Fuzzy Normed Algebras Generated by Homomorphisms

1. Introduction and Preliminaries

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].

The concept of fuzzy normed algebra is different from the notion of fuzzy normed linear space in one step that had to be done.

In this paper we will consider A as a normed algebra over the field $\mathbb{F}\in \left\{\mathbb{C},\mathbb{R}\right\}$. Also let $Hom(A,\mathbb{F})$ be the set of all algebra homomorphisms from A into $\mathbb{F}$.

Given a normed algebra A, we will introduce some different classes of fuzzy normed algebras. 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 in the field of fuzzy normed algebras.

2. Algebra Fuzzy Norms Generated by Homomorphisms

Definition 2.1. 

[1] Let X be a linear space and let $N:X\times \mathbb{R}⟶[0,1]$ be a function such that for all $x,y\in X$ and for all $s,t\in \mathbb{R}$,

• $N(x,t)=0$ for all $t\le 0,$
• $N(x,t)=1$ for all $t>0$ if and only if $x=0,$
• $N(cx,t)=N(x,\frac{t}{\left|c\right|})$ for all $c\ne 0,$
• $N(x+y,s+t)\ge min\left(N\right(x,s),N(y,t\left)\right),$
• For each fixed $x\in X,N(x,.):\mathbb{R}⟶[0,1]$ is an increasing function, and $\underset{t⟶\infty }{lim}N(x,t)=1.$
  Then $(X,N)$ is called a fuzzy normed linear space.

Definition 2.2. 

[13] Let A be an algebra and let $N:A\times \mathbb{R}⟶[0,1]$ be a function such that for all $a,b\in A$ and for all $s,t\in \mathbb{R}$,

• $N(a,t)=0$ for all $t\le 0,$
• $N(a,t)=1$ for all $t>0$ if and only if $a=0,$
• $N(\alpha a,t)=N(a,\frac{t}{\left|\alpha \right|})$ for all $\alpha \ne 0,$
• $N(a+b,s+t)\ge min\left(N\right(a,s),N(b,t\left)\right),$
• For each fixed $a\in A,N(a,.):\mathbb{R}⟶[0,1]$ is an increasing function, and $\underset{t⟶\infty }{lim}N(a,t)=1,$
• $N(ab,st)\ge N(a,s)N(b,t)$.
  Then $(A,N)$ is called a fuzzy normed algebra.

Example 2.3. 

If A is a normed algebra, then $N:A\times \mathbb{R}⟶[0,1]$ defined by

$$N(a,t)=\left\{\begin{array}{cc}0\hfill & t\le \parallel a\parallel \hfill \\ 1\hfill & t>\parallel a\parallel \hfill \end{array}\right.$$

is an algebra fuzzy norm on A.

Proposition 2.4. 

Let A be a normed algebra over the field $\mathbb{F}$ and $\phi \in Hom(A,\mathbb{F})$. Define $N_{\phi }:A\times \mathbb{R}⟶[0,1]$ by

$$N_{\phi }(a,t)=\left\{\begin{array}{cc}0\hfill & t\le \parallel a\parallel +\left|\phi \right(a\left)\right|\hfill \\ \frac{t-\parallel a\parallel -\left|\phi \right(a\left)\right|}{t+\parallel a\parallel +\left|\phi \right(a\left)\right|}\hfill & t>\parallel a\parallel +\left|\phi \right(a\left)\right|.\hfill \end{array}\right.$$

Then $N_{\phi }$ is an algebra fuzzy norm on A.

Proof. 

We only prove parts (2), (4), (5), and (6) of Definition 2.2.

(2): If $a=0$, then clearly $N_{\phi }(a,t)=1$ for all $t>0$. For the converse let $N_{\phi }(a,t)=1$ for all $t>0$ and suppose by contradiction that $a\ne 0$.

If $t=\parallel a\parallel +\left|\phi \right(a\left)\right|$, then $N_{\phi }(a,t)=0$ that is a contradiction.

(4): Let $a,b\in A$ and $s,t\in \mathbb{R}$. If $s+t\le 0$, then $s\le 0$ or $t\le 0$. So $N_{\phi }(a,s)=0$ or $N_{\phi }(b,t)=0$.

Hence $0=N_{\phi }(a+b,s+t)\ge min(N_{\phi }(a,s),N_{\phi }(b,t))=0$.

If $s+t>0$ and $s+t\le \parallel a+b\parallel +\left|\phi \right(a+b\left)\right|$, then $s\le \parallel a\parallel +\left|\phi \right(a\left)\right|$ or $t\le \parallel b\parallel +\left|\phi \right(b\left)\right|$. Therefore

$$0=N_{\phi }(a+b,s+t)\ge min(N_{\phi }(a,s),N_{\phi }(b,t))=0.$$

If $s+t>\parallel a+b\parallel +\left|\phi \right(a+b\left)\right|$, then $N_{\phi }(a+b,s+t)=\frac{s+t-\parallel a+b\parallel -\left|\phi \right(a)+\phi (b\left)\right|}{s+t+\parallel a+b\parallel +\left|\phi \right(a)+\phi (b\left)\right|}$.

In this case if $s\le \parallel a\parallel +\left|\phi \right(a\left)\right|$ or $t\le \parallel b\parallel +\left|\phi \right(b\left)\right|$, then clearly

$$\frac{s+t-\parallel a+b\parallel -\left|\phi \right(a)+\phi (b\left)\right|}{s+t+\parallel a+b\parallel +\left|\phi \right(a)+\phi (b\left)\right|}\ge min(N_{\phi }(a,s),N_{\phi }(b,t))=0.$$

If $s>\parallel a\parallel +\left|\phi \right(a\left)\right|$ and $t>\parallel b\parallel +\left|\phi \right(b\left)\right|$, then $N_{\phi }(a,s)=\frac{s-\parallel a\parallel -\left|\phi \right(a\left)\right|}{s+\parallel a\parallel +\left|\phi \right(a\left)\right|}$ and $N_{\phi }(b,t)=\frac{t-\parallel b\parallel -\left|\phi \right(b\left)\right|}{t+\parallel b\parallel +\left|\phi \right(b\left)\right|}$.

One can easily verify that if $N_{\phi }(a,s)\le N_{\phi }(b,t)$, then

$$s(\parallel b\parallel +|\phi \left(b\right)\left|\right)\le t(\parallel a\parallel +|\phi \left(a\right)\left|\right).$$

Also if $N_{\phi }(b,t)\le N_{\phi }(a,s)$, then $t(\parallel a\parallel +|\phi \left(a\right)\left|\right)\le s(\parallel b\parallel +|\phi \left(b\right)\left|\right)$.

A straightforward calculation reveals that

if $s(\parallel b\parallel +|\phi \left(b\right)\left|\right)\le t(\parallel a\parallel +|\phi \left(a\right)\left|\right)$, then

$$\begin{array}{cc}\hfill N_{\phi }(a+b,s+t)& =\frac{s+t-\parallel a+b\parallel -\left|\phi \right(a)+\phi (b\left)\right|}{s+t+\parallel a+b\parallel +\left|\phi \right(a)+\phi (b\left)\right|}\hfill \\ \hfill & \ge \frac{s-\parallel a\parallel -\left|\phi \right(a\left)\right|}{s+\parallel a\parallel +\left|\phi \right(a\left)\right|}\hfill \\ \hfill & =min(N_{\phi }(a,s),N_{\phi }(b,t)).\hfill \end{array}$$

Also if $t(\parallel a\parallel +|\phi \left(a\right)\left|\right)\le s(\parallel b\parallel +|\phi \left(b\right)\left|\right)$, then

$$\begin{array}{cc}\hfill N_{\phi }(a+b,s+t)& =\frac{s+t-\parallel a+b\parallel -\left|\phi \right(a)+\phi (b\left)\right|}{s+t+\parallel a+b\parallel +\left|\phi \right(a)+\phi (b\left)\right|}\hfill \\ \hfill & \ge \frac{t-\parallel b\parallel -\left|\phi \right(b\left)\right|}{t+\parallel b\parallel +\left|\phi \right(b\left)\right|}\hfill \\ \hfill & =min(N_{\phi }(a,s),N_{\phi }(b,t)).\hfill \end{array}$$

Therefore $N_{\phi }(a+b,s+t)\ge min(N_{\phi }(a,s),N_{\phi }(b,t))$ for all $a,b\in A$ and $s,t\in \mathbb{R}$.

(5): Let $a\in A$ and $s\le t$. If $N_{\phi }(a,s)=0$, then clearly $N_{\phi }(a,s)\le N_{\phi }(a,t)$. If $N_{\phi }(a,s)\ne 0$, then $N_{\phi }(a,s)=\frac{s-\parallel a\parallel -\left|\phi \right(a\left)\right|}{s+\parallel a\parallel +\left|\phi \right(a\left)\right|}$ and $s>\parallel a\parallel +\left|\phi \right(a\left)\right|$. Hence $t>\parallel a\parallel +\left|\phi \right(a\left)\right|$ and $N_{\phi }(a,t)=\frac{t-\parallel a\parallel -\left|\phi \right(a\left)\right|}{t+\parallel a\parallel +\left|\phi \right(a\left)\right|}$. Since $s\le t$,

$$s(\parallel a\parallel +|\phi \left(a\right)\left|\right)\le t(\parallel a\parallel +|\phi \left(a\right)\left|\right).$$

Therefore a straightforward calculation reveals that

$$N_{\phi }(a,s)=\frac{s-\parallel a\parallel -\left|\phi \right(a\left)\right|}{s+\parallel a\parallel +\left|\phi \right(a\left)\right|}\le \frac{t-\parallel a\parallel -\left|\phi \right(a\left)\right|}{t+\parallel a\parallel +\left|\phi \right(a\left)\right|}=N_{\phi }(a,t).$$

This shows that $N_{\phi }(a,.)$ is an increasing function for all $a\in A$.

Also $\underset{t\to \infty }{lim}{N}_{\phi }(a,t)=\underset{t\to \infty }{lim}\frac{t-\parallel a\parallel -\left|\phi \right(a\left)\right|}{t+\parallel a\parallel +\left|\phi \right(a\left)\right|}=1$.

(6): Let $a,b\in A$ and $s,t\in \mathbb{R}$. If $st\le 0$, then $s\le 0$ or $t\le 0$. So $N_{\phi }(a,s)=0$ or $N_{\phi }(b,t)=0$.

Hence $0=N_{\phi }(ab,st)\ge N_{\phi }(a,s)N_{\phi }(b,t)=0$.

If $st>0$ and $st\le \parallel ab\parallel +\left|\phi \right(ab\left)\right|$, then $s\le \parallel a\parallel +\left|\phi \right(a\left)\right|$ or $t\le \parallel b\parallel +\left|\phi \right(b\left)\right|$. Therefore

$$0=N_{\phi }(ab,st)\ge N_{\phi }(a,s)N_{\phi }(b,t)=0.$$

Let $st>\parallel ab\parallel +\left|\phi \right(ab\left)\right|$, $s>\parallel a\parallel +\left|\phi \right(a\left)\right|$ and $t>\parallel b\parallel +\left|\phi \right(b\left)\right|$. It is easy but tedious to show that the inequality

$$N_{\phi }(ab,st)\ge N_{\phi }(a,s)N_{\phi }(b,t)$$

(2.1)

is equivalent to

$$\begin{array}{cc}\hfill st\left(s(\parallel b\parallel +|\phi \left(b\right)\left|\right)+t(\parallel a\parallel +|\phi \left(a\right)|\right)& \ge st\left|\phi \right(a\left)\right|\left|\phi \right(b\left)\right|+st\parallel ab\parallel \hfill \\ \hfill & +\parallel ab\parallel \parallel a\parallel \parallel b\parallel +\parallel ab\parallel \parallel a\parallel \left|\phi \right(b\left)\right|\hfill \\ \hfill & +\parallel ab\parallel \parallel b\parallel \left|\phi \right(a\left)\right|+\parallel ab\parallel \left|\phi \right(a\left)\right|\left|\phi \right(b\left)\right|\hfill \\ \hfill & +\parallel a\parallel \parallel b\parallel \left|\phi \left(a\right)\right|\left|\phi \left(b\right)\right|+{\parallel a\parallel \left|\phi \left(a\right)\right|\left|\phi \left(b\right)\right|}^2\hfill \\ \hfill & +{\parallel b\parallel \left|\phi \left(b\right)\right|\left|\phi \left(a\right)\right|}^2+{\left|\phi \left(a\right)\right|}^2{\left|\phi \left(b\right)\right|}^2.\hfill \end{array}$$

So by hypothesis, we have

$$\begin{array}{cc}\hfill & st\left(s(\parallel b\parallel +|\phi \left(b\right)\left|\right)+t(\parallel a\parallel +|\phi \left(a\right)|\right)\hfill \\ \hfill & \ge st\left((\parallel a\parallel +|\phi \left(a\right)\left|\right)(\parallel b\parallel +|\phi \left(b\right)\left|\right)+(\parallel b\parallel +|\phi \left(b\right)\left|\right)(\parallel a\parallel +|\phi \left(a\right)\left|\right)\right)\hfill \\ \hfill & =2st\left((\parallel a\parallel +|\phi \left(a\right)\left|\right)(\parallel b\parallel +|\phi \left(b\right)\left|\right)\right)\hfill \\ \hfill & =(st+st)\left(\parallel a\parallel \parallel b\parallel +\parallel a\parallel \left|\phi \right(b\left)\right|+\parallel b\parallel \left|\phi \right(a\left)\right|+\left|\phi \right(a\left)\right|\left|\phi \right(b\left)\right|\right)\hfill \\ \hfill & =st\left(\parallel a\parallel \parallel b\parallel +\parallel a\parallel \left|\phi \right(b\left)\right|+\parallel b\parallel \left|\phi \right(a\left)\right|+\left|\phi \right(a\left)\right|\left|\phi \right(b\left)\right|\right)\hfill \\ \hfill & +st\left(\parallel a\parallel \parallel b\parallel +\parallel a\parallel \left|\phi \right(b\left)\right|+\parallel b\parallel \left|\phi \right(a\left)\right|+\left|\phi \right(a\left)\right|\left|\phi \right(b\left)\right|\right)\hfill \\ \hfill & \ge st\left(\parallel a\parallel \parallel b\parallel +\left|\phi \right(a\left)\right|\left|\phi \right(b\left)\right|\right)\hfill \\ \hfill & +\left(\parallel ab\parallel +\left|\phi \right(a\left)\phi \right(b\left)\right|\right)\left(\parallel a\parallel \parallel b\parallel +\parallel a\parallel \left|\phi \right(b\left)\right|+\parallel b\parallel \left|\phi \right(a\left)\right|+\left|\phi \right(a\left)\right|\left|\phi \right(b\left)\right|\right)\hfill \\ \hfill & \ge st\left(\parallel ab\parallel +\left|\phi \right(a\left)\right|\left|\phi \right(b\left)\right|\right)\hfill \\ \hfill & +\left(\parallel ab\parallel +\left|\phi \right(a\left)\phi \right(b\left)\right|\right)\left(\parallel a\parallel \parallel b\parallel +\parallel a\parallel \left|\phi \right(b\left)\right|+\parallel b\parallel \left|\phi \right(a\left)\right|+\left|\phi \right(a\left)\right|\left|\phi \right(b\left)\right|\right)\hfill \\ \hfill & =st\left|\phi \right(a\left)\right|\left|\phi \right(b\left)\right|+st\parallel ab\parallel +\parallel ab\parallel \parallel a\parallel \parallel b\parallel +\parallel ab\parallel \parallel a\parallel \left|\phi \right(b\left)\right|\hfill \\ \hfill & +\parallel ab\parallel \parallel b\parallel \left|\phi \right(a\left)\right|+\parallel ab\parallel \left|\phi \right(a\left)\right|\left|\phi \right(b\left)\right|+\parallel a\parallel \parallel b\parallel \left|\phi \right(a\left)\right|\left|\phi \right(b\left)\right|\hfill \\ \hfill & +{\parallel a\parallel \left|\phi \left(a\right)\right|\left|\phi \left(b\right)\right|}^2+{\parallel b\parallel \left|\phi \left(b\right)\right|\left|\phi \left(a\right)\right|}^2+{\left|\phi \left(a\right)\right|}^2{\left|\phi \left(b\right)\right|}^2.\hfill \end{array}$$

This shows that inequality 2.1 holds for all $a,b\in A$ and $s,t\in \mathbb{R}$. Hence $N_{\phi }$ is an algebra fuzzy norm on A. □

Proposition 2.5. 

Let A be a normed algebra, $\phi :A⟶\mathbb{F}$ be a continuous homomorphism, and $ϵ>0$ be an element such that $(ϵ+\parallel \phi \parallel )\ge 1$. Define $N_{\phi ,ϵ}:A\times \mathbb{R}⟶[0,1]$ by

$$N_{\phi ,ϵ}(a,t)=\left\{\begin{array}{cc}0\hfill & t\le \parallel a\parallel (ϵ+\parallel \phi \parallel )\hfill \\ \frac{t-\left|\phi \right(a\left)\right|}{t+\left|\phi \right(a\left)\right|}\hfill & t>\parallel a\parallel (ϵ+\parallel \phi \parallel ).\hfill \end{array}\right.$$

Then $N_{\phi ,ϵ}$ is an algebra fuzzy norm on A.

Proof. 

We only prove parts (2), (4), (5), and (6) of Definition 2.2.

(2) : If $a=0$, then clearly $N_{\phi ,ϵ}(a,t)=1$ for all $t>0$. For the converse let $N_{\phi ,ϵ}(a,t)=1$ for all $t>0$ and suppose by contradiction that $a\ne 0$.

If $t=\parallel a\parallel (ϵ+\parallel \phi \parallel )$, then $N_{\phi ,ϵ}(a,t)=0$ that is a contradiction.

(4) : Let $a,b\in A$ and $s,t\in \mathbb{R}$. If $s+t\le 0$, then $s\le 0$ or $t\le 0$. So $N_{\phi ,ϵ}(a,s)=0$ or $N_{\phi ,ϵ}(b,t)=0$.

Hence $0=N_{\phi ,ϵ}(a+b,s+t)\ge min(N_{\phi ,ϵ}(a,s),N_{\phi ,ϵ}(b,t))=0$.

If $s+t>0$ and $s+t\le \parallel a+b\parallel (ϵ+\parallel \phi \parallel )$, then $s\le \parallel a\parallel (ϵ+\parallel \phi \parallel )$ or $t\le \parallel b\parallel (ϵ+\parallel \phi \parallel )$. Therefore

$$0=N_{\phi ,ϵ}(a+b,s+t)\ge min(N_{\phi ,ϵ}(a,s),N_{\phi ,ϵ}(b,t))=0.$$

If $s+t>\parallel a+b\parallel (ϵ+\parallel \phi \parallel )$, then $N_{\phi ,ϵ}(a+b,s+t)=\frac{s+t-\left|\phi \right(a)+\phi (b\left)\right|}{s+t+\left|\phi \right(a)+\phi (b\left)\right|}$.

In this case if $s\le \parallel a\parallel (ϵ+\parallel \phi \parallel )$ or $t\le \parallel b\parallel (ϵ+\parallel \phi \parallel )$, then clearly

$$\frac{s+t-\left|\phi \right(a)+\phi (b\left)\right|}{s+t+\left|\phi \right(a)+\phi (b\left)\right|}\ge min(N_{\phi ,ϵ}(a,s),N_{\phi ,ϵ}(b,t))=0.$$

If $s>\parallel a\parallel (ϵ+\parallel \phi \parallel )$ and $t>\parallel b\parallel (ϵ+\parallel \phi \parallel )$, then $N_{\phi ,ϵ}(a,s)=\frac{s-\left|\phi \right(a\left)\right|}{s+\left|\phi \right(a\left)\right|}$ and $N_{\phi ,ϵ}(b,t)=\frac{t-\left|\phi \right(b\left)\right|}{t+\left|\phi \right(b\left)\right|}$.

One can easily verify that if $N_{\phi ,ϵ}(a,s)\le N_{\phi ,ϵ}(b,t)$, then $2s\left|\phi \right(b\left)\right|\le 2t\left|\phi \right(a\left)\right|$. Also if $N_{\phi ,ϵ}(b,t)\le N_{\phi ,ϵ}(a,s)$, then $2t\left|\phi \right(a\left)\right|\le 2s\left|\phi \right(b\left)\right|$.

A straightforward calculation reveals that if $2s\left|\phi \right(b\left)\right|\le 2t\left|\phi \right(a\left)\right|$, then

$$\begin{array}{cc}\hfill N_{\phi ,ϵ}(a+b,s+t)& =\frac{s+t-\left|\phi \right(a)+\phi (b\left)\right|}{s+t+\left|\phi \right(a)+\phi (b\left)\right|}\hfill \\ \hfill & \ge \frac{s-\left|\phi \right(a\left)\right|}{s+\left|\phi \right(a\left)\right|}\hfill \\ \hfill & =min(N_{\phi ,ϵ}(a,s),N_{\phi ,ϵ}(b,t)).\hfill \end{array}$$

Also if $2t\left|\phi \right(a\left)\right|\le 2s\left|\phi \right(b\left)\right|$, then

$$\begin{array}{cc}\hfill N_{\phi ,ϵ}(a+b,s+t)& =\frac{s+t-\left|\phi \right(a)+\phi (b\left)\right|}{s+t+\left|\phi \right(a)+\phi (b\left)\right|}\hfill \\ \hfill & \ge \frac{t-\left|\phi \right(b\left)\right|}{t+\left|\phi \right(b\left)\right|}\hfill \\ \hfill & =min(N_{\phi ,ϵ}(a,s),N_{\phi ,ϵ}(b,t)).\hfill \end{array}$$

Therefore $N_{\phi ,ϵ}(a+b,s+t)\ge min(N_{\phi ,ϵ}(a,s),N_{\phi ,ϵ}(b,t))$ for all $a,b\in A$ and $s,t\in \mathbb{R}$.

(5) : Let $a\in A$ and $s\le t$. If $N_{\phi ,ϵ}(a,s)=0$, then clearly $N_{\phi ,ϵ}(a,s)\le N_{\phi ,ϵ}(a,t)$. If $N_{\phi ,ϵ}(a,s)\ne 0$, then $N_{\phi ,ϵ}(a,s)=\frac{s-\left|\phi \right(a\left)\right|}{s+\left|\phi \right(a\left)\right|}$ and $s>\parallel a\parallel (ϵ+\parallel \phi \parallel )$. Hence $t>\parallel a\parallel (ϵ+\parallel \phi \parallel )$ and $N_{\phi ,ϵ}(a,t)=\frac{t-\left|\phi \right(a\left)\right|}{t+\left|\phi \right(a\left)\right|}$.

Since $s\le t$, $2s\left|\phi \right(a\left)\right|\le 2t\left|\phi \right(a\left)\right|$. Therefore a straightforward calculation reveals that $N_{\phi ,ϵ}(a,s)=\frac{s-\left|\phi \right(a\left)\right|}{s+\left|\phi \right(a\left)\right|}\le \frac{t-\left|\phi \right(a\left)\right|}{t+\left|\phi \right(a\left)\right|}=N_{\phi ,ϵ}(a,t)$. This shows that $N_{\phi ,ϵ}(a,.)$ is an increasing function for all $a\in A$. Also

$$\underset{t\to \infty }{lim}{N}_{\phi ,ϵ}(a,t)=\underset{t\to \infty }{lim}\frac{t-\left|\phi \right(a\left)\right|}{t+\left|\phi \right(a\left)\right|}=1.$$

(6): Let $a,b\in A$ and $s,t\in \mathbb{R}$. If $st\le 0$, then $s\le 0$ or $t\le 0$. So $N_{\phi ,ϵ}(a,s)=0$ or $N_{\phi ,ϵ}(b,t)=0$.

Hence $0=N_{\phi ,ϵ}(ab,st)\ge N_{\phi ,ϵ}(a,s)N_{\phi ,ϵ}(b,t)=0$.

Let $st>0$ and $st\le \parallel ab\parallel (ϵ+\parallel \phi \parallel )$. The condition $ϵ+\parallel \phi \parallel \ge 1$ implies that

$s\le \parallel a\parallel (ϵ+\parallel \phi \parallel )$ or $t\le \parallel b\parallel (ϵ+\parallel \phi \parallel )$. Therefore

$$0=N_{\phi ,ϵ}(ab,st)\ge N_{\phi ,ϵ}(a,s)N_{\phi ,ϵ}(b,t)=0.$$

Finally let $st>\parallel ab\parallel (ϵ+\parallel \phi \parallel )$, $s>\parallel a\parallel (ϵ+\parallel \phi \parallel )$ and $t>\parallel b\parallel (ϵ+\parallel \phi \parallel )$. It is easy to see that the inequality

$$N_{\phi ,ϵ}(ab,st)\ge N_{\phi ,ϵ}(a,s)N_{\phi ,ϵ}(b,t)$$

(2.2)

is equivalent to $st\left|\phi \left(a\right)\right|\left|\phi \left(b\right)\right|+{\left|\phi \left(a\right)\right|}^2{\left|\phi \left(b\right)\right|}^2\le s^{2}t\left|\phi \left(b\right)\right|+s{t}^2\left|\phi \left(a\right)\right|$.

Since $s>\parallel a\parallel (ϵ+\parallel \phi \parallel )\ge \parallel a\parallel \parallel \phi \parallel \ge \left|\phi \right(a\left)\right|$ and $t>\parallel b\parallel (ϵ+\parallel \phi \parallel )\ge \parallel b\parallel \parallel \phi \parallel \ge \left|\phi \right(b\left)\right|$, we have

$$\begin{array}{cc}\hfill st\left|\phi \left(a\right)\right|\left|\phi \left(b\right)\right|+{\left|\phi \left(a\right)\right|}^2{\left|\phi \left(b\right)\right|}^2& =st\left|\phi \left(a\right)\right|\left|\phi \left(b\right)\right|+{\left|\phi \left(b\right)\right|}^2\left|\phi \left(a\right)\right|\left|\phi \left(a\right)\right|\hfill \\ \hfill & \le st\left(s\right)\left|\phi \left(b\right)\right|+t^{2}s\left|\phi \left(a\right)\right|\hfill \\ \hfill & =s^{2}t\left|\phi \left(b\right)\right|+s{t}^2\left|\phi \left(a\right)\right|.\hfill \end{array}$$

This shows that inequality 2.2 holds for all $a,b\in A$ and $s,t\in \mathbb{R}$. □

Remark 2.6.

Note that the condition $ϵ+\parallel \phi \parallel \ge 1$ in Proposition 2.5 is necessary. Indeed, for $A=\mathbb{R}$, $\phi =0$, and $0<ϵ<1$ we have $ϵ+\parallel \phi \parallel =ϵ<1$. Also $N_{0,ϵ}(1·1,\sqrt{ϵ}·\sqrt{ϵ})=N_{0,ϵ}(1,ϵ)=0$ and $N_{0,ϵ}(1,\sqrt{ϵ})=1$. Since $ϵ\le \left|1\right|(ϵ+0)$ and $\sqrt{ϵ}>\left|1\right|(ϵ+0)$. So the inequality $N_{0,ϵ}(1·1,\sqrt{ϵ}·\sqrt{ϵ})\ge N_{0,ϵ}(1,\sqrt{ϵ})N_{0,ϵ}(1,\sqrt{ϵ})$ dose not hold. Hence $N_{0,ϵ}$ is not an algebra fuzzy norm .

Proposition 2.7. 

Let A be a normed algebra and $\psi \in Hom(A,\mathbb{F})$. Then the maps

$$\begin{array}{cc}\hfill N_{\psi }^{\left(1\right)}& :A\times \mathbb{R}⟶[0,1]\hfill \\ \hfill N_{\psi }^{\left(1\right)}(a,t)& =\left\{\begin{array}{cc}0\hfill & t\le \parallel a\parallel +\left|\psi \right(a\left)\right|\hfill \\ \frac{t}{t+\parallel a\parallel +\left|\psi \right(a\left)\right|}\hfill & t>\parallel a\parallel +\left|\psi \right(a\left)\right|,\hfill \end{array}\right.\hfill \end{array}$$

and

$$\begin{array}{cc}\hfill N_{\psi }^{\left(2\right)}& :A\times \mathbb{R}⟶[0,1]\hfill \\ \hfill N_{\psi }^{\left(2\right)}(a,t)& =\left\{\begin{array}{cc}0\hfill & t\le \left|\psi \right(a\left)\right|\hfill \\ \frac{t}{t+\parallel a\parallel +\left|\psi \right(a\left)\right|}\hfill & t>\left|\psi \right(a\left)\right|,\hfill \end{array}\right.\hfill \end{array}$$

are algebra fuzzy norms on A.

Also if $ker\psi =\left\{0\right\}$, then the map

$$\begin{array}{cc}\hfill N_{\psi }^{\left(3\right)}& :A\times \mathbb{R}⟶[0,1]\hfill \\ \hfill N_{\psi }^{\left(3\right)}(a,t)& =\left\{\begin{array}{cc}0\hfill & t\le \left|\psi \right(a\left)\right|\hfill \\ \frac{t}{t+\left|\psi \right(a\left)\right|}\hfill & t>\left|\psi \right(a\left)\right|,\hfill \end{array}\right.\hfill \end{array}$$

is an algebra fuzzy norm on A.

Proof. 

At the first we will prove that $N_{\psi }^{\left(2\right)}$ is an algebra fuzzy norm. At the end we will only prove part (6) of Definition 2.2 for $N_{\psi }^{\left(1\right)}$ and $N_{\psi }^{\left(3\right)}$. The proof of the other parts are similar. Note that for investigation of the fuzzy norm properties in the case $N_{\psi }^{\left(3\right)}$, the condition $ker\psi =\left\{0\right\}$ will be used in part (2) of Definition 2.2, since $\left|\psi \right(a\left)\right|=0$ implies that $a=0$.

(1): If $t\le 0$, then clearly $N_{\psi }^{\left(2\right)}(a,t)=0$ for all $a\in A$.

(2): If $a=0$, then $\psi \left(a\right)=0$ and consequently for all $t>0=\left|\psi \right(a\left)\right|$, $N_{\psi }^{\left(2\right)}(a,t)=\frac{t}{t+0+0}=1$. For the converse let $N_{\psi }^{\left(2\right)}(a,t)=1$ for all $t>0$. At the first we will show that $\psi \left(a\right)=0$. Suppose by contradiction that $\psi \left(a\right)\ne 0$.

If $t=\left|\psi \right(a\left)\right|$, then $N_{\psi }^{\left(2\right)}(a,t)=0$ that is a contradiction. This shows that $\psi \left(a\right)=0$. So by hypothesis, for all $t>0=\left|\psi \right(a\left)\right|$, $\frac{t}{t+\parallel a\parallel +0}=N_{\psi }^{\left(2\right)}(a,t)=1$. It follows that $\parallel a\parallel =0$ that implies $a=0$.

(3): If $\alpha \ne 0$, then

$$\begin{array}{cc}\hfill N_{\psi }^{\left(2\right)}(\alpha a,t)& =\left\{\begin{array}{cc}0\hfill & t\le \left|\psi \right(\alpha a\left)\right|\hfill \\ \frac{t}{t+\parallel \alpha a\parallel +\left|\psi \right(\alpha a\left)\right|}\hfill & t>\left|\psi \right(\alpha a\left)\right|\hfill \end{array}\right.\hfill \\ \hfill & =\left\{\begin{array}{cc}0\hfill & \frac{t}{\left|\alpha \right|}\le \left|\psi \left(a\right)\right|\hfill \\ \frac{\frac{t}{\left|\alpha \right|}}{\frac{t}{\left|\alpha \right|}+\parallel a\parallel +\left|\psi \left(a\right)\right|}\hfill & \frac{t}{\left|\alpha \right|}>\left|\psi \left(a\right)\right|\hfill \end{array}\right.\hfill \\ \hfill & =N_{\psi }^{\left(2\right)}\left(a,\frac{t}{\left|\alpha \right|}\right),a\in A,t\in \mathbb{R}.\hfill \end{array}$$

(4): Let $a,b\in A$ and $s,t\in \mathbb{R}$. If $s+t\le \left|\psi \right(a+b\left)\right|=\left|\psi \right(a)+\psi (b\left)\right|$, then $s\le \left|\psi \right(a\left)\right|$ or $t\le \left|\psi \right(b\left)\right|$. So

$$0=N_{\psi }^{\left(2\right)}(a+b,s+t)\ge min(N_{\psi }^{\left(2\right)}(a,s),N_{\psi }^{\left(2\right)}(b,t))=0.$$

Let $s+t>\left|\psi \right(a)+\psi (b\left)\right|$ and $s\le \left|\psi \right(a\left)\right|$ or $t\le \left|\psi \right(b\left)\right|$. Then clearly,

$$\begin{array}{cc}\hfill N_{\psi }^{\left(2\right)}(a+b,s+t)& =\frac{s+t}{s+t+\parallel a+b\parallel +\left|\psi \right(a)+\psi (b\left)\right|}\hfill \\ \hfill & >0\hfill \\ \hfill & =min(N_{\psi }^{\left(2\right)}(a,s),N_{\psi }^{\left(2\right)}(b,t)).\hfill \end{array}$$

Let $s+t>\left|\psi \right(a)+\psi (b\left)\right|$ and $s>\left|\psi \right(a\left)\right|$ and $t>\left|\psi \right(b\left)\right|$. Then $N_{\psi }^{\left(2\right)}(a+b,s+t)=\frac{s+t}{s+t+\parallel a+b\parallel +\left|\psi \right(a)+\psi (b\left)\right|}$ and

$N_{\psi }^{\left(2\right)}(a,s)=\frac{s}{s+\parallel a\parallel +\left|\psi \right(a\left)\right|}$ and $N_{\psi }^{\left(2\right)}(b,t)=\frac{t}{t+\parallel b\parallel +\left|\psi \right(b\left)\right|}$.

If $\frac{s}{s+\parallel a\parallel +\left|\psi \right(a\left)\right|}\le \frac{t}{t+\parallel b\parallel +\left|\psi \right(b\left)\right|}$, then

$$s(\parallel b\parallel +|\psi \left(b\right)\left|\right)\le t(\parallel a\parallel +|\psi \left(a\right)\left|\right).$$

(2.3)

If $\frac{t}{t+\parallel b\parallel +\left|\psi \right(b\left)\right|}\le \frac{s}{s+\parallel a\parallel +\left|\psi \right(a\left)\right|}$, then

$$t(\parallel a\parallel +|\psi \left(a\right)\left|\right)\le s(\parallel b\parallel +|\psi \left(b\right)\left|\right).$$

(2.4)

In the case where $N_{\psi }^{\left(2\right)}(a,s)\le N_{\psi }^{\left(2\right)}(b,t)$, inequality 2.3 implies that

$$\begin{array}{cc}\hfill \frac{s+t}{s+t+\parallel a+b\parallel +\left|\psi \right(a)+\psi (b\left)\right|}& =N_{\psi }^{\left(2\right)}(a+b,s+t)\hfill \\ \hfill & \ge \frac{s}{s+\parallel a\parallel +\left|\psi \right(a\left)\right|}\hfill \\ \hfill & =min(N_{\psi }^{\left(2\right)}(a,s),N_{\psi }^{\left(2\right)}(b,t)).\hfill \end{array}$$

Also in the case where $N_{\psi }^{\left(2\right)}(b,t)\le N_{\psi }^{\left(2\right)}(a,s)$, inequality 2.4 implies that

$N_{\psi }^{\left(2\right)}(a+b,s+t)\ge min(N_{\psi }^{\left(2\right)}(a,s),N_{\psi }^{\left(2\right)}(b,t))$.

(5): Let $s\le t$. If $N_{\psi }^{\left(2\right)}(a,s)=0$, then clearly $N_{\psi }^{\left(2\right)}(a,s)\le N_{\psi }^{\left(2\right)}(a,t)$. Let $N_{\psi }^{\left(2\right)}(a,s)\ne 0$. Then $s>\left|\psi \right(a\left)\right|$. It follows that $t>\left|\psi \right(a\left)\right|$.

Since $s(\parallel a\parallel +|\psi \left(a\right)\left|\right)\le t(\parallel a\parallel +|\psi \left(a\right)\left|\right)$, $N_{\psi }^{\left(2\right)}(a,s)\le N_{\psi }^{\left(2\right)}(a,t)$. So $N_{\psi }^{\left(2\right)}(a,.)$ is an increasing function and also

$$\underset{t\to \infty }{lim}{N}_{\psi }^{\left(2\right)}(a,t)=\underset{t\to \infty }{lim}\frac{t}{t+\parallel a\parallel +\left|\psi \right(a\left)\right|}=1.$$

(6): Let $a,b\in A$ and $s,t\in \mathbb{R}$. If $st\le 0$, then $s\le 0$ or $t\le 0$. So $N_{\psi }^{\left(2\right)}(a,s)=0$ or $N_{\psi }^{\left(2\right)}(b,t)=0$.

Hence $0=N_{\psi }^{\left(2\right)}(ab,st)\ge N_{\psi }^{\left(2\right)}(a,s)N_{\psi }^{\left(2\right)}(b,t)=0$.

If $st>0$ and $st\le \left|\psi \right(ab\left)\right|$, then $s\le \left|\psi \right(a\left)\right|$ or $t\le \left|\psi \right(b\left)\right|$. Therefore

$$0=N_{\psi }^{\left(2\right)}(ab,st)\ge N_{\psi }^{\left(2\right)}(a,s)N_{\psi }^{\left(2\right)}(b,t)=0.$$

Let $st>\left|\psi \right(ab\left)\right|$, $s>\left|\psi \right(a\left)\right|$, and $t>\left|\psi \right(b\left)\right|$. Clearly

$$st+\parallel ab\parallel +\left|\psi \right(a\left)\right|\left|\psi \right(b\left)\right|\le (s+\parallel a\parallel +|\psi \left(a\right)\left|\right)(t+\parallel b\parallel +|\psi \left(b\right)\left|\right).$$

So $\frac{st}{st+\parallel ab\parallel +\left|\psi \right(a\left)\right|\left|\psi \right(b\left)\right|}\ge \frac{st}{(s+\parallel a\parallel +|\psi \left(a\right)\left|\right)(t+\parallel b\parallel +|\psi \left(b\right)\left|\right)}$. Hence

$$N_{\psi }^{\left(2\right)}(ab,st)\ge N_{\psi }^{\left(2\right)}(a,s)N_{\psi }^{\left(2\right)}(b,t).$$

(2.5)

Therefore inequality 2.5 holds for all $a,b\in A$ and $s,t\in \mathbb{R}$.

Now we will investigate part (6) of Definition 2.2 for $N_{\psi }^{\left(1\right)}$ and $N_{\psi }^{\left(3\right)}$.

$$\begin{array}{cc}\hfill N_{\psi }^{\left(1\right)}& :A\times \mathbb{R}⟶[0,1]\hfill \\ \hfill N_{\psi }^{\left(1\right)}(a,t)& =\left\{\begin{array}{cc}0\hfill & t\le \parallel a\parallel +\left|\psi \right(a\left)\right|\hfill \\ \frac{t}{t+\parallel a\parallel +\left|\psi \right(a\left)\right|}\hfill & t>\parallel a\parallel +\left|\psi \right(a\left)\right|,\hfill \end{array}\right.\hfill \end{array}$$

(6): Let $a,b\in A$ and $s,t\in \mathbb{R}$. If $st\le 0$, then $s\le 0$ or $t\le 0$. So $N_{\psi }^{\left(1\right)}(a,s)=0$ or $N_{\psi }^{\left(1\right)}(b,t)=0$.

Hence $0=N_{\psi }^{\left(1\right)}(ab,st)\ge N_{\psi }^{\left(1\right)}(a,s)N_{\psi }^{\left(1\right)}(b,t)=0$.

If $st>0$ and $st\le \parallel ab\parallel +\left|\psi \right(ab\left)\right|$, then $s\le \parallel a\parallel +\left|\psi \right(a\left)\right|$ or $t\le \parallel b\parallel +\left|\psi \right(b\left)\right|$. Therefore

$$0=N_{\psi }^{\left(1\right)}(ab,st)\ge N_{\psi }^{\left(1\right)}(a,s)N_{\psi }^{\left(1\right)}(b,t)=0.$$

Finally in the case where $st>\parallel ab\parallel +\left|\psi \right(a\left)\psi \right(b\left)\right|$, $s>\parallel a\parallel +\left|\psi \right(a\left)\right|$, and $t>\parallel b\parallel +\left|\psi \right(b\left)\right|$ clearly $st+\parallel ab\parallel +\left|\psi \right(a\left)\right|\left|\psi \right(b\left)\right|\le (s+\parallel a\parallel +|\psi \left(a\right)\left|\right)(t+\parallel b\parallel +|\psi \left(b\right)\left|\right)$. So $\frac{st}{st+\parallel ab\parallel +\left|\psi \right(a\left)\right|\left|\psi \right(b\left)\right|}\ge \frac{st}{(s+\parallel a\parallel +|\psi \left(a\right)\left|\right)(t+\parallel b\parallel +|\psi \left(b\right)\left|\right)}$. Hence

$$N_{\psi }^{\left(1\right)}(ab,st)\ge N_{\psi }^{\left(1\right)}(a,s)N_{\psi }^{\left(1\right)}(b,t).$$

(2.6)

Therefore inequality 2.6 holds for all $a,b\in A$ and $s,t\in \mathbb{R}$.

$$\begin{array}{cc}\hfill N_{\psi }^{\left(3\right)}& :A\times \mathbb{R}⟶[0,1]\hfill \\ \hfill N_{\psi }^{\left(3\right)}(a,t)& =\left\{\begin{array}{cc}0\hfill & t\le \left|\psi \right(a\left)\right|\hfill \\ \frac{t}{t+\left|\psi \right(a\left)\right|}\hfill & t>\left|\psi \right(a\left)\right|,\hfill \end{array}\right.\hfill \end{array}$$

(6): Let $a,b\in A$ and $s,t\in \mathbb{R}$. If $st\le 0$, then $s\le 0$ or $t\le 0$. So $N_{\psi }^{\left(3\right)}(a,s)=0$ or $N_{\psi }^{\left(3\right)}(b,t)=0$.

Hence $0=N_{\psi }^{\left(3\right)}(ab,st)\ge N_{\psi }^{\left(3\right)}(a,s)N_{\psi }^{\left(3\right)}(b,t)=0$.

If $st>0$ and $st\le \left|\psi \right(ab\left)\right|$, then $s\le \left|\psi \right(a\left)\right|$ or $t\le \left|\psi \right(b\left)\right|$. Therefore

$$0=N_{\psi }^{\left(3\right)}(ab,st)\ge N_{\psi }^{\left(3\right)}(a,s)N_{\psi }^{\left(3\right)}(b,t)=0.$$

Finally in the case $st>\left|\psi \right(ab\left)\right|$, $s>\left|\psi \right(a\left)\right|$, and $t>\left|\psi \right(b\left)\right|$ clearly

$st+\left|\psi \right(a\left)\right|\left|\psi \right(b\left)\right|\le (s+|\psi \left(a\right)\left|\right)(t+|\psi \left(b\right)\left|\right)$. So $\frac{st}{st+\left|\psi \right(a\left)\right|\left|\psi \right(b\left)\right|}\ge \frac{st}{(s+|\psi \left(a\right)\left|\right)(t+|\psi \left(b\right)\left|\right)}$ and consequently

$$N_{\psi }^{\left(3\right)}(ab,st)\ge N_{\psi }^{\left(3\right)}(a,s)N_{\psi }^{\left(3\right)}(b,t).$$

(2.7)

This shows that for all $a,b\in A$ and $s,t\in \mathbb{R}$ inequality 2.7 holds. □

Proposition 2.8. 

Let A be a normed algebra and $\eta \in Hom(A,\mathbb{F})$. Then the map

$$\begin{array}{cc}\hfill N_{\eta }^{\left(1\right)}& :A\times \mathbb{R}⟶[0,1]\hfill \\ \hfill N_{\eta }^{\left(1\right)}(a,t)& =\left\{\begin{array}{cc}0\hfill & t\le \parallel a\parallel +\left|\eta \right(a\left)\right|\hfill \\ \frac{t-\parallel a\parallel -\left|\eta \right(a\left)\right|}{t}\hfill & t>\parallel a\parallel +\left|\eta \right(a\left)\right|,\hfill \end{array}\right.\hfill \end{array}$$

is an algebra fuzzy norm on A.

Also if $ker\eta =\left\{0\right\}$, then the map

$$\begin{array}{cc}\hfill N_{\eta }^{\left(2\right)}& :A\times \mathbb{R}⟶[0,1]\hfill \\ \hfill N_{\eta }^{\left(2\right)}(a,t)& =\left\{\begin{array}{cc}0\hfill & t\le \left|\eta \right(a\left)\right|\hfill \\ \frac{t-\left|\eta \right(a\left)\right|}{t}\hfill & t>\left|\eta \right(a\left)\right|,\hfill \end{array}\right.\hfill \end{array}$$

is an algebra fuzzy norm on A.

Proof. 

We only prove part (6) of Definition 2.2 for $N_{\eta }^{\left(1\right)}$ and $N_{\eta }^{\left(2\right)}$.

$$\begin{array}{cc}\hfill N_{\eta }^{\left(1\right)}& :A\times \mathbb{R}⟶[0,1]\hfill \\ \hfill N_{\eta }^{\left(1\right)}(a,t)& =\left\{\begin{array}{cc}0\hfill & t\le \parallel a\parallel +\left|\eta \right(a\left)\right|\hfill \\ \frac{t-\parallel a\parallel -\left|\eta \right(a\left)\right|}{t}\hfill & t>\parallel a\parallel +\left|\eta \right(a\left)\right|,\hfill \end{array}\right.\hfill \end{array}$$

(6): Let $a,b\in A$ and $s,t\in \mathbb{R}$. If $st\le 0$, then $s\le 0$ or $t\le 0$. So $N_{\eta }^{\left(1\right)}(a,s)=0$ or $N_{\eta }^{\left(1\right)}(b,t)=0$. Hence $0=N_{\eta }^{\left(1\right)}(ab,st)\ge N_{\eta }^{\left(1\right)}(a,s)N_{\eta }^{\left(1\right)}(b,t)=0$.

If $st>0$ and $st\le \parallel ab\parallel +\left|\eta \right(a\left)\eta \right(b\left)\right|$, then $s\le \parallel a\parallel +\left|\eta \right(a\left)\right|$ or $t\le \parallel b\parallel +\left|\eta \right(b\left)\right|$. Therefore

$$0=N_{\eta }^{\left(1\right)}(ab,st)\ge N_{\eta }^{\left(1\right)}(a,s)N_{\eta }^{\left(1\right)}(b,t)=0.$$

Finally in the case where $st>\parallel ab\parallel +\left|\eta \right(ab\left)\right|$, $s>\parallel a\parallel +\left|\eta \right(a\left)\right|$, and

$t>\parallel b\parallel +\left|\eta \right(b\left)\right|$ we will show that the inequality

$$N_{\eta }^{\left(1\right)}(ab,st)\ge N_{\eta }^{\left(1\right)}(a,s)N_{\eta }^{\left(1\right)}(b,t)$$

(2.8)

holds. It is easy to see that inequality 2.8 is equivalent to

$$s(\parallel b\parallel +|\eta \left(b\right)\left|\right)+t(\parallel a\parallel +|\eta \left(a\right)\left|\right)\ge \parallel a\parallel \parallel b\parallel +\parallel ab\parallel +\parallel a\parallel \left|\eta \right(b\left)\right|+\parallel b\parallel \left|\eta \right(a\left)\right|+2\left|\eta \right(a\left)\right|\left|\eta \right(b\left)\right|.$$

In this case

$$s(\parallel b\parallel +|\eta \left(b\right)\left|\right)\ge (\parallel a\parallel +|\eta \left(a\right)\left|\right)(\parallel b\parallel +|\eta \left(b\right)\left|\right)$$

and

$$t(\parallel a\parallel +|\eta \left(a\right)\left|\right)\ge (\parallel b\parallel +|\eta \left(b\right)\left|\right)(\parallel a\parallel +|\eta \left(a\right)\left|\right).$$

Hence

$$s(\parallel b\parallel +|\eta \left(b\right)\left|\right)+t(\parallel a\parallel +|\eta \left(a\right)\left|\right)\ge 2(\parallel a\parallel \parallel b\parallel +\parallel a\parallel |\eta \left(b\right)|+\parallel b\parallel |\eta \left(a\right)|+|\eta \left(a\right)\left|\right|\eta \left(b\right)\left|\right).$$

Therefore

$$\begin{array}{cc}\hfill s(\parallel b\parallel +|\eta \left(b\right)\left|\right)+t(\parallel a\parallel +|\eta \left(a\right)\left|\right)& \ge (\parallel a\parallel \parallel b\parallel +\parallel a\parallel |\eta \left(b\right)|+\parallel b\parallel |\eta \left(a\right)|+2|\eta \left(a\right)\left|\right|\eta \left(b\right)\left|\right)\hfill \\ \hfill & +(\parallel a\parallel \parallel b\parallel )+(\parallel a\parallel |\eta \left(b\right)|+\parallel b\parallel |\eta \left(a\right)\left|\right)\hfill \\ \hfill & \ge (\parallel a\parallel \parallel b\parallel +\parallel ab\parallel +\parallel a\parallel |\eta \left(b\right)|+\parallel b\parallel |\eta \left(a\right)|+2|\eta \left(a\right)\left|\right|\eta \left(b\right)\left|\right)\hfill \\ \hfill & +(\parallel a\parallel |\eta \left(b\right)|+\parallel b\parallel |\eta \left(a\right)\left|\right)\hfill \\ \hfill & \ge \parallel a\parallel \parallel b\parallel +\parallel ab\parallel +\parallel a\parallel \left|\eta \right(b\left)\right|+\parallel b\parallel \left|\eta \right(a\left)\right|+2\left|\eta \right(a\left)\right|\left|\eta \right(b\left)\right|.\hfill \end{array}$$

So inequality 2.8 holds for all $a,b\in A$ and $s,t\in \mathbb{R}$.

$$\begin{array}{cc}\hfill N_{\eta }^{\left(2\right)}& :A\times \mathbb{R}⟶[0,1]\hfill \\ \hfill N_{\eta }^{\left(2\right)}(a,t)& =\left\{\begin{array}{cc}0\hfill & t\le \left|\eta \right(a\left)\right|\hfill \\ \frac{t-\left|\eta \right(a\left)\right|}{t}\hfill & t>\left|\eta \right(a\left)\right|,\hfill \end{array}\right.\hfill \end{array}$$

(6): Let $a,b\in A$ and $s,t\in \mathbb{R}$. If $st\le 0$, then $s\le 0$ or $t\le 0$. So $N_{\eta }^{\left(2\right)}(a,s)=0$ or $N_{\eta }^{\left(2\right)}(b,t)=0$.

Hence $0=N_{\eta }^{\left(2\right)}(ab,st)\ge N_{\eta }^{\left(2\right)}(a,s)N_{\eta }^{\left(2\right)}(b,t)=0$.

If $st>0$ and $st\le \left|\eta \right(a\left)\eta \right(b\left)\right|$, then $s\le \left|\eta \right(a\left)\right|$ or $t\le \left|\eta \right(b\left)\right|$. Therefore

$$0=N_{\eta }^{\left(2\right)}(ab,st)\ge N_{\eta }^{\left(2\right)}(a,s)N_{\eta }^{\left(2\right)}(b,t)=0.$$

Let $st>\left|\eta \right(ab\left)\right|$, $s>\left|\eta \right(a\left)\right|$, and $t>\left|\eta \right(b\left)\right|$. It is easy to see that the inequality

$$N_{\eta }^{\left(2\right)}(ab,st)\ge N_{\eta }^{\left(2\right)}(a,s)N_{\eta }^{\left(2\right)}(b,t)$$

(2.9)

is equivalent to $s\left|\eta \right(b\left)\right|+t\left|\eta \right(a\left)\right|\ge 2\left|\eta \right(a\left)\right|\left|\eta \right(b\left)\right|$. So in this case we have,

$$s\left|\eta \right(b\left)\right|\ge \left|\eta \right(a\left)\right|\left|\eta \right(b\left)\right|$$

and

$$t\left|\eta \right(a\left)\right|\ge \left|\eta \right(b\left)\right|\left|\eta \right(a\left)\right|.$$

Hence

$$s\left|\eta \right(b\left)\right|+t\left|\eta \right(a\left)\right|\ge 2\left|\eta \right(a\left)\right|\left|\eta \right(b\left)\right|.$$

This shows that for all $a,b\in A$ and $s,t\in \mathbb{R}$ inequality 2.9 holds. □

3. The separate continuity of $N_{\phi }$, $N_{\phi ,ϵ}$, $N_{\psi }^{\left(i\right)}$, $N_{\eta }^{\left(j\right)}$, $1\le i\le 3$, $1\le j\le 2$

In this section we characterize separate continuity of the algebra fuzzy norms $N_{\phi }$, $N_{\phi ,ϵ}$, $N_{\psi }^{\left(i\right)}$, $N_{\eta }^{\left(j\right)}$, $1\le i\le 3$, $1\le j\le 2$, where $ϵ>0$ and $\phi ,\psi ,\eta$ are continuous homomorphisms from A into $\mathbb{F}$.

Theorem 3.1. 

Let A be a normed algebra and$\phi :A⟶\mathbb{F}$be a continuous homomorphism. If

$N_{\phi }:A\times \mathbb{R}⟶[0,1]$is defined by

$$N_{\phi }(a,t)=\left\{\begin{array}{cc}0\hfill & t\le \parallel a\parallel +\left|\phi \right(a\left)\right|\hfill \\ \frac{t-\parallel a\parallel -\left|\phi \right(a\left)\right|}{t+\parallel a\parallel +\left|\phi \right(a\left)\right|}\hfill & t>\parallel a\parallel +\left|\phi \right(a\left)\right|,\hfill \end{array}\right.$$

then the map

$$\begin{array}{cc}\hfill N_{\phi }(.,t):& A⟶[0,1]\hfill \\ \hfill & a⟶N_{\phi }(a,t),\hfill \end{array}$$

is continuous for all$t\in \mathbb{R}$.

Also the map

$$\begin{array}{cc}\hfill N_{\phi }(a,.):& \mathbb{R}⟶[0,1]\hfill \\ \hfill & t⟶N_{\phi }(a,t),\hfill \end{array}$$

is continuous for all$a\in A$except$a=0$.

Proof. 

If $t\le 0$, then $N_{\phi }(a,t)=0$ for all $a\in A$. So $N_{\phi }(.,t):A⟶[0,1]$ is a constant function and consequently is continuous on A for all $t\le 0$.

For a fixed $t>0$, let $a\in A$ and ${\left\{a_n\right\}}_{n=1}^{\infty }$ be a sequence such that $a_n⟶a$ as $n⟶\infty$. If $t=\parallel a\parallel +\left|\phi \right(a\left)\right|$, then for all subsequences ${\left\{a_{n_k}\right\}}_{k=1}^{\infty }\subseteq {\left\{a_n\right\}}_{n=1}^{\infty }$ satisfying $t\le \parallel a_{n_k}\parallel +|\phi \left(a_{n_k}\right)|$, $k\in \mathbb{N}$, we have

$$\underset{k\to \infty }{lim}{N}_{\phi }(x_{n_k},t)=\underset{k\to \infty }{lim}0=0=N_{\phi }(a,t).$$

Also for all subsequences ${\left\{a_{n_k}\right\}}_{k=1}^{\infty }\subseteq {\left\{a_n\right\}}_{n=1}^{\infty }$ satisfying $t>\parallel a_{n_k}\parallel +|\phi \left(a_{n_k}\right)|,k\in \mathbb{N}$, we have

$$\begin{array}{cc}\hfill \underset{k\to \infty }{lim}{N}_{\phi }(a_{n_k},t)& =\underset{k\to \infty }{lim}\frac{t-\parallel a_{n_k}\parallel -|\phi \left(a_{n_k}\right)|}{t+\parallel a_{n_k}\parallel +|\phi \left(a_{n_k}\right)|}\hfill \\ \hfill & =\frac{t-t}{t+t}\hfill \\ \hfill & =0\hfill \\ \hfill & =N_{\phi }(a,t).\hfill \end{array}$$

If $t<\parallel a\parallel +\left|\phi \right(a\left)\right|$, then there exists an $N\in \mathbb{N}$ such that $t<\parallel a_n\parallel +|\phi \left(a_n\right)|$ for all $n\ge N$. So $\underset{n\to \infty }{lim}{N}_{\phi }(a_n,t)=\underset{n\to \infty }{lim}0=0=N_{\phi }(a,t)$.

If $t>\parallel a\parallel +\left|\phi \right(a\left)\right|$, then there exists an $N\in \mathbb{N}$ such that $t>\parallel a_n\parallel +|\phi \left(a_n\right)|$ for all $n\ge N$. So

$$\begin{array}{cc}\hfill \underset{n\to \infty }{lim}{N}_{\phi }(a_n,t)& =\underset{n\to \infty }{lim}\frac{t-\parallel a_n\parallel -|\phi \left(a_n\right)|}{t+\parallel a_n\parallel +|\phi \left(a_n\right)|}\hfill \\ \hfill & =\frac{t-\parallel a\parallel -\left|\phi \right(a\left)\right|}{t+\parallel a\parallel +\left|\phi \right(a\left)\right|}\hfill \\ \hfill & =N_{\phi }(a,t).\hfill \end{array}$$

Hence for each $t>0$, $N_{\phi }(.,t)$ is continuous at every $a\in A$. So $N_{\phi }(.,t)$ is continuous on A for all $t\in \mathbb{R}.$

For any fixed $a\ne 0$, let $t\in \mathbb{R}$ and $t_n⟶t$ as $n⟶\infty$. If $t=\parallel a\parallel +\left|\phi \right(a\left)\right|$, then for all subsequences ${\left\{t_{n_k}\right\}}_{k=1}^{\infty }\subseteq {\left\{t_n\right\}}_{n=1}^{\infty }$ satisfying $t_{n_k}\le \parallel a\parallel +\left|\phi \left(a\right)\right|$ we have

$$\underset{k\to \infty }{lim}{N}_{\phi }(a,t_{n_k})=\underset{k\to \infty }{lim}0=0=N_{\phi }(a,t).$$

Also for all subsequences ${\left\{t_{n_k}\right\}}_{k=1}^{\infty }\subseteq {\left\{t_n\right\}}_{n=1}^{\infty }$ satisfying $t_{n_k}>\parallel a\parallel +\left|\phi \left(a\right)\right|$ we have

$$\begin{array}{cc}\hfill \underset{k\to \infty }{lim}{N}_{\phi }(a,t_{n_k})& =\underset{k\to \infty }{lim}\frac{t_{n_k}-\parallel a\parallel -\left|\phi \left(a\right)\right|}{t_{n_k}+\parallel a\parallel +\left|\phi \left(a\right)\right|}\hfill \\ \hfill & =\frac{t-\parallel a\parallel -\left|\phi \right(a\left)\right|}{t+\parallel a\parallel +\left|\phi \right(a\left)\right|}\hfill \\ & =\frac{t-t}{t+t}\hfill \\ & =0\hfill \\ \hfill & =N_{\phi }(a,t).\hfill \end{array}$$

If $t<\parallel a\parallel +\left|\phi \right(a\left)\right|$, then there exists an $N\in \mathbb{N}$ such that $t_n<\parallel a\parallel +\left|\phi \left(a\right)\right|$ for all $n\ge N$. So $\underset{n\to \infty }{lim}{N}_{\phi }(a,t_n)=\underset{n\to \infty }{lim}0=0=N_{\phi }(a,t)$.

If $t>\parallel a\parallel +\left|\phi \right(a\left)\right|$, then there exists an $N\in \mathbb{N}$ such that $t_n>\parallel a\parallel +\left|\phi \left(a\right)\right|$ for all $n\ge N$. So

$$\begin{array}{cc}\hfill \underset{n\to \infty }{lim}{N}_{\phi }(a,t_n)& =\underset{n\to \infty }{lim}\frac{t_n-\parallel a\parallel -\left|\phi \left(a\right)\right|}{t_n+\parallel a\parallel +\left|\phi \left(a\right)\right|}\hfill \\ \hfill & =\frac{t-\parallel a\parallel -\left|\phi \right(a\left)\right|}{t+\parallel a\parallel +\left|\phi \right(a\left)\right|}\hfill \\ \hfill & =N_{\phi }(a,t).\hfill \end{array}$$

It follows that for any fixed $a\ne 0$, $t_n⟶t$ as $n⟶\infty$, implies that $\underset{n⟶\infty }{lim}{N}_{\phi }(a,t_n)=N_{\phi }(a,t)$. Hence $N_{\phi }(a,.):\mathbb{R}⟶[0,1]$ is continuous for all $a\ne 0$.

We shall show that $N_{\phi }(0,.):\mathbb{R}⟶[0,1]$ is not continuous at $t=0$.

Since

$$N_{\phi }(0,t)=\left\{\begin{array}{cc}0\hfill & t\le 0\hfill \\ 1\hfill & t>0,\hfill \end{array}\right.$$

$\underset{t⟶0^{+}}{lim}{N}_{\phi }(0,t)=1\ne N_{\phi }(0,0)=0$. This shows that $N_{\phi }(0,.):\mathbb{R}⟶[0,1]$ is not continuous on $\mathbb{R}$. □

Theorem 3.2. 

Let A be a normed algebra,$\phi :A⟶\mathbb{F}$be a continuous homomorphism, and$ϵ>0$be an element such that$(ϵ+\parallel \phi \parallel )\ge 1$. If

$N_{\phi ,ϵ}:A\times \mathbb{R}⟶[0,1]$is defined by

$$N_{\phi ,ϵ}(a,t)=\left\{\begin{array}{cc}0\hfill & t\le \parallel a\parallel (ϵ+\parallel \phi \parallel )\hfill \\ \frac{t-\left|\phi \right(a\left)\right|}{t+\left|\phi \right(a\left)\right|}\hfill & t>\parallel a\parallel (ϵ+\parallel \phi \parallel ),\hfill \end{array}\right.$$

then

• the map

  $$\begin{array}{cc}\hfill N_{\phi ,ϵ}(.,t):& A⟶[0,1]\hfill \\ \hfill & a⟶N_{\phi ,ϵ}(a,t),\hfill \end{array}$$

  is continuous for all$t\le 0$.
• if$t>0$, then$N_{\phi ,ϵ}(.,t)$is continuous at every$a\in A\setminus S$, where$S=\{a\in A|\parallel a\parallel =\frac{t}{ϵ+\parallel \phi \parallel }\}$.
• the map$N_{\phi ,ϵ}(a,.)$is continuous at every$t\in \mathbb{R}\setminus T$, where$T=\{t\in \mathbb{R}|t=\parallel a\parallel (ϵ+\parallel \phi \parallel )\}$.

Proof. 

• If $t\le 0$, then $N_{\phi ,ϵ}(a,t)=0$ for all $a\in A$. So $N_{\phi ,ϵ}(.,t)$ is a constant function on A that is continuous.
• If $t>0$ and $a\in S$, then $\parallel a\parallel =\frac{t}{ϵ+\parallel \phi \parallel }$ and $N_{\phi ,ϵ}(a,t)=0$. Set $a_n=\frac{t-\frac{t}{2n}}{\parallel a\parallel (ϵ+\parallel \phi \parallel )}a$ for all $n\in \mathbb{N}$. So $a_n⟶a$ as $n⟶\infty$, and
  $\parallel a_n\parallel =\frac{t-\frac{t}{2n}}{\parallel a\parallel (ϵ+\parallel \phi \parallel )}\parallel a\parallel <\frac{t}{(ϵ+\parallel \phi \parallel )}$ for all $n\in \mathbb{N}$. This shows that
  $t>\parallel a_n\parallel (ϵ+\parallel \phi \parallel )$ for all $n\in \mathbb{N}$.
  Hence

  $$\begin{array}{cc}\hfill \underset{n\to \infty }{lim}{N}_{\phi ,ϵ}(a_n,t)& =\underset{n\to \infty }{lim}\frac{t-\left|\phi \right(a_n\left)\right|}{t+\left|\phi \right(a_n\left)\right|}\hfill \\ \hfill & =\frac{t-\left|\phi \right(a\left)\right|}{t+\left|\phi \right(a\left)\right|}\hfill \\ \hfill & \ne N_{\phi ,ϵ}(a,t)=0,\hfill \end{array}$$

  since, $t=\parallel a\parallel (ϵ+\parallel \phi \parallel )>\parallel a\parallel \parallel \phi \parallel \ge \left|\phi \right(a\left)\right|$. Therefore $N_{\phi ,ϵ}(.,t)$ is discontinuous at every $a\in S$.
  Let $a\notin S$ and let $a_n⟶a$ as $n⟶\infty$. So $t>\parallel a\parallel (ϵ+\parallel \phi \parallel )$ or $t<\parallel a\parallel (ϵ+\parallel \phi \parallel )$. If $t>\parallel a\parallel (ϵ+\parallel \phi \parallel )$, then there exists an $N\in \mathbb{N}$ such that $t>\parallel a_n\parallel (ϵ+\parallel \phi \parallel )$ for all $n\ge N$. Hence

  $$\begin{array}{cc}\hfill \underset{n\to \infty }{lim}{N}_{\phi ,ϵ}(a_n,t)& =\underset{n\to \infty }{lim}\frac{t-\left|\phi \right(a_n\left)\right|}{t+\left|\phi \right(a_n\left)\right|}\hfill \\ \hfill & =\frac{t-\left|\phi \right(a\left)\right|}{t+\left|\phi \right(a\left)\right|}\hfill \\ \hfill & =N_{\phi ,ϵ}(a,t).\hfill \end{array}$$
  If $t<\parallel a\parallel (ϵ+\parallel \phi \parallel )$, then there exists an $N\in \mathbb{N}$ such that
  $t<\parallel a_n\parallel (ϵ+\parallel \phi \parallel )$ for all $n\ge N$. So

  $$\underset{n\to \infty }{lim}{N}_{\phi ,ϵ}(a_n,t)=\underset{n\to \infty }{lim}0=0=N_{\phi ,ϵ}(a,t).$$
  Consequently in the case where $t>0$, $N_{\phi ,ϵ}(.,t)$ is continuous at every point $a\in A\setminus S$.
• Let $t\in T$. So $t=\parallel a\parallel (ϵ+\parallel \phi \parallel )$ and $N_{\phi ,ϵ}(a,t)=0$.
  Set $t_n=(\parallel a\parallel +\frac{1}{n})(ϵ+\parallel \phi \parallel )$ for all $n\in \mathbb{N}$. Clearly $t_n⟶t$ as $n⟶\infty$, and $t_n>\parallel a\parallel (ϵ+\parallel \phi \parallel )$ for all $n\in \mathbb{N}$. Hence

  $$\begin{array}{cc}\hfill \underset{n\to \infty }{lim}{N}_{\phi ,ϵ}(a,t_n)& =\underset{n\to \infty }{lim}\frac{t_n-\left|\phi \left(a\right)\right|}{t_n+\left|\phi \left(a\right)\right|}\hfill \\ \hfill & =\left\{\begin{array}{cc}1\hfill & a=0\hfill \\ \frac{t-\left|\phi \right(a\left)\right|}{t+\left|\phi \right(a\left)\right|}\hfill & a\ne 0.\hfill \end{array}\right.\hfill \end{array}$$
  It follows that $\underset{n\to \infty }{lim}{N}_{\phi ,ϵ}(a,t_n)\ne N_{\phi ,ϵ}(a,t)=0$. Note that if $t=\parallel a\parallel (ϵ+\parallel \phi \parallel )$ and $a\ne 0$, then $t-\left|\phi \right(a\left)\right|\ne 0$, since

  $$t=\parallel a\parallel (ϵ+\parallel \phi \parallel )>\parallel a\parallel \parallel \phi \parallel \ge \left|\phi \right(a\left)\right|.$$
  We shall show that $N_{\phi ,ϵ}(a,.)$ is continuous at every $t\in \mathbb{R}\setminus T$.
  Let $t\in \mathbb{R}\setminus T$ and let $t_n⟶t$ as $n⟶\infty$. Then $t<\parallel a\parallel (ϵ+\parallel \phi \parallel )$ or $t>\parallel a\parallel (ϵ+\parallel \phi \parallel )$. If $t<\parallel a\parallel (ϵ+\parallel \phi \parallel )$, then there exists an $N\in \mathbb{N}$ such that $t_n<\parallel a\parallel (ϵ+\parallel \phi \parallel )$ for all $n\ge N$. Hence

  $$\underset{n\to \infty }{lim}{N}_{\phi ,ϵ}(a,t_n)=\underset{n\to \infty }{lim}0=0=N_{\phi ,ϵ}(a,t).$$
  If $t>\parallel a\parallel (ϵ+\parallel \phi \parallel )$, then there exists an $N\in \mathbb{N}$ such that
  $t_n>\parallel a\parallel (ϵ+\parallel \phi \parallel )$ for all $n\ge N$. So

  $$\begin{array}{cc}\hfill \underset{n\to \infty }{lim}{N}_{\phi ,ϵ}(a,t_n)& =\underset{n\to \infty }{lim}\frac{t_n-\left|\phi \left(a\right)\right|}{t_n+\left|\phi \left(a\right)\right|}\hfill \\ \hfill & =\frac{t-\left|\phi \right(a\left)\right|}{t+\left|\phi \right(a\left)\right|}\hfill \\ \hfill & =N_{\phi ,ϵ}(a,t).\hfill \end{array}$$
  This shows that $N_{\phi ,ϵ}(a,.)$ is continuous at every $t\in \mathbb{R}\setminus T$.

□

Theorem 3.3. 

Let A be a normed algebra and$\psi :A⟶\mathbb{F}$be a continuous homomorphism. If

$N_{\psi }^{\left(1\right)}:A\times \mathbb{R}⟶[0,1]$and$N_{\psi }^{\left(2\right)}:A\times \mathbb{R}⟶[0,1]$are defined by

$$\begin{array}{cc}\hfill N_{\psi }^{\left(1\right)}(a,t)& =\left\{\begin{array}{cc}0\hfill & t\le \parallel a\parallel +\left|\psi \right(a\left)\right|\hfill \\ \frac{t}{t+\parallel a\parallel +\left|\psi \right(a\left)\right|}\hfill & t>\parallel a\parallel +\left|\psi \right(a\left)\right|,\hfill \end{array}\right.\hfill \end{array}$$

$$\begin{array}{cc}\hfill N_{\psi }^{\left(2\right)}(a,t)& =\left\{\begin{array}{cc}0\hfill & t\le \left|\psi \right(a\left)\right|\hfill \\ \frac{t}{t+\parallel a\parallel +\left|\psi \right(a\left)\right|}\hfill & t>\left|\psi \right(a\left)\right|,\hfill \end{array}\right.\hfill \end{array}$$

and in the case where$ker\psi =\left\{0\right\}$, $N_{\psi }^{\left(3\right)}:A\times \mathbb{R}⟶[0,1]$is defined by

$$\begin{array}{cc}\hfill N_{\psi }^{\left(3\right)}(a,t)& =\left\{\begin{array}{cc}0\hfill & t\le \left|\psi \right(a\left)\right|\hfill \\ \frac{t}{t+\left|\psi \right(a\left)\right|}\hfill & t>\left|\psi \right(a\left)\right|,\hfill \end{array}\right.\hfill \end{array}$$

then

• the maps$N_{\psi }^{\left(1\right)}(.,t)$, $N_{\psi }^{\left(2\right)}(.,t)$, and$N_{\psi }^{\left(3\right)}(.,t)$are continuous on A for all$t\le 0$.
• if$t>0$, then the map$N_{\psi }^{\left(1\right)}(.,t)$is continuous at every
  $a\in A\setminus S_1$, where$S_1=\{a\in A|t=\parallel a\parallel +\left|\psi \left(a\right)\right|\}$.
• if$t>0$, then the maps$N_{\psi }^{\left(2\right)}(.,t)$ and $N_{\psi }^{\left(3\right)}(.,t)$are continuous at every$a\in A\setminus S_2$, where$S_2=\{a\in A|t=\left|\psi \left(a\right)\right|\}$.
• for$a\in A$, the map$N_{\psi }^{\left(1\right)}(a,.)$is continuous at every$t\in \mathbb{R}\setminus T_1$, where$T_1=\{t\in \mathbb{R}|t=\parallel a\parallel +\left|\psi \left(a\right)\right|\}$.
• for$a\ne 0$, the map$N_{\psi }^{\left(2\right)}(a,.)$is continuous at every$t\in \mathbb{R}\setminus T_2$, where$T_2=\{t>0|t=\left|\psi \left(a\right)\right|\}$. Also the map$N_{\psi }^{\left(2\right)}(0,.)$is continuous at every$t\in \mathbb{R}$except$t=0$.
• for$a\in A$, the map$N_{\psi }^{\left(3\right)}(a,.)$is continuous at every$t\in \mathbb{R}\setminus T_3$, where$T_3=\{t\in \mathbb{R}|t=\left|\psi \left(a\right)\right|\}$.

Proof. 

• It is obvious.
• Let $t>0$ and $a\in S_1$. So $t=\parallel a\parallel +\left|\psi \right(a\left)\right|$. Set $a_n=\frac{t-\frac{t}{2n}}{\parallel a\parallel +\left|\psi \right(a\left)\right|}a$ for all $n\in \mathbb{N}$. Obviously $a_n⟶a$ and so $\psi \left(a_n\right)⟶\psi \left(a\right)$ as $n⟶\infty$. Also

  $$\begin{array}{cc}\hfill \parallel a_n\parallel & =\frac{t-\frac{t}{2n}}{\parallel a\parallel +\left|\psi \right(a\left)\right|}\parallel a\parallel \hfill \\ \hfill & <\frac{t}{\parallel a\parallel +\left|\psi \right(a\left)\right|}\parallel a\parallel \hfill \\ \hfill & =\parallel a\parallel ,n\in \mathbb{N},\hfill \end{array}$$

  and

  $$\begin{array}{cc}\hfill \left|\psi \right(a_n\left)\right|& =\frac{t-\frac{t}{2n}}{\parallel a\parallel +\left|\psi \right(a\left)\right|}\left|\psi \left(a\right)\right|\hfill \\ \hfill & \le \frac{t}{\parallel a\parallel +\left|\psi \right(a\left)\right|}\left|\psi \left(a\right)\right|\hfill \\ \hfill & =\left|\psi \right(a\left)\right|,n\in \mathbb{N}.\hfill \end{array}$$
  So $\parallel a_n\parallel +|\psi \left(a_n\right)|<\parallel a\parallel +|\psi \left(a\right)|=t$ for all $n\in \mathbb{N}$. It follows that

  $$\begin{array}{cc}\hfill \underset{n\to \infty }{lim}{N}_{\psi }^{\left(1\right)}(a_n,t)& =\underset{n\to \infty }{lim}\frac{t}{t+\parallel a_n\parallel +|\psi \left(a_n\right)|}\hfill \\ \hfill & =\frac{t}{t+t}\hfill \\ \hfill & =\frac{1}{2}\hfill \\ \hfill & \ne N_{\psi }^{\left(1\right)}(a,t)\hfill \\ \hfill & =0.\hfill \end{array}$$
  This shows that $N_{\psi }^{\left(1\right)}(.,t)$ is discontinuous at every $a\in S_1$. Now let $a\in A\setminus S_1$ and ${\left\{z_n\right\}}_{n=1}^{\infty }$ be a sequence such that $z_n⟶a$ as $n⟶\infty$. So $t>\parallel a\parallel +\left|\psi \right(a\left)\right|$ or $t<\parallel a\parallel +\left|\psi \right(a\left)\right|$. If $t>\parallel a\parallel +\left|\psi \right(a\left)\right|$, then there exists an $N\in \mathbb{N}$ such that $t>\parallel z_n\parallel +|\psi \left(z_n\right)|$ for all $n\ge N$. Hence

  $$\begin{array}{cc}\hfill \underset{n\to \infty }{lim}{N}_{\psi }^{\left(1\right)}(z_n,t)& =\underset{n\to \infty }{lim}\frac{t}{t+\parallel z_n\parallel +|\psi \left(z_n\right)|}\hfill \\ \hfill & =\frac{t}{t+\parallel a\parallel +\left|\psi \right(a\left)\right|}\hfill \\ \hfill & =N_{\psi }^{\left(1\right)}(a,t).\hfill \end{array}$$
  If $t<\parallel a\parallel +\left|\psi \right(a\left)\right|$, then there exists an $N\in \mathbb{N}$ such that $t<\parallel z_n\parallel +|\psi \left(z_n\right)|$ for all $n\ge N$. So

  $$\underset{n\to \infty }{lim}{N}_{\psi }^{\left(1\right)}(z_n,t)=\underset{n\to \infty }{lim}0=0=N_{\psi }^{\left(1\right)}(a,t).$$
  This shows that $N_{\psi }^{\left(1\right)}(.,t)$ is continuous at every $a\in A\setminus S_1$.
• Let $a\in S_2$. So $t=\left|\psi \right(a\left)\right|$, $a\ne 0$ and $N_{\psi }^{\left(2\right)}(a,t)=N_{\psi }^{\left(3\right)}(a,t)=0$. Set $a_n=(1-\frac{1}{2n})a$ for all $n\in \mathbb{N}$. Clearly $a_n⟶a$, $\psi \left(a_n\right)⟶\psi \left(a\right)$ as $n⟶\infty$. Also $|\psi \left(a_n\right)|=(1-\frac{1}{2n})|\psi \left(a\right)|<|\psi \left(a\right)|=t$ for all $n\in \mathbb{N}$. Hence

  $$\begin{array}{cc}\hfill \underset{n\to \infty }{lim}{N}_{\psi }^{\left(2\right)}(a_n,t)& =\underset{n\to \infty }{lim}\frac{t}{t+\parallel a_n\parallel +|\psi \left(a_n\right)|}\hfill \\ \hfill & =\frac{t}{t+\parallel a\parallel +t}\hfill \\ \hfill & =\frac{t}{2t+\parallel a\parallel }\hfill \\ \hfill & \ne 0\hfill \\ \hfill & =N_{\psi }^{\left(2\right)}(a,t).\hfill \end{array}$$

  and

  $$\begin{array}{cc}\hfill \underset{n\to \infty }{lim}{N}_{\psi }^{\left(3\right)}(a_n,t)& =\underset{n\to \infty }{lim}\frac{t}{t+\left|\psi \right(a_n\left)\right|}\hfill \\ \hfill & =\frac{t}{t+t}\hfill \\ \hfill & =\frac{1}{2}\hfill \\ \hfill & \ne 0\hfill \\ \hfill & =N_{\psi }^{\left(3\right)}(a,t).\hfill \end{array}$$
  Hence $N_{\psi }^{\left(2\right)}(.,t)$ and $N_{\psi }^{\left(3\right)}(.,t)$ are discontinuous at every $a\in S_2$.
  Now let $a\notin S_2$ and $z_n⟶a$ as $n⟶\infty$. Then $t<\left|\psi \right(a\left)\right|$ or $t>\left|\psi \right(a\left)\right|$. If $t<\left|\psi \right(a\left)\right|$, then there exists an $N\in \mathbb{N}$ such that $t<\left|\psi \right(z_n\left)\right|$ for all $n\ge N$. Hence

  $$\underset{n\to \infty }{lim}{N}_{\psi }^{\left(2\right)}(z_n,t)=0=N_{\psi }^{\left(2\right)}(a,t),$$

  and

  $$\underset{n\to \infty }{lim}{N}_{\psi }^{\left(3\right)}(z_n,t)=0=N_{\psi }^{\left(3\right)}(a,t).$$
  Also if $t>\left|\psi \right(a\left)\right|$, then there exists an $N\in \mathbb{N}$ such that
  $t>\left|\psi \right(z_n\left)\right|$ for all $n\ge N$. So

  $$\begin{array}{cc}\hfill \underset{n\to \infty }{lim}{N}_{\psi }^{\left(2\right)}(z_n,t)& =\underset{n\to \infty }{lim}\frac{t}{t+\parallel z_n\parallel +|\psi \left(z_n\right)|}\hfill \\ \hfill & =\frac{t}{t+\parallel a\parallel +\left|\psi \right(a\left)\right|}\hfill \\ \hfill & =N_{\psi }^{\left(2\right)}(a,t),\hfill \end{array}$$

  and

  $$\begin{array}{cc}\hfill \underset{n\to \infty }{lim}{N}_{\psi }^{\left(3\right)}(z_n,t)& =\underset{n\to \infty }{lim}\frac{t}{t+\left|\psi \right(z_n\left)\right|}\hfill \\ \hfill & =\frac{t}{t+\left|\psi \right(a\left)\right|}\hfill \\ \hfill & =N_{\psi }^{\left(3\right)}(a,t).\hfill \end{array}$$
  Hence $N_{\psi }^{\left(2\right)}(.,t)$ and $N_{\psi }^{\left(3\right)}(.,t)$ are continuous at every $a\in A\setminus S_2$.
• Let $t\in T_1$. Then $t=\parallel a\parallel +\left|\psi \right(a\left)\right|$ and $N_{\psi }^{\left(1\right)}(a,t)=0$. Set
  $t_n=\parallel a\parallel +\left|\psi \left(a\right)\right|+\frac{1}{n}$ for all $n\in \mathbb{N}$. Clearly $t_n>\parallel a\parallel +\left|\psi \left(a\right)\right|$ for all $n\in \mathbb{N}$ and $\underset{n\to \infty }{lim}{t}_n=t$. So

  $$\begin{array}{cc}\hfill \underset{n\to \infty }{lim}{N}_{\psi }^{\left(1\right)}(a,t_n)& =\underset{n\to \infty }{lim}\frac{t_n}{t_n+\parallel a\parallel +\left|\psi \left(a\right)\right|}\hfill \\ \hfill & =\left\{\begin{array}{cc}1\hfill & a=0\hfill \\ \frac{t}{t+t}=\frac{1}{2}\hfill & a\ne 0.\hfill \end{array}\right.\hfill \end{array}$$
  It follows that $\underset{n\to \infty }{lim}{N}_{\psi }^{\left(1\right)}(a,t_n)\ne N_{\psi }^{\left(1\right)}(a,t)=0$. This shows that $N_{\psi }^{\left(1\right)}(a,.)$ is discontinuous at every $t\in T_1$. One can easily verify that if $t\in \mathbb{R}\setminus T_1$, then $N_{\psi }^{\left(1\right)}(a,.)$ is continuous at t.
• Let $a\ne 0$. If $t\in T_2$, then $t=\left|\psi \right(a\left)\right|>0$ and $N_{\psi }^{\left(2\right)}(a,t)=0$. Set $t_n=(1+\frac{1}{n})\left|\psi \left(a\right)\right|$ for all $n\in \mathbb{N}$. Clearly $t_n⟶t$ as $n⟶\infty$, and $t_n>\left|\psi \left(a\right)\right|$ for all $n\in \mathbb{N}$. So

  $$\begin{array}{cc}\hfill \underset{n\to \infty }{lim}{N}_{\psi }^{\left(2\right)}(a,t_n)& =\underset{n\to \infty }{lim}\frac{t_n}{t_n+\parallel a\parallel +\left|\psi \left(a\right)\right|}\hfill \\ \hfill & =\frac{t}{2t+\parallel a\parallel }\hfill \\ \hfill & \ne 0\hfill \\ \hfill & =N_{\psi }^{\left(2\right)}(a,t).\hfill \end{array}$$
  This shows that $N_{\psi }^{\left(2\right)}(a,.)$ is discontinuous at every $t\in T_2$.
  Let $t\in \mathbb{R}\setminus T_2$. Then $t\le 0$ or $t\ne \left|\psi \right(a\left)\right|$. In the case where $t<0$ or $t\ne \left|\psi \right(a\left)\right|$, the continuity of $N_{\psi }^{\left(2\right)}(a,.)$ at t, can be obviously verified. If $t=0$ and $t_n⟶0$ as $n⟶\infty$, then for all subsequences ${\left\{t_{n_k}\right\}}_{k=1}^{\infty }\subseteq {\left\{t_n\right\}}_{n=1}^{\infty }$ satisfying $t_{n_k}\le \left|\psi \left(a\right)\right|$ we have, $\underset{k⟶\infty }{lim}{N}_{\psi }^{\left(2\right)}(a,t_{n_k})=0=N_{\psi }^{\left(2\right)}(a,0)$. Also for all subsequences ${\left\{t_{n_k}\right\}}_{k=1}^{\infty }\subseteq {\left\{t_n\right\}}_{n=1}^{\infty }$ satisfying $t_{n_k}>\left|\psi \left(a\right)\right|$ we have,

  $$\underset{k\to \infty }{lim}{N}_{\psi }^{\left(2\right)}(a,t_{n_k})=\underset{k\to \infty }{lim}\frac{t_{n_k}}{t_{n_k+\parallel a\parallel +\left|\psi \left(a\right)\right|}}=\frac{0}{0+\parallel a\parallel +\left|\psi \right(a\left)\right|}=0=N_{\psi }^{\left(2\right)}(a,0).$$
  So $N_{\psi }^{\left(2\right)}(a,.)$ is continuous at $t=0$.
  Clearly the map $N_{\psi }^{\left(2\right)}(0,.)$ is continuous at every $t\in \mathbb{R}$ except $t=0$.
• Inspired by part (5), the proof is obvious.

□

Theorem 3.4. 

Let A be a normed algebra and$\eta :A⟶\mathbb{F}$be a continuous homomorphism. If

$N_{\eta }^{\left(1\right)}:A\times \mathbb{R}⟶[0,1]$is defined by

$$\begin{array}{cc}\hfill N_{\eta }^{\left(1\right)}(a,t)& =\left\{\begin{array}{cc}0\hfill & t\le \parallel a\parallel +\left|\eta \right(a\left)\right|\hfill \\ \frac{t-\parallel a\parallel -\left|\eta \right(a\left)\right|}{t}\hfill & t>\parallel a\parallel +\left|\eta \right(a\left)\right|,\hfill \end{array}\right.\hfill \end{array}$$

and in the case where$ker\eta =\left\{0\right\}$, $N_{\eta }^{\left(2\right)}:A\times \mathbb{R}⟶[0,1]$is defined by

$$\begin{array}{cc}\hfill N_{\eta }^{\left(2\right)}(a,t)& =\left\{\begin{array}{cc}0\hfill & t\le \left|\eta \right(a\left)\right|\hfill \\ \frac{t-\left|\eta \right(a\left)\right|}{t}\hfill & t>\left|\eta \right(a\left)\right|,\hfill \end{array}\right.\hfill \end{array}$$

then

• the map$N_{\eta }^{\left(1\right)}(.,t)$is continuous for all$t\in \mathbb{R}$.
• the map$N_{\eta }^{\left(1\right)}(a,.)$is continuous for all$a\in A$except$a=0$.
• the map$N_{\eta }^{\left(2\right)}(.,t)$is continuous for all$t\in \mathbb{R}$.
• the map$N_{\eta }^{\left(2\right)}(a,.)$is continuous for all$a\in A$except$a=0$.

Proof. 

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