Stability of general A-cubic functional equations in modular spaces

Abstract

In this paper, by using fixed point theory, we investigate the generalized Hyers–Ulam stability of an $\alpha$-cubic functional equation in modular spaces.

1 Introduction and preliminaries

In 1940, Ulam [23] asked the first question on the stability problem. In 1941, Hyers [9] solved the problem of Ulam. This result was generalized by Aoki [1] for additive mappings and by Rassias [20] for linear mappings by considering an unbounded Cauchy difference. In 1994, a further generalization was obtained by Găvruta [8]. Rassias [16,17,18,19] generalized Hyers result. During the last two decades, a number of papers and research monographs have been published on various generalizations and applications of the generalized Hyers-Ulam stability to a number of functional equations and mappings (see [2, 5,6,7, 11, 15, 16, 21]). We also refer the readers to the books: Czerwik [3] and Hyers, Isac and Rassias [10].

The theory of modulars on linear spaces and the corresponding theory of modular linear spaces were founded by Nakano [12] and were intensively developed by Amemiya, Koshi, Shimogaki, Yamamuro [14, 25] and others.

Definition 1.1

Let X be a vector space over a field K ($\mathbb {R}$ or $\mathbb {C}$). A generalized functional $\rho : X\longrightarrow [0,\infty ]$ is called a modular if for arbitrary $x, y \in X,$ $\rho$ satisfies:

• (a) $\rho (x) = 0$ if and only if $x = 0$,

• (b) $\rho (a x) = \rho (x)$ for every scalar a with $|a| = 1,$

• (c) $\rho (a x+b y)\le \rho (x) + \rho (y)$, whenever $a,b \ge 0$ and $a+b =1$.

If we replace (c) by

($c'$) $\rho (a x+b y)\le a\rho (x) +b\rho (y)$, whenever $a,b \ge 0$ and $a+b =1,$ then the modular $\rho$ is called convex. A modular $\rho$ defines a corresponding modular space, i.e., the vector space $X_{\rho }$ given by:

$$\begin{aligned} X_{\rho }=\{x\in X|\,\rho (\lambda x)\longrightarrow 0\, as \,\lambda \longrightarrow 0\}. \end{aligned}$$

Definition 1.2

Let $\{x_{n}\}$ and x be in $X_{\rho }$. Then

• (i) The sequence $\{x_{n}\}$, with $x_{n} \in X_{\rho }$, is $\rho$-convergent to x if $\rho (x_{n}-x)\rightarrow 0$ as $n\rightarrow \infty$.

• (ii) The sequence $\{x_{n}\}$, with $x_{n} \in X_{\rho }$, is called $\rho$-Cauchy if $\rho (x_{n}-x_{m})\rightarrow 0$ as $n,m\rightarrow \infty$.

• (iii) A subset S of $X_{\rho }$ is called $\rho$-complete if and only if any $\rho$-Cauchy sequence is $\rho$-convergent to an element of S.

Fatou property The modular $\rho$ has the Fatou property if and only if $\rho (x)\le \displaystyle \liminf _{n\rightarrow \infty }\rho (x_{n})$ whenever the sequence $\{x_{n}\}$ is $\rho$-convergent to x. A function modular is said to satisfy the $\Delta _{\alpha }$-condition $(\alpha \in \mathbb {N}$,$\alpha \ge 2)$ if there exists $\kappa >0$ such that $\rho (\alpha x)\le \kappa \rho (x),$ for all $x \in X_{\rho }$.

Remark

$\Delta _{\alpha }$-condition implies $\Delta _{2}$-condition.

Definition 1.3

Let $X_{\rho }$ be a modular space and C be a nonempty subset of $X_{\rho }$. The self-map $T:C\rightarrow C$ is said to be quasicontraction if there exists $k < 1$ such that

$$\begin{aligned} \rho \big (Tx-Ty\big )\le k\max \Big \{\rho (x-y),\rho (x-Ty), \rho (y-Tx),\rho (x-Tx),\rho (y-Ty)\Big \}, \end{aligned}$$

for any $x, y\in C.$

Definition 1.4

Given a modular space $X_{\rho }$, a nonempty subset $C\subseteq X_{\rho }$, and a mapping $T:C\rightarrow C$, the orbit of T around a point x is the set

$$\begin{aligned} O(T) := \{x, Tx, T^{2}x,\ldots \}, \end{aligned}$$

the quantity

$$\begin{aligned} \delta _{\rho }(T) := \sup \{\rho (u-v) | u,v\in O(T)\}, \end{aligned}$$

is then associated to T and is called the orbital diameter of T at x. In particular, if $\delta _{\rho }(T)<\infty ,$ one says that T has a bounded orbit at x.

Theorem 1.5

([13]) Let $X_{\rho }$ be a modular space such that $\rho$ satisfies the Fatou property and let $C\subseteq X_{\rho }$ be a $\rho$-complete subset. If $T:C\rightarrow C$ is a quasicontraction and T has a bounded orbit at $x_{0}$, then the sequence $\{T^{n}x_{0}\}$ is $\rho$-convergent to a point $\omega \in C.$

Stability of quadratic and generalized Jensen functional equation in modular spaces have been investigated in [22] and [24].

In this paper, we investigate the generalized Hyers–Ulam stability of the $\alpha$- cubic functional equation

$$\begin{aligned}&f(\alpha x+y)+f(\alpha x-y)+f(x+\alpha y)-f(x-\alpha y)\nonumber \\&\quad =2\alpha f(x+y)+2\alpha (\alpha ^2-1)[f(x)+f(y)], \end{aligned}$$

(1.1)

with $\alpha \in {\mathbb {N}}, \alpha \ne 1$ via the extensive studies of fixed point theory in modular spaces.

2 Stability of $\alpha$-cubic functional equation (1.1)

Throughout this section, we assume that $\rho$ is a convex modular on $\rho$-complete modular space $X_{\rho }$ with the Fatou property such that satisfies the $\Delta _{\alpha }$-condition with $0<\kappa \le \alpha$. In addition, let V be a linear space. For convenience, we use the following abbreviation for a given function $f:V\longrightarrow X_{\rho }$:

$$\begin{aligned} D_{\alpha }f(x,y):= & {} f(\alpha x+y)+f(\alpha x-y)+f(x+\alpha y)-f(x-\alpha y)\\&-2\alpha f(x+y)-2\alpha (\alpha ^2-1)[f(x)+f(y)] \end{aligned}$$

with $\alpha \in {\mathbb {N}}, \alpha \ne 1$ and for all $x,y\in V.$ We shall need the following lemmas:

Lemma 2.1

If a mapping $f:X\rightarrow Y$ satisfies the functional equation

$$\begin{aligned} f(x+\alpha y)-f(x-\alpha y)=\alpha [f(x+y)-f(x-y)]+2\alpha (\alpha ^{2}-1)f(y), \end{aligned}$$

(2.1)

with $\alpha \in {\mathbb {N}}, \alpha \ne 1$ and for all $x, y\in X,$ then f is cubic.

Proof

Replacing (x, y) with (0, 0) in (2.1), we get $2\alpha (\alpha ^{2}-1)f(0)=0$ with $\alpha \in {\mathbb {N}}, \alpha \ne 1$. Therefore $f(0)=0$. Replacing (x, y) with (0, x) and $(0,-x)$ in (2.1), we get, respectively, equations:

$$\begin{aligned} f(\alpha x)-f(-\alpha x)= & {} \alpha [(2\alpha ^{2}-1)f(x)-f(-x)],\nonumber \\ f(-\alpha x)-f(\alpha x)= & {} \alpha [(2\alpha ^{2}-1)f(-x)-f(x)]. \end{aligned}$$

(2.2)

By adding these two equations, one can obtain $f(-x)=-f(x)$. By using (2.2) and $f(-x)=-f(x)$, we get $f(\alpha x)=\alpha ^{3}f(x)$ with $\alpha \in {\mathbb {N}}, \alpha \ne 1$ and for all $x\in X$ (See [4]). $\square$

Lemma 2.2

If a mapping $f:X\rightarrow Y$ satisfies (1.1) for all $x, y\in X,$ then f is cubic.

Proof

Replacing (x, y) with (0, 0) in (1.1), we get $f(0)=0$. Replacing (x, y) with (x, 0) in (1.1), we get,

$$\begin{aligned} f(\alpha x)=\alpha ^{3} f(x), \end{aligned}$$

(2.3)

for all $x\in X.$ By setting $x=0$ and using (2.3), we get $f(-y)=-f(y)$ for all $y\in X,$ that is f is odd. Replacing (x, y) with $(x,-y)$ in (1.1) and using oddness of f, we get,

$$\begin{aligned}&f(\alpha x-y)+f(\alpha x+y)+f(x-\alpha y)-f(x+\alpha y)\nonumber \\&\quad =2\alpha f(x-y)+2\alpha (\alpha ^2-1)[f(x)-f(y)] \end{aligned}$$

(2.4)

for all $x,y\in X.$ It follows from (1.1) and (2.4) that

$$\begin{aligned} f(x+\alpha y)-f(x-\alpha y)=\alpha [f(x+y)-f(x-y)]+2\alpha (\alpha ^2-1)f(y), \end{aligned}$$

(2.5)

for all $x,y\in X.$ It follows from Lemma 2.1 that f is cubic. $\square$

Theorem 2.3

Let $\varphi : V^{2}\longrightarrow [0, +\infty )$ be a function such that

$$\begin{aligned} \lim _{n\rightarrow \infty }\frac{1}{\alpha ^{3n}}\varphi (\alpha ^{n}x,\alpha ^{n}y)=0, \end{aligned}$$

(2.6)

and

$$\begin{aligned} \varphi (\alpha x,\alpha y)\le \alpha ^{3} L\varphi (x,y), \end{aligned}$$

(2.7)

for all $x,y\in V$ with $L<1.$ Suppose that $f:V\longrightarrow X_{\rho }$ satisfies the condition

$$\begin{aligned} \rho \big (D_{\alpha }f(x,y)\big )\le \varphi (x,y), \end{aligned}$$

(2.8)

for all $x,y\in V$ and $f(0)=0$. Then there exists a unique cubic mapping $C_{\alpha }: V\longrightarrow X_{\rho }$ such that

$$\begin{aligned} \rho \big (C_{\alpha }(x)-f(x)\big )\le \frac{1}{\alpha ^{3}(1- L)}\varphi (x,0), \end{aligned}$$

(2.9)

for all $x\in V$.

Proof

We consider the set

$$\begin{aligned} M= \{g:V\rightarrow X_{\rho }\} \end{aligned}$$

and define the function $\overline{\rho }$ on M as follows,

$$\begin{aligned} \overline{\rho }(g)=:\inf \{c>0:\, \rho (g(x))\le c\varphi (x,0),\;\, \forall x\in V\} . \end{aligned}$$

We show that $\overline{\rho }$ is a convex modular on M. It is also easy to verify that $\overline{\rho }$ satisfies the axioms (a) and (b) of a modular. We will next show that $\overline{\rho }$ is convex, and hence $(c')$ is satisfied. Let $\epsilon >0$ be given. Then there exist real constants $c_{1}>0$ and $c_{2}>0$ such that

$$\begin{aligned} \overline{\rho }(g)\le c_{1}\le \overline{\rho }(g)+\epsilon ,\,\; \overline{\rho }(h)\le c_{2}\le \overline{\rho }(h)+\epsilon . \end{aligned}$$

Also

$$\begin{aligned} \rho (g(x))\le c_{1}\varphi (x,0),\,\;\rho (h(x))\le c_{2}\varphi (x,0). \end{aligned}$$

for all $x\in V.$ If $a+b=1$ and $a,b\ge 0$, then we get

$$\begin{aligned} \rho (a g(x) +b h(x))\le & {} a\rho (g(x)) +b \rho (h(x)) \\\le & {} (c_{1}a+c_{2}b )\varphi (x,0), \end{aligned}$$

so we get

$$\begin{aligned} \overline{\rho }(a g+b h)\le a\overline{\rho }(g)+b \overline{\rho }(h)+(a+b)\epsilon , \end{aligned}$$

This concludes that $\overline{\rho }$ is a convex modular on M. Now we show that $M_{\overline{\rho }}$ is $\overline{\rho }$-complete. Let $\{g_{n}\}$ be a $\overline{\rho }$-Cauchy sequence in $M_{\overline{\rho }}$ and let $\epsilon > 0$ be given. There exists a positive integer $n_{0}\in \mathbb {N}$ such that

$$\begin{aligned} \overline{\rho }(g_{n}-g_{m})<\epsilon , \end{aligned}$$

(2.10)

for all $n,m\ge n_{0}$. We have

$$\begin{aligned} \rho (g_{n}(x)-g_{m}(x))\le \epsilon \varphi (x,0) \end{aligned}$$

(2.11)

for all $x\in V$ and $n,m\ge n_{0}$. Therefore if x is any given point of V, $\{g_{n}(x)\}$ is a $\rho$-Cauchy sequence in $X_{\rho }$. Since $X_{\rho }$ is $\rho$-complete, so $\{g_{n}(x)\}$ is convergent in $X_{\rho }$, for each $x\in V$. Hence, we can define a function $g:V\rightarrow X_{\rho }$ by:

$$\begin{aligned} g(x):=\lim _{n\rightarrow \infty }\displaystyle g_{n}(x), \end{aligned}$$

(2.12)

for all $x\in V.$ Since $\rho$ satisfies the Fatou property, it follows from (2.11) that

$$\begin{aligned} \rho (g_{n}(x)-g(x))\le \liminf _{m\rightarrow \infty }\displaystyle \rho (g_{n}(x)-g_{m}(x))\le \epsilon \varphi (x,0) , \end{aligned}$$

(2.13)

so

$$\begin{aligned} \overline{\rho }(g_{n}-g)\le \epsilon , \end{aligned}$$

(2.14)

for all $n\ge n_{0}$. Thus, $\{g_{n}\}$ is $\overline{\rho }$-converges, so that $M_{\overline{\rho }}$ is $\overline{\rho }$-complete.

Now we show that $\overline{\rho }$ satisfies the Fatou property. Suppose that $\{g_{n}\}$ is a sequence in $M_{\overline{\rho }}$ which is $\overline{\rho }$- convergent to an element $g\in M_{\overline{\rho }}$. Let $\epsilon > 0$ be given. For each $n\in \mathbb {N}$, let $c_{n}$ be a constant such that

$$\begin{aligned} \overline{\rho }(g_{n})\le c_{n}\le \overline{\rho }(g_{n})+\epsilon . \end{aligned}$$

(2.15)

so

$$\begin{aligned} \rho (g_{n}(x))\le c_{n}\varphi (x,0), \end{aligned}$$

(2.16)

for all $x\in V$. Since $\rho$ satisfies the Fatou property, we have

$$\begin{aligned} \rho (g(x))\le & {} \liminf _{n\rightarrow \infty }\displaystyle \rho (g_{n}(x))\\\le & {} \liminf _{n\rightarrow \infty }\displaystyle c_{n}\varphi (x,0)\\\le & {} \left[ \liminf _{n\rightarrow \infty } \overline{\rho }(g_{n})+\epsilon \right] \varphi (x,0) \end{aligned}$$

Thus, we have

$$\begin{aligned} \overline{\rho }(g)\le \liminf _{n\rightarrow \infty }\displaystyle \overline{\rho } (g_{n})+\epsilon . \end{aligned}$$

So $\overline{\rho }$ satisfies the Fatou property. We consider the function $\tau :M_{\overline{\rho }}\rightarrow M_{\overline{\rho }}$ defined by:

$$\begin{aligned} \tau g(x)=\frac{1}{\alpha ^{3}}g(\alpha x), \end{aligned}$$

for all $x\in V$ and $g\in M_{\overline{\rho }}$. Let $g,h\in M_{\overline{\rho }}$ and let $c \in [0,1]$ be an arbitrary constant with $\overline{\rho }(g-h)<c$. From the definition of $\overline{\rho }$, we have $\rho (g(x)-h(x))\le c\varphi (x,0)$ for all $x\in V$. By (2.7) and the last inequality, we get

$$\begin{aligned} \rho \big (\frac{g(\alpha x)}{\alpha ^{3}}-\frac{h(\alpha x)}{\alpha ^{3}}\big )\le & {} \frac{1}{\alpha ^{3}}\rho (g(\alpha x)-h(\alpha x))\\\le & {} \frac{1}{\alpha ^{3}}c\varphi (\alpha x,0)\\\le & {} cL\varphi (x,0), \end{aligned}$$

for all $x\in V$. Hence, $\overline{\rho }(\tau g-\tau h)\le L\overline{\rho }(g-h),$ for all $g,h\in M_{\overline{\rho }}$, that is, $\tau$ is a $\overline{\rho }$-contraction. Next, we show that $\tau$ has a bounded orbit at f. Letting $y =0$ in (2.8), we get

$$\begin{aligned} \rho \left( \frac{f(\alpha x)}{\alpha ^{3}}-f(x)\right)\le & {} \frac{1}{\alpha ^{3}}\varphi (x,0), \end{aligned}$$

(2.17)

for all $x\in V$. Replacing x with $\alpha x$ in (2.17), we get

$$\begin{aligned} \rho \left( \frac{f(\alpha ^{2} x)}{\alpha ^{3}}-f(\alpha x)\right)\le & {} \frac{1}{\alpha ^{3}}\varphi (\alpha x,0), \end{aligned}$$

(2.18)

By using (2.17) and (2.18), we get

$$\begin{aligned} \rho \left( \frac{f(\alpha ^{2} x)}{\alpha ^{6}}-f(x)\right)\le & {} \rho (\frac{f(\alpha ^{2}x)}{\alpha ^{6}}-\frac{f(\alpha x)}{\alpha ^{3}})+\rho (\frac{f(\alpha x)}{\alpha ^{3}}-f(x))\nonumber \\\le & {} \frac{1}{\alpha ^{6}}\varphi (\alpha x,0)+\frac{1}{\alpha ^{3}}\varphi (x,0), \qquad \qquad \end{aligned}$$

(2.19)

for all $x\in V$. By induction, we can easily see that

$$\begin{aligned} \rho (\frac{f(\alpha ^{n}x)}{\alpha ^{3n}}-f(x))\le & {} \sum _{i=1}^{n}\frac{1}{\alpha ^{3i}}\varphi (\alpha ^{i-1}x,0) \nonumber \\\le & {} \frac{1}{L\alpha ^{3}}\varphi (x,0)\sum \limits _{i=1}^{n}L^{i}\nonumber \\\le & {} \frac{1}{\alpha ^{3}(1-L)}\varphi (x,0), \end{aligned}$$

(2.20)

for all $x\in V$. It follows from inequality (2.20) that

$$\begin{aligned} \rho \left( \frac{f(\alpha ^{n}x)}{\alpha ^{3n}}-\frac{f(\alpha ^{k}x)}{\alpha ^{3k}}\right)\le & {} \frac{1}{2}\rho \left( 2\frac{f(\alpha ^{n}x)}{\alpha ^{3n}}-2f(x)\right) +\frac{1}{2}\rho \left( 2\frac{f(\alpha ^{k}x)}{\alpha ^{3k}}-2f(x)\right) \\\le & {} \frac{\kappa }{2}\rho \left( \frac{f(\alpha ^{n}x)}{\alpha ^{3n}}-f(x)\right) +\frac{\kappa }{2}\rho \left( \frac{f(\alpha ^{k}x)}{\alpha ^{3k}}-f(x)\right) \\\le & {} \frac{\kappa }{\alpha ^{3}(1-L)}\varphi (x,0), \end{aligned}$$

for every $x\in V$ and $n,k \in \mathbb {N}$, By the definition of $\overline{\rho }$, we conclude that

$$\begin{aligned} \overline{\rho }(\tau ^{n} f-\tau ^{k} f)\le \frac{\kappa }{\alpha ^{3}(1-L)}, \end{aligned}$$

which implies the boundedness of an orbit of $\tau$ at f. It follows from Theorem 1.5 that, the sequence $\{\tau ^{n}f\}$ $\overline{\rho }$-converges to $C_{\alpha }\in M_{\overline{\rho }}$. Now, by the $\overline{\rho }$-contractivity of $\tau$, we have

$$\begin{aligned} \overline{\rho }(\tau ^{n} f-\tau C_{\alpha })\le L\overline{\rho }(\tau ^{n-1} f-C_{\alpha }). \end{aligned}$$

Passing to the limit $n\rightarrow \infty$ and applying the Fatou property of $\overline{\rho }$, we obtain that

$$\begin{aligned} \overline{\rho }(\tau C_{\alpha }-C_{\alpha })\le & {} \liminf _{n\rightarrow \infty }\overline{\rho }(\tau C_{\alpha }-\tau ^{n} f)\\\le & {} L\liminf _{n\rightarrow \infty }\overline{\rho }(C_{\alpha }-\tau ^{n-1} f)=0. \end{aligned}$$

Therefore, $C_{\alpha }$ is a fixed point of $\tau$. Letting $x = \alpha ^{n}x$ and $y = \alpha ^{n}y$ in (2.8), we get

$$\begin{aligned} \rho \big (D_{\alpha }f(\alpha ^{n}x,\alpha ^{n}y)\big )\le \varphi (\alpha ^{n}x_,\alpha ^{n}y), \end{aligned}$$

for all $x,y\in V$. Therefore

$$\begin{aligned} \rho \big (\frac{1}{\alpha ^{3n}}D_{\alpha }f(\alpha ^{n}x,\alpha ^{n}y)\big )\le \frac{1}{\alpha ^{3n}}\varphi (\alpha ^{n}x_,\alpha ^{n}y), \end{aligned}$$

(2.21)

Employing the limit $n\rightarrow \infty$, we get

$$\begin{aligned} D_{\alpha }C_{\alpha }(x,y)=0, \end{aligned}$$

for all $x,y\in V$. It follows from Lemma 2.2, that $C_{\alpha }$ is cubic. By using (2.20), we get (2.9).

To prove the uniqueness of $C_{\alpha },$ let $C:V\rightarrow X_{\rho }$ be another cubic mapping satisfying (2.9). Then, C is a fixed point of $\tau$.

$$\begin{aligned} \overline{\rho }\big (C_{\alpha }-C\big )= \overline{\rho }\big (\tau C_{\alpha }-\tau C\big )\le L \overline{\rho }\big (C_{\alpha }-C\big ), \end{aligned}$$

which implies that $\overline{\rho }\big (C_{\alpha }-C\big )=0$ or $C_{\alpha }=C$. This completes the proof.

Corollary 2.4

Let X be a Banach space, $\varphi : V^{2}\longrightarrow [0, +\infty )$ be a function such that

$$\begin{aligned} \lim _{n\rightarrow \infty }\frac{1}{\alpha ^{3n}}\varphi (\alpha ^{n}x,\alpha ^{n}y)=0, \end{aligned}$$

and

$$\begin{aligned} \varphi (\alpha x,\alpha y)\le L\alpha ^{3}\varphi (x,y), \end{aligned}$$

for all $x,y\in V$ with $L<1.$ Suppose that $f:V\longrightarrow X$ satisfies the following condition

$$\begin{aligned} \big \Vert D_{\alpha }f(x,y)\big \Vert \le \varphi (x,y), \end{aligned}$$

$x,y\in V$ and $f(x)=0.$ Then there exists a unique cubic mapping $C_{\alpha }: V\longrightarrow X$ such that

$$\begin{aligned} \big \Vert C_{\alpha }(x)-f(x)\big \Vert \le \frac{1}{\alpha ^{3}(1-L)}\varphi (x,0), \end{aligned}$$

for all $x\in V$.

Proof

It is known that every normed space is modular space with the modular $\rho (x) = \Vert x\Vert$ and satisfies the $\Delta _{\alpha }$-condition with $\kappa = \alpha$. $\square$

Theorem 2.5

Let $\varphi : V^{2}\longrightarrow [0, +\infty )$ be a function such that

$$\begin{aligned} \lim _{n\rightarrow \infty }\kappa ^{3n}\varphi \left( \frac{x}{\alpha ^{n}},\frac{y}{\alpha ^{n}}\right) =0, \end{aligned}$$

(2.22)

and

$$\begin{aligned} \varphi \left( \frac{x}{\alpha },\frac{y}{\alpha }\right) \le \frac{L}{\alpha ^{3}}\varphi (x,y), \end{aligned}$$

(2.23)

for all $x,y\in V$ with $L<1.$ Suppose that $f:V\longrightarrow X_{\rho }$ satisfies the condition

$$\begin{aligned} \rho \big (D_{\alpha }f(x,y)\big )\le \varphi (x,y), \end{aligned}$$

(2.24)

for all $x,y\in V$ and $f(0)=0.$ Then there exists a unique mapping $C_{\alpha }: V\longrightarrow X_{\rho }$ such that

$$\begin{aligned} \rho \big (C_{\alpha }(x)-f(x)\big )\le \frac{L}{\alpha ^{3}(1- L)}\varphi (x,0), \end{aligned}$$

(2.25)

for all $x\in V$.

Proof

We consider the set

$$\begin{aligned} M=\{g:V\rightarrow X_{\rho }\} \end{aligned}$$

and define the function $\overline{\rho }$ on M as follows,

$$\begin{aligned} \overline{\rho }(g)=:\inf \{c>0:\, \rho (g(x))\le c\varphi (x,0),\;\, \forall x\in V\} . \end{aligned}$$

Similar to the proof of Theorem 2.3, we have:

• 1. $\overline{\rho }$ is a convex modular on M, 

• 2. $M_{\overline{\rho }}$ is $\overline{\rho }$-complete.

• 3. $\overline{\rho }$ satisfies the Fatou property.

Now, we consider the function $\tau :M_{\overline{\rho }}\rightarrow M_{\overline{\rho }}$ defined by:

$$\begin{aligned} \tau g(x)=\alpha ^{3}g\left( \frac{x}{\alpha }\right) , \end{aligned}$$

for all $x\in V$ and $g\in M_{\overline{\rho }}.$ Let $g,h\in M_{\overline{\rho }}$ and let $c \in [0,1]$ be an arbitrary constant with $\overline{\rho }(g-h)<c$. From the definition of $\overline{\rho }$, we have $\rho (g(x)-h(x))\le c\varphi (x,0)$ for all $x\in V$. By the assumption and the last inequality, we get

$$\begin{aligned} \rho \left( \alpha ^{3}g\left( \frac{x}{\alpha }\right) -\alpha ^{3}h\left( \frac{x}{\alpha }\right) \right)\le & {} \kappa ^{3}\rho \left( g\left( \frac{x}{\alpha }\right) -g\left( \frac{x}{\alpha }\right) \right) \\\le & {} \kappa ^{3} c\varphi \left( \frac{x}{\alpha },0\right) \\\le & {} cL\varphi (x,0), \end{aligned}$$

for all $x\in V$. Hence, $\overline{\rho }(\tau g-\tau h)\le L\overline{\rho }(g-h),$ for all $g,h\in \mathfrak {M_{\overline{\rho }}}$ that is, $\tau$ is a $\overline{\rho }$-contraction. Next, we show that $\tau$ has a bounded orbit at f. Letting $y=0$ in (2.24), we get

$$\begin{aligned} \rho (\alpha ^{3}f(x)-f(\alpha x))&\le \varphi (x,0), \end{aligned}$$

(2.26)

for all $x\in V$. Replacing x with $\frac{x}{\alpha }$ in (2.26), we get

$$\begin{aligned} \rho \left( \alpha ^{3}f\left( \frac{x}{\alpha }\right) -f(x)\right)\le & {} \varphi (\frac{x}{\alpha },0), \end{aligned}$$

(2.27)

for all $x\in V$. Replacing x with $\frac{x}{\alpha }$ in (2.27), we get

$$\begin{aligned} \rho \left( \alpha ^{3}f\left( \frac{x}{\alpha ^{2}}\right) -f\left( \frac{x}{\alpha }\right) \right)\le & {} \varphi (\frac{x}{\alpha ^{2}},0), \end{aligned}$$

(2.28)

for all $x\in V$. By using (2.26), (2.27) and (2.28), we get

$$\begin{aligned} \rho (\alpha ^{6}f(\frac{x}{\alpha ^{2}})-f(x))\le & {} \rho (\alpha ^{6}f(\frac{x}{\alpha ^{2}})- \alpha ^{3}f(\frac{x}{\alpha }))+\rho (\alpha ^{3}f(\frac{x}{\alpha })- f(x))\nonumber \\\le & {} \kappa ^{3}\rho (\alpha ^{3}f(\frac{x}{\alpha ^{2}})- f(\frac{x}{\alpha }))+\rho (\alpha ^{3}f(\frac{x}{\alpha })-f(x))\nonumber \\\le & {} \alpha ^{3}\varphi (\frac{x}{\alpha ^{2}},0)+\varphi (\frac{x}{\alpha },0), \end{aligned}$$

(2.29)

for all $x\in V$. By induction, we can easily see that

$$\begin{aligned} \rho \big (\alpha ^{3n}f(\frac{x}{\alpha ^{n}})-f(x)\big )\le & {} \frac{1}{\alpha ^{3}}\sum _{i=1}^{n}\alpha ^{3i}\varphi (\frac{x}{\alpha ^{i}},0)\nonumber \\\le & {} \frac{1}{\alpha ^{3}}\varphi (x,0)\sum _{i=1}^{n}L^{i}\nonumber \\\le & {} \frac{L}{\alpha ^{3}(1-L)}\varphi (x,0), \end{aligned}$$

(2.30)

for all $x\in V$. It follows from inequality (2.30) that

$$\begin{aligned} \rho \big (\alpha ^{3n}f(\frac{x}{\alpha ^{n}})-\alpha ^{3k}f(\frac{x}{\alpha ^{k}})\big )\le & {} \frac{1}{2}\rho (2\alpha ^{3n}f(\frac{x}{\alpha ^{n}})-2f(x))+\frac{1}{2}\rho (2\alpha ^{3k}f(\frac{x}{\alpha ^{k}})-2f(x))\nonumber \\\le & {} \frac{kL}{\alpha ^{3}(1-L)}\varphi (x,0), \end{aligned}$$

(2.31)

for every $x\in V$ and $n,k \in \mathbb {N}$, By the definition of $\overline{\rho }$, we conclude that

$$\begin{aligned} \overline{\rho }(\tau ^{n} f-\tau ^{k} f)\le \frac{kL}{\alpha ^{3}(1-L)}, \end{aligned}$$

which implies the boundedness of an orbit of $\tau$ at f. It follows from Theorem 1.5 that, the sequence $\{\tau ^{n}f\}$ $\overline{\rho }$-converges to $C_{\alpha }\in M_{\overline{\rho }}$. Now, by the $\overline{\rho }$-contractivity of $\tau$, we have

$$\begin{aligned} \overline{\rho }(\tau ^{n} f-\tau C_{\alpha })\le L\overline{\rho }(\tau ^{n-1} f-C_{\alpha }). \end{aligned}$$

Employing the limit $n\rightarrow \infty$ and applying the Fatou property of $\overline{\rho }$, we obtain that

$$\begin{aligned} \overline{\rho }(\tau C_{\alpha }-C_{\alpha })\le & {} \liminf _{n\rightarrow \infty }\overline{\rho }(\tau C_{\alpha }-\tau ^{n} f)\\\le & {} L\liminf _{n\rightarrow \infty }\overline{\rho }(C_{\alpha }-\tau ^{n-1} f)=0. \end{aligned}$$

Therefore, $C_{\alpha }$ is a fixed point of $\tau$. Letting $x = \displaystyle \frac{x}{\alpha ^{n}}$ and $y=\displaystyle \frac{y}{\alpha ^{n}}$ in (2.24), we get

$$\begin{aligned} \rho \big (D_{\alpha }f(\frac{x}{\alpha ^{n}},\frac{y}{\alpha ^{n}})\big )\le \varphi (\frac{x}{\alpha ^{n}},\frac{y}{\alpha ^{n}}), \end{aligned}$$

for all $x,y\in V$. Therefore

$$\begin{aligned} \rho \left( \alpha ^{3n}D_{\alpha }f\left( \frac{x}{\alpha ^{n}},\frac{y}{\alpha ^{n}}\right) \right) \le \kappa ^{3n}\varphi \left( \frac{x}{\alpha ^{n}},\frac{y}{\alpha ^{n}}\right) , \end{aligned}$$

Passing to the limit $n\rightarrow \infty$, we get

$$\begin{aligned} D_{\alpha }C_{\alpha }(x,y)=0 \end{aligned}$$

for all $x,y\in V$. It follows from Lemma 2.2 that $C_{\alpha }$ is cubic. By using (2.30), we get (2.25). $\square$

Corollary 2.6

Let X be a Banach space, $\varphi : V^{2}\longrightarrow [0, +\infty )$ be a function such that

$$\begin{aligned} \lim _{n\rightarrow \infty }\alpha ^{3n}\varphi \left( \frac{x}{\alpha ^{n}},\frac{y}{\alpha ^{n}}\right) =0, \end{aligned}$$

and

$$\begin{aligned} \varphi \left( \frac{x}{\alpha },\frac{y}{\alpha }\right) \le \frac{L}{\alpha ^{3}}\varphi (x,y), \end{aligned}$$

for all $x,y\in V$ with $L<1.$ Suppose that $f:V\longrightarrow X$ satisfies the condition

$$\begin{aligned} \big \Vert D_{\alpha }f(x,y)\big \Vert \le \varphi (x,y), \end{aligned}$$

for all $x,y\in V$ and $f(0)=0.$ Then there exists a unique cubic mapping $C_{\alpha }: V\longrightarrow X$ such that

$$\begin{aligned} \big \Vert C_{\alpha }(x)-f(x)\big \Vert \le \frac{L}{\alpha ^{3}(1-L)}\varphi (x,0), \end{aligned}$$

for all $x\in V$.

Proof

It is known that every normed space is modular space with the modular $\rho (x) = \Vert x\Vert$ and satisfies the $\Delta _{\alpha }$-condition with $\kappa = \alpha$. $\square$

Remark 2.7

In Corollaries 2.4 and 2.6, by replacing $\varphi$ with:

$$\begin{aligned} \varphi (x,y)= & {} \Vert x\Vert ^{p}+\Vert y\Vert ^{p},\\ \varphi (x,y)= & {} \Vert x\Vert ^{p}\Vert y\Vert ^{q},\\ \varphi (x,y)= & {} \Vert x\Vert ^{p}+\Vert y\Vert ^{p}+\Vert x\Vert ^{r}\Vert y\Vert ^{s}, \end{aligned}$$

under suitable conditions, it is possible to obtain some corollaries.