Certain properties of difference operator and stability of Fréchet functional equation

Abstract

Let M be a commutative monoid, and let B be a Banach space. We give a new recursive method to obtain a Gǎvruţa-type stability result for the functional equation

$${\mathrm{\Delta }}_y^{n+1}f(x):=\sum _{k=0}^{n+1}{(-1)}^{n+1+k}\left(\begin{array}{c}n+1\\ k\end{array}\right)f(x+ky)=0$$

via algebraic manipulations of the forward difference operator.

1 Motivation

One of the best-known classes of functional equations, the Fréchet functional equations, consists of functional equations equivalent to the equation

$$ \Delta _{y_{1}}\Delta _{y_{2}}\cdots \Delta _{y_{n+1}}f(0)=0, $$

(1)

studied by Fréchet [1] in 1909. In this paper, we focus on the equation

$$ \Delta ^{n+1}_{y}f(x)=0, $$

(2)

for which the stability result relies on its equivalence with the equation

$$ \Delta _{y_{1}}\Delta _{y_{2}}\cdots \Delta _{y_{n+1}}f(x)=0. $$

(3)

The stability problem of functional equations originated from a problem posed by S. Ulam regarding “almost additive” functions that satisfy

$$ \bigl\Vert f(x+y)-f(x)-f(y) \bigr\Vert \leq \epsilon $$

for a fixed ϵ. This inequality was studied (under stricter assumptions than those for the posed problem) by Hyers [2] in 1941. The result was that such a function can be approximated by an additive function. Hence the problems in this sense, where the assigned bound is a constant, became known as Hyers–Ulam-type stability. Years later, Aoki [3] and Rassias [4] presented stability results where the bound is a power function of x and y. Thus the stability in this sense became known as the Aoki–Rassias-type stability. Later Gǎvruţa [5] generalized this so that the bound (which can also be called the control function) is a function with specific properties.

The solutions of (1), (2), and (3) are called generalized polynomials of degree at most n. It is well known that a generalized polynomial p is constructed from diagonalization of multiadditive functions [6]. The Hyers–Ulam stability of (2) and (3) has been studied in [6–9]. The Gǎvruţa-type stability of (1) has been studied by Dăianu [10]. Dăianu used an equivalence theorem, which is more general than that presented by Kuczma [11], to obtain a stability result for (2) under the assumption that M is $(n+1)!$-divisible. In this paper, we present a result in which divisibility of M is not required. Instead, we require $(n+1)!$-divisibility of B, which is readily true since B is a Banach space.

2 The forward difference operator

Let M be a commutative monoid, let be B be a Banach space, and let $\mathbb{N}$ be the set of positive integers. For a function $f:M\to B$ and $y\in M$, define $\Delta _{y}f:M\to B$ by

$$ \Delta _{y}f(x)=f(x+y)-f(x) $$

for $x\in M$. Also, define its iterations

$$ \Delta _{y_{1}}\Delta _{y_{2}}\cdots \Delta _{y_{n}}f= \Delta _{y_{1}} ( \Delta _{y_{2}}\Delta _{y_{3}}\cdots \Delta _{y_{n}}f ) \quad \text{and}\quad \Delta _{y}^{n}f= \underbrace{\Delta _{y}\Delta _{y}\cdots \Delta _{y}}_{n\text{ terms}}f $$

for $y,y_{1},y_{2},\dots ,y_{n}\in M$. Observe that $\Delta _{y_{1}}\Delta _{y_{2}}f=\Delta _{y_{2}}\Delta _{y_{1}}f$, so the ordering of $y_{i}$ is interchangable, and

$${\mathrm{\Delta }}_y^{n}f(x)=\sum _{k=0}^n{(-1)}^{n-k}\left(\begin{array}{c}n\\ k\end{array}\right)f(x+ky)$$

and

$$\begin{aligned}& \Delta _{y_{1}} \Delta _{y_{2}}\cdots \Delta _{y_{n}}f(x)=\sum _{ \epsilon _{1},\epsilon _{2},\dots ,\epsilon _{n}=0}^{1}(-1)^{n+ \epsilon _{1}+\epsilon _{2}+\cdots +\epsilon _{n}}f \Biggl( x+\sum _{i=1}^{n} \epsilon _{i}y_{i} \Biggr) . \end{aligned}$$

We will establish some algebraic manipulation of the forward difference operator Δ. We begin with a modified version of Kuczma’s theorem.

Lemma 2.1

Let$n\in \mathbb{N}$and$f:M\to B$. Then

$$ \Delta _{y_{1}}\Delta _{2y_{2}}\Delta _{3y_{3}}\cdots \Delta _{ny_{n}}f(x)= \sum _{\epsilon _{1},\epsilon _{2},\dots ,\epsilon _{n}=0}^{1}(-1)^{ \epsilon _{1}+\epsilon _{2}+\cdots +\epsilon _{n}}\Delta ^{n}_{b_{{ \epsilon _{1},\epsilon _{2},\dots ,\epsilon _{n}}}}f(x+a_{\epsilon _{1}, \epsilon _{2},\dots ,\epsilon _{n}}) $$

(4)

for all$x,y_{1},y_{2},\dots ,y_{n}\in X$, where

$$ a_{\epsilon _{1},\epsilon _{2},\dots ,\epsilon _{n}}=\sum_{i=1}^{n}i \epsilon _{i}y_{i}\quad \textit{and}\quad b_{\epsilon _{1},\epsilon _{2}, \dots ,\epsilon _{n}}= \sum_{i=1}^{n}(1-\epsilon _{i})y_{i} . $$

Proof

Recall that

$$\begin{array}{rl}& {\mathrm{\Delta }}_{b_{{ϵ}_1,{ϵ}_2,\dots ,{ϵ}_n}}^{n}f(x+a_{{ϵ}_1,{ϵ}_2,\dots ,{ϵ}_n})\\ & =\sum _{k=0}^n{(-1)}^{n-k}\left(\begin{array}{c}n\\ k\end{array}\right)f(x+\sum _{i=1}^{n}i{ϵ}_i{y}_i+k\sum _{i=1}^n(1-{ϵ}_i)y_i).\end{array}$$

We will show that each of these terms cancels out in the sum on the right-hand side of (4), except for $k=0$. For $k\neq 0$, we observe that $\epsilon _{k}$ is absent in the term where $i=k$. So changing $\epsilon _{k}$ does not change the argument of this term.

Consider another term in the sum on the right-hand side of (4), where only $\epsilon _{k}$ differs from this term. Then the kth term in its expansion cancels the term we mentioned before.

Hence every term where $k\neq 0$ has its negative in the sum in (4). Then

$$\begin{aligned} &\sum_{\epsilon _{1},\epsilon _{2},\dots ,\epsilon _{n}=0}^{1}(-1)^{ \epsilon _{1}+\epsilon _{2}+\cdots +\epsilon _{n}} \Delta ^{n}_{b_{{ \epsilon _{1},\epsilon _{2},\dots ,\epsilon _{n}}}}f(x+a_{\epsilon _{1}, \epsilon _{2},\dots ,\epsilon _{n}}) \\ &\quad =\sum_{\epsilon _{1},\epsilon _{2},\dots ,\epsilon _{n}=0}^{1}(-1)^{n+ \epsilon _{1}+\epsilon _{2}+\cdots +\epsilon _{n}}f \Biggl( x+\sum_{i=1}^{n} \epsilon _{i}iy_{i} \Biggr) \\ &\quad = \Delta _{y_{1}}\Delta _{2y_{2}}\Delta _{3y_{3}}\cdots \Delta _{ny_{n}}f(x) . \end{aligned}$$

 □

Lemma 2.2

Let$n,m\in \mathbb{N}$and$f:M\to B$. Then

$$ \Delta ^{n}_{y}f(x+my)=\Delta ^{n}_{y}f(x)+ \sum_{k=0}^{m-1}\Delta ^{n+1}_{y}f(x+ky) $$

for all$x,y\in X$.

Proof

Since $\Delta ^{n+1}_{y}f(x)=\Delta ^{n}_{y}f(x+y)-\Delta ^{n}_{y}f(x)$ for all $x,y\in M$,

$$\begin{aligned} \Delta ^{n}_{y}f(x+my) =& \Delta ^{n}_{y}f \bigl(x+(m-1)y\bigr)+\Delta ^{n+1}_{y}f\bigl(x+(m-1)y\bigr) \\ =& \Delta ^{n}_{y}f\bigl(x+(m-2)y\bigr)+\Delta _{y}^{n+1}f\bigl(x+(m-2)y\bigr)+\Delta _{y}^{n+1}f\bigl(x+(m-1)y\bigr) \\ \vdots& \\ =& \Delta ^{n}_{y}f(x)+\sum _{k=0}^{m-1}\Delta ^{n+1}_{y}f(x+ky). \end{aligned}$$

(5)

 □

In the following theorems, for convenience, we let $\sum_{i=0}^{-1}a_{i}=0$ for any sequence $(a_{i})$.

Lemma 2.3

Let$n\in \mathbb{N}$and$f:M\to B$. Then

$$ n!\Delta ^{n}_{y}f(x)=\Delta _{2y,3y,\dots ,(n+1)y}f(x)-\sum _{l_{1}=0}^{1} \sum _{l_{2}=0}^{2}\dots \sum _{l_{n}=0}^{n}\sum_{s=0}^{l_{2}+l_{3}+ \cdots +l_{n}-1} \Delta ^{n+1}_{y}f(x+sy) $$

for all$x,y\in X$.

Proof

By Lemma 2.2 it is sufficient to show that

$$ \Delta _{2y,3y,\ldots ,(n+1)y}f(x)=\sum_{l_{1}=0}^{1} \sum_{l_{2}=0}^{2} \dots \sum _{l_{n}=0}^{n}\Delta ^{n}_{y}f \bigl(x+(l_{1}+l_{2}+\cdots +l_{n})y\bigr). $$

(6)

For the case $n=1$, we have

$$\begin{aligned} \Delta _{2y}f(x)&=f(x+2y)-f(x) \\ &=f(x+2y)-f(x+y)+f(x+y)-f(x) =\Delta _{y}f(x+y)+\Delta _{y}f(x). \end{aligned}$$

Suppose that (6) is true for $n=k$. Since the order of $\Delta _{iy}$ is interchangable,

$$\begin{aligned} \Delta _{2y,3y,\dots ,(k+2)y}f(x)&=\Delta _{2y,3y,\dots ,(k+1)y} ( \Delta _{(k+2)y}f ) (x) \\ &=\sum_{l_{1}=0}^{1}\sum _{l_{2}=0}^{2}\dots \sum _{l_{k}=0}^{k} \Delta ^{n}_{y} ( \Delta _{(k+2)y}f ) \bigl(x+(l_{1}+l_{2}+ \cdots +l_{k})y\bigr) \\ &=\sum_{l_{1}=0}^{1}\sum _{l_{2}=0}^{2}\dots \sum _{l_{k}=0}^{k} \Delta _{(k+2)y} \bigl( \Delta ^{k}_{y}f \bigr) \bigl(x+(l_{1}+l_{2}+ \cdots +l_{k})y\bigr) \\ &=\sum_{l_{1}=0}^{1}\sum _{l_{2}=0}^{2}\dots \sum _{l_{k}=0}^{k} \sum_{l_{k+1}=0}^{k+1} \Delta _{y} \bigl( \Delta ^{k}_{y}f \bigr) \bigl(x+(l_{1}+l_{2}+ \cdots +l_{k})y+l_{k+1}y \bigr) . \end{aligned}$$

 □

The following theorem follows from Lemmas 2.3 and 2.1.

Theorem 2.4

Let$n\in \mathbb{N}$and$f:M\to B$. Then there exist$m\in \mathbb{N}$and nonnegative integers$a_{i}$, $b_{i}$, $c_{i}$, $d_{i}$, $e_{i}$for$i\in \{ 1,2,\dots ,m\}$such that

$$ n!\Delta _{y_{1}}\Delta ^{n}_{y_{2}}f(x)=\sum _{i=1}^{m}(-1)^{a_{i}} \Delta ^{n+1}_{b_{i}y_{1}+c_{i}y_{2}}f(x+d_{i}y_{1}+e_{i}y_{2}) $$

for$x,y_{1},y_{2}\in M$. To be precise,

$$\begin{aligned} n!\Delta _{y_{1}}\Delta ^{n}_{y_{2}}f(x)={}& \sum _{\epsilon _{1}, \epsilon _{2},\dots ,\epsilon _{n+1}=0}^{1}(-1)^{\epsilon _{1}+ \epsilon _{2}+\cdots +\epsilon _{n+1}}\Delta ^{n+1}_{b_{{\epsilon _{1}, \epsilon _{2},\dots ,\epsilon _{n+1}}}}f(x+a_{\epsilon _{1},\epsilon _{2}, \dots ,\epsilon _{n+1}}) \\ &{}-\sum_{l_{1}=0}^{1}\sum _{l_{2}=0}^{2}\dots \sum _{l_{n}=0}^{n} \sum_{s=0}^{l_{1}+l_{2}+\cdots +l_{n}-1} \Delta ^{n+1}_{y_{2}}f(x+y_{1}+sy_{2}) \\ &{}+\sum_{l_{1}=0}^{1}\sum _{l_{2}=0}^{2}\dots \sum _{l_{n}=0}^{n} \sum_{s=0}^{l_{1}+l_{2}+\cdots +l_{n}-1} \Delta ^{n+1}_{y_{2}}f(x+sy_{2}), \end{aligned}$$

where

$$ a_{\epsilon _{1},\epsilon _{2},\dots ,\epsilon _{n+1}}=\epsilon _{1}y_{1}+ \sum _{i=2}^{n+1}i\epsilon _{i}y_{2} \quad \textit{and}\quad b_{\epsilon _{1}, \epsilon _{2},\dots ,\epsilon _{n+1}}=(1-\epsilon _{1})y_{1}+ \sum_{i=2}^{n+1}(1- \epsilon _{i})y_{2} . $$

3 The stability of $\Delta ^{n+1}_{y}f(x)=0$

We recall a theorem from the author’s dissertation [12].

Theorem 3.1

Let$n\in \mathbb{N}$and$f:M\to B$. Then

$$2^n{\mathrm{\Delta }}_y^{n}f(x)={\mathrm{\Delta }}_{2y}^{n}f(x)-\sum _{i=1}^n\left(\begin{array}{c}n\\ i\end{array}\right)\sum _{k=0}^{i-1}{\mathrm{\Delta }}_y^{n+1}f(x+ky)$$

for$x,y\in M$.

Next, we introduce the class of preferred control functions. Consider the following properties of a function $\varphi :M\to [0,\infty )$.

• P1 $\varphi (x+y)\leq \varphi (x)+\varphi (y)$ for all $x,y\in M$.

• P2 For each $x\in M$, either there exists $N\in \mathbb{N}$ such that $\varphi (2^{k}x)=0$ for every $k>N$, or $\lim_{k\to \infty } \frac{\varphi (2^{k+1}x)}{\varphi (2^{k}x)} <2$.

We take the notation of limit in P2 more loosely than normal. We allow it to not actually converge, as long as every of its limit points are in $[0,2)$.

Also note that P2 implies that $\sum^{\infty }_{k=0}\frac{\varphi (2^{k}x)}{2^{k}}$ converges. The next proposition states that the set of these functions forms a convex cone under pointwise addition and scalar multiplication.

Proposition 3.2

Let$\varphi _{1},\varphi _{2}:M\to [0,\infty )$satisfy P1 and P2. Then, for all$c_{1},c_{2}\in [0,\infty )$, $c_{1}\varphi _{1}+c_{2}\varphi _{2}$also satisfies P1 and P2.

Proof

The case $c_{1}c_{2}=0$ is straightforward, so we omit it. Firstly, it is clear that $c_{1}\varphi _{1}+c_{2}\varphi _{2}$ satisfies P1. Let $x\in M$. We will consider the cases depending on whether $N_{1}$ and $N_{2}$ exist such that $\varphi _{1}(2^{k}x)=0$ for $k>N_{1}$ and $\varphi _{2}(2^{l}x)=0$ for $l>N_{2}$.

• If both such $N_{1}$ and $N_{2}$ exist, then $c_{1}\varphi _{1}(2^{k}x)+c_{2}\varphi _{2}(2^{k}x)=0$ whenever $k>\max \{ N_{1},N_{2}\}$.

• If only one exists, then without loss of generality we assume that $N_{1}$ exists but $N_{2}$ does not. Then

  $$ \lim_{k\to \infty } \frac{c_{1}\varphi _{1}(2^{k+1}x)+c_{2}\varphi _{2}(2^{k+1}x)}{c_{1}\varphi _{1}(2^{k}x)+c_{2}\varphi _{2}(2^{k}x)} =\lim_{k\to \infty } \frac{c_{2}\varphi _{2}(2^{k+1}x)}{c_{2}\varphi _{2}(2^{k}x)} < 2. $$

• If there are no such $N_{1}$ and $N_{2}$, then there exist $N'_{1},N'_{2}\in \mathbb{N}$ and $r\in (0,2)$ such that $\varphi _{1}(2^{k+1}x)< r\varphi _{1}(2^{k}x)$ for $k>N'_{1}$ and $\varphi _{2}(2^{l+1}x)< r\varphi _{2}(2^{l}x)$ for $l>N'_{2}$. Thus

  $$ c_{1}\varphi _{1}\bigl(2^{k+1}x \bigr)+c_{2}\varphi _{2}\bigl(2^{k+1}x \bigr)< rc_{1} \varphi _{1}\bigl(2^{k}x \bigr)+rc_{2}\varphi _{2}\bigl(2^{k}x\bigr) $$

  for $k>\max \{ N'_{1},N'_{2}\}$. This implies that $\lim_{k\to \infty } \frac{c_{1}\varphi _{1}(2^{k+1}x)+c_{2}\varphi _{2}(2^{k+1}x)}{c_{1}\varphi _{1}(2^{k}x)+c_{2}\varphi _{2}(2^{k}x)} \leq r<2$,

and we conclude that $c_{1}\varphi _{1}+c_{2}\varphi _{2}$ satisfies P2. □

Let $C=\{ \varphi :M\to [0,\infty ) | \varphi \text{ satisfy P1 and P2} \}$. For all $\varphi :M\to [0,\infty )$ and $n\in \mathbb{N}$, we define $\lambda _{n}\varphi :M\to [0,\infty )$ by

$$ \lambda _{n}\varphi (x)=\sum^{\infty }_{k=0} \frac{\varphi (2^{k}x)}{2^{kn}} . $$

The next theorem shows that $\lambda _{n}(C)\subseteq C$.

Theorem 3.3

Let$\varphi \in C$and$n\in \mathbb{N}$. Then$\lambda _{n}\varphi \in C$.

Proof

It is clear that $\lambda _{n}\varphi$ satisfies P1. We will consider P2 for $\lambda _{n}\varphi$. Let $x\in M$.

If there exists $N\in \mathbb{N}$ such that $\varphi (2^{k}x)=0$ for all $k>N$, then it is straightforward to show that $\lambda _{n}\varphi (2^{k}x)=0$ for every $k>N$.

If no such N exists, then $\varphi (2^{k}x)>0$ for all $k\in \mathbb{N}$. This implies that $\lambda _{n}\varphi (2^{k}x)>0$ for all $k\in \mathbb{N}$. Since φ satisfies P2, there exist $N\in \mathbb{N}$ and $r\in (0,2)$ such that $\varphi (2^{k+1}x)< r\varphi (2^{k}x)$ whenever $k>N$. So,

$$ \frac{\lambda _{n}\varphi (2^{k+1}x)}{\varphi (2^{k}x)} =\sum^{ \infty }_{i=0} \frac{\varphi (2^{k+i+1}x)}{2^{in}\varphi (2^{k}x)} < \sum^{\infty }_{i=0} \frac{r^{i+1}}{2^{in}} =r\sum^{\infty }_{i=0} \biggl( \frac{r}{2^{n}} \biggr) ^{i}=\frac{2^{n}r}{2^{n}-r} . $$

Also note that $\lambda _{n}\varphi (2^{k}x)=\varphi (2^{k}x)+\frac{1}{2^{n}} \lambda _{n}\varphi (2^{k+1}x)$. Thus

$$ \frac{\lambda _{n}\varphi (2^{k}x)}{\lambda _{n}\varphi (2^{k+1}x)} = \frac{\varphi (2^{k}x)}{\lambda _{n}\varphi (2^{k+1}x)} + \frac{1}{2^{n}} > \frac{2^{n}-r}{2^{n}r} +\frac{1}{2^{n}} =\frac{1}{r} . $$

Hence we have $\frac{\lambda _{n}\varphi (2^{k+1}x)}{\lambda _{n}\varphi (2^{k}x)} \leq r<2$ whenever $k>N$. This completes the proof. □

Now we establish our main theorems.

Theorem 3.4

Let$n\in \mathbb{N}$, $f:M\to B$, $\theta \in [0,\infty )$, and$\varphi _{1},\varphi _{2}\in C$. If

$$ \bigl\Vert \Delta ^{n+1}_{y}f(x) \bigr\Vert \leq \theta +\varphi _{1}(x)+\varphi _{2}(y) $$

for all$x,y\in M$, then there exists$\varphi _{3}\in C$such that

$$ \bigl\vert \Delta _{y_{1}}\Delta ^{n}_{y_{2}}f(x) \bigr\vert \leq \frac{n2^{n}}{2^{n}-1} \theta +\frac{n2^{n-1}\varphi _{1}(y_{1})}{2^{n}-1} + \frac{n2^{n}}{2^{n}-1} \varphi _{1}(x)+\varphi _{3}(y_{2}) $$

for all$x,y_{1},y_{2}\in M$, where$\varphi _{3}$is defined by

$$ \varphi _{3}:=\frac{n(n-1)}{4} \lambda _{n}\varphi _{1}+n\lambda _{n} \varphi _{2}. $$

Proof

According to Theorem 2.4, there exist $m\in \mathbb{N}$ and nonnegative integers $a_{i}$, $b_{i}$, $c_{i}$, $d_{i}$, $e_{i}$ for $i\in \{ 1,2,\dots ,m\}$ such that

$$\begin{aligned} \bigl\Vert \Delta _{y_{1}}\Delta ^{n}_{y_{2}}f(x) \bigr\Vert &=\frac{1}{n!} \Biggl\Vert \sum _{i=1}^{m}(-1)^{a_{i}}\Delta ^{n+1}_{b_{i}y_{1}+c_{i}y_{2}}f(x+d_{i}y_{1}+e_{i}y_{2}) \Biggr\Vert \\ &\leq \frac{1}{n!} \sum_{i=1}^{m} \bigl( \theta +\varphi _{1}(x+d_{i}y_{1}+e_{i}y_{2})+ \varphi _{2}(b_{i}y_{1}+c_{i}y_{2}) \bigr). \end{aligned}$$

(7)

Denote the right-hand side of (7) by $\alpha _{0}(x,y_{1},y_{2})$. Since $\varphi _{1},\varphi _{2}\in C$,

$$ \lim_{k\to \infty }\frac{\alpha _{0}(x,y_{1},2^{k}y_{2})}{2^{kn}} =0. $$

For each nonnegative integer k, let

$$\begin{array}{rcl}{\alpha }_{k+1}(x,y_1,y_2)& =& \frac{{\alpha }_k(x,y_1,2y_2)}{2^n}\\ & & +\frac{1}{2^n}\sum _{i=1}^n\left(\begin{array}{c}n\\ i\end{array}\right)\sum _{k=0}^{i-1}(2\theta +2{\phi }_1(x)+{\phi }_1(y_1)+2k{\phi }_1(y_2)+2{\phi }_2(y_2)).\end{array}$$

We can see that if $\| \Delta _{y_{1}}\Delta ^{n}_{y_{2}}f(x)\| \leq \alpha _{k}(x,y_{1},y_{2})$, then by Theorem 3.1

$$\begin{array}{rl}& \parallel {\mathrm{\Delta }}_{y_1}{\mathrm{\Delta }}_{y_2}^{n}f(x)\parallel \\ & =\parallel \frac{1}{2^n}{\mathrm{\Delta }}_{y_1}{\mathrm{\Delta }}_{2y_2}^{n}f(x)-\frac{1}{2^n}\sum _{i=1}^n\left(\begin{array}{c}n\\ i\end{array}\right)\sum _{k=0}^{i-1}{\mathrm{\Delta }}_{y_1}{\mathrm{\Delta }}_{y_2}^{n+1}f(x+k{y}_2)\parallel \\ & \le \frac{{\alpha }_k(x,y_1,2y_2)}{2^n}+\frac{1}{2^n}\sum _{i=1}^n\left(\begin{array}{c}n\\ i\end{array}\right)\sum _{k=0}^{i-1}\parallel {\mathrm{\Delta }}_{y_2}^{n+1}f(x+y_1+k{y}_2)-{\mathrm{\Delta }}_{y_2}^{n+1}f(x+k{y}_2)\parallel \\ & \le \frac{{\alpha }_k(x,y_1,2y_2)}{2^n}+\frac{1}{2^n}\sum _{i=1}^n\left(\begin{array}{c}n\\ i\end{array}\right)\sum _{k=0}^{i-1}(2\theta +{\phi }_1(x+y_1+k{y}_2)+{\phi }_1(x+k{y}_2)+2{\phi }_2(y_2))\\ & \le \frac{{\alpha }_k(x,y_1,2y_2)}{2^n}+\frac{1}{2^n}\sum _{i=1}^n\left(\begin{array}{c}n\\ i\end{array}\right)\sum _{k=0}^{i-1}(2\theta +2{\phi }_1(x)+{\phi }_1(y_1)+2k{\phi }_1(y_2)+2{\phi }_2(y_2))\\ & ={\alpha }_{k+1}(x,y_1,y_2).\end{array}$$

Hence $\| \Delta _{y_{1}}\Delta ^{n}_{y_{2}}f(x)\| \leq \alpha _{k}(x,y_{1},y_{2})$ for any nonnegative integer k. Observe that, for $m\geq 1$,

$$\begin{array}{rl}{\alpha }_m(x,y_1,y_2)=& \frac{{\alpha }_0(x,y_1,2^m{y}_2)}{2^{mn}}\\ & +\frac{1}{2^n}\sum _{j=0}^{m-1}\sum _{i=1}^n\left(\begin{array}{c}n\\ i\end{array}\right)\sum _{k=0}^{i-1}(\frac{2\theta +2{\phi }_1(x)+{\phi }_1(y_1)}{2^{jn}}+\frac{2k{\phi }_1(2^j{y}_2)+2{\phi }_2(2^j{y}_2)}{2^{jn}})\\ =& \frac{{\alpha }_0(x,y_1,2^m{y}_2)}{2^{mn}}\\ & +\frac{1}{2^n}\sum _{i=1}^n\left(\begin{array}{c}n\\ i\end{array}\right)\sum _{k=0}^{i-1}\sum _{j=0}^{m-1}(\frac{2\theta +2{\phi }_1(x)+{\phi }_1(y_1)}{2^{jn}}+\frac{2k{\phi }_1(2^j{y}_2)+2{\phi }_2(2^j{y}_2)}{2^{jn}}).\end{array}$$

It follows that

$$\begin{array}{rl}& \underset{m\to \mathrm{\infty }}{lim}{a}_m(a,y_1,y_2)\\ & =0+\frac{1}{2^n}\sum _{i=1}^n\left(\begin{array}{c}n\\ i\end{array}\right)\sum _{k=0}^{i-1}\sum _{j=0}^{\mathrm{\infty }}(\frac{2\theta +2{\phi }_1(x)+{\phi }_1(y_1)}{2^{jn}}+\frac{2k{\phi }_1(2^j{y}_2)+2{\phi }_2(2^j{y}_2)}{2^{jn}})\\ & =\frac{1}{2^n}\sum _{i=1}^n\left(\begin{array}{c}n\\ i\end{array}\right)\sum _{k=0}^{i-1}(\frac{2^n}{2^n-1}(2\theta +2{\phi }_1(x)+{\phi }_1(y_1))+2k{\lambda }_n{\phi }_1(y_2)+2{\lambda }_n{\phi }_2(y_2))\\ & =\frac{n{2}^{n-1}}{2^n-1}(2\theta +2{\phi }_1(x)+{\phi }_1(y_1))+\frac{n(n-1)}{4}{\lambda }_n{\phi }_1(y_2)+n{\lambda }_n{\phi }_2(y_2).\end{array}$$

Hence

$$\begin{aligned} \bigl\Vert \Delta _{y_{1}}\Delta ^{n}_{y_{2}}f(x) \bigr\Vert \leq {}&\lim_{m\to \infty } \alpha _{m}(x,y_{1},y_{2}) \\ = {}&\frac{n2^{n}}{2^{n}-1} \biggl(\theta +\varphi _{1}(x)+ \frac{\varphi _{1}(y_{1})}{2} \biggr)+\frac{n(n-1)}{4} \lambda _{n} \varphi _{1}(y_{2})+n \lambda _{n}\varphi _{2}(y_{2}). \end{aligned}$$

 □

Let $\varLambda _{n,k}\varphi = ( \prod^{n}_{k+1} \frac{2^{k}}{2^{k}-1} ) \lambda _{1}\lambda _{2}\cdots \lambda _{k} \varphi$ for $k< n$ and $\varLambda _{n,n}=\lambda _{1}\lambda _{2}\cdots \lambda _{n}\varphi$. Using Theorem 3.4 inductively, we get the following theorem.

Theorem 3.5

Let$n\in \mathbb{N}$, $\theta \in [0,\infty )$, $\varphi _{1},\varphi _{2}\in C$, and$f:M\to B$. If$|\Delta ^{n+1}_{y}f(x)|\leq \theta +\varphi _{1}(x)+\varphi _{2}(y)$for all$x,y\in M$, then there exists$\varphi _{3}\in C$such that

$$ \bigl\Vert \Delta _{y_{1}}\Delta _{y_{2}}\cdots \Delta _{y_{n+1}}f(x) \bigr\Vert \leq n! \Biggl( \prod ^{n}_{k=1}\frac{2^{k}}{2^{k}-1} \Biggr) \Biggl( \theta +\varphi _{1}(x)+\sum^{n}_{i=1} \frac{\varphi _{1} (y_{i})}{2} \Biggr) +\varphi _{3}(y_{n+1}) $$

for all$x,y_{1},y_{2}\in M$, where$\varphi _{3}$is defined by

$$ \varphi _{3}:=n!\varLambda _{n,n}\varphi _{2}+n!\sum^{n}_{i=1} \frac{i-1}{4} \varLambda _{n,i}\varphi _{1}. $$

Proof

Let $y_{1}\in M$. By Theorem 3.4 there exists $\varphi '_{2}\in C$ such that

$$ \bigl\Vert \Delta _{y_{1}}\Delta ^{n}_{y_{2}}f(x) \bigr\Vert \leq \frac{n2^{n}}{2^{n}-1} \biggl(\theta +\varphi _{1}(x)+ \frac{\varphi _{1}(y_{1})}{2} \biggr)+\frac{n(n-1)}{4} \lambda _{n}\varphi _{1}+n \lambda _{n}\varphi _{2}. $$

Let $f_{y_{1}}=\Delta _{y_{1}}f$, $\theta _{y_{1}}=\frac{n2^{n}}{2^{n}-1} (\theta + \frac{\varphi _{1}(y_{1})}{2} )$, $\psi _{1}=\frac{n2^{n}}{2^{n}-1} \varphi _{1}$, and $\psi _{2}=\frac{n(n-1)}{4} \lambda _{n}\varphi _{1}+n\lambda _{n} \varphi _{2}$. Since the order of $\Delta _{y_{i}}$ can be interchanged without affecting the value on the left-hand side, we have

$$\begin{aligned} \bigl\Vert \Delta ^{n}_{y_{2}}f_{y_{1}}(x) \bigr\Vert ={}& \bigl\Vert \Delta ^{n}_{y_{2}}\Delta _{y_{1}}f(x) \bigr\Vert \\ \leq {}& \frac{n2^{n}}{2^{n}-1} \biggl(\theta +\varphi _{1}(x)+ \frac{\varphi _{1}(y_{1})}{2} \biggr)+\frac{n(n-1)}{4} \lambda _{n} \varphi _{1}(y_{2})+n \lambda _{n}\varphi _{2}(y_{2}) \\ ={}& \theta '_{y_{1}}+\psi _{1}(x)+\psi _{2}(y_{2}) . \end{aligned}$$

Since $y_{1}$ is currently fixed and $\psi _{1},\psi _{2}\in C$, the theorem is true by induction on n. □

Now we apply this to the result of Theorem 4.4 in [10]. For all $\varphi :M^{n+1}\to [0,\infty )$ and $n\in \mathbb{N}$, define $r_{n}\varphi ,R_{n}\varphi :M^{n}\to \mathbb{R}^{*}$ by

$$\begin{aligned} &r_{n}\varphi (x_{1},x_{2},\dots ,x_{n}) \\ &\quad = \varphi (2x_{1},2x_{2}, \dots ,2x_{n-1},x_{n},x_{n})+2\varphi (2x_{1},2x_{2},\dots ,2x_{n-2},x_{n},x_{n-1},x_{n-1})+ \cdots \\ &\qquad {}+2^{n-2}\varphi (2x_{1},x_{n},x_{n-1}, \dots ,x_{3},x_{2},x_{2})+2^{n-1} \varphi (x_{n},x_{n-1},\dots ,x_{2},x_{1},x_{1}), \\ &R_{n}\varphi (x_{1},x_{2},\dots ,x_{n})= \sum^{\infty }_{k=0} \frac{r_{n}\varphi (2^{k}x_{1},2^{k}x_{2},\dots ,2^{k}x_{n})}{2^{n(k+1)}} . \end{aligned}$$

Also, let

$$\begin{aligned} D^{+}_{n}={}& \Biggl\{ \bigl(\varphi ,\varphi '\bigr) | \varphi :M^{n+1}\to \mathbb{R}^{*}, \sum^{\infty }_{k=0}2^{-n(k+1)}\varphi \bigl(2^{k}z\bigr)< \infty ,z\in M^{n+1}, \\ &{} \varphi ':M^{n}\to [0,\infty ),\varphi '(y)\geq R_{n} \varphi (y),\text{ and }\lim _{k\to \infty }2^{-nk}\varphi ' \bigl(2^{k}y\bigr)=0, y\in M^{n} \Biggr\} . \end{aligned}$$

We restate the theorem as follows.

Theorem 3.6

Let$n\in \mathbb{N}$, and let$\varphi _{1},\varphi _{2},\dots ,\varphi _{n+1}:M^{i}\to [0,\infty )$for$i\in \{ 1,2,\dots ,n+1\}$be such that$(\varphi _{i+1},\varphi _{i})\in D^{+}_{i}$for$1\leq i\leq n$. If$f:M\to B$satisfies

$$ \bigl\Vert \Delta _{y_{1}}\Delta _{y_{2}}\cdots \Delta _{y_{n+1}}f(0) \bigr\Vert \leq \varphi _{n+1}(y_{1},y_{2}, \dots ,y_{n+1}) $$

for all$y_{1},y_{2},\dots ,y_{n+1}\in M$, then there exists a generalized polynomial$p:M\to B$of degree at mostnsuch that

$$ \bigl\Vert f(x)-p(x) \bigr\Vert \leq \varphi _{1}(x) $$

for all$x\in M$and$p(0)=f(0)$.

If we let

$$ \varphi (x_{1},x_{2},\dots ,x_{n},x_{n+1})= \theta +\sum^{n+1}_{i=1} \varphi _{i} (x_{i}) $$

with $\varphi _{1},\varphi _{2},\dots ,\varphi _{n+1}\in C$, then

$$\begin{aligned} r_{n}(x_{1},x_{2},\dots ,x_{n}) \leq {}& \bigl(2^{n}-1\bigr)\theta + \bigl( \bigl(2^{n}-2 \bigr) \varphi _{1}(x_{1})+2^{n-1}\varphi _{n}(x_{1})+2^{n-1}\varphi _{n+1}(x_{1}) \bigr) \\ &{} +\bigl(2^{n-1}-2\bigr)\varphi _{2}(x_{2})+2^{n-1} \varphi _{n-1}(x_{2})+2^{n-2} \varphi _{n}(x_{2})+2^{n-2}\varphi _{n+1}(x_{2}) \\ &{} \vdots \\ &{} +\varphi _{n+1}(x_{n})+\sum ^{n}_{i=1}2^{n-i}\varphi _{i}(x_{n}). \end{aligned}$$

So

$$\begin{aligned} R_{n}\varphi (y_{1},y_{2},\dots ,y_{n})\leq{}& 2^{n}\theta + \biggl( \frac{2^{n}-2}{2^{n}} \lambda _{n}\varphi _{1}(x_{1})+ \frac{1}{2} \lambda _{n}\varphi _{n}(x_{1})+ \frac{1}{2} \lambda _{n}\varphi _{n+1}(x_{1}) \biggr) \\ &{} +\frac{2^{n-1}-2}{2^{n}} \lambda _{n}\varphi _{2}(x_{2})+ \frac{1}{2} \lambda _{n}\varphi _{n-1}(x_{2})+ \frac{1}{4} \lambda _{n} \varphi _{n}(x_{2})+ \frac{1}{4} \lambda _{n}\varphi _{n+1}(x_{2}) \\ &{} \vdots \\ &{} +\frac{1}{2^{n}} \lambda _{n}\varphi _{n+1}(x_{n})+ \sum^{n}_{i=1} \frac{1}{2^{i}} \lambda _{n}\varphi _{i}(x_{n}). \end{aligned}$$

(8)

Let $\varPsi (y_{1},y_{2},\dots ,y_{n})$ be the right-hand side of (8). Then $(\varphi ,\varPsi )\in D^{+}_{n}$ and Ψ can be used to produce the next pair, resulting in a stability chain. We have the following result.

Theorem 3.7

Let$n\in \mathbb{N}$, $\theta \in [0,\infty )$, $\varphi _{1},\varphi _{2}\in C$, and$f:M\to B$. If

$$ \bigl\Vert \Delta ^{n+1}_{y}f(x) \bigr\Vert \leq \theta +\varphi _{1}(x)+\varphi _{2}(y) $$

for all$x,y\in M$, then there exist a generalized polynomial$p:M\to B$of degree at mostnand$\varphi _{3}\in C$such that

$$ \bigl\Vert f(x)-p(x) \bigr\Vert \leq \bigl(2^{\frac{n(n+1)}{2} }\bigr) \Biggl( n!\prod^{n}_{i=1} \frac{2^{n}}{2^{n}-1} \Biggr) \theta +\varphi _{3}(x) $$

for all$x\in M$.

A direct corollary of this theorem is the Aoki–Rassias stability:

$$ \bigl\Vert \Delta ^{n+1}f(x) \bigr\Vert \leq \theta +c_{1} \bigl\vert x^{p} \bigr\vert +c_{2} \vert y \vert ^{p} $$

for $0< p<1$ when M is either $\mathbb{N} \cup \{ 0\}$ or the set of all integers. In this case, $\varphi _{1}(x)=|x|^{p}$ and

$$ \lambda _{n}\varphi _{1}(x)=\sum ^{\infty }_{k=0} \frac{ \vert 2^{k}x \vert ^{p}}{2^{kn}} =\sum ^{\infty }_{k=0} \frac{ \vert 2^{k}x \vert ^{p}}{2^{kn}} = \vert x \vert ^{p}\sum^{\infty }_{k=0} \frac{1}{2^{k(n-p)}} =\frac{2^{n-p}}{2^{n-p}-1} \vert x \vert ^{p}. $$

Theorem 3.8

Let$n\in \mathbb{N}$, $\theta ,c_{1},c_{2}\in [0,\infty )$, $p\in (0,1)$, and$f:\mathbb{N} \cup \{ 0\} \to B$. If

$$ \bigl\Vert \Delta ^{n+1}_{y}f(x) \bigr\Vert \leq \theta +c_{1} \vert x \vert ^{p}+c_{2} \vert y \vert ^{p} $$

for all$x,y\in \mathbb{N} \cup \{ 0\}$, then there exist$M_{n}\in [0,\infty )$and a polynomial$p:\mathbb{N} \cup \{ 0\} \to B$of degree at mostnsuch that

$$ \bigl\Vert f(x)-p(x) \bigr\Vert \leq \bigl(2^{\frac{n(n+1)}{2} }\bigr) \Biggl( n!\prod^{n}_{i=1} \frac{2^{n}}{2^{n}-1} \Biggr) \theta +M_{n} \vert x \vert ^{p} $$

for all$x\in \mathbb{N} \cup \{ 0\}$.