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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

Δyn+1f(x):=k=0n+1(1)n+1+k(n+1k)f(x+ky)=0

via algebraic manipulations of the forward difference operator.

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 [69]. 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.

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

Δynf(x)=k=0n(1)nk(nk)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

Δbϵ1,ϵ2,,ϵnnf(x+aϵ1,ϵ2,,ϵn)=k=0n(1)nk(nk)f(x+i=1niϵiyi+ki=1n(1ϵi)yi).

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

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

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

2nΔynf(x)=Δ2ynf(x)i=1n(ni)k=0i1Δyn+1f(x+ky)

for\(x,y\in M\).

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

  1. P1

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

  2. 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

αk+1(x,y1,y2)=αk(x,y1,2y2)2n+12ni=1n(ni)k=0i1(2θ+2φ1(x)+φ1(y1)+2kφ1(y2)+2φ2(y2)).

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

Δy1Δy2nf(x)=12nΔy1Δ2y2nf(x)12ni=1n(ni)k=0i1Δy1Δy2n+1f(x+ky2)αk(x,y1,2y2)2n+12ni=1n(ni)k=0i1Δy2n+1f(x+y1+ky2)Δy2n+1f(x+ky2)αk(x,y1,2y2)2n+12ni=1n(ni)k=0i1(2θ+φ1(x+y1+ky2)+φ1(x+ky2)+2φ2(y2))αk(x,y1,2y2)2n+12ni=1n(ni)k=0i1(2θ+2φ1(x)+φ1(y1)+2kφ1(y2)+2φ2(y2))=αk+1(x,y1,y2).

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 \(m\geq 1\),

αm(x,y1,y2)=α0(x,y1,2my2)2mn+12nj=0m1i=1n(ni)k=0i1(2θ+2φ1(x)+φ1(y1)2jn+2kφ1(2jy2)+2φ2(2jy2)2jn)=α0(x,y1,2my2)2mn+12ni=1n(ni)k=0i1j=0m1(2θ+2φ1(x)+φ1(y1)2jn+2kφ1(2jy2)+2φ2(2jy2)2jn).

It follows that

limmam(a,y1,y2)=0+12ni=1n(ni)k=0i1j=0(2θ+2φ1(x)+φ1(y1)2jn+2kφ1(2jy2)+2φ2(2jy2)2jn)=12ni=1n(ni)k=0i1(2n2n1(2θ+2φ1(x)+φ1(y1))+2kλnφ1(y2)+2λnφ2(y2))=n2n12n1(2θ+2φ1(x)+φ1(y1))+n(n1)4λnφ1(y2)+nλnφ2(y2).

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\(x\in \mathbb{N} \cup \{ 0\} \).

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Acknowledgements

The author would like to show his gratitude to the reviewers for their beneficial remarks. They were also generous enough to understand many mistypes the author made, including \([0,1]\), where it should be \([0,\infty )\) at many places.

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Correspondence to Teerapol Sukhonwimolmal.

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Sukhonwimolmal, T. Certain properties of difference operator and stability of Fréchet functional equation. Adv Differ Equ 2020, 108 (2020). https://doi.org/10.1186/s13662-020-02561-9

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MSC

  • 39B52
  • 39B82

Keywords

  • Fréchet functional equation
  • Stability
  • Difference operator