Preprint
Article

On Fuzzy Near Best Approximation

Altmetrics

Downloads

97

Views

32

Comments

0

This version is not peer-reviewed

Submitted:

16 September 2023

Posted:

19 September 2023

You are already at the latest version

Alerts
Abstract
Given a fuzzy normed space $ \left( X,N \right) $‎, ‎we will introduce the notion of fuzzy near best approximation within a relative distance $ \rho \geq 0 $‎. ‎Some basic properties are characterized and also many examples for illustration are presented‎.
Keywords: 
Subject: Computer Science and Mathematics  -   Analysis

1. INTRODUCTION AND PRELIMINARIES

Let X be a normed linear space and Y be a subset of X. If x X , then the distance of x from Y is denoted by d ( x , Y ) that is
d x , Y = inf x y s . t . y Y .
An element y Y is said to be a best approximation to x X from Y if x y = d ( x , Y ) . The set of all best approximations to x X from Y is denoted by P Y ( x ) . If for any x X , P Y ( x ) , then we say that Y is proximinal in X. Also if for any x X , P Y ( x ) is singleton, therefore Y is a Chebyshev subset of X. A sequence y n n Y is called a minimizing sequence for x X if lim n x y n = d x , Y [4].
An element y n Y is said to be a near best approximation to x within a relative distance ρ 0 if,
x y n ( 1 + ρ ) x x b = ( 1 + ρ ) d x , Y
where x b is a best approximation to x from Y[5]. The set of all near best approximations to x X from Y is denoted by P Y n ( x ) .
Let X i i I be a family of linear spaces. Then the algebraic direct sum of the spaces X i , i.e.,
i I X i = x = x i i I | x i = 0 for all but finitely many i I
with the pointwise vector-space operations as follows, is a linear space,
x + y = x i + y i i I
and
α x = α x i i I
for all x , y i I X i and α C or R [1].
Also let X and Y be linear spaces over C or R . Then the algebraic tensor product of X and Y is denoted by X Y . If X and Y are the dual spaces of X and Y respectively, then for all x X and y Y , the map x y : X × Y C or R defined by
x y f , g = f x g y , f X , g Y
is a bilinear map. For the basic properties concerning the tensor product of linear spaces, we refer the reader to [2].
Definition 1.1.
[7]Let X be a linear space. A function N : X × R [ 0 , 1 ] is said to be fuzzy norm on X if for all x , y X and all s , t R :
1
- N x , t = 0 for t 0 .
2
- N x , t = 1 for every t R + if and only if x = 0 .
3
- N c x , t = N x , t c for every c 0 and t R .
4
- N x + y , s + t min N x , s , N y , t for every s , t R .
5
- N x , . is non-decreasing on R and lim t N x , t = 1 .
Note that by part 3 of Definition 1.1, N x , t = N x , t for all x X and t R .
Definition 1.2.
[6]Let Y be a nonempty subset of a fuzzy normed space X , N . For x X and t R , let
d Y , x , t = sup N y x , t , y Y .
An element y 0 Y is said to be a fuzzy best approximation to x from Y if
N y 0 x , t = d Y , x , t ,
for all t R . The set of all fuzzy best approximations to x from Y is denoted by P Y f x .
Definition 1.3.
Let X be a linear space and Y be a subset of X. Also let N : X × R [ 0 , 1 ] be a fuzzy norm on X. An element y 0 Y is said to be a fuzzy near best approximation to x from Y within a relative distance ρ 0 if,
N x y 0 , t N x Y , t 1 + ρ
for all t R , where
N x Y , t 1 + ρ = sup N x y , t 1 + ρ , y Y .
The set of all fuzzy near best approximations to x from Y within the relative distance ρ is denoted by P Y f n ρ x .
Remark 1.4.
Trivially the notion of fuzzy best approximation is nothing else than fuzzy near best approximation within the relative distance ρ = 0 .
Proposition 1.5.
Let X be a linear space and Y be a nonempty subset of X. Also let x , y X , z Y , ρ 0 and α R . Then
1-
If P Y f n ρ x , then P α Y f n ρ α x = α P Y f n ρ x
2-
If α 0 , then P Y f n ρ α x = α P Y α f n ρ x
3-
P Y f n ρ x + z = P Y z f n ρ x + z
4-
P Y + y f n ρ x + y = P Y f n ρ x + y .
Proof. 
We’ll prove the first part. The rest of the parts are easily verified. In the case where α = 0 , the equality trivially holds. For α 0 , let z 0 P α Y f n ρ α x . Then
N α x z 0 , t N α x α Y , t 1 + ρ , t R .
Therefore
N x z 0 α , t α N x Y , t α 1 + ρ , t R .
Replacing t by α t , we have
N x z 0 α , t N x Y , t 1 + ρ , t R .
So z 0 α P Y f n ρ x . Hence z 0 α P Y f n ρ x .
Conversely, let z 0 α P Y f n ρ x . Then z 0 α P Y f n ρ x . Therefore
N x z 0 α , t N x Y , t 1 + ρ , t R .
So
N α x z 0 , α t N x Y , t 1 + ρ , t R .
Replacing t by t α , we have
N α x z 0 , t N x Y , t α 1 + ρ = N α x α Y , t 1 + ρ , t R .
Then z 0 P α Y f n ρ α x . So we can conclude that P α Y f n ρ α x = α P Y f n ρ x . □
Proposition 1.6.
Let X , N be a fuzzy normed linear space and Y be a subset of X. Then every fuzzy best approximation to x X from Y is a fuzzy near best approximation to x from Y within every relative distance ρ 0 .
Proof. 
Let y 0 P Y f x . So N x y 0 , t = N x Y , t for all t R . We shall show that N x y 0 , t N x Y , t 1 + ρ for all t R and every ρ 0 . Let y Y and t R . So by Definition 1.1 part 5,
N x y , t 1 + ρ N x y , t N x Y , t = N x y 0 , t .
It follows that N x y , t 1 + ρ N x y 0 , t for all t R and y Y . Hence
sup N x y , t 1 + ρ , y Y N x y 0 , t
for all t R . Therefore
N x Y , t 1 + ρ N x y 0 , t
for all t R , providing y 0 P Y f n ρ x . □
Proposition 1.7.
Let X , N be a fuzzy normed linear space, Y be a subset of X , x X , and ρ 0 . If Y is convex, then P Y f x and P Y f n ρ ( x ) are convex.
Proof. 
Let y 1 , y 2 P Y f n ρ ( x ) and 0 < α < 1 . So by Definition 1.1, for all t R we have
N x α y 1 + 1 α y 2 , t = N α x + 1 α x α y 1 + 1 α y 2 , α t + 1 α t = N α x y 1 + 1 α x y 2 , α t + 1 α t min N α x y 1 , α t , N 1 α x y 2 , 1 α t = min N x y 1 , t , N x y 2 , t N x Y , t 1 + ρ .
Hence α y 1 + 1 α y 2 P Y f n ρ ( x ) . □
The next example shows that the notion of fuzzy near best approximation is different from the notion of fuzzy best approximation.
Example 1.8.
Suppose that X = R , Y = [ 1 , 1 . 5 ] , x = 0 and ρ = 0 . 5 . Also let
N x , t = 0 t x 1 t > x
be a fuzzy norm on X. Then, P Y 0 = P Y f 0 = 1 and P Y f n 0 . 5 0 = Y . Generally for 0 ρ 0 . 5 , P Y f n ρ 0 = [ 1 , 1 + ρ ] . Indeed, if 0 < t 1 , then N 1 , t = 0 and for all y Y , t y . So N y , t = 0 for all y Y . It follows that N Y , t = 0 . Hence N 1 , t = N Y , t for all 0 < t 1 . If t > 1 , then N 1 , t = 1 and so N Y , t = 1 . Therefore N 1 , t = N Y , t for all t R . This shows that 1 P Y f 0 . If 1 < y 0 1 . 5 , then for t = 1 + y 0 1 2 , N y 0 , t = 0 and N 1 , t = 1 . It follows that 0 = N y 0 , t N Y , t = 1 for t = 1 + y 0 1 2 . So y 0 P Y f 0 . Hence P Y f 0 = 1 .
Now we will show that P Y f n 0 . 5 0 = [ 1 , 1 . 5 ] = Y . Let y 0 [ 1 , 1 . 5 ] . If t y 0 , then N y 0 , t = 0 and 2 3 t 2 3 y 0 2 3 3 2 = 1 . So N y , 2 3 t = 0 for all y [ 1 , 1 . 5 ] . It follows that N Y , 2 3 t = 0 . Hence N y 0 , t N Y , 2 3 t for all t y 0 . If t > y 0 , then N y 0 , t = 1 N Y , 2 3 t . Therefore N y 0 , t N Y , 2 3 t for all t R . Hence y 0 P Y f n 0 . 5 0 . This shows that P Y f n 0 . 5 0 = [ 1 , 1 . 5 ] = Y .
Example 1.9.
Suppose that X = R , Y = 0 , 1 , x = 0 and ρ 0 . Also let
N x , t = 0 t x 1 t > x
be a fuzzy norm on X. Then P Y 0 = P Y f 0 = P Y f n ρ 0 = . Indeed, since P Y 0 P Y f 0 P Y f n ρ 0 , it’s enough to prove that P Y f n ρ 0 = . Let y 0 Y be an arbitrary element. Choose n N such that 1 n < y 0 1 + ρ . So if t = y 0 , then
0 = N y 0 , t < N 1 n , t 1 + ρ = 1 .
It follows that
0 = N y 0 , t < N Y , t 1 + ρ = 1 .
Hence y 0 P Y f n ρ 0 and so P Y f n ρ 0 = .
Proposition 1.10.
Let X be a linear space, Y be a subset of X and N : X × R [ 0 , 1 ] be a fuzzy norm on X. For any ρ 0 , x Y , P Y f n ρ x = x .
Proof. 
As 1 = N 0 , t = N x x , t N x Y , t 1 + ρ for all t > 0 , we can conclude that N x x , t N x Y , t 1 + ρ for all t R . Hence x P Y f n ρ x .
Now let y 0 P Y f n ρ x . Then for all t > 0
N x y 0 , t N x Y , t 1 + ρ N x x , t 1 + ρ = 1 .
Therefore N x y 0 , t = 1 for all t > 0 . Hence x y 0 = 0 . Then y 0 = x . □
Theorem 1.11
Let X be a linear space, Y be a subset of X and N : X × R [ 0 , 1 ] be a fuzzy norm on X. For x X , if ρ 1 ρ 2 , then P Y f n ρ 1 x P Y f n ρ 2 x .
Proof. 
If y 0 P Y f n ρ 1 x , then
N x y 0 , t N x Y , t 1 + ρ 1 .
Since for all t R and y Y
N x y , t 1 + ρ 1 N x y , t 1 + ρ 2 ,
N x Y , t 1 + ρ 1 N x Y , t 1 + ρ 2 .
Therefore
N x y 0 , t N x Y , t 1 + ρ 2 .
It follows that y 0 P Y f n ρ 2 x . □
Theorem 1.12
Let X be a linear space, Y be a subset of X and N : X × R [ 0 , 1 ] be a fuzzy norm on X. If x X and ρ 1 ρ 2 ρ 3 such that ρ m ρ as m , then P Y f n ρ x m = 1 P Y f n ρ m x . In particular, if N z , . is lower semicontinuous at every z x Y , then P Y f n ρ x = m = 1 P Y f n ρ m x .
Proof. 
By Theorem 1.11, since ρ m ρ for all m N ,
P Y f n ρ x P Y f n ρ 1 x P Y f n ρ x P Y f n ρ 2 x P Y f n ρ x P Y f n ρ m x , m N .
Then P Y f n ρ x m = 1 P Y f n ρ m x .
If y 0 m = 1 P Y f n ρ m x and N z , . is lower semicontinuous for every z x Y , then for all m N and for all t R we have
N x y 0 , t N x Y , t 1 + ρ m N x y , t 1 + ρ m , y Y .
Since N x y , . is lower semicontinuous for all x y and t 1 + ρ m t 1 + ρ for every m N ,
N x y 0 , t lim m N x y , t 1 + ρ m = N x y , t 1 + ρ , y Y .
So m = 1 P Y f n ρ m x P Y f n ρ x .
N x y 0 , t sup N x y , t 1 + ρ , y Y = N x Y , t 1 + ρ .
Hence m = 1 P Y f n ρ m x P Y f n ρ x .
We use the following lemma in the proof of Proposition 2.1 and Theorem 2.5.
Lemma 1.13.
Let 0 a i 1 and 0 b i 1 for all 1 i m . Then
min min a i , b i | 1 i m = min min a i | 1 i m , min b i | 1 i m .
Proof. 
It’s obvious. □

2. Fuzzy Near Best Approximation On Direct Sum And Tensor Product Of Linear Spaces

Proposition 2.1.
Let X i , N i i I be a family of fuzzy normed spaces. Then i I X i , N is a fuzzy normed space, where N : i I X i × R [ 0 , 1 ] is defined by
N x i i I , t = inf N i x i , t | i I .
Proof. 
To prove the above proposition, we only prove conditions 4 and 5 of Definition 1.1 and we leave the rest to the reader. To prove the fourth part, suppose that x i i I , y i i I i I X i and s , t R . So x i = 0 and y i = 0 for all but finitely many i k I , 1 k m . Clearly if s + t 0 , then the inequality
N x i i I + y i i I , s + t min N x i i I , s , N y i i I , t
holds. Also if s + t > 0 , s 0 or t 0 , then obviously inequality 2.1 holds.
Let s + t > 0 , s > 0 and t > 0 . Then
N x i i I + y i i I , s + t = N x i + y i i I , s + t inf N i x i + y i , s + t | i I inf min N i x i , s , N i y i , t | i I = inf min N i k x i k , s , N i k y i k , t , 1 | 1 k m = min min N i k x i k , s , N i k y i k , t , 1 | 1 k m = min min N i k x i k , s , N i k y i k , t | 1 k m = min min N i k x i k , s | 1 k m , min N i k y i k , t | 1 k m = min min N i x i , s | i I , min N i y i , t | i I = min inf N i x i , s | i I , inf N i y i , t | i I = min N x i i I , s , N y i i I , t .
To prove the fifth part, suppose that x i i I i I X i . So x i = 0 for all but finitely many i k I , 1 k m . Hence
lim t N x i i I , t = lim t inf N i x i , t | i I = lim t inf N i k x i k , t , 1 | 1 k m = lim t min N i k x i k , t , 1 | 1 k m = lim t min N i k x i k , t | 1 k m = min lim t N i k x i k , t | 1 k m = min 1 = 1 .
Since N i x i , . is increasing for all x i X i , N x i i I , . is increasing for all x i i I . □
Corollary 2.2.
Let N i : X i × R [ 0 , 1 ] be a fuzzy norm on X i for i = 1 , 2 , , n . Then N : X 1 × X 2 × × X n × R [ 0 , 1 ] defined by
N x 1 , x 2 , , x n , t = min N 1 x 1 , t , N 2 x 2 , t , , N n x n , t
is a fuzzy norm on X 1 × X 2 × × X n .
Proposition 2.3.
Let X i , N i i = 1 n be a finite family of fuzzy normed spaces and N : X 1 × X 2 × × X n × R [ 0 , 1 ] is defined by
N x 1 , x 2 , , x n , t = min N 1 x 1 , t , N 2 x 2 , t , , N n x n , t .
Also let Y i X i , x i X i and ρ i 0 for all 1 i n . Then
P Y 1 f n ρ 1 x 1 × P Y 2 f n ρ 2 x 2 × × P Y n f n ρ n x n P Y 1 × Y 2 × × Y n f n max ρ i x 1 , x 2 , , x n .
Proof. 
Let y 1 , y 2 , , y n P Y 1 f n ρ 1 x 1 × P Y 2 f n ρ 2 x 2 × × P Y n f n ρ n x n . Then y 1 P Y 1 f n ρ 1 x 1 , y 2 P Y 2 f n ρ 2 x 2 , ⋯, y n P Y n f n ρ n x n . Therefore
N 1 x 1 y 1 , t N 1 x 1 z 1 , t 1 + ρ 1 N 1 x 1 z 1 , t 1 + max 1 i n ρ i N 2 x 2 y 2 , t N 2 x 2 z 2 , t 1 + ρ 2 N 2 x 2 z 2 , t 1 + max 1 i n ρ i N n x n y n , t N n x n z n , t 1 + ρ n N n x n z n , t 1 + max 1 i n ρ i
for every z i Y i and t R . Then
min N 1 x 1 y 1 , t , N 2 x 2 y 2 , t , , N n x n y n , t min N 1 1 , N 2 2 , , N n n ,
where N k k = N k x k z k , t 1 + max 1 i n ρ i , 1 k n . Therefore
N x 1 , x 2 , , x n y 1 , y 2 , , y n , t sup z 1 , z 2 , , z n Y 1 × Y 2 × × Y n N x 1 , x 2 , , x n z 1 , z 2 , , z n , t 1 + max 1 i n ρ i .
Hence
N x 1 , x 2 , , x n y 1 , y 2 , , y n , t N x 1 , x 2 , , x n Y 1 × Y 2 × × Y n , t 1 + max 1 i n ρ i .
In the next example, we will show that the converse of inclusion 2 is not true in general.
Example 2.4.
Let X 1 = X 2 = R , Y 1 = Y 2 = [ 1 , 3 ] , and N i : X i × R [ 0 , 1 ] is defined by
N i α , t = 0 t α 1 t > α
for i = 1 , 2 . Assume that N x , y , t = min N 1 x , t , N 2 y , t , ρ = 1 , x 1 = 0 and x 2 = 1 2 . Then we have
P Y 1 f n 1 0 = [ 1 , 2 ] P Y 2 f n 1 1 2 = [ 1 , 3 2 ] P Y 1 f n 1 0 × P Y 2 f n 1 1 2 = [ 1 , 2 ] × [ 1 , 3 2 ] .
It is easy to see that
N x , y , t = 0 t max x , y 1 t > max x , y
for all x , y , t R . Also a sufficient effort can be applied to show that
P Y 1 × Y 2 f n 1 0 , 1 2 = [ 1 , 2 ] × [ 1 , 5 2 ] .
Theorem 2.5.
Let X , N 1 and Y , N 2 be fuzzy normed spaces. Also let B X and B Y be the bases of X and Y respectively. Define N : X Y × R [ 0 , 1 ] by
N j = 1 n α j x j y j , t = min N 1 α j x j , t , N 2 α j y j , t | 1 j n ,
where x j B X , y j B Y , n N , α j C and t R . Then X Y , N is a fuzzy normed space.
Proof. (1) : At the first we will prove that N is well defined. Let z , s , w , t X Y × R . So z = w and s = t . Hence there exists an n N , x j j = 1 n B X , y j j = 1 n B Y and α j , β j C such that z = j = 1 n α j x j y j and w = j = 1 n β j x j y j . It follows that α j = β j for all 1 j n . Therefore
min N 1 α j x j , s , N 2 α j y j , s | 1 j n = min N 1 β j x j , t , N 2 β j y j , t | 1 j n ,
providing N z , s = N w , t .
In the sequel we will prove the parts 2, 3, 4 and 5 of Definition 1.1.
(2) : Let z = j = 1 n α j x j y j and N z , t = 1 for all t > 0 . So
min N 1 α j x j , t , N 2 α j y j , t | 1 j n = 1
for all t > 0 . It follows that N 1 α j x j , t = N 2 α j y j , t = 1 for all 1 j n and for all t > 0 . Hence α j x j = 0 and α j y j = 0 for all 1 j n . Therefore α j = 0 for all 1 j n . This shows that z = 0 0 . Also for all t > 0 , since 0 0 = 0 x y for all x B X and y B Y ,
N 0 0 , t = N 0 x y , t = min N 1 0 , t , N 2 0 , t = min 1 = 1 .
(3) : Let c 0 and j = 1 n α j x j y j X Y . So
N c j = 1 n α j x j y j , t = N j = 1 n c α j x j y j , t = min N 1 c α j x j , t , N 2 c α j y j , t | 1 j n = min N 1 α j x j , t c , N 2 α j y j , t c | 1 j n = N j = 1 n α j x j y j , t c ,
for all t R .
(4) : Let z , w X Y and s , t R . So z = j = 1 n α j x j y j and w = j = 1 n β j x j y j , where n N , α j , β j C , x j j = 1 n B X and y j j = 1 n B Y . Hence
N j = 1 n α j x j y j + j = 1 n β j x j y j , s + t = N j = 1 n α j + β j x j y j , s + t = min N 1 α j + β j x j , s + t , N 2 α j + β j y j , s + t | 1 j n = min N 1 α j x j + β j x j , s + t , N 2 α j y j + β j y j , s + t | 1 j n min min N 1 α j x j , s , N 1 β j x j , t , min N 2 α j y j , s , N 2 β j y j , t | 1 j n = min min A j , B j , min C j , D j | 1 j n = min min A j , C j | 1 j n , min B j , D j | 1 j n = min N j = 1 n α j x j y j , s , N j = 1 n β j x j y j , t ,
where
A j = N 1 α j x j , s , B j = N 1 β j x j , t , C j = N 2 α j y j , s , D j = N 2 β j y j , t .
(5) : Let s t and z = j = 1 n α j x j y j . Clearly N 1 α j x j , s N 1 α j x j , t and N 2 α j y j , s N 2 α j y j , t for all 1 j n . So
min N 1 α j x j , s , N 2 α j y j , s | 1 j n min N 1 α j x j , t , N 2 α j y j , t | 1 j n .
It follows that
N j = 1 n α j x j y j , s N j = 1 n α j x j y j , t .
Hence N z , . : R [ 0 , 1 ] is increasing for all z X Y . Also
lim t N z , t = lim t min N 1 α j x j , t , N 2 α j y j , t | 1 j n = min lim t N 1 α j x j , t , lim t N 2 α j y j , t | 1 j n = min 1 = 1 .
Theorem 2.6.
Let X and Y be linear spaces and N : X Y × R [ 0 , 1 ] be a fuzzy norm. Then for all x X 0 and y Y 0 , the maps N x : Y × R [ 0 , 1 ] and N y : X × R [ 0 , 1 ] , where N x z , t = N x z , t and N y w , t = N w y , t , are fuzzy norms on Y and X respectively.
Proof. 
We only prove that N x : Y × R [ 0 , 1 ] is a fuzzy norm for all x X 0 .
(1) : Let z Y and t 0 . So N x z , t = N x z , t = 0 .
(2) : If z = 0 , then N x 0 , t = N x 0 , t = N 0 0 , t = 1 for all t > 0 . Also if N x z , t = 1 for all t > 0 , then N x z , t = 1 for all t > 0 . It follows that x z = 0 0 . Since x 0 , there exists f X such that f x 0 . Let g Y be an arbitrary element. As x z is a bilinear map on X × Y , x z f , g = 0 0 f , g . So f x g z = 0 for all g Y . Since f x 0 , g z = 0 for all g Y . It follows that z = 0 .
(3) : Let c 0 and z Y . So
N x c z , t = N x c z , t = N c x z , t = N x z , t c = N x z , t c
for all z Y and t R .
(4) : Let z 1 , z 2 Y and s , t R . So
N x z 1 + z 2 , s + t = N x z 1 + z 2 , s + t = N x z 1 + x z 2 , s + t min N x z 1 , s , N x z 2 , t = min N x z 1 , s , N x z 2 , t .
(5) : Let s t and z Y . So N x z , s N x z , t . It follows that N x z , s N x z , t . Hence N x z , . : R [ 0 , 1 ] is increasing and
lim t N x z , t = lim t N x z , t = 1
for all z Y .
Therefore N x : Y × R [ 0 , 1 ] is a fuzzy norm. □
Example 2.7.
Let X = Y = R 2 be linear spaces over R with the bases B X = B Y = e 1 = 1 , 0 , e 2 = 0 , 1 . Also let N 1 = N 2 : R 2 × R [ 0 , 1 ] are defined by
N i j = 1 2 α j e j , t = 0 t max α 1 , α 2 1 t > max α 1 , α 2
for i = 1 , 2 . Clearly N 1 and N 2 are fuzzy norms on R 2 . According to Theorem 2.5, if N : X Y × R [ 0 , 1 ] is defined by
N α 1 e 1 e 1 + α 2 e 1 e 2 + α 3 e 2 e 1 + α 4 e 2 e 2 , t = min N 1 α 1 e 1 , t , N 1 α 2 e 1 , t , N 1 α 3 e 2 , t , N 1 α 4 e 2 , t N 2 α 1 e 1 , t , N 2 α 2 e 2 , t , N 2 α 3 e 1 , t , N 2 α 4 e 2 , t ,
then the equalities
N 2 α 1 e 1 , t = N 1 α 1 e 1 , t , N 2 α 2 e 2 , t = N 1 α 2 e 1 , t , N 2 α 3 e 1 , t = N 1 α 3 e 2 , t , N 2 α 4 e 2 , t = N 1 α 4 e 2 , t
imply that
N α 1 e 1 e 1 + α 2 e 1 e 2 + α 3 e 2 e 1 + α 4 e 2 e 2 = min N 1 α 1 e 1 , t , N 1 α 2 e 1 , t , N 1 α 3 e 2 , t , N 1 α 4 e 2 , t = 0 t max α i , 1 i 4 1 t > max α i , 1 i 4 .
Let x 0 = 1 5 e 1 , y 0 = 2 e 2 , K 1 = B X , K 2 = B Y , K = B X Y = B X B Y and ρ = 1 10 . Also let P K 1 f n 1 10 x 0 and P K 2 f n 1 10 y 0 be the set of all fuzzy near best approximations to x 0 and y 0 within the relative distance ρ = 1 10 with respect to the fuzzy norms N 1 and N 2 respectively. If P K f n 1 10 x 0 y 0 be the set of all fuzzy near best approximations to x 0 y 0 within the relative distance ρ = 1 10 with respect to the fuzzy norm N, then a straightforward calculation reveals that
P K 1 f n 1 10 x 0 = e 2 P K 2 f n 1 10 y 0 = e 1 P K f n 1 10 x 0 y 0 = e 1 e 2 e 2 e 1 .
This example shows that there is no a relation between P K f n 1 10 x 0 y 0 and P K 1 f n 1 10 x 0 P K 2 f n 1 10 y 0 .

References

  1. Alexander C. R. Belton. Functional Analysis, Hyperlinked and revised edition, (2004).
  2. F.F. Bonsall and J. Duncan. Complete Normed Algebras, Springer-Verlag, Berlin Heidelberg New York, (1973).
  3. T. Bag and S.K. Samanta. Some fixed point theorems in fuzzy normed linear spaces. Inform. Sci., 2007, 177, 3271–3289. [CrossRef]
  4. F. Deutschi. Best Approximation in Inner Product Spaces, Canad. Math. Soc., (2000).
  5. F.D. Kovaca and F.E. Levisb. Extended best Lp-approximation is near-best approximation in Lq, p − 1 ≤ q, Journal of Approximation Theory., 284 (2022), 105819. [CrossRef]
  6. H. Mazaheri, Z. Bizhanzadeh, S. M. Moosavi, M. A. Dehghan. Fuzzy farthest points and fuzzy best approximation points in fuzzy normed spaces, Theory of Approximation and Applications., Vol. 13 No.1, (2019), 11-25.
  7. I. Sadeqi, F. Solaty Kia. Fuzzy normed linear space and its topological structure Solitons and Fractals., 40 (2009), 2576–2589. [CrossRef]
Disclaimer/Publisher’s Note: The statements, opinions and data contained in all publications are solely those of the individual author(s) and contributor(s) and not of MDPI and/or the editor(s). MDPI and/or the editor(s) disclaim responsibility for any injury to people or property resulting from any ideas, methods, instructions or products referred to in the content.
Copyright: This open access article is published under a Creative Commons CC BY 4.0 license, which permit the free download, distribution, and reuse, provided that the author and preprint are cited in any reuse.
Prerpints.org logo

Preprints.org is a free preprint server supported by MDPI in Basel, Switzerland.

Subscribe

© 2024 MDPI (Basel, Switzerland) unless otherwise stated