1 Introduction

The concept of random normed spaces and their properties are discussed in [2]. 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 [3–23].

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 [24] 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.