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# Approximation of the multiplicatives on random multi-normed space

Journal of Inequalities and Applications20172017:204

https://doi.org/10.1186/s13660-017-1478-9

• Received: 15 May 2017
• Accepted: 23 August 2017
• Published:

## Abstract

In this paper, we consider random multi-normed spaces introduced by Dales and Polyakov (Multi-Normed Spaces, 2012). Next, by the fixed point method, we approximate the multiplicatives on these spaces.

## Keywords

• approximation
• homomorphisms
• random multi-normed space
• dual random multi-normed space

• 39A10
• 39B52
• 39B72
• 46L05
• 47H10
• 46B03
• 54E40
• 54E35
• 54H25

## 1 Introduction

The concept of random normed spaces and their properties are discussed in . Also, the concept of multi-normed spaces was introduced by Dales and Polyakov. In this paper we combine the mentioned concepts and introduce random multi-normed spaces. Next, we get an approximation for homomorphisms in these spaces. For more results and applications, one can see .

### Definition 1.1

Let $$(E,\mu,\ast)$$ be a random normed space. is a continuous t-norm. A multi-random norm on $$\{ E^{k},k\in\mathbb{N} \}$$ is sequence $$\{N_{k}\}$$ such that $$N_{k}$$ is a random norm on $$E^{k}$$ ($$k\in\mathbb{N}$$), $$\mu_{x}^{1}(t)= \mu_{x}(t)$$ for each $$x\in E$$ and $$t\in\mathbb{R}$$ and the following axioms are satisfied for each $$k\in\mathbb{N}$$ with $$k\geq2$$:
1. (NF1)

$$\mu_{A_{\sigma}(x)}^{k}(t)=\mu_{x}^{k}(t)$$, for each $$\sigma\in\sigma_{k}$$, $$x\in E^{k}$$, $$t\in\mathbb{R}$$,

2. (NF2)

$$\mu_{M_{\alpha}(x)}^{k}(t)\geq \mu_{\max_{i\in\mathbb{N}_{k}} \vert \alpha_{i} \vert x}^{k}(t)$$, for each $$\alpha= (\alpha_{1},\ldots,\alpha_{k})\in\mathbb{R}^{k}$$, $$x\in E^{k}$$, $$t\in\mathbb{R}$$,

3. (NF3)

$$\mu_{(x_{1},\ldots,x_{k},0)}^{k+1}(t)=\mu_{(x_{1},\ldots,x_{k})} ^{k}(t)$$, for each $$x_{1},\ldots,x_{k}\in E$$ and $$t\in\mathbb{R}$$,

4. (NF4)

$$\mu_{(x_{1},\ldots,x_{k},x_{k})}^{k+1}(t)=\mu_{(x_{1},\ldots,x _{k})}^{k}(t)$$, for each $$x_{1},\ldots,x_{k}\in E$$ and $$t\in\mathbb{R}$$.

In this case $$\{(E^{k},\mu^{k},\ast),k\in\mathbb{N} \}$$ is called a random multi-normed space. Moreover, if axiom (NF4) is replaced by the following axiom:
1. (DF4)

$$\mu_{(x_{1},\ldots,x_{k},x_{k})}^{k+1}(t)=\mu_{(x_{1},\ldots,2x _{k})}^{k}(t)$$, for each $$x_{1},\ldots,x_{k}\in E$$ and $$t\in\mathbb{R}$$,

then $$\{\mu^{k} \}$$ is called a dual random multi-normed and $$\{(E ^{k},\mu^{k},\ast), k\in\mathbb{N} \}$$ is called a dual random multi-normed space.

## 2 Approximation of the multiplicatives

We apply fixed point theory  to get an approximation for multiplicatives. A metric d on non-empty set ϒ with range $$[0,\infty]$$ is called a generalized metric.

### Lemma 2.1

[25, 26]

Let $$k\in\mathbb{N}$$, and let E and F be linear spaces such that $$(F^{k},\mu^{k},*)$$ is a complete random multi-normed space. Let there exist $$0\leq M<1$$, $$\lambda> 0$$, and a function $$\psi: E^{k} \longrightarrow[0,\infty)$$ such that
\begin{aligned} \psi(\lambda x_{1},\ldots,\lambda x_{k})\leq \lambda M\psi (x_{1},\ldots,x _{k}) \quad(x_{1}, \ldots,x_{k}\in E). \end{aligned}
(2.1)
We set $$\Upsilon:= \{ \eta:E \longrightarrow F:\eta(0)=0 \}$$, and define $$d:\Upsilon\times\Upsilon$$ on $$[0,\infty]$$ by
\begin{aligned}& d(\eta,\zeta) \\& \quad = \inf\biggl\{ c> 0: \mu_{(\eta(x_{1})-\zeta(x_{1}),\ldots,\eta(x_{k})-\zeta(x_{k}))}(ct) \geq\frac{t}{t+\psi(x_{1},\ldots,x_{k})}, x_{1},\ldots,x_{k} \in E \biggr\} . \end{aligned}

Then $$(\Upsilon,d)$$ is a complete generalized metric space, and the mapping $$J: \Upsilon\longrightarrow\Upsilon$$ defined by $$(Jg)(x):=\frac{g( \lambda x)}{\lambda}$$ ($$x\in\Upsilon$$) is a strictly contractive mapping.

### Theorem 2.2

Let E be a linear space and let $$((F^{n},\mu ^{n},*):n\in\mathbb{N})$$ be a complete random multi-normed space. Let $$k\in\mathbb{N}$$ and let there exist $$0\leq M_{0} < 1$$ and a function $$\varphi:E^{2k} \longrightarrow [0,\infty)$$ satisfying
$$\varphi(2x_{1},2y_{1},\ldots,2x_{k},2y_{k}) \leq2M_{0}\varphi(x_{1},y _{1}, \ldots,x_{k},y_{k})$$
(2.2)
for all $$x_{1},y_{1},\ldots,x_{k},y_{k}\in E$$. Suppose that $$f:E \longrightarrow F$$ is a mapping with $$f(0)=0$$ and
\begin{aligned}& \mu^{k}_{(f(\lambda x_{1}+\lambda y_{1})-\lambda f(x_{1})-\lambda f(y _{1}),\ldots,f(\lambda x_{k}+\lambda y_{k})-\lambda f(x_{k})-\lambda f(y _{k}))}(t) \\& \quad \geq \frac{t}{t+\varphi(x_{1},y_{1},\ldots,x_{k},y_{k})}, \end{aligned}
(2.3)
\begin{aligned}& \mu^{k}_{(f(x_{1}y_{1})-f(x_{1})f(y_{1}),\ldots ,f(x_{k}y_{k})-f(x_{k})f(y _{k}))}(t)\geq\frac{t}{t+\varphi(x_{1},y_{1},\ldots,x_{k},y_{k})}, \end{aligned}
(2.4)
for all $$\lambda\in\mathbb{T}:=\{\lambda\in\mathbb{C}: \vert \lambda \vert =1\}$$ and $$x_{1},y_{1},\ldots,x_{k},y_{k}\in E$$, $$t> 0$$.
Then
$$H(x):=\lim_{n\rightarrow\infty}2^{n}f\biggl( \frac{x}{2^{n}}\biggr)$$
(2.5)
exists for any $$x_{1},\ldots,x_{k} \in E$$ and defines a random homomorphism $$H:E \longrightarrow F$$ such that
\begin{aligned}& \mu_{(f(x_{1})-H(x_{1}),\ldots,f(x_{k})-H(x_{k}))}(t) \geq \frac{(1-M _{0})t}{(1-M_{0})t+M_{0}\psi(x_{1},\ldots,x_{k})}, \end{aligned}
(2.6)
\begin{aligned}& \psi(x_{1},\ldots,x_{k})= \varphi\biggl( \frac{x_{1}}{2},\frac {x_{1}}{2},\ldots,\frac{x _{k}}{2},\frac{x_{k}}{2} \biggr), \end{aligned}
(2.7)
for all $$x_{1},\ldots,x_{k} \in E$$ and $$t> 0$$.

### Proof

Let $$x_{1}=\frac{x_{1}}{2},\ldots,x_{k}=\frac{x_{k}}{2}$$, $$y_{1}=\frac{y _{1}}{2},\ldots,y_{k}=\frac{y_{k}}{2}$$ in (2.2). We get
$$\varphi(x_{1},y_{1},\ldots,x_{k},y_{k}) \leq2M_{0}\varphi\biggl( \frac{x_{1}}{2},\frac{y_{1}}{2},\ldots, \frac{x_{k}}{2},\frac{y_{k}}{2}\biggr),$$
(2.8)
since f is odd, $$f(0)=0$$. So $$\mu_{f(0)}(\frac{t}{2})=1$$. Letting $$\lambda=1$$ and $$y= x$$, we get
$$\mu^{k}_{(f(2x_{1})-2f(x_{1}),\ldots,f(2x_{k})-2f(x_{k}))}(t) \geq\frac{t}{t+ \varphi(x_{1},x_{1},\ldots,x_{k},x_{k})}$$
(2.9)
for all $$x_{1},y_{1},\ldots,x_{k},y_{k}\in E$$. Consider the following set:
$$s=:\{g:E \longrightarrow F \}$$
and introduce the generalized metric on s:
\begin{aligned}& d(g,h) \\& \quad= \inf\biggl\{ \nu\in\mathbb{R}_{+} :\mu^{k}_{(g(x_{1})-h(x_{1}),\ldots,g(x _{k})-h(x_{k}))}( \nu t)\geq\frac{t}{t+\varphi(x_{1},\ldots,x_{k})}, x _{1},\ldots,x_{k}\in E,t> 0 \biggr\} , \end{aligned}
where, as usual, $$\inf\phi=+\infty$$. It is easy to show $$(s,d)$$ is complete. Now, we consider the linear mapping $$J:s \longrightarrow s$$ such that
$$J\bigl(g(x)\bigr) :=2g \biggl( \frac{x}{2} \biggr)$$
for all $$x \in E$$. Let $$g,h \in s$$ be given such that $$d(g,h)=\varepsilon$$. Then we have
$$\mu^{k}_{(g(x_{1})-h(x_{1}),\ldots,g(x_{k})-h(x_{k}))}(\varepsilon t) \geq\frac{t}{t+\varphi(x_{1},x_{1},\ldots,x_{k},x_{k})},$$
for all $$x_{1},\ldots,x_{k} \in E$$ and all $$t> 0$$ and hence we have
\begin{aligned} \mu^{k}_{(Jg(x_{1})-Jh(x_{1}),\ldots,Jg(x_{k})-Jh(x_{k}))} ( M_{0} \varepsilon t ) &=\mu^{k}_{2g(\frac{x_{1}}{2})-2h(\frac{x_{1}}{2}),\ldots,2g( \frac{x_{k}}{2})-2h(\frac{x_{k}}{2}))} ( M_{0}\varepsilon t ) \\ &=\mu^{k}_{g(\frac{x_{1}}{2})-h(\frac{x_{1}}{2}),\ldots,g( \frac{x_{k}}{2})-h(\frac{x_{k}}{2}))} \biggl( \frac{M_{0}}{2} \varepsilon t \biggr) \\ &\geq\frac{\frac{M_{0}}{2}t}{\frac{M_{0}}{2}+\varphi ( \frac{x _{1}}{2},\frac{x_{1}}{2},\ldots,\frac{x_{k}}{2},\frac{x_{k}}{2} ) } \\ &\geq\frac{\frac{M_{0}}{2}t}{\frac{M_{0}}{2}+\frac{M_{0}}{2}\varphi (x_{1},x_{1},\ldots,x_{k},x_{k})} \\ &=\frac{t}{t+\varphi(x_{1},x_{1},\ldots,x_{k},x_{k})} \end{aligned}
for all $$x_{1},\ldots,x_{k} \in E$$ and $$t> 0$$. Then $$d(g,h)=\varepsilon$$ implies that $$d(J{g},J{h})\leq M_{0}\varepsilon$$. This means that
$$d(J{g},J{h})\leq M_{0}\varepsilon$$
for all $$g,h\in s$$. It follows that
$$\mu_{(f(x_{1})-2f(\frac{x_{1}}{2}),\ldots,f(x_{k})-2f(\frac{x_{k}}{2}))} \biggl( \frac{M_{0}}{2}t \biggr) \geq\frac{t}{t+\varphi (x_{1},x_{1},\ldots,x _{k},x_{k})}$$
for all $$x_{1},\ldots,x_{k} \in E$$ and $$t> 0$$. So $$d(f,J{f})\leq\frac{M _{0}}{2}$$.
Now, there exists a mapping $$H: E \longrightarrow F$$ satisfying the following:
1. (1)
H is a fixed point of J, i.e.,
$$H \biggl( \frac{x}{2} \biggr) =\frac{1}{2}H(x)$$
(2.10)
for all $$x \in E$$. Since $$f:E \longrightarrow E$$ is odd, $$H:E \longrightarrow F$$ is an odd mapping. The mapping H is a unique fixed point of J in the set
$$M=\bigl\{ g\in s:d(f,g)< \infty\bigr\} .$$
This implies that H is a unique mapping satisfying (2.10) such that there exists a $$\nu\in(0,\infty)$$ satisfying
$$\mu^{k}_{(f(x_{1})-H(x_{1}),\ldots,f(x_{k})-H(x_{k}))}(\nu t)\geq\frac{t}{t+ \varphi(x_{1},\ldots,x_{k})}$$
for all $$x_{1},\ldots,x_{k} \in E$$,

2. (2)
$$d(J^{n}f,H)\rightarrow0$$ as $$n\rightarrow\infty$$. This implies that
$$\lim_{n\rightarrow\infty}2^{n}f \biggl( \frac{x}{2_{n}} \biggr) =H(x)$$
for all $$x \in E$$,

3. (3)
$$d(f,H)\leq\frac{1}{1-M_{0}}d(f,Jf)$$, which implies
$$d(f,H)\leq\frac{M_{0}}{2-2M_{0}}.$$

Put $$\lambda=1$$ in (2.3). Then
\begin{aligned}& \mu^{k}_{(2^{n}(f(\frac{x_{1}}{2^{n}}+\frac{y_{1}}{2^{n}})-f(\frac{x _{1}}{2^{n}})-f(\frac{y_{1}}{2^{n}})),\ldots,2^{n}(f(\frac {x_{k}}{2^{n}}+\frac{y _{k}}{2^{n}})-f(\frac{x_{k}}{2^{n}})-f(\frac{y_{k}}{2^{n}})))}(t) \\& \quad \geq \frac{\frac{t}{2^{n}}}{\frac{t}{2^{n}}+ \frac{M_{0}^{n}}{2^{n}}\varphi(x_{1},y_{1},\ldots,x_{k},y_{k})} \end{aligned}
for all $$x_{1},\ldots,x_{k},y_{1},\ldots,y_{k} \in E$$, $$t> 0$$ and $$n\geq1$$. Since
$$\lim_{n\rightarrow\infty} \frac{\frac{t}{2^{n}}}{\frac{t}{2^{n}}+\frac{M _{0}^{n}}{2^{n}}\varphi(x_{1},y_{1},\ldots,x_{k},y_{k})}=1$$
for all $$x_{1},\ldots,x_{k},y_{1},\ldots,y_{k} \in E$$, $$t> 0$$. It follows that
$$\mu^{k}_{(H(x_{1}+y_{1})-H(x_{1})-H(y_{1}),\ldots,H(x_{k}+y_{k})-H(x_{k})-H(y _{k}))}(t)= 1$$
for all $$x_{1},\ldots,x_{k},y_{1},\ldots,y_{k} \in E$$, $$t> 0$$. So mapping $$H:E \longrightarrow F$$ is Cauchy additive.
Let $$y_{1}=x_{1},\ldots,y_{k}=x_{k}$$ in (2.3). Then we have
$$\mu^{k}_{2^{n}(f(\frac{\beta x_{1}}{2^{n}})-f( \frac{\beta x_{1}}{2^{n}}),\ldots,f(\frac{\beta x_{k}}{2^{n}})-f(\frac{ \beta x_{k}}{2^{n}}))}\bigl(2^{n}t\bigr)\geq \frac{t}{t+\varphi( \frac{x_{1}}{2^{n}},\frac{x_{1}}{2^{n}},\ldots,\frac{x_{k}}{2^{n}},\frac{x _{k}}{2^{n}})}$$
for all $$\lambda,\beta\in\mathbb{T}$$, $$\lambda=\frac{\beta}{2}$$, $$x_{1},\ldots,x_{k} \in E$$, $$t> 0$$ and $$n\geq1$$. So we have
$$\mu^{k}_{2^{n}(f(\frac{\beta x_{1}}{2^{n}})-f( \frac{\beta x_{1}}{2^{n}}),\ldots,f(\frac{\beta x_{k}}{2^{n}})-f(\frac{ \beta x_{k}}{2^{n}}))}(t)\geq\frac{\frac{t}{2^{n}}}{\frac {t}{2^{n}}+\frac{M _{0}^{n}}{2^{n}}\varphi(x_{1},x_{1},\ldots,x_{k},x_{k})}$$
for all $$\beta\in\mathbb{T}$$, $$x_{1},\ldots,x_{k} \in E$$, $$t> 0$$ and $$n\geq1$$. We have
$$\lim_{n\rightarrow\infty}\frac{\frac{t}{2^{n}}}{\frac{t}{2^{n}}+\frac{M _{0}^{n}}{2^{n}}\varphi(x_{1},x_{1},\ldots,x_{k},x_{k})}=1$$
for all $$x_{1},\ldots,x_{k} \in E$$, $$t> 0$$, and
$$\mu^{k}_{(H(\beta x_{1})-\beta H(x_{1}),\ldots,H(\beta x_{k})-\beta H(x _{k}))}(t)=1$$
for all $$\beta\in\mathbb{T}$$, $$x_{1},\ldots,x_{k} \in E$$, $$t> 0$$. Thus, the additive mapping $$H:E \longrightarrow F$$ is $$\mathbb{R}$$-linear. From (2.4), we have
\begin{aligned}& \mu^{k}_{(4^{n}f(\frac{x_{1}}{2^{n}}\frac{y_{1}}{2^{n}})-2^{n}f(\frac{x _{1}}{2^{n}})2^{n}f(\frac{y_{1}}{2^{n}}),\ldots,4^{n} f( \frac{x_{k}}{2^{n}}\frac{y_{k}}{2^{n}})-2^{n}f(\frac {x_{k}}{2^{n}})2^{n}f(\frac{y _{k}}{2^{n}}))}\bigl(4^{n}t\bigr) \\& \quad \geq\frac{t}{t+\varphi(\frac{x_{1}}{2^{n}},\ldots, \frac{x_{k}}{2^{n}},\frac{y_{1}}{2^{n}},\ldots,\frac{y_{k}}{2^{n}})} \end{aligned}
for all $$x_{1},\ldots,x_{k} \in E$$, $$t> 0$$ and $$n\geq1$$.
Then we have
\begin{aligned}& \mu^{k}_{(4^{n}f(\frac{x_{1}}{2^{n}}\frac{y_{1}}{2^{n}})-2^{n}f(\frac{x _{1}}{2^{n}})2^{n}f(\frac{y_{1}}{2^{n}}),\ldots,4^{n} f( \frac{x_{k}}{2^{n}}\frac{y_{k}}{2^{n}})-2^{n}f(\frac {x_{k}}{2^{n}})2^{n}f(\frac{y _{k}}{2^{n}}))}\bigl(4^{n}t\bigr) \\& \quad \geq\frac{\frac{t}{4^{n}}}{\frac{t}{4^{n}}+\frac{M_{0}^{n}}{t^{n}} \varphi(x_{1},\ldots,x_{k},y_{1},\ldots,y_{k})} \end{aligned}
for all $$x_{1},\ldots,x_{k} \in E$$, $$t> 0$$ and $$n\geq1$$.
Since
$$\lim_{n\rightarrow\infty} \frac{\frac{t}{4^{n}}}{\frac{t}{4^{n}}+\frac{M _{0}^{n}}{t^{n}}\varphi(x_{1},\ldots,x_{k},y_{1},\ldots,y_{k})}=1$$
for all $$x_{1},\ldots,x_{k} \in E$$, $$t> 0$$, we have
$$\mu^{k}_{(H(x_{1}y_{1})-H(x_{1})H(y_{1}),\ldots,H(x_{k}y_{k})-H(x_{k})H(y _{k}))}(t)=1$$
for all $$x_{1},\ldots,x_{k},y_{1},\ldots,y_{k} \in E$$, $$t> 0$$. Thus, the mapping $$H:E \longrightarrow F$$ is multiplicative. Therefore, there exists a unique random homomorphism $$H:E\longrightarrow F$$ satisfying (2.6), and this completes the proof. □

## 3 Approximation in dual random multi-normed space

The following lemma is an immediate result of the definition of random multi-normed space.

### Lemma 3.1

Let $$\{(E^{k},\mu^{k},\ast),k\in\mathbb{N}\}$$ be a dual random multi-normed space, $$k,n\in\mathbb{N}$$, $$x_{1},x_{2},\ldots ,x_{k},x_{k+1},\ldots,x _{k+n}\in\mathbb{E}$$ and $$\lambda_{1},\ldots,\lambda_{k}$$ be real numbers of absolute value 1. Then we have:
1. (i)

$$\mu_{(\lambda_{1}x_{1},\ldots,\lambda_{k}x_{k})}^{k}(t)= \mu_{(x_{1},\ldots,x_{k})}^{k}(t)$$,

2. (ii)

$$\mu_{(x_{1},\ldots,x_{k})}^{k}(t)\geq \mu_{(x_{1},\ldots,x_{k},x_{k+1})}^{k+1}(t)$$,

3. (iii)

$$\mu_{(x_{1},\ldots,x_{k},x_{k+1},\ldots,x_{k+n})}^{k+n}(t) \geq T_{M}(\mu_{(x_{1},\ldots,x_{k})}^{k}(\alpha t), \mu_{(x_{k+1},\ldots,x_{k+n})}^{n}(\beta t))$$, where $$\alpha,\beta \geq0$$ and $$\alpha+ \beta=1$$,

4. (iv)

$$\min_{i\in{\mathbb{N}_{k}}} \mu_{x_{i}}(t)\geq \mu_{(x_{1},\ldots,x_{k})}^{k}(t)\geq\min_{i\in\mathbb{N}_{k}} \mu_{x _{i}}(\alpha_{i}t)$$,

where $$\alpha_{1},\ldots,\alpha_{k} \geq0$$ and $$\sum_{i=1}^{k}\alpha _{i}=1$$. In particular, we have
$$\mu_{(x_{1},\ldots,x_{k})}^{k}(t)\geq\min_{i\in\mathbb{N}_{k}} \mu_{kx _{i}}(t).$$

### Theorem 3.2

Let E be a linear space, and $$\{(E^{k},\mu^{k},\ast),k\in \mathbb{N}\}$$ be a random multi space. Let $$\alpha\in(0,1)$$ and $$f:E\longrightarrow F$$ is a mapping satisfying $$f(0)=0$$ and
$$\mu_{ ( f(\frac{x_{1}+y_{1}}{2})-\frac{f(x_{1})}{2}- \frac{f(y_{1})}{2},\ldots,f(\frac{x_{k}+y_{k}}{2})-\frac {f(x_{k})}{2}-\frac{f(y _{k})}{2} ) }^{k} \biggl( \frac{t}{s} \biggr) \geq1-\frac{\alpha}{t},$$
(3.1)
where $$x_{1},\ldots,x_{k},y_{1},\ldots,y_{k}\in E$$ and $$t,s\in\mathbb{N}$$ with the greatest common divisor $$(t,s)=1$$.
Then there exists a unique additive mapping $$T:E\longrightarrow F$$ such that
$$\mu_{(f(x_{1})-T(x_{1}),\ldots,f(x_{k})-T(x_{k}))}^{k} \biggl( \frac {2t}{s} \biggr) \geq1-\frac{\alpha}{t}$$
(3.2)
for all $$x_{1},\ldots,x_{k}\in E$$ and $$t,s\in\mathbb{N}$$ with $$(t,s)=1$$.

### Proof

Replacing $$x_{1},\ldots,x_{k}$$ and $$y_{1},\ldots,y_{k}$$ by $$2x_{1},\ldots,2x _{k}$$ and $$0,\ldots,0$$ in (3.1), respectively, yields
$$\mu_{(2f(x_{1})-f(2x_{1}),\ldots,2f(x_{k})-f(2x_{k}))}^{k} \biggl( \frac {2t}{s} \biggr) \geq1-\frac{\alpha}{t}.$$
(3.3)
Replacing $$x_{1},\ldots,x_{k}$$, t, s by $$2x_{1},\ldots,2x_{k}$$, 2t, 2s, respectively, in (3.3) and repeating this process for n-time ($$n\in\mathbb{N}$$), it follows that
$$\mu_{ ( \frac{f(2^{n-1}x_{1})}{2^{n-1}}- \frac{f(2^{n}x_{1})}{2^{n}},\ldots,\frac{f(2^{n-1}x_{k})}{2^{n-1}}-\frac {f(2^{n}x _{k})}{2^{n}} ) }^{k} \biggl( \frac{t}{2^{n-1}s} \biggr) \geq1-\frac{ \alpha}{2^{n-1}t}$$
(3.4)
for $$n,m\in\mathbb{N}$$ with $$n> m$$. Using (3.4) and (RN2) we get
$$\mu_{ ( \frac{f(2^{m}x_{1})}{2^{m}}-\frac {f(2^{n}x_{1})}{2^{n}},\ldots,\frac{f(2^{m}x _{k})}{2^{m}}-\frac{f(2^{n}x_{k})}{2^{n}} ) }^{k} \Biggl( \sum_{i=m} ^{n-1}2^{-i}\frac{t}{s} \Biggr) \geq1-\frac{\alpha}{2^{m}t}.$$
Then
$$\mu_{ ( \frac{f(2^{m}x_{1})}{2^{m}}-\frac {f(2^{n}x_{1})}{2^{n}},\ldots,\frac{f(2^{m}x _{k})}{2^{m}}-\frac{f(2^{n}x_{k})}{2^{n}} ) }^{k} \biggl( \frac {t}{s} \biggr) \geq1-\frac{\alpha}{2^{m}t}$$
(3.5)
for $$x\in E$$. Then, replacing $$x_{1},\ldots,x_{k}$$ by $$x,2x,\ldots,2^{k-1}x$$ in (3.5), we have
\begin{aligned}[b] & \mu_{ ( \frac{f(2^{m}x)}{2^{m}}-\frac{f(2^{n}x)}{2^{n}},\ldots, \frac{f(2^{m+k-1}x)}{2^{m+k-1}}-\frac{f(2^{n+k-1}x)}{2^{n+k-1}} ) } ^{k} \biggl( \frac{t}{s} \biggr) \\ & \quad \geq 1-\frac{\alpha}{2^{m}t} \\ & \quad \geq 1-\frac{\alpha}{2^{m}}. \end{aligned}
(3.6)
Let $$\varepsilon> 0$$ be given. Then there exists $$n_{0} \in \mathbb{N}$$ such that $$\frac{\alpha}{2^{n_{0}}}< \varepsilon$$. Now we substitute m, n with n, $$n+p$$ ($$p\in\mathbb{N}$$), respectively, in (3.6), for each $$n\geq n_{0}$$, and we get
\begin{aligned} \mu_{ ( \frac{f(2^{n}x)}{2^{n}}-\frac{f(2^{n+p}x)}{2^{n+p}},\ldots, \frac{f(2^{n+k-1}x)}{2^{n+k-1}}-\frac {f(2^{n+p+k-1}x)}{2^{n+p+k-1}} ) } ^{k} \biggl( \frac{t}{s} \biggr) & \geq 1-\frac{\alpha}{2^{n}t} \\ &> 1-\varepsilon. \end{aligned}
By Lemma 3.1, we have
$$\mu_{\frac{f(2^{n}x)}{2^{n}}-\frac{f(2^{n+p}x)}{2^{n+p}}} \biggl( \frac {t}{s} \biggr) > 1- \varepsilon$$
(3.7)
for all $$n> n_{0}$$ and $$p\in\mathbb{N}$$. The density of rational numbers in $$\mathbb{R}$$ is useful in checking correctness of (3.6) with positive real number r instead of $$\frac{t}{s}$$. Then we have
$$\mu_{\frac{f(2^{n}x)}{2^{n}}-\frac{f(2^{n+p}x)}{2^{n+p}}}(r)> 1- \varepsilon$$
for each $$x\in E$$, $$r\in\mathbb{R^{+}}$$, $$n\geq n_{0}$$ and $$p\in\mathbb{N}$$. Then $$\{\frac{f(2^{n}x)}{2^{n}}\}$$ is a Cauchy sequence, so it is convergent in the random multi-Banach space $$\{(E^{k},\mu^{k},\ast),k\in\mathbb{N}\}$$. Setting $$T(x):= \lim_{n\rightarrow\infty}\frac{f(2^{n}x)}{2^{n}}$$ and applying again Lemma 3.1, for each $$r> 0$$, we have
$$\mu_{ ( \frac{f(2^{n}x_{1})}{2^{n}}-T(x_{1}),\ldots,\frac{f(2^{n}x _{k})}{2^{n}}-T(x_{k}) ) }^{k}(r)\geq\min_{i\in\mathbb{N}_{k}} \mu_{\frac{f(2^{n}x_{i})}{2^{n}}-T(x_{i})} \biggl( \frac{r}{k} \biggr) ,$$
and
$$\lim_{n\rightarrow\infty} \frac{f(2^{n}x_{k})}{2^{n}} = T(x_{k}).$$
We put $$m= 0$$ in (3.5), and we get
$$\mu_{ ( f(x_{1})-\frac{f(2^{n}x_{1})}{2^{n}},\ldots ,f(x_{k})-\frac{f(2^{n}x _{k})}{2^{n}} ) }^{k} \biggl( \frac{t}{s} \biggr) \geq1-\frac{\alpha }{t}.$$
(3.8)
Then
\begin{aligned} &\mu_{f(x_{1})-T(x_{1}),\ldots,f(x_{k})-T(x_{k})}^{k} \biggl( \frac {2t}{s} \biggr) \\ & \quad \geq T_{M} \biggl( \mu_{ ( f(x_{1})-\frac{f(2^{n}x_{1})}{2^{n}},\ldots ,f(x_{k})-\frac{f(2^{n}x _{k})}{2^{n}} ) }^{k} \biggl( \frac{t}{s} \biggr) , \\ & \quad\quad \mu_{ ( \frac{f(2^{n}x_{1})}{2^{n}}-T(x_{1}),\ldots,\frac{f(2^{n}x _{k})}{2^{n}}-T(x_{k}) ) }^{k} \biggl( \frac{t}{s} \biggr) \biggr) \\ & \quad \geq 1-\frac{\alpha}{t} \end{aligned}
(3.9)
by (3.8) and when $$n\rightarrow\infty$$, which implies that (3.2).
Now, we show that T is additive. Let $$x,y\in E$$ and replace $$x_{1},\ldots,x_{k}$$ by $$2^{n}x$$, $$y_{1},\ldots,y_{k}$$ by $$2^{n}y$$, and t by $$2^{n}t$$ in (3.1). We get
$$\mu_{ ( f(2^{n}\frac{x+y}{2})-\frac{f(2^{n}x)}{2}- \frac{f(2^{n}y)}{2},\ldots,f(2^{n}\frac{x+y}{2})-\frac{f(2^{n}x)}{2}- \frac{f(2^{n}y)}{2} ) }^{k} \biggl( \frac{2^{n}t}{s} \biggr) \geq 1- \frac{ \alpha}{2^{n}t}.$$
Using (NF4), we conclude that
$$\mu_{\frac{f2^{n}(\frac{x+y}{2})}{2^{n}}-\frac{1}{2} \frac{f(2^{n}x)}{2^{n}}-\frac{1}{2}\frac{f(2^{n}y)}{2^{n}}} \biggl( \frac {t}{s} \biggr) \geq1- \frac{\alpha}{2^{n}t}.$$
(3.10)
On the other hand, we obtain that
\begin{aligned} \mu_{T(\frac{x+y}{2})-\frac{1}{2}T(x)-\frac{1}{2}T(y)} \biggl( \frac {4t}{s} \biggr) \geq& T_{M} \biggl( \mu_{T(\frac{x+y}{2})- \frac{f(2^{n}(\frac{x+y}{2}))}{2^{n}}} \biggl( \frac{t}{s} \biggr) , \\ & \mu_{\frac{T(x)}{2}-\frac{1}{2}\frac{f(2^{n}x)}{2^{n}}}\biggl(\frac{t}{s}\biggr), \\ & \mu_{\frac{T(y)}{2}-\frac{1}{2}\frac{f(2^{n}y)}{2^{n}}} \biggl( \frac {t}{s} \biggr) , \\ & \mu_{\frac{f(2^{n}(\frac{x+y}{2}))}{2^{n}}-\frac{1}{2} \frac{f(2^{n})}{2^{n}x}-\frac{1}{2}\frac{f(2^{n}y)}{2^{n}}} \biggl( \frac {t}{s} \biggr) \biggr) \\ \geq& 1-\frac{\alpha}{2^{n}} \end{aligned}
(3.11)
for each $$x,y\in E$$, $$t,s\in\mathbb{N}$$ with $$(t,s)=1$$. Utilizing again the density of $$\mathbb{Q}$$ in $$\mathbb{R}$$, we find that (3.11) remains true if $$\frac{4t}{s}$$ is substituted with a positive real number r.
Consequently,
$$\mu_{T(\frac{x+y}{2})-\frac{1}{2}T(x)-\frac{1}{2}T(y)}(r)\geq1-\frac{ \alpha}{2^{n}}$$
for each $$x,y\in E$$ and $$r\in\mathbb{R}$$. Letting $$n\rightarrow \infty$$ reveals that T complies with Jensen, and using the fact that $$T(0)=0$$, we conclude that T is additive [27, Theorem 6].
It remains to show the uniqueness of T. Suppose that $$T'$$ is another additive mapping satisfying (3.2). Then, for each $$t,s\in \mathbb{N}$$, sufficiently large n in $$\mathbb{N}$$ and $$x\in E$$,
\begin{aligned} \mu_{T'(x)-T(x)} \biggl( \frac{t}{s} \biggr) &= \mu_{\frac{T'(2^{n}x)}{2^{n}}-\frac{T(2^{n}x)}{2^{n}}} \biggl( \frac {t}{s} \biggr) \\ &\geq T_{M} \biggl( \mu_{T'(2^{n}x)-f(2^{n}x)} \biggl( \frac {2^{n-1}t}{s} \biggr) , \mu_{T(2^{n}x)-f(2^{n}x)} \biggl( \frac{2^{n-1}t}{s} \biggr) \biggr) \\ &\geq1-\frac{\alpha}{2^{n-2}t} \\ &\geq1-\frac{\alpha}{2^{n-2}}. \end{aligned}

This inequality holds for each $$r\in\mathbb{R}^{+}$$ instead of $$\frac{t}{s}$$, too. Therefore, for each $$r\in\mathbb{R}^{+}$$, $$n\in\mathbb{N}$$, $$\mu_{T'(x)-T(x)}(r)\geq1-\frac{\alpha}{2^{n-2}}$$, letting $$n\rightarrow\infty$$, it follows that $$T= T'$$. □

## 4 Conclusion

In this paper, we consider multi-Banach spaces, approximate by multiplicatives, and provide some controlled mappings, which are stable by control functions.

## Declarations

### Acknowledgements

The authors are grateful to the reviewer(s) for their valuable comments and suggestions. 