 →𝐹 spaces. Suppose that and are Proofs of (b) and (c) are similar to the proof of (a). vectorially continuous functions. Then the function ℎ from 𝑋×𝑌to 𝐸×𝐹defined by ℎ(𝑥, 𝑦) = (𝑓(𝑥), 𝑔(𝑦)) for all 𝑥∈𝑋, 𝑦∈𝑌is vectorially continuous with the product 3. Fundamental Vector Valued vector metric 𝜋 and the biabsolute valued vector metric |⋅|. Function Classes 𝐸 𝑑 𝐹 The last three examples inspire the following results. Definition 14. The -valued vector metric and -valued vector metric 𝜌 on 𝑋 are said to be (𝐸, 𝐹)-equivalent if for Corollary 11. (a) If 𝑓 : (𝑋,𝑑,𝐸) → (𝑌,𝜂,𝐺) and 𝑔: any 𝑥∈𝑋and any sequence (𝑥𝑛) in 𝑋 (𝑋, 𝜌, 𝐹) → (𝑍, 𝜉,𝐻) are vectorially continuous functions, 𝑑,𝐸 𝜌,𝐹 then the function ℎ from 𝑋 to 𝑌×𝑍defined by ℎ(𝑥) = 𝑥𝑛 󳨀󳨀→𝑥 iff 𝑥𝑛 󳨀󳨀→𝑥. (5) (𝑓(𝑥), 𝑔(𝑥)) for all 𝑥∈𝑋is vectorially continuous with the Lemma 15. For any two 𝐸-valued vector metrics 𝑑 and 𝜌 on 𝐸×𝐹-valued double vector metric 𝛿 and the 𝐺×𝐻-valued 𝑋, the following statements are equivalent. product vector metric 𝜋. (b) Let 𝐺 be a Riesz space. If 𝑓:(𝑋,𝑑,𝐸)→𝐺and 𝑔: (a) There exist some 𝛼, 𝛽 >0 in R such that 𝛼𝑑(𝑥, 𝑦)≤ (𝑌,𝜌,𝐹) →𝐺 are vectorially continuous functions, then the 𝜌(𝑥, 𝑦) ≤ 𝛽𝑑(𝑥, 𝑦) for all 𝑥, 𝑦 ∈𝑋. 4 Journal of Function Spaces

(b) There exist two positive and 𝜎-order continuous oper- Remark 18. Vectorial continuity is invariant under equivalent ators 𝑇 and 𝑆 from 𝐸 to itself such that 𝜌(𝑥, 𝑦) ≤ vector metrics. 𝑇(𝑑(𝑥, 𝑦)) and 𝑑(𝑥, 𝑦) ≤ 𝑆(𝜌(𝑥, 𝑦)) for all 𝑥, 𝑦 ∈𝑋. Let us show how an isometry is defined between two Proof. Let 𝑇 and 𝑆 be two operators defined as 𝑇(𝑎) = 𝛽𝑎 vector metric spaces. −1 and 𝑆(𝑎) =𝛼 𝑎 for all 𝑎∈𝐸. If (a) holds, then 𝑇 and (𝑋, 𝑑, 𝐸) (𝑌,𝜌,𝐹) 𝑆 are positive and 𝜎-order continuous operators and satisfy Definition 19. Let and be vector metric 𝑓:𝑋 →𝑌 𝜌(𝑥, 𝑦) ≤ 𝑇(𝑑(𝑥, 𝑦)) and 𝑑(𝑥, 𝑦) ≤ 𝑆(𝜌(𝑥, 𝑦)) for all 𝑥, 𝑦 ∈ spaces. A function is said to be a vector isometry 𝑇 :𝐸 → 𝐹 𝑋. Conversely, (a) holds since every (𝜎-)order continuous if there exists a linear operator 𝑓 satisfying the operator is order bounded [1,1.54]. following two conditions: 𝑇 (𝑑(𝑥, 𝑦)) = 𝜌(𝑓(𝑥), 𝑓(𝑦)) 𝑥, 𝑦 ∈𝑋 Now, we give the following result for the equivalence of (i) 𝑓 for all , vector metrics. (ii) 𝑇𝑓(𝑎) = 0 implies 𝑎=0for all 𝑎∈𝐸. 𝑓 𝑇 Theorem 16. An 𝐸-valued vector metric 𝑑 and an 𝐹-valued If the function isonto,andtheoperator 𝑓 is a lattice vector metric 𝜌 on 𝑋 are (𝐸, 𝐹)-equivalent if there exist positive homomorphism, then the vector metric spaces (𝑋,𝑑,𝐸)and (𝑌,𝜌,𝑇 (𝐸)) and 𝜎-order continuous two operators 𝑇:𝐸, →𝐹 𝑆:𝐹 →𝐸 𝑓 arecalledvectorisometric. such that Remark 20. A vector isometry is a vectorial distance preserv- 𝜌 (𝑥,) 𝑦 ≤𝑇(𝑑 (𝑥,)) 𝑦 ,𝑑(𝑥,) 𝑦 ≤𝑆(𝜌 (𝑥,)) 𝑦 (6) ing one-to-one function.

2 for all 𝑥, 𝑦 ∈𝑋. Example 21. Let 𝑑 be R-valued vector metric and let 𝜌 be R - R 2 valued vector metric on defined as Example 17. Suppose that the ordering of R is coordinate- 󵄨 󵄨 󵄨 󵄨 󵄨 󵄨 wise. 𝑑 (𝑥,) 𝑦 =𝑎󵄨𝑥−𝑦󵄨 ,𝜌(𝑥,) 𝑦 = (𝑏 󵄨𝑥−𝑦󵄨 ,𝑐󵄨𝑥−𝑦󵄨) , 2 (10) (a) Let 𝑑 and 𝜌 be R-valued and R -valued vector metrics on R, respectively, defined as where 𝑏, 𝑐 ≥0 and 𝑎, 𝑏 + 𝑐. >0 Consider the identity 2 󵄨 󵄨 󵄨 󵄨 󵄨 󵄨 mapping 𝐼 on R and the operator 𝑇𝐼 : R → R defined as 𝑑 (𝑥,) 𝑦 =𝑎󵄨𝑥−𝑦󵄨 ,𝜌(𝑥,) 𝑦 = (𝑏 󵄨𝑥−𝑦󵄨 ,𝑐󵄨𝑥−𝑦󵄨) , −1 𝑇𝐼(𝑥) = 𝑎 (𝑏𝑥, 𝑐𝑥) for all 𝑥∈R. Then, the identity mapping (7) 𝐼 is a vector isometry. So, the vector metric spaces (R,𝑑,R) and (R,𝜌,{(𝑥,𝑦):𝑐𝑥=𝑏𝑦;𝑥,𝑦∈R}) are vector isometric. where 𝑏, 𝑐 ≥0 and 𝑎, 𝑏 + 𝑐. >0 Consider the two 2 −1 operators 𝑇:R → R ; 𝑇(𝑥) =𝑎 (𝑏𝑥, 𝑐𝑥) and 2 −1 Definition 22. Let (𝑋, 𝑑, 𝐸) and (𝑌,𝜌,𝐹) be vector metric 𝑆:R → R; 𝑆(𝑥, 𝑦) =𝑎𝑏 𝑥 for all 𝑥, 𝑦 ∈ R. 𝑓:𝑋 →𝑌 𝑇 𝑆 𝜎 spaces. A function is said to be a vector homeo- Then, the operators and are positive and -order morphism if 𝑓 is one-to-one and vectorially continuous and continuous, and (6) is satisfied. Hence, the metrics 𝑑 𝑓(𝑋) 𝜌 (R, R2) R has a vectorially continuous inverse on .Ifthefunction and are -equivalent on . 𝑓 is onto, then the vector metric spaces 𝑋 and 𝑌 are called 2 (b) Let 𝑑 and 𝜌 be R-valued and R -valued vector metrics vector homeomorphic. 2 on R , respectively, defined as Remark 23. A vector homeomorphism is one-to-one func- 󵄨 󵄨 󵄨 󵄨 𝑑(𝑥,𝑦)=𝑎󵄨𝑥1 −𝑦1󵄨 +𝑏󵄨𝑥2 −𝑦2󵄨 , tion that preserves vectorial convergence of sequences. (8) 󵄨 󵄨 󵄨 󵄨 𝜌(𝑥,𝑦)=(𝑐󵄨𝑥1 −𝑦1󵄨 ,𝑒󵄨𝑥2 −𝑦2󵄨), By Theorem 7, we can develop another characterization resultforvectorhomeomorphisms. where 𝑥=(𝑥1,𝑥2), 𝑦=(𝑦1,𝑦2),and𝑎, 𝑏, 𝑐,.Let 𝑒>0 2 2 Theorem 24. 𝑇:R → R and 𝑆:R → R be two operators An onto vector homeomorphism is a one-to-one −1 −1 defined as 𝑇(𝑥) = (𝑐𝑎 𝑥, 𝑒𝑏 𝑥) and 𝑆(𝑥, 𝑦) = function that preserves vector closed sets. −1 −1 𝑎𝑐 𝑥+𝑏𝑒 𝑦.Then,theoperators𝑇 and 𝑆 are positive Proof. Let 𝑓:𝑋 →𝑌be an (𝐸, 𝐹)-homeomorphism. Since and 𝜎-order continuous. The condition (6) is satisfied. −1 2 𝑓 𝑓 So, the vector metrics 𝑑 and 𝜌 are (R, R )-equivalent isaone-to-onefunctionanditsinverse is vectorially 2 continuous, by Theorem 7 for every 𝐸-closed set 𝐴 in 𝑋, on R . On the other hand, if 𝜂 is another R-valued −1 −1 2 𝑓(𝐴) = (𝑓 ) (𝐴) is 𝐹-closed in 𝑌. vector metric on R defined as 󵄨 󵄨 󵄨 󵄨 The following example illustrates a relationship between 𝜂 (𝑥,) 𝑦 = max {𝑎 󵄨𝑥1 −𝑦1󵄨 ,𝑏󵄨𝑥2 −𝑦2󵄨} , (9) 󵄨 󵄨 󵄨 󵄨 vectorial equivalence and vector homeomorphism.

where 𝑥=(𝑥1,𝑥2), 𝑦=(𝑦1,𝑦2),and𝑎, 𝑏,andthe >0 𝑑 𝜌 (𝐸, 𝐹) −1 −1 Example 25. Let and be two -equivalent vector operator 𝑆 is defined as 𝑆(𝑥, 𝑦) = max{𝑎𝑐 𝑥, 𝑏𝑒 𝑦}, 𝑋 (𝑋, 𝑑, 𝐸) 2 metrics on . Then the vector metric spaces and then the vector metrics 𝜂 and 𝜌 are (R, R )-equivalent (𝑋, 𝜌, 𝐹) are vector homeomorphic under the identity map- 2 on R . ping. On the other hand, if any two vector metric spaces Journal of Function Spaces 5

(𝑋, 𝑑, 𝐸) and (𝑌,𝜌,𝐹) are vector homeomorphic under a exist elements 𝑐𝑛 =𝑎𝑛 ∨𝑏𝑛 in 𝐸 such that 𝑑(𝑥𝑛,𝑥𝑛+𝑝)≤𝑎𝑛 <𝑐𝑛 function 𝑓, then the vector metrics 𝑑 and 𝛿 defined as implies 𝜌(𝑓(𝑥𝑛), 𝑓(𝑥𝑛+𝑝)) < (1/𝑛)𝑏 for all 𝑛 and 𝑝.Since𝐹 is Archimedean, (1/𝑛)𝑏. ↓0 Hence, the sequence 𝑓(𝑥𝑛) is 𝐹- 𝛿 (𝑥,) 𝑦 =𝜌(𝑓 (𝑥) ,𝑓(𝑦)) (11) Cauchy. for all 𝑥, 𝑦 ∈𝑋 are (𝐸, 𝐹)-equivalent vector metrics on 𝑋. Example 30. (a) For a vector isometry 𝑓 between two vector metric spaces (𝑋, 𝑑, 𝐸) and (𝑌,𝜌,𝐹),thefunction𝑓 is 𝑇 𝜎 4. Extension Theorems on Continuity vectorial uniformly continuous if 𝑓 is positive and -order continuous. 𝑦 (𝑋,𝑑,𝐸) If 𝑋 and 𝑌 are vector metric spaces, 𝐴⊆𝑋,and𝑓:𝐴 →𝑌 (b) For an element in a vector metric space , 𝑓 :𝑋 → 𝐸 𝑓 (𝑥) = 𝑑(𝑥, 𝑦) is vectorially continuous, then we might ask whether there the function 𝑦 defined by 𝑦 for all 𝑥∈𝑋 exists a vectorial continuous extension 𝑔 of 𝑓.Below,wedeal is vectorial uniformly continuous. 𝐴 (𝑋,𝑑,𝐸) with some simple extension techniques. (c) For a subset of a vector metric space where 𝐸 is Dedekind complete, the function 𝑓𝐴 :𝑋 →defined 𝐸 Theorem 26. (𝑋, 𝑑, 𝐸) (𝑌,𝜌,𝐹) Let and be vector metric by 𝑓𝐴(𝑥) = 𝑑(𝑥, 𝐴)= inf{𝑑(𝑥, 𝑦) : 𝑦 ∈𝐴} for all 𝑥∈𝑋is spaces, and let 𝑓 and 𝑔 be vectorially continuous functions from vectorial uniformly continuous. 𝑋 to 𝑌.Thentheset{𝑥 ∈ 𝑋 : 𝑓(𝑥) = 𝑔(𝑥)} is an 𝐸-closed subset of 𝑋. The following theorem enables us to establish an extension property for the functions between vector metric spaces. Proof. Let 𝐵 = {𝑥 ∈ 𝑋 : 𝑓(𝑥) =𝑔(𝑥)}.Suppose(𝑥𝑛)⊆𝐵 𝑑,𝐸 Theorem 31. Let 𝐴 be 𝐸-dense subset of a vector metric space 𝑥 󳨀󳨀→𝑥 𝑓 𝑔 and 𝑛 .Since and are vectorially continuous, there (𝑋,𝑑,𝐸)and let (𝑌,𝜌,𝐹)be an 𝐹-complete vector metric space (𝑎 ) (𝑏 ) 𝑎 ↓0𝑏 ↓0 exist sequences 𝑛 and 𝑛 such that 𝑛 , 𝑛 and where 𝐹 is Archimedean. If 𝑓:𝐴is →𝑌 a topological 𝜌(𝑓(𝑥 ), 𝑓(𝑥)) ≤𝑎 𝜌(𝑔(𝑥 ), 𝑔(𝑥)) ≤𝑏 𝑛 𝑛 𝑛, 𝑛 𝑛 for all .Then uniformly continuous function, then 𝑓 has a unique vectorially 𝑋 𝜌(𝑓(𝑥) ,𝑔(𝑥))≤𝜌(𝑓(𝑥 ),𝑓(𝑥))+𝜌(𝑓(𝑥 ),𝑔(𝑥 )) continuous extension to which is also topological uniformly 𝑛 𝑛 𝑛 continuous.

+ 𝜌 (𝑔𝑛 (𝑥 ),𝑔(𝑥))≤𝑎𝑛 +𝑏𝑛 Proof. Let 𝑥∈𝑋. Then there exists a sequence (𝑥𝑛) in 𝐴 such (12) 𝑑,𝐸 that 𝑥𝑛 󳨀󳨀→𝑥. Since the function 𝑓 is vectorial uniformly 𝑛 𝑓(𝑥) = 𝑔(𝑥) 𝑥∈𝐵 𝐵 for all .So ;thatis, .Hence,theset is an continuous on 𝐴 by Theorem 29,thesequence(𝑓(𝑥𝑛)) is 𝐹- 𝐸-closed subset of 𝑋. Cauchy in 𝐹-complete vector metric space 𝑌.Hence,there The following corollary is a consequence of Theorem 26. 𝜌,𝐹 exists an element 𝑦∈𝑌such that 𝑓(𝑥𝑛) 󳨀󳨀→𝑦.Define 𝑔 𝑓 𝑋 𝑔(𝑥) =𝑦 Corollary 27. (𝑋, 𝑑, 𝐸) (𝑌,𝜌,𝐹) an extension of on by . This extension is Let and be vector metric 𝑔 𝑥 𝑓 𝑔 well-defined; that is, the value of at is independent of the spaces and let and be vectorially continuous functions from (𝑥 ) 𝐸 𝑥 𝑋 𝑌 {𝑥 ∈ 𝑋 : 𝑓(𝑥) = 𝑔(𝑥)} 𝐸 𝑋 particular sequence 𝑛 chosen -convergent to . We need to .Iftheset is -dense in ,then 𝑔 𝑋 𝑓=𝑔. to show that is topological uniformly continuous on . Let 𝑎>0in 𝐹.Choose𝑏>0in 𝐸 such that for Definition 28. Let (𝑋,𝑑,𝐸) and (𝑌,𝜌,𝐹) be vector metric 𝑥,𝑦∈𝐴𝑑(𝑥,𝑦)<𝑏implies 𝜌(𝑓(𝑥), 𝑓(𝑦)).Let <𝑎 spaces. 𝑥, 𝑦 ∈𝑋 satisfy 𝑑(𝑥, 𝑦).Choosetwosequences <𝑏 (𝑥𝑛) 𝑑,𝐸 𝑑,𝐸 𝑓:𝑋 →𝑌 and (𝑦𝑛) in 𝐴 such that 𝑥𝑛 󳨀󳨀→𝑥and 𝑦𝑛 󳨀󳨀→𝑦.Then, (a) A function is said to be topological 𝑜 uniformly continuous on 𝑋 if for every 𝑏>0in 𝐹 𝑑(𝑥𝑛,𝑦𝑛) →󳨀 𝑑(𝑥, 𝑦) in 𝐸.Fix𝑛0 such that 𝑑(𝑥𝑛,𝑦𝑛)<𝑏 there exist some 𝑎 in 𝐸 such that for all 𝑥, 𝑦 ∈𝑋, for all 𝑛>𝑛0.Then𝜌(𝑓(𝑥𝑛), 𝑓(𝑦𝑛)) < 𝑎 for all 𝑛>𝑛0.By 𝜌(𝑓(𝑥), 𝑓(𝑦)) <𝑏 whenever 𝑑(𝑥, 𝑦). <𝑎 the vectorial uniform continuity of 𝑓, 𝑓(𝑥𝑛) and 𝑓(𝑦𝑛) are 𝐹- 𝑌 𝑌 𝐹 (b) A function 𝑓:𝑋is →𝑌 said to be vectorial Cauchy sequences in .Since is -complete, there exist two 𝑋 𝐸 𝜌,𝐹 𝜌,𝐹 uniformly continuous on if for every -Cauchy points 𝑢 and V in 𝑌 such that 𝑓(𝑥𝑛) 󳨀󳨀→𝑢and 𝑓(𝑦𝑛) 󳨀󳨀→ V. sequence (𝑥𝑛) the sequence 𝑓(𝑥𝑛) is 𝐹-Cauchy. By the definition of the function 𝑔,wehave𝑔(𝑥) =𝑢 and 𝑜 𝑔(𝑦) = V.Then𝜌(𝑔(𝑥𝑛), 𝑔(𝑦𝑛)) →󳨀 𝜌(𝑔(𝑥), 𝑔(𝑦)) in 𝐹,and Theorem 29. Let (𝑋, 𝑑, 𝐸) and (𝑌,𝜌, 𝐹) be vector metric spaces therefore, 𝜌(𝑔(𝑥), 𝑔(𝑦)).Thisshowsthat ≤𝑏 𝑔 is a topological where 𝐹 is Archimedean. If a function 𝑓:𝑋is →𝑌 uniformly continuous function on 𝑋. topological uniformly continuous, then 𝑓 is vectorial uniformly continuous. 5. Vectorially Continuous Function Spaces Proof. Suppose that (𝑥𝑛) is an 𝐸-Cauchy sequence. Then there exists a sequence (𝑎𝑛) in 𝐸 such that 𝑎𝑛 ↓0and 𝑑(𝑥𝑛,𝑥𝑛+𝑝)≤ Definition 32. Let 𝑋 be any nonempty set and let (𝑌,𝜌,𝐹)be 𝑎𝑛 for all 𝑛 and 𝑝.Let𝑏 be any nonzero positive element in a vector metric space. Then a sequence (𝑓𝑛) of functions from 𝐹. By topological uniform continuity of 𝑓, there exist some 𝑋 to 𝑌 is said to be uniformly 𝐹-convergent to a function 𝑓: 𝑏𝑛 =𝑏𝑛((1/𝑛)𝑏) in 𝐸 such that the inequality 𝑑(𝑥𝑛,𝑥𝑛+𝑝)<𝑏𝑛 𝑋→𝑌, if there exists a sequence (𝑎𝑛) in 𝐹 such that 𝑎𝑛 ↓0 implies 𝜌(𝑓(𝑥𝑛), 𝑓(𝑥𝑛+𝑝)) < (1/𝑛)𝑏 for all 𝑛 and 𝑝. Then there and 𝜌(𝑓𝑛(𝑥), 𝑓(𝑥))𝑛 ≤𝑎 holds for all 𝑥∈𝑋and all 𝑛∈N. 6 Journal of Function Spaces

Now we give the uniform limit theorem in vector metric is vectorial bounded function. This argument gives us the spaces. following result.

Theorem 33. Let (𝑓𝑛) be a sequence of vectorially continuous Corollary 37. 𝐶𝑜(𝑋, 𝐹) (𝑋, 𝑑, 𝐸) The subset V described above is a vector functions between two vector metric spaces and metric space equipped with the 𝐹-valued uniform vector metric (𝑌,𝜌,𝐹).If(𝑓𝑛) is uniformly 𝐹-convergent to 𝑓, then the defined as 𝑑∞(𝑓,𝑔) = sup𝑥∈𝑋|𝑓(𝑥) − 𝑔(𝑥)|. function 𝑓 is vectorially continuous. 𝑑,𝐸 Conflict of Interests Proof. Let (𝑥𝑛)⊆𝑋such that 𝑥𝑛 󳨀󳨀→𝑥in 𝑋.Since(𝑓𝑛) is 𝐹 𝑓 (𝑎 ) 𝐹 uniformly -convergent to ,thereisasequence 𝑛 in The author declares that there is no conflict of interests 𝑎 ↓0 𝜌(𝑓 (𝑥), 𝑓(𝑥)) ≤𝑎 𝑛∈N such that 𝑛 and 𝑛 𝑛 for all .For regarding the publication of this paper. each 𝑘∈N,thereisasequence(𝑏𝑛) in 𝐹 such that 𝑏𝑛 ↓0and 𝜌(𝑓𝑘(𝑥𝑛), 𝑓𝑘(𝑥)) ≤𝑛 𝑏 for all 𝑛∈N by the vectorial continuity References of 𝑓𝑘.Notethatfor𝑘=𝑛 [1] C. D. Aliprantis and O. Burkinshaw, Positive Operators, 𝜌(𝑓(𝑥𝑛),𝑓(𝑥))≤𝜌(𝑓(𝑥𝑛),𝑓𝑛 (𝑥𝑛)) + 𝜌 (𝑓 (𝑥) ,𝑓𝑛 (𝑥)) Springer,Dordrecht,TheNetherlands,2006.

+𝜌(𝑓𝑛 (𝑥𝑛),𝑓𝑛 (𝑥))≤2𝑎𝑛 +𝑏𝑛. [2]W.A.J.LuxemburgandA.C.Zaanen,Riesz Space I,North- (13) Holland, Amsterdam, The Netherland, 1971. [3] C. C¸ evik and I. Altun, “Vector metric spaces and some proper- 𝜌,𝐹 ties,” Topological Methods in Nonlinear Analysis,vol.34,no.2, This implies 𝑓(𝑥𝑛) 󳨀󳨀→ 𝑓(𝑥). pp. 375–382, 2009. Let 𝐴 be a nonempty subset of a vector metric space [4] I. Altun and C. C¸ evik, “Some common fixed point theorems in (𝑋, 𝑑, 𝐸). 𝐸-diameter of 𝐴, 𝑑(𝐴), is defined as sup{𝑑(𝑥, 𝑦) : vector metric spaces,” Filomat,vol.25,no.1,pp.105–113,2011. 𝑥, 𝑦 .Theset∈ 𝐴} 𝐴 is called 𝐸-bounded if there exists an [5] Z. Pales´ and I.-R. Petre, “Iterative fixed point theorems in 𝐸- element 𝑎>0in 𝐸 such that 𝑑(𝑥, 𝑦) ≤𝑎 for all 𝑥, 𝑦. ∈𝐴 metric spaces,” Acta Mathematica Hungarica,vol.140,no.1-2, Every 𝐸-bounded subset of 𝑋 has an 𝐸-diameter whenever 𝐸 pp. 134–144, 2013. is Dedekind complete. [6] I.-R. Petre, “Fixed point theorems in vector metric spaces for single-valued operators,” in Annals of the Tiberiu Popoviciu Definition 34. Afunction𝑓:𝑋between →𝑌 two Seminar of Functional Equations, Approximation and Convexity, vector metric spaces (𝑋, 𝑑, 𝐸) and (𝑌,𝜌, 𝐹) is called vectorial vol. 9, pp. 59–80, 2011. bounded if 𝑓 maps 𝐸-bounded subsets of 𝑋 to 𝐹-bounded [7] I.-R. Petre, “Fixed points for 𝜑-contractions in E-Banach subsets of 𝑌. spaces,” Fixed Point Theory,vol.13,no.2,pp.623–640,2012. [8]P.P.Zabrejko,“𝐾-metric and 𝐾-normed linear spaces: survey,” Theorem 35. Afunction𝑓:𝑋between →𝑌 two vector Universitat de Barcelona: Collectanea Mathematica,vol.48,no. metric spaces (𝑋,𝑑,𝐸) and (𝑌,𝜌,𝐹) is vectorial bounded if 4–6, pp. 825–859, 1997. there exists a positive operator 𝑇:𝐸such →𝐹 that 𝜌(𝑓(𝑥), 𝑓(𝑦)) ≤ 𝑇(𝑑(𝑥, 𝑦)) for all 𝑥, 𝑦 ∈𝑋.

Let 𝐶V(𝑋, 𝐹) and 𝐶𝑡(𝑋, 𝐹) be the collections of all vectorially continuous and topologically continuous functions between a vector metric space (𝑋, 𝑑, 𝐸) and a Riesz space 𝐹, respectively. By Theorem 4, 𝐶𝑡(𝑋, 𝐹) ⊆𝐶 V(𝑋, 𝐹) whenever 𝐹 is Archimedean.

Theorem 36. The spaces 𝐶V(𝑋, 𝐹) and 𝐶𝑡(𝑋, 𝐹) are Riesz spaces with the natural partial ordering defined as 𝑓≤𝑔 whenever 𝑓(𝑥) ≤ 𝑔(𝑥) for all 𝑥∈𝑋.

Consider an 𝐸-bounded vector metric space 𝑋 and a 𝑜 Dedekind complete Riesz space 𝐹.Let𝐶V(𝑋, 𝐹) be a subset of 𝑜 𝐶V(𝑋, 𝐹) such that, for any 𝑓 in 𝐶V(𝑋, 𝐹),thereisapositive operator 𝑇:𝐸satisfying →𝐹 |𝑓(𝑥) − 𝑓(𝑦)| ≤ 𝑇(𝑑(𝑥, 𝑦)) for all 𝑥, 𝑦. ∈𝑋 Since the Birkhoff inequality1 ([ , 1.9(2)]; [2, 12.4(ii)]) 󵄨 󵄨 󵄨 󵄨 󵄨 󵄨 󵄨𝑓∨𝑔(𝑥) −𝑓∨𝑔(𝑦)󵄨 ≤ 󵄨𝑓 (𝑥) −𝑓(𝑦)󵄨 + 󵄨𝑔 (𝑥) −𝑔(𝑦)󵄨 (14) 𝑜 holds for all 𝑥, 𝑦, ∈𝑋 then the subset 𝐶V(𝑋, 𝐹) is a Riesz 𝑜 subspace of 𝐶V(𝑋, 𝐹).ByTheorem 35,every𝑓∈𝐶V(𝑋, 𝐹) Copyright of Journal of Function Spaces is the property of Hindawi Publishing Corporation and its content may not be copied or emailed to multiple sites or posted to a listserv without the copyright holder's express written permission. However, users may print, download, or email articles for individual use.