On Fuzzy Near Best Approximation

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On Fuzzy Near Best Approximation

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Submitted:

16 September 2023

Posted:

19 September 2023

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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

MSC: 46A32; 41A50; 41A17; 41A65

1. INTRODUCTION AND PRELIMINARIES

Let X be a normed linear space and Y be a subset of X. If

x \in 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 \in Y\} .

An element

y \in Y

is said to be a best approximation to

x \in X

from Y if

∥x - y∥ = d (x, Y)

. The set of all best approximations to

x \in X

from Y is denoted by

P_{Y} (x)

. If for any

x \in X

P_{Y} (x) \neq \emptyset

, then we say that Y is proximinal in X. Also if for any

x \in X

P_{Y} (x)

is singleton, therefore Y is a Chebyshev subset of X. A sequence

{\{y_{n}\}}_{n} \subseteq Y

is called a minimizing sequence for

x \in X

{lim}_{n \to \infty} ∥x - y_{n}∥ = d (x, Y)

[4].

An element

y^{n} \in Y

is said to be a near best approximation to x within a relative distance

ρ \geq 0

if,

∥x - y^{n}∥ \leq (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 \in X

from Y is denoted by

P_{Y}^{n} (x)

Let

{\{X_{i}\}}_{i \in I}

be a family of linear spaces. Then the algebraic direct sum of the spaces

X_{i}

, i.e.,

\sum_{i \in I} X_{i} = \{x = {(x_{i})}_{i \in I} | x_{i} = 0 for all but finitely many i \in I\}

with the pointwise vector-space operations as follows, is a linear space,

x + y = {(x_{i} + y_{i})}_{i \in I}

and

α x = {(α x_{i})}_{i \in I}

for all

x, y \in \sum_{i \in I} X_{i}

and

α \in C or R

[1].

Also let X and Y be linear spaces over

C

R

. Then the algebraic tensor product of X and Y is denoted by

X \otimes Y

. If

X^{'}

and

Y^{'}

are the dual spaces of X and Y respectively, then for all

x \in X

and

y \in Y

, the map

x \otimes y : X^{'} \times Y^{'} ⟶ C (or R)

defined by

(x \otimes y) (f, g) = f (x) g (y), f \in X^{'}, g \in 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 \times R ⟶ [0, 1]

is said to be fuzzy norm on X if for all

x, y \in X

and all

s, t \in R

1: - $N (x, t) = 0$ for $t \leq 0$ .
2: - $N (x, t) = 1$ for every $t \in R^{+}$ if and only if $x = 0$ .
3: - $N (c x, t) = N (x, \frac{t}{|c|})$ for every $c \neq 0$ and $t \in R$ .
4: - $N (x + y, s + t) \geq min \{N (x, s), N (y, t)\}$ for every $s, t \in R$ .
5: - $N (x, .)$ is non-decreasing on $R$ and ${lim}_{t \to \infty} N (x, t) = 1$ .

Note that by part 3 of Definition 1.1,

N (- x, t) = N (x, t)

for all

x \in X

and

t \in R

Definition 1.2.

[6]Let Y be a nonempty subset of a fuzzy normed space

(X, N)

. For

x \in X

and

t \in R

, let

d (Y, x, t) = sup \{N (y - x, t), y \in Y\} .

An element

y_{0} \in 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 \in 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 \times R ⟶ [0, 1]

be a fuzzy norm on X. An element

y_{0} \in Y

is said to be a fuzzy near best approximation to x from Y within a relative distance

ρ \geq 0

if,

N (x - y_{0}, t) \geq N (x - Y, \frac{t}{1 + ρ})

for all

t \in R

, where

N (x - Y, \frac{t}{1 + ρ}) = sup \{N (x - y, \frac{t}{1 + ρ}), y \in 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 \in X

z \in Y, ρ \geq 0

and

α \in R

. Then

1-: If $P_{Y}^{f n (ρ)} (x) \neq \emptyset$ , then $P_{α Y}^{f n (ρ)} (α x) = α P_{Y}^{f n (ρ)} (x)$
2-: If $α \neq 0$ , then $P_{Y}^{f n (ρ)} (α x) = α P_{\frac{Y}{α}}^{f n (ρ)} (x)$
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 \neq 0

, let

z_{0} \in P_{α Y}^{f n (ρ)} (α x)

. Then

N (α x - z_{0}, t) \geq N (α x - α Y, \frac{t}{1 + ρ}), (\forall t \in R) .

Therefore

N (x - \frac{z_{0}}{α}, \frac{t}{|α|}) \geq N (x - Y, \frac{t}{|α| (1 + ρ)}), (\forall t \in R) .

Replacing t by

|α| t

, we have

N (x - \frac{z_{0}}{α}, t) \geq N (x - Y, \frac{t}{1 + ρ}), (\forall t \in R) .

\frac{z_{0}}{α} \in P_{Y}^{f n (ρ)} (x)

. Hence

z_{0} \in α P_{Y}^{f n (ρ)} (x)

Conversely, let

z_{0} \in α P_{Y}^{f n (ρ)} (x)

. Then

\frac{z_{0}}{α} \in P_{Y}^{f n (ρ)} (x)

. Therefore

N (x - \frac{z_{0}}{α}, t) \geq N (x - Y, \frac{t}{1 + ρ}), (\forall t \in R) .

N (α x - z_{0}, |α| t) \geq N (x - Y, \frac{t}{1 + ρ}), (\forall t \in R) .

Replacing t by

\frac{t}{|α|}

, we have

\begin{matrix} N (α x - z_{0}, t) & \geq N (x - Y, \frac{t}{|α| (1 + ρ)}) \\ = N (α x - α Y, \frac{t}{1 + ρ}), (\forall t \in R) . \end{matrix}

Then

z_{0} \in 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 \in X

from Y is a fuzzy near best approximation to x from Y within every relative distance

ρ \geq 0

Proof.

Let

y_{0} \in P_{Y}^{f} (x)

. So

N (x - y_{0}, t) = N (x - Y, t)

for all

t \in R

. We shall show that

N (x - y_{0}, t) \geq N (x - Y, \frac{t}{1 + ρ})

for all

t \in R

and every

ρ \geq 0

. Let

y \in Y

and

t \in R

. So by Definition 1.1 part 5,

N (x - y, \frac{t}{1 + ρ}) \leq N (x - y, t) \leq N (x - Y, t) = N (x - y_{0}, t) .

It follows that

N (x - y, \frac{t}{1 + ρ}) \leq N (x - y_{0}, t)

for all

t \in R

and

y \in Y

. Hence

\begin{matrix} sup \{N (x - y, \frac{t}{1 + ρ}), y \in Y\} \leq N (x - y_{0}, t) \end{matrix}

for all

t \in R

. Therefore

\begin{matrix} N (x - Y, \frac{t}{1 + ρ}) \leq N (x - y_{0}, t) \end{matrix}

for all

t \in R

, providing

y_{0} \in 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 \in X,

and

ρ \geq 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} \in P_{Y}^{f n (ρ)} (x)

and

0 < α < 1

. So by Definition 1.1, for all

t \in R

we have

\begin{matrix} 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) \\ \geq 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)\} \\ \geq N (x - Y, \frac{t}{1 + ρ}) . \end{matrix}

Hence

α y_{1} + (1 - α) y_{2} \in 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

\begin{matrix} N (x, t) = \{\begin{matrix} 0 & t \leq |x| \\ 1 & t > |x| \end{matrix} \end{matrix}

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 \leq ρ \leq 0.5

P_{Y}^{f n (ρ)} (0) = [1, 1 + ρ]

. Indeed, if

0 < t \leq 1

, then

N (1, t) = 0

and for all

y \in Y

t \leq y

. So

N (y, t) = 0

for all

y \in Y

. It follows that

N (Y, t) = 0

. Hence

N (1, t) = N (Y, t)

for all

0 < t \leq 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 \in R

. This shows that

1 \in P_{Y}^{f} (0)

. If

1 < y_{0} \leq 1.5

, then for

t = 1 + \frac{y_{0} - 1}{2}

N (y_{0}, t) = 0

and

N (1, t) = 1

. It follows that

0 = N (y_{0}, t) \neq N (Y, t) = 1

for

t = 1 + \frac{y_{0} - 1}{2}

. So

y_{0} \notin 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} \in [1, 1.5]

. If

t \leq y_{0}

, then

N (y_{0}, t) = 0

and

\frac{2}{3} t \leq \frac{2}{3} y_{0} \leq \frac{2}{3} (\frac{3}{2}) = 1

. So

N (y, \frac{2}{3} t) = 0

for all

y \in [1, 1.5]

. It follows that

N (Y, \frac{2}{3} t) = 0

. Hence

N (y_{0}, t) \geq N (Y, \frac{2}{3} t)

for all

t \leq y_{0}

. If

t > y_{0}

, then

N (y_{0}, t) = 1 \geq N (Y, \frac{2}{3} t)

. Therefore

N (y_{0}, t) \geq N (Y, \frac{2}{3} t)

for all

t \in R

. Hence

y_{0} \in 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 \geq 0

. Also let

\begin{matrix} N (x, t) = \{\begin{matrix} 0 & t \leq |x| \\ 1 & t > |x| \end{matrix} \end{matrix}

be a fuzzy norm on X. Then

P_{Y} (0) = P_{Y}^{f} (0) = P_{Y}^{f n (ρ)} (0) = \emptyset

. Indeed, since

P_{Y} (0) \subseteq P_{Y}^{f} (0) \subseteq P_{Y}^{f n (ρ)} (0)

, it’s enough to prove that

P_{Y}^{f n (ρ)} (0) = \emptyset

. Let

y_{0} \in Y

be an arbitrary element. Choose

n \in N

such that

\frac{1}{n} < \frac{y_{0}}{1 + ρ}

. So if

t = y_{0}

, then

0 = N (y_{0}, t) < N (\frac{1}{n}, \frac{t}{1 + ρ}) = 1 .

It follows that

0 = N (y_{0}, t) < N (Y, \frac{t}{1 + ρ}) = 1 .

Hence

y_{0} \notin P_{Y}^{f n (ρ)} (0)

and so

P_{Y}^{f n (ρ)} (0) = \emptyset

Proposition 1.10.

Let X be a linear space, Y be a subset of X and

N : X \times R ⟶ [0, 1]

be a fuzzy norm on X. For any

ρ \geq 0

x \in Y

P_{Y}^{f n (ρ)} (x) = \{x\}

Proof.

1 = N (0, t) = N (x - x, t) \geq N (x - Y, \frac{t}{1 + ρ})

for all

t > 0

, we can conclude that

N (x - x, t) \geq N (x - Y, \frac{t}{1 + ρ})

for all

t \in R

. Hence

x \in P_{Y}^{f n (ρ)} (x)

Now let

y_{0} \in P_{Y}^{f n (ρ)} (x)

. Then for all

t > 0

\begin{matrix} N (x - y_{0}, t) & \geq N (x - Y, \frac{t}{1 + ρ}) \\ \geq N (x - x, \frac{t}{1 + ρ}) \\ = 1 . \end{matrix}

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 \times R ⟶ [0, 1]

be a fuzzy norm on X. For

x \in X

, if

ρ_{1} \leq ρ_{2}

, then

P_{Y}^{f n (ρ_{1})} (x) \subseteq P_{Y}^{f n (ρ_{2})} (x)

Proof.

y_{0} \in P_{Y}^{f n (ρ_{1})} (x)

, then

N (x - y_{0}, t) \geq N (x - Y, \frac{t}{1 + ρ_{1}}) .

Since for all

t \in R

and

y \in Y

N (x - y, \frac{t}{1 + ρ_{1}}) \geq N (x - y, \frac{t}{1 + ρ_{2}}),

N (x - Y, \frac{t}{1 + ρ_{1}}) \geq N (x - Y, \frac{t}{1 + ρ_{2}}) .

Therefore

N (x - y_{0}, t) \geq N (x - Y, \frac{t}{1 + ρ_{2}}) .

It follows that

y_{0} \in 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 \times R ⟶ [0, 1]

be a fuzzy norm on X. If

x \in X

and

ρ_{1} \geq ρ_{2} \geq ρ_{3} \geq \dots

such that

ρ_{m} ⟶ ρ

m ⟶ \infty

, then

P_{Y}^{f n (ρ)} (x) \subseteq ⋂_{m = 1}^{\infty} P_{Y}^{f n (ρ_{m})} (x)

. In particular, if

N (z, .)

is lower semicontinuous at every

z \in x - Y

, then

P_{Y}^{f n (ρ)} (x) = ⋂_{m = 1}^{\infty} P_{Y}^{f n (ρ_{m})} (x)

Proof.

By Theorem 1.11, since

ρ_{m} \geq ρ

for all

m \in N

\begin{matrix} P_{Y}^{f n (ρ)} (x) \subseteq P_{Y}^{f n (ρ_{1})} (x) \\ P_{Y}^{f n (ρ)} (x) \subseteq P_{Y}^{f n (ρ_{2})} (x) \\ ⋮ \\ P_{Y}^{f n (ρ)} (x) \subseteq P_{Y}^{f n (ρ_{m})} (x), m \in N . \end{matrix}

Then

P_{Y}^{f n (ρ)} (x) \subseteq ⋂_{m = 1}^{\infty} P_{Y}^{f n (ρ_{m})} (x)

y_{0} \in ⋂_{m = 1}^{\infty} P_{Y}^{f n (ρ_{m})} (x)

and

N (z, .)

is lower semicontinuous for every

z \in x - Y

, then for all

m \in N

and for all

t \in R

we have

\begin{matrix} N (x - y_{0}, t) \geq & N (x - Y, \frac{t}{1 + ρ_{m}}) \\ \geq & N (x - y, \frac{t}{1 + ρ_{m}}), y \in Y . \end{matrix}

Since

N (x - y, .)

is lower semicontinuous for all

x - y

and

\frac{t}{1 + ρ_{m}} \leq \frac{t}{1 + ρ}

for every

m \in N

\begin{matrix} N (x - y_{0}, t) \geq & lim_{m ⟶ \infty} N (x - y, \frac{t}{1 + ρ_{m}}) \\ = & N (x - y, \frac{t}{1 + ρ}), y \in Y . \end{matrix}

⋂_{m = 1}^{\infty} P_{Y}^{f n (ρ_{m})} (x) \subset P_{Y}^{f n (ρ)} (x)

\begin{matrix} N (x - y_{0}, t) \geq & sup \{N (x - y, \frac{t}{1 + ρ}), y \in Y\} \\ = & N (x - Y, \frac{t}{1 + ρ}) . \end{matrix}

Hence

⋂_{m = 1}^{\infty} P_{Y}^{f n (ρ_{m})} (x) \subseteq 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 \leq a_{i} \leq 1

and

0 \leq b_{i} \leq 1

for all

1 \leq i \leq m

. Then

\begin{matrix} min & \{min \{a_{i}, b_{i}\} | 1 \leq i \leq m\} \\ = min & \{min \{a_{i} | 1 \leq i \leq m\}, min \{b_{i} | 1 \leq i \leq m\}\} . \end{matrix}

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 \in I}

be a family of fuzzy normed spaces. Then

(\sum_{i \in I} X_{i}, N)

is a fuzzy normed space, where

N : (\sum_{i \in I} X_{i}) \times R ⟶ [0, 1]

is defined by

N ({(x_{i})}_{i \in I}, t) = inf \{N_{i} (x_{i}, t) | i \in 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 \in I}, {(y_{i})}_{i \in I} \in \sum_{i \in I} X_{i}

and

s, t \in R

. So

x_{i} = 0

and

y_{i} = 0

for all but finitely many

i_{k} \in I, 1 \leq k \leq m

. Clearly if

s + t \leq 0

, then the inequality

N ({(x_{i})}_{i \in I} + {(y_{i})}_{i \in I}, s + t) \geq min \{N ({(x_{i})}_{i \in I}, s), N ({(y_{i})}_{i \in I}, t)\}

(1)

holds. Also if

s + t > 0

s \leq 0

t \leq 0

, then obviously inequality 2.1 holds.

Let

s + t > 0, s > 0

and

t > 0

. Then

\begin{matrix} N ({(x_{i})}_{i \in I} + {(y_{i})}_{i \in I}, s + t) \\ = N ({(x_{i} + y_{i})}_{i \in I}, s + t) \\ inf \{N_{i} (x_{i} + y_{i}, s + t) | i \in I\} \\ \geq inf \{min \{N_{i} (x_{i}, s), N_{i} (y_{i}, t)\} | i \in I\} \\ = inf \{min \{N_{i_{k}} (x_{i_{k}}, s), N_{i_{k}} (y_{i_{k}}, t)\}, 1 | 1 \leq k \leq m\} \\ = min \{min \{N_{i_{k}} (x_{i_{k}}, s), N_{i_{k}} (y_{i_{k}}, t)\}, 1 | 1 \leq k \leq m\} \\ = min \{min \{N_{i_{k}} (x_{i_{k}}, s), N_{i_{k}} (y_{i_{k}}, t)\} | 1 \leq k \leq m\} \\ = min \{min \{N_{i_{k}} (x_{i_{k}}, s) | 1 \leq k \leq m\}, min \{N_{i_{k}} (y_{i_{k}}, t) | 1 \leq k \leq m\}\} \\ = min \{min \{N_{i} (x_{i}, s) | i \in I\}, min \{N_{i} (y_{i}, t) | i \in I\}\} \\ = min \{inf \{N_{i} (x_{i}, s) | i \in I\}, inf \{N_{i} (y_{i}, t) | i \in I\}\} \\ = min \{N ({(x_{i})}_{i \in I}, s), N ({(y_{i})}_{i \in I}, t)\} . \end{matrix}

To prove the fifth part, suppose that

{(x_{i})}_{i \in I} \in \sum_{i \in I} X_{i}

. So

x_{i} = 0

for all but finitely many

i_{k} \in I

1 \leq k \leq m

. Hence

\begin{matrix} lim_{t ⟶ \infty} N ({(x_{i})}_{i \in I}, t) \\ = lim_{t ⟶ \infty} \{inf \{N_{i} (x_{i}, t) | i \in I\}\} \\ = lim_{t ⟶ \infty} \{inf \{N_{i_{k}} (x_{i_{k}}, t), 1 | 1 \leq k \leq m\}\} \\ = lim_{t ⟶ \infty} \{min \{N_{i_{k}} (x_{i_{k}}, t), 1 | 1 \leq k \leq m\}\} \\ = lim_{t ⟶ \infty} \{min \{N_{i_{k}} (x_{i_{k}}, t) | 1 \leq k \leq m\}\} \\ = min \{lim_{t ⟶ \infty} N_{i_{k}} (x_{i_{k}}, t) | 1 \leq k \leq m\} \\ = min \{1\} \\ = 1 . \end{matrix}

Since

N_{i} (x_{i}, .)

is increasing for all

x_{i} \in X_{i}

N ({(x_{i})}_{i \in I}, .)

is increasing for all

{(x_{i})}_{i \in I}

. □

Corollary 2.2.

Let

N_{i} : X_{i} \times R ⟶ [0, 1]

be a fuzzy norm on

X_{i}

for

i = 1, 2, \dots, n

. Then

N : X_{1} \times X_{2} \times \dots \times X_{n} \times R ⟶ [0, 1]

defined by

N ((x_{1}, x_{2}, \dots, x_{n}), t) = min \{N_{1} (x_{1}, t), N_{2} (x_{2}, t), \dots, N_{n} (x_{n}, t)\}

is a fuzzy norm on

X_{1} \times X_{2} \times \dots \times 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} \times X_{2} \times \dots \times X_{n} \times R ⟶ [0, 1]

is defined by

N ((x_{1}, x_{2}, \dots, x_{n}), t) = min \{N_{1} (x_{1}, t), N_{2} (x_{2}, t), \dots, N_{n} (x_{n}, t)\} .

Also let

Y_{i} \subseteq X i

x_{i} \in X_{i}

and

ρ_{i} \geq 0

for all

1 \leq i \leq n

. Then

P_{Y_{1}}^{f n (ρ_{1})} (x_{1}) \times P_{Y_{2}}^{f n (ρ_{2})} (x_{2}) \times \dots \times P_{Y_{n}}^{f n (ρ_{n})} (x_{n}) \subseteq P_{Y_{1} \times Y_{2} \times \dots \times Y_{n}}^{f n (max \{ρ_{i}\})} (x_{1}, x_{2}, \dots, x_{n}) .

(2)

Proof.

Let

(y_{1}, y_{2}, \dots, y_{n}) \in P_{Y_{1}}^{f n (ρ_{1})} (x_{1}) \times P_{Y_{2}}^{f n (ρ_{2})} (x_{2}) \times \dots \times P_{Y_{n}}^{f n (ρ_{n})} (x_{n})

. Then

y_{1} \in P_{Y_{1}}^{f n (ρ_{1})} (x_{1})

y_{2} \in P_{Y_{2}}^{f n (ρ_{2})} (x_{2})

, ⋯,

y_{n} \in P_{Y_{n}}^{f n (ρ_{n})} (x_{n})

. Therefore

\begin{matrix} N_{1} (x_{1} - y_{1}, t) \geq N_{1} (x_{1} - z_{1}, \frac{t}{1 + ρ_{1}}) \geq N_{1} (x_{1} - z_{1}, \frac{t}{1 + {max}_{1 \leq i \leq n} \{ρ_{i}\}}) \\ N_{2} (x_{2} - y_{2}, t) \geq N_{2} (x_{2} - z_{2}, \frac{t}{1 + ρ_{2}}) \geq N_{2} (x_{2} - z_{2}, \frac{t}{1 + {max}_{1 \leq i \leq n} \{ρ_{i}\}}) \\ ⋮ \\ N_{n} (x_{n} - y_{n}, t) \geq N_{n} (x_{n} - z_{n}, \frac{t}{1 + ρ_{n}}) \geq N_{n} (x_{n} - z_{n}, \frac{t}{1 + {max}_{1 \leq i \leq n} \{ρ_{i}\}}) \end{matrix}

for every

z_{i} \in Y_{i}

and

t \in R

. Then

\begin{matrix} min & \{N_{1} (x_{1} - y_{1}, t), N_{2} (x_{2} - y_{2}, t), \dots, N_{n} (x_{n} - y_{n}, t)\} \\ \geq min & \{N_{1}^{(1)}, N_{2}^{(2)}, \dots, N_{n}^{(n)}\}, \end{matrix}

where

N_{k}^{(k)} = N_{k} (x_{k} - z_{k}, \frac{t}{1 + {max}_{1 \leq i \leq n} \{ρ_{i}\}}), 1 \leq k \leq n

. Therefore

\begin{matrix} N ((x_{1}, x_{2}, \dots, x_{n}) - (y_{1}, y_{2}, \dots, y_{n}), t) \\ \geq sup_{(z_{1}, z_{2}, \dots, z_{n}) \in Y_{1} \times Y_{2} \times \dots \times Y_{n}} \{N ((x_{1}, x_{2}, \dots, x_{n}) - (z_{1}, z_{2}, \dots, z_{n}), \frac{t}{1 + {max}_{1 \leq i \leq n} \{ρ_{i}\}})\} . \end{matrix}

Hence

\begin{matrix} N ((x_{1}, x_{2}, \dots, x_{n}) - (y_{1}, y_{2}, \dots, y_{n}), t) \\ \geq N ((x_{1}, x_{2}, \dots, x_{n}) - Y_{1} \times Y_{2} \times \dots \times Y_{n}, \frac{t}{1 + {max}_{1 \leq i \leq n} \{ρ_{i}\}}) . \end{matrix}

□

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} \times R ⟶ [0, 1]

is defined by

\begin{matrix} N_{i} (α, t) = \{\begin{matrix} 0 & t \leq |α| \\ 1 & t > |α| \end{matrix} \end{matrix}

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} = \frac{1}{2}

. Then we have

\begin{matrix} P_{Y_{1}}^{f n (1)} (0) = [1, 2] \\ P_{Y_{2}}^{f n (1)} (\frac{1}{2}) = [1, \frac{3}{2}] \\ P_{Y_{1}}^{f n (1)} (0) \times P_{Y_{2}}^{f n (1)} (\frac{1}{2}) = [1, 2] \times [1, \frac{3}{2}] . \end{matrix}

It is easy to see that

\begin{matrix} N ((x, y), t) = \{\begin{matrix} 0 & t \leq max (|x|, |y|) \\ 1 & t > max (|x|, |y|) \end{matrix} \end{matrix}

for all

x, y, t \in R

. Also a sufficient effort can be applied to show that

P_{Y_{1} \times Y_{2}}^{f n (1)} (0, \frac{1}{2}) = [1, 2] \times [1, \frac{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 \otimes Y) \times R ⟶ [0, 1]

N (\sum_{j = 1}^{n} α_{j} x_{j} \otimes y_{j}, t) = min \{N_{1} (α_{j} x_{j}, t), N_{2} (α_{j} y_{j}, t) | 1 \leq j \leq n\},

where

x_{j} \in B_{X}, y_{j} \in B_{Y}, n \in N, α_{j} \in C

and

t \in R

. Then

(X \otimes 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) \in (X \otimes Y) \times R

. So

z = w

and

s = t

. Hence there exists an

n \in N, {\{x_{j}\}}_{j = 1}^{n} \subseteq B_{X}, {\{y_{j}\}}_{j = 1}^{n} \subseteq B_{Y}

and

α_{j}, β_{j} \in C

such that

z = \sum_{j = 1}^{n} α_{j} x_{j} \otimes y_{j}

and

w = \sum_{j = 1}^{n} β_{j} x_{j} \otimes y_{j}

. It follows that

α_{j} = β_{j}

for all

1 \leq j \leq n

. Therefore

\begin{matrix} min & \{N_{1} (α_{j} x_{j}, s), N_{2} (α_{j} y_{j}, s) | 1 \leq j \leq n\} \\ = min & \{N_{1} (β_{j} x_{j}, t), N_{2} (β_{j} y_{j}, t) | 1 \leq j \leq n\}, \end{matrix}

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 = \sum_{j = 1}^{n} α_{j} x_{j} \otimes 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 \leq j \leq 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 \leq j \leq n

and for all

t > 0

. Hence

α_{j} x_{j} = 0

and

α_{j} y_{j} = 0

for all

1 \leq j \leq n

. Therefore

α_{j} = 0

for all

1 \leq j \leq n

. This shows that

z = 0 \otimes 0

. Also for all

t > 0

, since

0 \otimes 0 = 0 x \otimes y

for all

x \in B_{X}

and

y \in B_{Y}

\begin{matrix} N (0 \otimes 0, t) \\ = N (0 x \otimes y, t) \\ = min \{N_{1} (0, t), N_{2} (0, t)\} \\ = min \{1\} \\ = 1 . \end{matrix}

(3) : Let

c \neq 0

and

\sum_{j = 1}^{n} α_{j} x_{j} \otimes y_{j} \in X \otimes Y

. So

\begin{matrix} N (c \sum_{j = 1}^{n} α_{j} x_{j} \otimes y_{j}, t) \\ = N (\sum_{j = 1}^{n} c α_{j} x_{j} \otimes y_{j}, t) \\ = min \{N_{1} (c α_{j} x_{j}, t), N_{2} (c α_{j} y_{j}, t) | 1 \leq j \leq n\} \\ = min \{N_{1} (α_{j} x_{j}, \frac{t}{|c|}), N_{2} (α_{j} y_{j}, \frac{t}{|c|}) | 1 \leq j \leq n\} \\ = N (\sum_{j = 1}^{n} α_{j} x_{j} \otimes y_{j}, \frac{t}{|c|}), \end{matrix}

for all

t \in R

(4) : Let

z, w \in X \otimes Y

and

s, t \in R

. So

z = \sum_{j = 1}^{n} α_{j} x_{j} \otimes y_{j}

and

w = \sum_{j = 1}^{n} β_{j} x_{j} \otimes y_{j}

, where

n \in N, α_{j}, β_{j} \in C, {\{x_{j}\}}_{j = 1}^{n} \subseteq B_{X}

and

{\{y_{j}\}}_{j = 1}^{n} \subseteq B_{Y}

. Hence

\begin{matrix} N & (\sum_{j = 1}^{n} α_{j} x_{j} \otimes y_{j} + \sum_{j = 1}^{n} β_{j} x_{j} \otimes y_{j}, s + t) \\ = N & (\sum_{j = 1}^{n} (α_{j} + β_{j}) x_{j} \otimes 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 \leq j \leq 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 \leq j \leq n\} \\ \geq 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 \leq j \leq n\} \\ = min & \{min (A_{j}, B_{j}), min (C_{j}, D_{j}) | 1 \leq j \leq n\} \\ = min & \{min \{A_{j}, C_{j} | 1 \leq j \leq n\}, min \{B_{j}, D_{j} | 1 \leq j \leq n\}\} \\ = min & \{N (\sum_{j = 1}^{n} α_{j} x_{j} \otimes y_{j}, s), N (\sum_{j = 1}^{n} β_{j} x_{j} \otimes y_{j}, t)\}, \end{matrix}

where

\begin{matrix} 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) . \end{matrix}

(5) : Let

s \leq t

and

z = \sum_{j = 1}^{n} α_{j} x_{j} \otimes y_{j}

. Clearly

N_{1} (α_{j} x_{j}, s) \leq N_{1} (α_{j} x_{j}, t)

and

N_{2} (α_{j} y_{j}, s) \leq N_{2} (α_{j} y_{j}, t)

for all

1 \leq j \leq n

. So

\begin{matrix} min & \{N_{1} (α_{j} x_{j}, s), N_{2} (α_{j} y_{j}, s) | 1 \leq j \leq n\} \\ \leq min & \{N_{1} (α_{j} x_{j}, t), N_{2} (α_{j} y_{j}, t) | 1 \leq j \leq n\} . \end{matrix}

It follows that

N (\sum_{j = 1}^{n} α_{j} x_{j} \otimes y_{j}, s) \leq N (\sum_{j = 1}^{n} α_{j} x_{j} \otimes y_{j}, t) .

Hence

N (z, .) : R ⟶ [0, 1]

is increasing for all

z \in X \otimes Y

. Also

\begin{matrix} lim_{t ⟶ \infty} N (z, t) \\ = lim_{t ⟶ \infty} min \{N_{1} (α_{j} x_{j}, t), N_{2} (α_{j} y_{j}, t) | 1 \leq j \leq n\} \\ = min \{lim_{t ⟶ \infty} N_{1} (α_{j} x_{j}, t), lim_{t ⟶ \infty} N_{2} (α_{j} y_{j}, t) | 1 \leq j \leq n\} \\ = min \{1\} \\ = 1 . \end{matrix}

□

Theorem 2.6.

Let X and Y be linear spaces and

N : (X \otimes Y) \times R ⟶ [0, 1]

be a fuzzy norm. Then for all

x \in X ∖ \{0\}

and

y \in Y ∖ \{0\}

, the maps

N_{x} : Y \times R ⟶ [0, 1]

and

N_{y} : X \times R ⟶ [0, 1]

, where

N_{x} (z, t) = N (x \otimes z, t)

and

N_{y} (w, t) = N (w \otimes y, t)

, are fuzzy norms on Y and X respectively.

Proof.

We only prove that

N_{x} : Y \times R ⟶ [0, 1]

is a fuzzy norm for all

x \in X ∖ \{0\}

(1) : Let

z \in Y

and

t \leq 0

. So

N_{x} (z, t) = N (x \otimes z, t) = 0

(2) : If

z = 0

, then

N_{x} (0, t) = N (x \otimes 0, t) = N (0 \otimes 0, t) = 1

for all

t > 0

. Also if

N_{x} (z, t) = 1

for all

t > 0

, then

N (x \otimes z, t) = 1

for all

t > 0

. It follows that

x \otimes z = 0 \otimes 0

. Since

x \neq 0

, there exists

f \in X^{'}

such that

f (x) \neq 0

. Let

g \in Y^{'}

be an arbitrary element. As

x \otimes z

is a bilinear map on

X^{'} \times Y^{'}

(x \otimes z) (f, g) = (0 \otimes 0) (f, g)

. So

f (x) g (z) = 0

for all

g \in Y^{'}

. Since

f (x) \neq 0

g (z) = 0

for all

g \in Y^{'}

. It follows that

z = 0

(3) : Let

c \neq 0

and

z \in Y

. So

\begin{matrix} N_{x} & (c z, t) \\ = N & (x \otimes c z, t) \\ = N & (c x \otimes z, t) \\ = N & (x \otimes z, \frac{t}{|c|}) \\ = N_{x} & (z, \frac{t}{|c|}) \end{matrix}

for all

z \in Y

and

t \in R

(4) : Let

z_{1}, z_{2} \in Y

and

s, t \in R

. So

\begin{matrix} N_{x} & (z_{1} + z_{2}, s + t) \\ = N & (x \otimes (z_{1} + z_{2}), s + t) \\ = N & (x \otimes z_{1} + x \otimes z_{2}, s + t) \\ \geq min & (N (x \otimes z_{1}, s), N (x \otimes z_{2}, t)) \\ = min & (N_{x} (z_{1}, s), N_{x} (z_{2}, t)) . \end{matrix}

(5) : Let

s \leq t

and

z \in Y

. So

N (x \otimes z, s) \leq N (x \otimes z, t)

. It follows that

N_{x} (z, s) \leq N_{x} (z, t)

. Hence

N_{x} (z, .) : R ⟶ [0, 1]

is increasing and

\begin{matrix} lim_{t ⟶ \infty} N_{x} (z, t) \\ = & lim_{t ⟶ \infty} N (x \otimes z, t) \\ = & 1 \end{matrix}

for all

z \in Y

Therefore

N_{x} : Y \times 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} \times R ⟶ [0, 1]

are defined by

\begin{matrix} N_{i} (\sum_{j = 1}^{2} α_{j} e_{j}, t) = \{\begin{matrix} 0 & t \leq max (|α_{1}|, |α_{2}|) \\ 1 & t > max (|α_{1}|, |α_{2}|) \end{matrix} \end{matrix}

for

i = 1, 2

. Clearly

N_{1}

and

N_{2}

are fuzzy norms on

R^{2}

. According to Theorem 2.5, if

N : (X \otimes Y) \times R ⟶ [0, 1]

is defined by

\begin{matrix} N (α_{1} e_{1} \otimes e_{1} + α_{2} e_{1} \otimes e_{2} + α_{3} e_{2} \otimes e_{1} + α_{4} e_{2} \otimes e_{2}, t) \\ = min (\begin{matrix} 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) \end{matrix}), \end{matrix}

then the equalities

\begin{matrix} 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) \end{matrix}

imply that

\begin{matrix} N (α_{1} e_{1} \otimes e_{1} + α_{2} e_{1} \otimes e_{2} + α_{3} e_{2} \otimes e_{1} + α_{4} e_{2} \otimes 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)\} \\ = \{\begin{matrix} 0 & t \leq max (|α_{i}|, 1 \leq i \leq 4) \\ 1 & t > max (|α_{i}|, 1 \leq i \leq 4) . \end{matrix} \end{matrix}

Let

x_{0} = - \frac{1}{5} e_{1}, y_{0} = - 2 e_{2}, K_{1} = B_{X}, K_{2} = B_{Y}, K = B_{X \otimes Y} = B_{X} \otimes B_{Y}

and

ρ = \frac{1}{10}

. Also let

P_{K_{1}}^{f n (\frac{1}{10})} (x_{0})

and

P_{K_{2}}^{f n (\frac{1}{10})} (y_{0})

be the set of all fuzzy near best approximations to

x_{0}

and

y_{0}

within the relative distance

ρ = \frac{1}{10}

with respect to the fuzzy norms

N_{1}

and

N_{2}

respectively. If

P_{K}^{f n (\frac{1}{10})} (x_{0} \otimes y_{0})

be the set of all fuzzy near best approximations to

x_{0} \otimes y_{0}

within the relative distance

ρ = \frac{1}{10}

with respect to the fuzzy norm N, then a straightforward calculation reveals that

\begin{matrix} P_{K_{1}}^{f n (\frac{1}{10})} (x_{0}) & = \{e_{2}\} \\ P_{K_{2}}^{f n (\frac{1}{10})} (y_{0}) & = \{e_{1}\} \\ P_{K}^{f n (\frac{1}{10})} (x_{0} \otimes y_{0}) & = \{e_{1} \otimes e_{2}\} \neq \{e_{2} \otimes e_{1}\} . \end{matrix}

This example shows that there is no a relation between

P_{K}^{f n (\frac{1}{10})} (x_{0} \otimes y_{0})

and

P_{K_{1}}^{f n (\frac{1}{10})} (x_{0}) \otimes P_{K_{2}}^{f n (\frac{1}{10})} (y_{0})

References

Alexander C. R. Belton. Functional Analysis, Hyperlinked and revised edition, (2004).
F.F. Bonsall and J. Duncan. Complete Normed Algebras, Springer-Verlag, Berlin Heidelberg New York, (1973).
T. Bag and S.K. Samanta. Some fixed point theorems in fuzzy normed linear spaces. Inform. Sci., 2007, 177, 3271–3289. [CrossRef]
F. Deutschi. Best Approximation in Inner Product Spaces, Canad. Math. Soc., (2000).
F.D. Kovaca and F.E. Levisb. Extended best L^p-approximation is near-best approximation in L^q, p − 1 ≤ q, Journal of Approximation Theory., 284 (2022), 105819. [CrossRef]
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.
I. Sadeqi, F. Solaty Kia. Fuzzy normed linear space and its topological structure Solitons and Fractals., 40 (2009), 2576–2589. [CrossRef]

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On Fuzzy Near Best Approximation

Abstract

1. INTRODUCTION AND PRELIMINARIES

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

References

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