Fuzzy Normed Linear Spaces Generated by Linear Functionals

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Fuzzy Normed Linear Spaces Generated by Linear Functionals

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Submitted:

25 September 2023

Posted:

27 September 2023

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Abstract

For a nonzero normed linear space X‎, ‎we will define some different classes of fuzzy norms on X generated by linear and bounded linear functionals‎. ‎Also separate continuity of the elements of each class are investigated‎. ‎The aim of this research is to introduce a source of examples and counterexamples in the field of fuzzy normed spaces‎.

Keywords:

Subject: Computer Science and Mathematics - Analysis

1. Introduction

The idea of fuzzy norm on a linear space introduced by Katsaras [11] in 1984. Also a type of fuzzy metric introduced by O. Kaleva and S. Seikkala [10] in 1984. Felbin [7] in 1992 introduced an idea of fuzzy norm on a linear space, such that its corresponding fuzzy metric is of type of introduced by O. Kaleva and S. Seikkala. Another idea of a fuzzy norm on a linear space was introduced by Cheng and Mordeson [6] in 1994. Following Cheng and Mordeson, a definition of a fuzzy norm whose associated fuzzy metric is similar to Kramosil and Michalek type [12], was introduced by T. Bag and S. K. Samanta [1] in 2003. A large number of papers concerning fuzzy norms have been published by different authors such as papers [2,3,4,5,8,9].

In this paper we will consider X as a normed linear space over

C

R

. Also let

X^{'}

be the set of all linear functionals on X and let

X^{*}

be the set of all bounded linear functionals on X.

Given a normed linear space X, we will introduce some different classes of fuzzy normed linear spaces. Also separate continuity of the elements of each class are investigated. The results of this paper can be applied as a source of examples and counterexamples concerning fuzzy normed linear spaces.

2. FUZZY NORMS GENERATED BY LINEAR FUNCTIONALS

Definition 2.1.

[1] Let X be a linear space and let

N : X \times R ⟶ [0, 1]

be a function with the following properties

(1): $N (x, t) = 0$ for all $t \leq 0,$
(2): $N (x, t) = 1$ for all $t > 0$ if and only if $x = 0,$
(3): $N (c x, t) = N (x, \frac{t}{| c |})$ for all $c \neq 0,$
(4): $N (x + y, s + t) \geq min (N (x, s), N (y, t)),$
(5): For each fixed $x \in X, N (x, .) : R ⟶ [0, 1]$ is an increasing function, and $lim_{t ⟶ \infty} N (x, t) = 1 .$

Then $(X, N)$ is called a fuzzy normed linear space.

Example 2.2.

If X is a normed linear space, then

N : X \times R ⟶ [0, 1]

defined by

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

is a fuzzy norm on X.

In the sequel we present some different classes of fuzzy normed linear spaces generated by linear functionals.

Proposition 2.3.

Let X be a normed linear space and

f \in X^{'}

. Define

N_{f} : X \times R ⟶ [0, 1]

N_{f} (x, t) = \{\begin{matrix} 0 & t \leq ∥ x ∥ + | f (x) | \\ \frac{t - ∥ x ∥ - | f (x) |}{t + ∥ x ∥ + | f (x) |} & t > ∥ x ∥ + | f (x) | . \end{matrix}

Then

N_{f}

is a fuzzy norm on X.

Proof.

We only prove parts (2), (4) and (5) of Definition 2.1.

(2) : If

x = 0

, then clearly

N_{f} (x, t) = 1

for all

t > 0

. For the converse let

N_{f} (x, t) = 1

for all

t > 0

and suppose by contradiction that

x \neq 0

t = ∥ x ∥ + | f (x) |

, then

N_{f} (x, t) = 0

that is a contradiction.

(4) : Let

x, y \in X

and

s, t \in R

. If

s + t \leq 0

, then

s \leq 0

t \leq 0

. So

N_{f} (x, s) = 0

N_{f} (y, t) = 0

Hence

0 = N_{f} (x + y, s + t) \geq min (N_{f} (x, s), N_{f} (y, t)) = 0

s + t > 0

and

s + t \leq ∥ x + y ∥ + | f (x) + f (y) |

, then

s \leq ∥ x ∥ + | f (x) |

t \leq ∥ y ∥ + | f (y) |

. Therefore

0 = N_{f} (x + y, s + t) \geq min (N_{f} (x, s), N_{f} (y, t)) = 0 .

s + t > ∥ x + y ∥ + | f (x) + f (y) |

, then

N_{f} (x + y, s + t) = \frac{s + t - ∥ x + y ∥ - | f (x) + f (y) |}{s + t + ∥ x + y ∥ + | f (x) + f (y) |}

In this case if

s \leq ∥ x ∥ + | f (x) |

t \leq ∥ y ∥ + | f (y) |

, then clearly

\frac{s + t - ∥ x + y ∥ - | f (x) + f (y) |}{s + t + ∥ x + y ∥ + | f (x) + f (y) |} \geq min (N_{f} (x, s), N_{f} (y, t)) = 0 .

s > ∥ x ∥ + | f (x) |

and

t > ∥ y ∥ + | f (y) |

, then

N_{f} (x, s) = \frac{s - ∥ x ∥ - | f (x) |}{s + ∥ x ∥ + | f (x) |}

and

N_{f} (y, t) = \frac{t - ∥ y ∥ - | f (y) |}{t + ∥ y ∥ + | f (y) |}

One can easily verify that if

N_{f} (x, s) \leq N_{f} (y, t)

, then

s (∥ y ∥ + | f (y) |) \leq t (∥ x ∥ + | f (x) |) .

Also if

N_{f} (y, t) \leq N_{f} (x, s)

, then

t (∥ x ∥ + | f (x) |) \leq s (∥ y ∥ + | f (y) |)

A straightforward calculation reveals that

s (∥ y ∥ + | f (y) |) \leq t (∥ x ∥ + | f (x) |)

, then

\begin{matrix} N_{f} (x + y, s + t) & = \frac{s + t - ∥ x + y ∥ - | f (x) + f (y) |}{s + t + ∥ x + y ∥ + | f (x) + f (y) |} \\ \geq \frac{s - ∥ x ∥ - | f (x) |}{s + ∥ x ∥ + | f (x) |} \\ = min (N_{f} (x, s), N_{f} (y, t)) . \end{matrix}

Also if

t (∥ x ∥ + | f (x) |) \leq s (∥ y ∥ + | f (y) |)

, then

\begin{matrix} N_{f} (x + y, s + t) & = \frac{s + t - ∥ x + y ∥ - | f (x) + f (y) |}{s + t + ∥ x + y ∥ + | f (x) + f (y) |} \\ \geq \frac{t - ∥ y ∥ - | f (y) |}{t + ∥ y ∥ + | f (y) |} \\ = min (N_{f} (x, s), N_{f} (y, t)) . \end{matrix}

Therefore

N_{f} (x + y, s + t) \geq min (N_{f} (x, s), N_{f} (y, t))

for all

x, y \in X

and

s, t \in R

(5) : Let

x \in X

and

s \leq t

. If

N_{f} (x, s) = 0

, then clearly

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

. If

N_{f} (x, s) \neq 0

, then

N_{f} (x, s) = \frac{s - ∥ x ∥ - | f (x) |}{s + ∥ x ∥ + | f (x) |}

and

s > ∥ x ∥ + | f (x) |

. Hence

t > ∥ x ∥ + | f (x) |

and

N_{f} (x, t) = \frac{t - ∥ x ∥ - | f (x) |}{t + ∥ x ∥ + | f (x) |}

. Since

s \leq t

s (∥ x ∥ + | f (x) |) \leq t (∥ x ∥ + | f (x) |) .

Therefore a straightforward calculation reveals that

N_{f} (x, s) = \frac{s - ∥ x ∥ - | f (x) |}{s + ∥ x ∥ + | f (x) |} \leq \frac{t - ∥ x ∥ - | f (x) |}{t + ∥ x ∥ + | f (x) |} = N_{f} (x, t) .

This shows that

N_{f} (x, .)

is an increasing function for all

x \in X

Also

lim_{t \to \infty} N_{f} (x, t) = lim_{t \to \infty} \frac{t - ∥ x ∥ - | f (x) |}{t + ∥ x ∥ + | f (x) |} = 1

. □

Proposition 2.4.

Let X be a normed linear space,

f \in X^{*}

and

ϵ > 0

Define

N_{f, ϵ} : X \times R ⟶ [0, 1]

N_{f, ϵ} (x, t) = \{\begin{matrix} 0 & t \leq ∥ x ∥ (ϵ + ∥ f ∥) \\ \frac{t - | f (x) |}{t + | f (x) |} & t > ∥ x ∥ (ϵ + ∥ f ∥) . \end{matrix}

Then

N_{f, ϵ}

is a fuzzy norm on X.

Proof.

We only prove parts (2), (4) and (5) of Definition 2.1.

(2) : If

x = 0

, then clearly

N_{f, ϵ} (x, t) = 1

for all

t > 0

. For the converse let

N_{f, ϵ} (x, t) = 1

for all

t > 0

and suppose by contradiction that

x \neq 0

t = ∥ x ∥ (ϵ + ∥ f ∥)

, then

N_{f, ϵ} (x, t) = 0

that is a contradiction.

(4) : Let

x, y \in X

and

s, t \in R

. If

s + t \leq 0

, then

s \leq 0

t \leq 0

. So

N_{f, ϵ} (x, s) = 0

N_{f, ϵ} (y, t) = 0

Hence

0 = N_{f, ϵ} (x + y, s + t) \geq min (N_{f, ϵ} (x, s), N_{f, ϵ} (y, t)) = 0

s + t > 0

and

s + t \leq ∥ x + y ∥ (ϵ + ∥ f ∥)

, then

s \leq ∥ x ∥ (ϵ + ∥ f ∥)

t \leq ∥ y ∥ (ϵ + ∥ f ∥)

. Therefore

0 = N_{f, ϵ} (x + y, s + t) \geq min (N_{f, ϵ} (x, s), N_{f, ϵ} (y, t)) = 0 .

s + t > ∥ x + y ∥ (ϵ + ∥ f ∥)

, then

N_{f, ϵ} (x + y, s + t) = \frac{s + t - | f (x) + f (y) |}{s + t + | f (x) + f (y) |}

In this case if

s \leq ∥ x ∥ (ϵ + ∥ f ∥)

t \leq ∥ y ∥ (ϵ + ∥ f ∥)

, then clearly

\frac{s + t - | f (x) + f (y) |}{s + t + | f (x) + f (y) |} \geq min (N_{f, ϵ} (x, s), N_{f, ϵ} (y, t)) = 0 .

s > ∥ x ∥ (ϵ + ∥ f ∥)

and

t > ∥ y ∥ (ϵ + ∥ f ∥)

, then

N_{f, ϵ} (x, s) = \frac{s - | f (x) |}{s + | f (x) |}

and

N_{f, ϵ} (y, t) = \frac{t - | f (y) |}{t + | f (y) |}

One can easily verify that if

N_{f, ϵ} (x, s) \leq N_{f, ϵ} (y, t)

, then

2 s | f (y) | \leq 2 t | f (x) |

. Also if

N_{f, ϵ} (y, t) \leq N_{f, ϵ} (x, s)

, then

2 t | f (x) | \leq 2 s | f (y) |

A straightforward calculation reveals that if

2 s | f (y) | \leq 2 t | f (x) |

, then

\begin{matrix} N_{f, ϵ} (x + y, s + t) & = \frac{s + t - | f (x) + f (y) |}{s + t + | f (x) + f (y) |} \\ \geq \frac{s - | f (x) |}{s + | f (x) |} \\ = min (N_{f, ϵ} (x, s), N_{f, ϵ} (y, t)) . \end{matrix}

Also if

2 t | f (x) | \leq 2 s | f (y) |

, then

\begin{matrix} N_{f, ϵ} (x + y, s + t) & = \frac{s + t - | f (x) + f (y) |}{s + t + | f (x) + f (y) |} \\ \geq \frac{t - | f (y) |}{t + | f (y) |} \\ = min (N_{f, ϵ} (x, s), N_{f, ϵ} (y, t)) . \end{matrix}

Therefore

N_{f, ϵ} (x + y, s + t) \geq min (N_{f, ϵ} (x, s), N_{f, ϵ} (y, t))

for all

x, y \in X

and

s, t \in R

(5) : Let

x \in X

and

s \leq t

. If

N_{f, ϵ} (x, s) = 0

, then clearly

N_{f, ϵ} (x, s) \leq N_{f, ϵ} (x, t)

. If

N_{f, ϵ} (x, s) \neq 0

, then

N_{f, ϵ} (x, s) = \frac{s - | f (x) |}{s + | f (x) |}

and

s > ∥ x ∥ (ϵ + ∥ f ∥)

. Hence

t > ∥ x ∥ (ϵ + ∥ f ∥)

and

N_{f, ϵ} (x, t) = \frac{t - | f (x) |}{t + | f (x) |}

Since

s \leq t

2 s | f (x) | \leq 2 t | f (x) |

. Therefore a straightforward calculation reveals that

N_{f, ϵ} (x, s) = \frac{s - | f (x) |}{s + | f (x) |} \leq \frac{t - | f (x) |}{t + | f (x) |} = N_{f, ϵ} (x, t)

. This shows that

N_{f, ϵ} (x, .)

is an increasing function for all

x \in X

. Also

lim_{t \to \infty} N_{f, ϵ} (x, t) = lim_{t \to \infty} \frac{t - | f (x) |}{t + | f (x) |} = 1 .

□

Proposition 2.5.

Let X be a normed linear space and

g \in X^{'}

. Then the maps

\begin{matrix} N_{g}^{(1)} & : X \times R ⟶ [0, 1] \\ N_{g}^{(1)} (x, t) & = \{\begin{matrix} 0 & t \leq ∥ x ∥ + | g (x) | \\ \frac{t}{t + ∥ x ∥ + | g (x) |} & t > ∥ x ∥ + | g (x) |, \end{matrix} \end{matrix}

and

\begin{matrix} N_{g}^{(2)} & : X \times R ⟶ [0, 1] \\ N_{g}^{(2)} (x, t) & = \{\begin{matrix} 0 & t \leq | g (x) | \\ \frac{t}{t + ∥ x ∥ + | g (x) |} & t > | g (x) |, \end{matrix} \end{matrix}

are fuzzy norms on X.

Also if

ker g = {0}

, then the map

\begin{matrix} N_{g}^{(3)} & : X \times R ⟶ [0, 1] \\ N_{g}^{(3)} (x, t) & = \{\begin{matrix} 0 & t \leq | g (x) | \\ \frac{t}{t + | g (x) |} & t > | g (x) |, \end{matrix} \end{matrix}

is a fuzzy norm on X.

Proof.

At the first we will prove that

N_{g}^{(2)}

is a fuzzy norm. The proof of the other parts are similar. Note that for investigation of the fuzzy norm properties in the case

N_{g}^{(3)}

ker g = {0}

will be used in part (2) of Definition 2.1. Since

| g (x) | = 0

implies that

x = 0

(1): If $t \leq 0$ , then clearly $N_{g}^{(2)} (x, t) = 0$ for all $x \in X$ .
(2): If $x = 0$ , then $g (x) = 0$ and consequently for all $t > 0 = | g (x) |$ , $N_{g}^{(2)} (x, t) = \frac{t}{t + 0 + 0} = 1$ . For the converse let $N_{g}^{(2)} (x, t) = 1$ for all $t > 0$ . At the first we will show that $g (x) = 0$ . Suppose by contradiction that $g (x) \neq 0$ .

If $t = | g (x) |$ , then $N_{g}^{(2)} (x, t) = 0$ that is a contradiction. This shows that $g (x) = 0$ . So by hypothesis, for all $t > 0 = | g (x) |$ , $\frac{t}{t + ∥ x ∥ + 0} = N_{g}^{(2)} (x, t) = 1$ . It follows that $∥ x ∥ = 0$ that implies $x = 0$ .
(3): If $c \neq 0$ , then

$\begin{matrix} N_{g}^{(2)} (c x, t) & = \{\begin{matrix} 0 & t \leq | g (c x) | \\ \frac{t}{t + ∥ c x ∥ + | g (c x) |} & t > | g (c x) | \end{matrix} \\ = \{\begin{matrix} 0 & \frac{t}{| c |} \leq | g (x) | \\ \frac{\frac{t}{| c |}}{\frac{t}{| c |} + ∥ x ∥ + | g (x) |} & \frac{t}{| c |} > | g (x) | \end{matrix} \\ = N_{g}^{(2)} (x, \frac{t}{| c |}), x \in X, t \in R . \end{matrix}$
(4): Let $x, y \in X$ and $s, t \in R$ . If $s + t \leq | g (x + y) | = | g (x) + g (y) |$ , then $s \leq | g (x) |$ or $t \leq | g (y) |$ . So

$0 = N_{g}^{(2)} (x + y, s + t) \geq min (N_{g}^{(2)} (x, s), N_{g}^{(2)} (y, t)) = 0 .$

Let $s + t > | g (x) + g (y) |$ and $s \leq | g (x) |$ or $t \leq | g (y) |$ , then clearly,

$\begin{matrix} N_{g}^{(2)} (x + y, s + t) & = \frac{s + t}{s + t + ∥ x + y ∥ + | g (x) + g (y) |} \\ > 0 \\ = min (N_{g}^{(2)} (x, s), N_{g}^{(2)} (y, t)) . \end{matrix}$

Let $s + t > | g (x) + g (y) |$ and $s > | g (x) |$ and $t > | g (y) |$ . Then $N_{g}^{(2)} (x + y, s + t) = \frac{s + t}{s + t + ∥ x + y ∥ + | g (x) + g (y) |}$ and

$N_{g}^{(2)} (x, s) = \frac{s}{s + ∥ x ∥ + | g (x) |}$ and $N_{g}^{(2)} (y, t) = \frac{t}{t + ∥ y ∥ + | g (y) |}$ .

If $\frac{s}{s + ∥ x ∥ + | g (x) |} \leq \frac{t}{t + ∥ y ∥ + | g (y) |}$ , then

$s (∥ y ∥ + | g (y) |) \leq t (∥ x ∥ + | g (x) |) .$

(1)

If $\frac{t}{t + ∥ y ∥ + | g (y) |} \leq \frac{s}{s + ∥ x ∥ + | g (x) |}$ , then

$t (∥ x ∥ + | g (x) |) \leq s (∥ y ∥ + | g (y) |) .$

(2)

In the case where $N_{g}^{(2)} (x, s) \leq N_{g}^{(2)} (y, t)$ , inequality 1 implies that

$\begin{matrix} \frac{s + t}{s + t + ∥ x + y ∥ + | g (x) + g (y) |} & = N_{g}^{(2)} (x + y, s + t) \\ \geq \frac{s}{s + ∥ x ∥ + | g (x) |} \\ = min (N_{g}^{(2)} (x, s), N_{g}^{(2)} (y, t)) . \end{matrix}$

Also in the case where $N_{g}^{(2)} (y, t) \leq N_{g}^{(2)} (x, s)$ , inequality 2 implies that $N_{g}^{(2)} (x + y, s + t) \geq min (N_{g}^{(2)} (x, s), N_{g}^{(2)} (y, t))$ .
(5): Let $s \leq t$ . If $N_{g}^{(2)} (x, s) = 0$ , then clearly $N_{g}^{(2)} (x, s) \leq N_{g}^{(2)} (x, t)$ . Let $N_{g}^{(2)} (x, s) \neq 0$ . Then $s > | g (x) |$ . It follows that $t > | g (x) |$ .

Since $s (∥ x ∥ + | g (x) |) \leq t (∥ x ∥ + | g (x) |)$ , $N_{g}^{(2)} (x, s) \leq N_{g}^{(2)} (x, t)$ . So $N_{g}^{(2)} (x, .)$ is an increasing function and also

$lim_{t \to \infty} N_{g}^{(2)} (x, t) = lim_{t \to \infty} \frac{t}{t + ∥ x ∥ + | g (x) |} = 1 .$

□

Proposition 2.6.

Let X be a normed linear space and

h \in X^{'}

. Then the map

\begin{matrix} N_{h}^{(1)} & : X \times R ⟶ [0, 1] \\ N_{h}^{(1)} (x, t) & = \{\begin{matrix} 0 & t \leq ∥ x ∥ + | h (x) | \\ \frac{t - ∥ x ∥ - | h (x) |}{t} & t > ∥ x ∥ + | h (x) |, \end{matrix} \end{matrix}

is a fuzzy norm on X.

Also if

ker h = {0}

, then the map

\begin{matrix} N_{h}^{(2)} & : X \times R ⟶ [0, 1] \\ N_{h}^{(2)} (x, t) & = \{\begin{matrix} 0 & t \leq | h (x) | \\ \frac{t - | h (x) |}{t} & t > | h (x) |, \end{matrix} \end{matrix}

is a fuzzy norm on X.

Proof.

A similar argument used to Proposition 2.5 can be applied. □

3. The separate continuity of $N_{f}$ , $N_{f, ϵ}$ , $N_{g}^{(i)}$ , $N_{h}^{(j)}$ , $1 \leq i \leq 3$ , $1 \leq j \leq 2$

In this section we characterize separate continuity of the fuzzy norms

N_{f}

N_{f, ϵ}

N_{g}^{(i)}

N_{h}^{(j)}

1 \leq i \leq 3

1 \leq j \leq 2

, where

ϵ > 0

and

f, g, h \in X^{*}

Theorem 3.1.

Let X be a normed linear space and

f \in X^{*}

. If

N_{f} : X \times R ⟶ [0, 1]

is defined by

N_{f} (x, t) = \{\begin{matrix} 0 & t \leq ∥ x ∥ + | f (x) | \\ \frac{t - ∥ x ∥ - | f (x) |}{t + ∥ x ∥ + | f (x) |} & t > ∥ x ∥ + | f (x) |, \end{matrix}

then the map

\begin{matrix} N_{f} (., t) : & X ⟶ [0, 1] \\ x ⟶ N_{f} (x, t), \end{matrix}

is continuous for all

t \in R

Also the map

\begin{matrix} N_{f} (x, .) : & R ⟶ [0, 1] \\ t ⟶ N_{f} (x, t), \end{matrix}

is continuous for all

x \in X

except

x = 0

Proof.

t \leq 0

, then

N_{f} (x, t) = 0

for all

x \in X

. So

N_{f} (., t) : X ⟶ [0, 1]

is a constant function and consequently is continuous on X for all

t \leq 0

For a fixed

t > 0

, let

x \in X

and

{x_{n}}_{n = 1}^{\infty}

be a sequence such that

x_{n} ⟶ x

n ⟶ \infty

. If

t = ∥ x ∥ + | f (x) |

, then for all subsequences

{x_{n_{k}}}_{k = 1}^{\infty} \subseteq {x_{n}}_{n = 1}^{\infty}

satisfying

t \leq ∥ x_{n_{k}} ∥ + | f (x_{n_{k}}) |

k \in N

, we have

lim_{k \to \infty} N_{f} (x_{n_{k}}, t) = lim_{k \to \infty} 0 = 0 = N_{f} (x, t) .

Also for all subsequences

{x_{n_{k}}}_{k = 1}^{\infty} \subseteq {x_{n}}_{n = 1}^{\infty}

satisfying

t > ∥ x_{n_{k}} ∥ + | f (x_{n_{k}}) |, k \in N

, we have

\begin{matrix} lim_{k \to \infty} N_{f} (x_{n_{k}}, t) & = lim_{k \to \infty} \frac{t - ∥ x_{n_{k}} ∥ - | f (x_{n_{k}}) |}{t + ∥ x_{n_{k}} ∥ + | f (x_{n_{k}}) |} \\ = \frac{t - t}{t + t} \\ = 0 \\ = N_{f} (x, t) . \end{matrix}

t < ∥ x ∥ + | f (x) |

, then there exists an

N \in N

such that

t < ∥ x_{n} ∥ + | f (x_{n}) |

for all

n \geq N

. So

lim_{n \to \infty} N_{f} (x_{n}, t) = lim_{n \to \infty} 0 = 0 = N_{f} (x, t)

t > ∥ x ∥ + | f (x) |

, then there exists an

N \in N

such that

t > ∥ x_{n} ∥ + | f (x_{n}) |

for all

n \geq N

. So

\begin{matrix} lim_{n \to \infty} N_{f} (x_{n}, t) & = lim_{n \to \infty} \frac{t - ∥ x_{n} ∥ - | f (x_{n}) |}{t + ∥ x_{n} ∥ + | f (x_{n}) |} \\ = \frac{t - ∥ x ∥ - | f (x) |}{t + ∥ x ∥ + | f (x) |} \\ = N_{f} (x, t) . \end{matrix}

Hence for each

t > 0

N_{f} (., t)

is continuous at every

x \in X

. So

N_{f} (., t)

is continuous on X for all

t \in R .

For any fixed

x \neq 0

, let

t \in R

and

t_{n} ⟶ t

n ⟶ \infty

. If

t = ∥ x ∥ + | f (x) |

, then for all subsequences

{t_{n_{k}}}_{k = 1}^{\infty} \subseteq {t_{n}}_{n = 1}^{\infty}

satisfying

t_{n_{k}} \leq ∥ x ∥ + | f (x) |

we have

lim_{k \to \infty} N_{f} (x, t_{n_{k}}) = lim_{k \to \infty} 0 = 0 = N_{f} (x, t) .

Also for all subsequences

{t_{n_{k}}}_{k = 1}^{\infty} \subseteq {t_{n}}_{n = 1}^{\infty}

satisfying

t_{n_{k}} > ∥ x ∥ + | f (x) |

we have

\begin{matrix} lim_{k \to \infty} N_{f} (x, t_{n_{k}}) & = lim_{k \to \infty} \frac{t_{n_{k}} - ∥ x ∥ - | f (x) |}{t_{n_{k}} + ∥ x ∥ + | f (x) |} \\ = \frac{t - ∥ x ∥ - | f (x) |}{t + ∥ x ∥ + | f (x) |} \\ = \frac{t - t}{t + t} \\ = 0 \\ = N_{f} (x, t) . \end{matrix}

t < ∥ x ∥ + | f (x) |

, then there exists an

N \in N

such that

t_{n} < ∥ x ∥ + | f (x) |

for all

n \geq N

. So

lim_{n \to \infty} N_{f} (x, t_{n}) = lim_{n \to \infty} 0 = 0 = N_{f} (x, t)

t > ∥ x ∥ + | f (x) |

, then there exists an

N \in N

such that

t_{n} > ∥ x ∥ + | f (x) |

for all

n \geq N

. So

\begin{matrix} lim_{n \to \infty} N_{f} (x, t_{n}) & = lim_{n \to \infty} \frac{t_{n} - ∥ x ∥ - | f (x) |}{t_{n} + ∥ x ∥ + | f (x) |} \\ = \frac{t - ∥ x ∥ - | f (x) |}{t + ∥ x ∥ + | f (x) |} \\ = N_{f} (x, t) . \end{matrix}

It follows that for any fixed

x \neq 0

t_{n} ⟶ t

n ⟶ \infty

, implies that

lim_{n ⟶ \infty} N_{f} (x, t_{n}) = N_{f} (x, t)

. Hence

N_{f} (x, .) : R ⟶ [0, 1]

is continuous for all

x \neq 0

We shall show that

N_{f} (0, .) : R ⟶ [0, 1]

is not continuous at

t = 0

Since

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

lim_{t ⟶ 0^{+}} N_{f} (0, t) = 1 \neq N_{f} (0, 0) = 0

. This shows that

N_{f} (0, .) : R ⟶ [0, 1]

is not continuous on

R

. □

Theorem 3.2.

Let X be a normed linear space,

f \in X^{*}

and

ϵ > 0

. If

N_{f, ϵ} : X \times R ⟶ [0, 1]

is defined by

N_{f, ϵ} (x, t) = \{\begin{matrix} 0 & t \leq ∥ x ∥ (ϵ + ∥ f ∥) \\ \frac{t - | f (x) |}{t + | f (x) |} & t > ∥ x ∥ (ϵ + ∥ f ∥), \end{matrix}

then

(1): the map

$\begin{matrix} N_{f, ϵ} (., t) : & X ⟶ [0, 1] \\ x ⟶ N_{f, ϵ} (x, t), \end{matrix}$

is continuous for all $t \leq 0$ .
(2): if $t > 0$ , then $N_{f, ϵ} (., t)$ is continuous at every $x \in X ∖ S$ , where $S = {x \in X | ∥ x ∥ = \frac{t}{ϵ + ∥ f ∥}}$ .
(3): the map $N_{f, ϵ} (x, .)$ is continuous at every $t \in R ∖ T$ , where $T = {t \in R | t = ∥ x ∥ (ϵ + ∥ f ∥)}$ .

Proof.

(1): If $t \leq 0$ , then $N_{f, ϵ} (x, t) = 0$ for all $x \in X$ . So $N_{f, ϵ} (., t)$ is a constant function on X that is continuous.
(2): If $t > 0$ and $x \in S$ , then $∥ x ∥ = \frac{t}{ϵ + ∥ f ∥}$ and $N_{f, ϵ} (x, t) = 0$ . Set $x_{n} = \frac{t - \frac{t}{2 n}}{∥ x ∥ (ϵ + ∥ f ∥)} x$ for all $n \in N$ . So $x_{n} ⟶ x$ as $n ⟶ \infty$ , and

$∥ x_{n} ∥ = \frac{t - \frac{t}{2 n}}{∥ x ∥ (ϵ + ∥ f ∥)} ∥ x ∥ < \frac{t}{(ϵ + ∥ f ∥)}$ for all $n \in N$ . This shows that

$t > ∥ x_{n} ∥ (ϵ + ∥ f ∥)$ for all $n \in N$ .

Hence

$\begin{matrix} lim_{n \to \infty} N_{f, ϵ} (x_{n}, t) & = lim_{n \to \infty} \frac{t - | f (x_{n}) |}{t + | f (x_{n}) |} \\ = \frac{t - | f (x) |}{t + | f (x) |} \\ \neq N_{f, ϵ} (x, t) = 0, \end{matrix}$

since, $t = ∥ x ∥ (ϵ + ∥ f ∥) > ∥ x ∥ ∥ f ∥ \geq | f (x) |$ . Therefore $N_{f, ϵ} (., t)$ is discontinuous at every $x \in S$ .

Let $x \notin S$ and let $x_{n} ⟶ x$ as $n ⟶ \infty$ . So $t > ∥ x ∥ (ϵ + ∥ f ∥)$ or $t < ∥ x ∥ (ϵ + ∥ f ∥)$ . If $t > ∥ x ∥ (ϵ + ∥ f ∥)$ , then there exists an $N \in N$ such that $t > ∥ x_{n} ∥ (ϵ + ∥ f ∥)$ for all $n \geq N$ . Hence

$\begin{matrix} lim_{n \to \infty} N_{f, ϵ} (x_{n}, t) & = lim_{n \to \infty} \frac{t - | f (x_{n}) |}{t + | f (x_{n}) |} \\ = \frac{t - | f (x) |}{t + | f (x) |} \\ = N_{f, ϵ} (x, t) . \end{matrix}$

If $t < ∥ x ∥ (ϵ + ∥ f ∥)$ , then there exists an $N \in N$ such that

$t < ∥ x_{n} ∥ (ϵ + ∥ f ∥)$ for all $n \geq N$ . So

$lim_{n \to \infty} N_{f, ϵ} (x_{n}, t) = lim_{n \to \infty} 0 = 0 = N_{f, ϵ} (x, t) .$

Consequently in the case where $t > 0$ , $N_{f, ϵ} (., t)$ is continuous at every point $x \in X ∖ S$ .
(3): Let $t \in T$ . So $t = ∥ x ∥ (ϵ + ∥ f ∥)$ and $N_{f, ϵ} (x, t) = 0$ .

Set $t_{n} = (∥ x ∥ + \frac{1}{n}) (ϵ + ∥ f ∥)$ for all $n \in N$ . Clearly $t_{n} ⟶ t$ as $n ⟶ \infty$ , and $t_{n} > ∥ x ∥ (ϵ + ∥ f ∥)$ for all $n \in N$ . Hence

$\begin{matrix} lim_{n \to \infty} N_{f, ϵ} (x, t_{n}) & = lim_{n \to \infty} \frac{t_{n} - | f (x) |}{t_{n} + | f (x) |} \\ = \{\begin{matrix} 1 & x = 0 \\ \frac{t - | f (x) |}{t + | f (x) |} & x \neq 0 . \end{matrix} \end{matrix}$

It follows that $lim_{n \to \infty} N_{f, ϵ} (x, t_{n}) \neq N_{f, ϵ} (x, t) = 0$ . Note that if $t = ∥ x ∥ (ϵ + ∥ f ∥)$ and $x \neq 0$ , then $t - | f (x) | \neq 0$ . Since

$t = ∥ x ∥ (ϵ + ∥ f ∥) > ∥ x ∥ ∥ f ∥ \geq | f (x) | .$

We shall show that $N_{f, ϵ} (x, .)$ is continuous at every $t \in R ∖ T$ .

Let $t \in R ∖ T$ and let $t_{n} ⟶ t$ as $n ⟶ \infty$ . Then $t < ∥ x ∥ (ϵ + ∥ f ∥)$ or $t > ∥ x ∥ (ϵ + ∥ f ∥)$ . If $t < ∥ x ∥ (ϵ + ∥ f ∥)$ , then there exists an $N \in N$ such that $t_{n} < ∥ x ∥ (ϵ + ∥ f ∥)$ for all $n \geq N$ . Hence

$lim_{n \to \infty} N_{f, ϵ} (x, t_{n}) = lim_{n \to \infty} 0 = 0 = N_{f, ϵ} (x, t) .$

If $t > ∥ x ∥ (ϵ + ∥ f ∥)$ , then there exists an $N \in N$ such that

$t_{n} > ∥ x ∥ (ϵ + ∥ f ∥)$ for all $n \geq N$ . So

$\begin{matrix} lim_{n \to \infty} N_{f, ϵ} (x, t_{n}) & = lim_{n \to \infty} \frac{t_{n} - | f (x) |}{t_{n} + | f (x) |} \\ = \frac{t - | f (x) |}{t + | f (x) |} \\ = N_{f, ϵ} (x, t) . \end{matrix}$

This shows that $N_{f, ϵ} (x, .)$ is continuous at every $t \in R ∖ T$ .

□

Theorem .3.3

Let X be a normed linear space and

g \in X^{*}

. If

N_{g}^{(1)} : X \times R ⟶ [0, 1]

and

N_{g}^{(2)} : X \times R ⟶ [0, 1]

are defined by

\begin{matrix} N_{g}^{(1)} (x, t) & = \{\begin{matrix} 0 & t \leq ∥ x ∥ + | g (x) | \\ \frac{t}{t + ∥ x ∥ + | g (x) |} & t > ∥ x ∥ + | g (x) |, \end{matrix} \end{matrix}

\begin{matrix} N_{g}^{(2)} (x, t) & = \{\begin{matrix} 0 & t \leq | g (x) | \\ \frac{t}{t + ∥ x ∥ + | g (x) |} & t > | g (x) |, \end{matrix} \end{matrix}

and in the case where

ker g = {0}

N_{g}^{(3)} : X \times R ⟶ [0, 1]

is defined by

\begin{matrix} N_{g}^{(3)} (x, t) & = \{\begin{matrix} 0 & t \leq | g (x) | \\ \frac{t}{t + | g (x) |} & t > | g (x) |, \end{matrix} \end{matrix}

then

(1): the maps $N_{g}^{(1)} (., t)$ , $N_{g}^{(2)} (., t)$ and $N_{g}^{(3)} (., t)$ are continuous on X for all $t \leq 0$ .
(2): if $t > 0$ , then the map $N_{g}^{(1)} (., t)$ is continuous at every

$x \in X ∖ S_{1}$ , where $S_{1} = {x \in X | t = ∥ x ∥ + | g (x) |}$ .
(3): if $t > 0$ , then the maps $N_{g}^{(2)} (., t)$ and $N_{g}^{(3)} (., t)$ are continuous at every $x \in X ∖ S_{2}$ , where $S_{2} = {x \in X | t = | g (x) |}$ .
(4): for $x \in X$ , the map $N_{g}^{(1)} (x, .)$ is continuous at every $t \in R ∖ T_{1}$ , where $T_{1} = {t \in R | t = ∥ x ∥ + | g (x) |}$ .
(5): for $x \neq 0$ , the map $N_{g}^{(2)} (x, .)$ is continuous at every $t \in R ∖ T_{2}$ , where $T_{2} = {t > 0 | t = | g (x) |}$ . Also the map $N_{g}^{(2)} (0, .)$ is continuous at every $t \in R$ except $t = 0$ .
(6): for $x \in X$ , the map $N_{g}^{(3)} (x, .)$ is continuous at every $t \in R ∖ T_{3}$ , where $T_{3} = {t \in R | t = | g (x) |}$ .

Proof.

(1): It is obvious.
(2): Let $t > 0$ and $x \in S_{1}$ . So $t = ∥ x ∥ + | g (x) |$ . Set $x_{n} = \frac{t - \frac{t}{2 n}}{∥ x ∥ + | g (x) |} x$ for all $n \in N$ . Obviously $x_{n} ⟶ x$ and so $g (x_{n}) ⟶ g (x)$ as $n ⟶ \infty$ . Also

$\begin{matrix} ∥ x_{n} ∥ & = \frac{t - \frac{t}{2 n}}{∥ x ∥ + | g (x) |} ∥ x ∥ \\ < \frac{t}{∥ x ∥ + | g (x) |} ∥ x ∥ \\ = ∥ x ∥, n \in N, \end{matrix}$

and

$\begin{matrix} | g (x_{n}) | & = \frac{t - \frac{t}{2 n}}{∥ x ∥ + | g (x) |} | g (x) | \\ \leq \frac{t}{∥ x ∥ + | g (x) |} | g (x) | \\ = | g (x) |, n \in N . \end{matrix}$

So $∥ x_{n} ∥ + | g (x_{n}) | < ∥ x ∥ + | g (x) | = t$ for all $n \in N$ . It follows that

$\begin{matrix} lim_{n \to \infty} N_{g}^{(1)} (x_{n}, t) & = lim_{n \to \infty} \frac{t}{t + ∥ x_{n} ∥ + | g (x_{n}) |} \\ = \frac{t}{t + t} \\ = \frac{1}{2} \\ \neq N_{g}^{(1)} (x, t) \\ = 0 . \end{matrix}$

This shows that $N_{g}^{(1)} (., t)$ is discontinuous at every $x \in S_{1}$ . Now let $x \in X ∖ S_{1}$ and ${z_{n}}_{n = 1}^{\infty}$ be a sequence such that $z_{n} ⟶ x$ as $n ⟶ \infty$ . So $t > ∥ x ∥ + | g (x) |$ or $t < ∥ x ∥ + | g (x) |$ . If $t > ∥ x ∥ + | g (x) |$ , then there exists an $N \in N$ such that $t > ∥ z_{n} ∥ + | g (z_{n}) |$ for all $n \geq N$ . Hence

$\begin{matrix} lim_{n \to \infty} N_{g}^{(1)} (z_{n}, t) & = lim_{n \to \infty} \frac{t}{t + ∥ z_{n} ∥ + | g (z_{n}) |} \\ = \frac{t}{t + ∥ x ∥ + | g (x) |} \\ = N_{g}^{(1)} (x, t) . \end{matrix}$

If $t < ∥ x ∥ + | g (x) |$ , then there exists an $N \in N$ such that $t < ∥ z_{n} ∥ + | g (z_{n}) |$ for all $n \geq N$ . So

$lim_{n \to \infty} N_{g}^{(1)} (z_{n}, t) = lim_{n \to \infty} 0 = 0 = N_{g}^{(1)} (x, t) .$

This shows that $N_{g}^{(1)} (., t)$ is continuous at every $x \in X ∖ S_{1}$ .
(3): Let $x \in S_{2}$ . So $t = | g (x) |$ , $x \neq 0$ and $N_{g}^{(2)} (x, t) = N_{g}^{(3)} (x, t) = 0$ . Set $x_{n} = (1 - \frac{1}{2 n}) x$ for all $n \in N$ . Clearly $x_{n} ⟶ x$ , $g (x_{n}) ⟶ g (x)$ as $n ⟶ \infty$ . Also $| g (x_{n}) | = (1 - \frac{1}{2 n}) | g (x) | < | g (x) | = t$ for all $n \in N$ . Hence

$\begin{matrix} lim_{n \to \infty} N_{g}^{(2)} (x_{n}, t) & = lim_{n \to \infty} \frac{t}{t + ∥ x_{n} ∥ + | g (x_{n}) |} \\ = \frac{t}{t + ∥ x ∥ + t} \\ = \frac{t}{2 t + ∥ x ∥} \\ \neq 0 \\ = N_{g}^{(2)} (x, t) . \end{matrix}$

and

$\begin{matrix} lim_{n \to \infty} N_{g}^{(3)} (x_{n}, t) & = lim_{n \to \infty} \frac{t}{t + | g (x_{n}) |} \\ = \frac{t}{t + t} \\ = \frac{1}{2} \\ \neq 0 \\ = N_{g}^{(3)} (x, t) . \end{matrix}$

Hence $N_{g}^{(2)} (., t)$ and $N_{g}^{(3)} (., t)$ are discontinuous at every $x \in S_{2}$ .

Now let $x \notin S_{2}$ and $z_{n} ⟶ x$ as $n ⟶ \infty$ . Then $t < | g (x) |$ or $t > | g (x) |$ . If $t < | g (x) |$ , then there exists an $N \in N$ such that $t < | g (z_{n}) |$ for all $n \geq N$ . Hence

$lim_{n \to \infty} N_{g}^{(2)} (z_{n}, t) = 0 = N_{g}^{(2)} (x, t),$

and

$lim_{n \to \infty} N_{g}^{(3)} (z_{n}, t) = 0 = N_{g}^{(3)} (x, t) .$

Also if $t > | g (x) |$ , then there exists an $N \in N$ such that

$t > | g (z_{n}) |$ for all $n \geq N$ . So

$\begin{matrix} lim_{n \to \infty} N_{g}^{(2)} (z_{n}, t) & = lim_{n \to \infty} \frac{t}{t + ∥ z_{n} ∥ + | g (z_{n}) |} \\ = \frac{t}{t + ∥ x ∥ + | g (x) |} \\ = N_{g}^{(2)} (x, t), \end{matrix}$

and

$\begin{matrix} lim_{n \to \infty} N_{g}^{(3)} (z_{n}, t) & = lim_{n \to \infty} \frac{t}{t + | g (z_{n}) |} \\ = \frac{t}{t + | g (x) |} \\ = N_{g}^{(3)} (x, t) . \end{matrix}$

Hence $N_{g}^{(2)} (., t)$ and $N_{g}^{(3)} (., t)$ are continuous at every $x \in X ∖ S_{2}$ .
(4): Let $t \in T_{1}$ . Then $t = ∥ x ∥ + | g (x) |$ and $N_{g}^{(1)} (x, t) = 0$ . Set

$t_{n} = ∥ x ∥ + | g (x) | + \frac{1}{n}$ for all $n \in N$ . Clearly $t_{n} > ∥ x ∥ + | g (x) |$ for all $n \in N$ and $lim_{n \to \infty} t_{n} = t$ . So

$\begin{matrix} lim_{n \to \infty} N_{g}^{(1)} (x, t_{n}) & = lim_{n \to \infty} \frac{t_{n}}{t_{n} + ∥ x ∥ + | g (x) |} \\ = \{\begin{matrix} 1 & x = 0 \\ \frac{t}{t + t} = \frac{1}{2} & x \neq 0 . \end{matrix} \end{matrix}$

It follows that $lim_{n \to \infty} N_{g}^{(1)} (x, t_{n}) \neq N_{g}^{(1)} (x, t) = 0$ . This shows that $N_{g}^{(1)} (x, .)$ is discontinuous at every $t \in T_{1}$ . One can easily verify that if $t \in R ∖ T_{1}$ , then $N_{g}^{(1)} (x, .)$ is continuous at t.
(5): Let $x \neq 0$ . If $t \in T_{2}$ , then $t = | g (x) | > 0$ and $N_{g}^{(2)} (x, t) = 0$ . Set $t_{n} = (1 + \frac{1}{n}) | g (x) |$ for all $n \in N$ . Clearly $t_{n} ⟶ t$ as $n ⟶ \infty$ , and $t_{n} > | g (x) |$ for all $n \in N$ . So

$\begin{matrix} lim_{n \to \infty} N_{g}^{(2)} (x, t_{n}) & = lim_{n \to \infty} \frac{t_{n}}{t_{n} + ∥ x ∥ + | g (x) |} \\ = \frac{t}{2 t + ∥ x ∥} \\ \neq 0 \\ = N_{g}^{(2)} (x, t) . \end{matrix}$

This shows that $N_{g}^{(2)} (x, .)$ is discontinuous at every $t \in T_{2}$ .

Let $t \in R ∖ T_{2}$ . Then $t \leq 0$ or $t \neq | g (x) |$ . In the case where $t < 0$ or $t \neq | g (x) |$ , the continuity of $N_{g}^{(2)} (x, .)$ at t can be obviously verified. If $t = 0$ and $t_{n} ⟶ 0$ as $n ⟶ \infty$ , then for all subsequences ${t_{n_{k}}}_{k = 1}^{\infty} \subseteq {t_{n}}_{n = 1}^{\infty}$ satisfying $t_{n_{k}} \leq | g (x) |$ we have, $lim_{k ⟶ \infty} N_{g}^{(2)} (x, t_{n_{k}}) = 0 = N_{g}^{(2)} (x, 0)$ . Also for all subsequences ${t_{n_{k}}}_{k = 1}^{\infty} \subseteq {t_{n}}_{n = 1}^{\infty}$ satisfying $t_{n_{k}} > | g (x) |$ we have,

$lim_{k \to \infty} N_{g}^{(2)} (x, t_{n_{k}}) = lim_{k \to \infty} \frac{t_{n_{k}}}{t_{n_{k} + ∥ x ∥ + | g (x) |}} = \frac{0}{0 + ∥ x ∥ + | g (x) |} = 0 = N_{g}^{(2)} (x, 0) .$

So $N_{g}^{(2)} (x, .)$ is continuous at $t = 0$ .

Clearly the map $N_{g}^{(2)} (0, .)$ is continuous at every $t \in R$ except $t = 0$ .
(6): Inspired by part (5), the proof is obvious.

□

Theorem 3.4.

Let X be a normed linear space and

h \in X^{*}

. If

N_{h}^{(1)} : X \times R ⟶ [0, 1]

is defined by

\begin{matrix} N_{h}^{(1)} (x, t) & = \{\begin{matrix} 0 & t \leq ∥ x ∥ + | h (x) | \\ \frac{t - ∥ x ∥ - | h (x) |}{t} & t > ∥ x ∥ + | h (x) |, \end{matrix} \end{matrix}

and in the case where

ker h = {0}

N_{h}^{(2)} : X \times R ⟶ [0, 1]

is defined by

\begin{matrix} N_{h}^{(2)} (x, t) & = \{\begin{matrix} 0 & t \leq | h (x) | \\ \frac{t - | h (x) |}{t} & t > | h (x) |, \end{matrix} \end{matrix}

then

(1): the map $N_{h}^{(1)} (., t)$ is continuous for all $t \in R$ .
(2): the map $N_{h}^{(1)} (x, .)$ is continuous for all $x \in X$ except $x = 0$ .
(3): the map $N_{h}^{(2)} (., t)$ is continuous for all $t \in R$ .
(4): the map $N_{h}^{(2)} (x, .)$ is continuous for all $x \in X$ except $x = 0$ .

Proof.

An argument similar to the proofs of the previous theorems can be applied. □

4. Declarations

4.1. Ethical Approval

All of the authors consent to participate and consent to publish the manuscript.

4.2. Competing interests

The authors declare that they have no known competing financial interests or personal relationships that could have appeared to influence the work reported in this paper.

4.3. Authors’ contributions

Ali Reza Khoddami and Fatemeh Kouhsari prepared and reviewed the manuscript.

4.4. Funding

not applicable.

4.5. Availability of data and materials

not applicable.

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Fuzzy Normed Linear Spaces Generated by Linear Functionals

Abstract

1. Introduction

2. FUZZY NORMS GENERATED BY LINEAR FUNCTIONALS

3. The separate continuity of N f , N f , ϵ , N g ( i ) , N h ( j ) , 1 ≤ i ≤ 3 , 1 ≤ j ≤ 2

4. Declarations

4.1. Ethical Approval

4.2. Competing interests

4.3. Authors’ contributions

4.4. Funding

4.5. Availability of data and materials

References

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3. The separate continuity of $N_{f}$ , $N_{f, ϵ}$ , $N_{g}^{(i)}$ , $N_{h}^{(j)}$ , $1 \leq i \leq 3$ , $1 \leq j \leq 2$