Stability of a functional equation connected with Reynolds operator

Abstract

Let G be a commutative semigroup, $\mathbb{K}=\mathbb{R}$ or ℂ and $F:G\to {\mathbb{K}}^n$. Generalizing the stability of the functional equation $F(x\circ g(y))-F(x)F(y)=0$ with bounded difference (Najdecki in J. Inequal. Appl. 2007:79816, 2007), we prove the stability of the above functional equation with unbounded differences. We also give a more precise description for bounded components of $F=(f_1,f_2,\dots ,f_n)$.

MSC:39B82.

1 Main results

Throughout this paper, $〈G,\circ 〉$ is a commutative semigroup with an identity e, ℝ the set of real numbers, ℂ the set of complex numbers, $\mathbb{K}=\mathbb{R}$ or ℂ, $ϵ\ge 0$, and $g:G\to G$ and $\varphi :G\to [0,\mathrm{\infty })$ are given functions. For $(a_1,a_2,\dots ,a_n),(b_1,b_2,\dots ,b_n)\in {\mathbb{K}}^n$, we define $(a_1,a_2,\dots ,a_n)(b_1,b_2,\dots ,b_n)=(a_1{b}_1,a_2{b}_2,\dots ,a_n{b}_n)$. A function $\sigma :G\to G$ is said to be an involution if $\sigma (x\circ y)=\sigma (x)\circ \sigma (y)$ and $\sigma (\sigma (x))=x$ for all $x,y\in G$. A function $m:G\to {\mathbb{K}}^n$ is called an exponential function provided that $m(x\circ y)=m(x)m(y)$ for all $x,y\in G$.

Generalizing the result of Ger and Šemrl [1], Najdecki [2] proved the stability of the functional equation

$$F(x\circ g(y))-F(x)F(y)=0$$

(1.1)

in the class of functions $F:G\to {\mathbb{K}}^n$. The particular cases of (1.1) are the exponential equation $f(xy)=f(x)f(y)$ (see Aczél and Dhombres [3] and Baker [4]) and the equation

$$f(xf(y))=f(x)f(y)$$

(1.2)

for all $x,y\in \mathbb{K}\setminus \{0\}$, where $f:\mathbb{K}\setminus \{0\}\to \mathbb{K}\setminus \{0\}$ (see Brzdȩk [5], Brzdȩk, Najdecki and Xu [6] and Chudziak and Tabor [7] for related equations). As mentioned in [2, 5], (1.2) arises in averaging theory applied to the turbulent fluid motion and is connected with the Reynolds operator (see Marias [8]), the averaging operator and the multiplicatively symmetric operator (see [3]). Moreover, the equation (1.2) is connected with a description of some associative operations, i.e., the binary operation $\circ :(\mathbb{K}\setminus \{0\})\times (\mathbb{K}\setminus \{0\})\to \mathbb{K}\setminus \{0\}$ defined by $x\circ f(y)=xf(y)$ is associative if and only if f satisfies (1.2) (see [5] for more details). We also refer the reader to Belluot, Brzdȩk and Ciepliński [9] and Brzdȩk and Ciepliński [10] for some recent developments on the issues of stability and superstability for functional equations.

The main result of Najdecki [2] is the following.

Theorem 1.1 Let $F:G\to {\mathbb{K}}^n$, $F=(f_1,f_2,\dots ,f_n)$ satisfy

$$\parallel F(x\circ g(y))-F(x)F(y)\parallel \le ϵ$$

(1.3)

for all $x,y\in G$ with any norm $\parallel \cdot \parallel$ in ${\mathbb{K}}^n$. Then there exist ideals $I,J\subset {\mathbb{K}}^n$ such that ${\mathbb{K}}^n=I\oplus J$, PF is bounded and QF satisfies (1.1), where $P:{\mathbb{K}}^n\to I$, $Q:{\mathbb{K}}^n\to J$ are the natural projections.

In this paper, generalizing the above result we consider the functional inequalities

$$\parallel F(x\circ g(y))-F(x)F(y)\parallel \le \varphi (y),$$

(1.4)

$$\parallel F(x\circ g(y))-F(x)F(y)\parallel \le \varphi (x)$$

(1.5)

for all $x,y\in G$ with any norm $\parallel \cdot \parallel$ in ${\mathbb{K}}^n$ (see [6] for related results).

Throughout this paper we denote

$$\begin{array}{r}L=\{j:f_j\text{ is bounded},j=1,2,\dots ,n\},\\ K=\{j:f_j\text{ is unbounded},j=1,2,\dots ,n\},\end{array}$$

where $F=(f_1,f_2,\dots ,f_n)$.

Theorem 1.2 Let $F:G\to {\mathbb{K}}^n$, $F=(f_1,f_2,\dots ,f_n)$ satisfy (1.4) for all $x,y\in G$ with any norm $\parallel \cdot \parallel$ in ${\mathbb{K}}^n$. Assume that one of the following two conditions is fulfilled.

• (i) g is an involution,

• (ii) for each $j\in K$, there exists a sequence $x_n$, $n=1,2,3,\dots$ (possibly depending on j) such that

  $$\frac{|f_j(x_n)|}{1+\varphi (x_n)}\to \mathrm{\infty }\mathit{\text{as }}n\to \mathrm{\infty }.$$

  (1.6)

Then there exist ideals $I,J\subset {\mathbb{K}}^n$ such that ${\mathbb{K}}^n=I\oplus J$, PF is bounded and QF satisfies (1.1), where $P:{\mathbb{K}}^n\to I$, $Q:{\mathbb{K}}^n\to J$ are the natural projections. Moreover, $Q(F\circ g^{-1})$ is exponential provided g is bijective.

Remark The case (ii) of Theorem 1.2 includes Theorem 1.1.

Theorem 1.3 Let $F:G\to {\mathbb{K}}^n$, $F=(f_1,f_2,\dots ,f_n)$ satisfy (1.5) for all $x,y\in G$ with any norm $\parallel \cdot \parallel$ in ${\mathbb{K}}^n$. Assume that g is an involution. Then there exist ideals $I,J\subset {\mathbb{K}}^n$ such that ${\mathbb{K}}^n=I\oplus J$, PF is bounded, QF satisfies (1.1), where $P:{\mathbb{K}}^n\to I$, $Q:{\mathbb{K}}^n\to J$ are the natural projections.

If we replace $\parallel \cdot \parallel$ by the usual norm ${\parallel \cdot \parallel }_u$ on ${\mathbb{K}}^n$ defined by

$${\parallel (a_1,a_2,\dots ,a_n)\parallel }_u=\sqrt{{|a_1|}^2+{|a_2|}^2+\cdots +{|a_n|}^2},$$

we can estimate PF (in Theorem 1.2 and Theorem 1.3) as follows.

Theorem 1.4 The following two statements are valid.

• (a) If $F:G\to {\mathbb{K}}^n$, $F=(f_1,f_2,\dots ,f_n)$ satisfies (1.4), then PF satisfies

  $${\parallel PF(y)\parallel }_u\le \frac{\sqrt{|L|}}{2}(1+\sqrt{1+4\varphi (y)})$$

  (1.7)

for all $y\in G$, where $|L|$ denotes the number of the elements of L. In particular, if $|L|=1$ and G is a group, then PF satisfies either

$$\frac{1}{2}(1+\sqrt{1-4\varphi (y)})\le {\parallel PF(y)\parallel }_u\le \frac{1}{2}(1+\sqrt{1+4\varphi (y)})$$

(1.8)

for all $y\in B:=\{y\in G:\varphi (y)<\frac{1}{4}\}$, or

$${\parallel PF(y)\parallel }_u\le \frac{1}{2}(1-\sqrt{1-4\varphi (y)})$$

(1.9)

for all $y\in B$.

• (b) If $F:G\to {\mathbb{K}}^n$, $F=(f_1,f_2,\dots ,f_n)$ satisfies (1.5), then PF satisfies (1.7). In particular if G is a group, g is surjective and $|L|=1$, then PF satisfies (1.8) or (1.9).

2 Proofs

Let $g:G\to G$ and $\varphi :G\to [0,\mathrm{\infty })$ be given. We first consider the stability of the functional equation

$$f(x\circ g(y))-f(x)f(y)=0$$

(2.1)

in the class of functions $f:G\to \mathbb{K}$, i.e., we investigate both bounded and unbounded functions $f:G\to \mathbb{K}$ satisfying the functional inequalities

$$|f(x\circ g(y))-f(x)f(y)|\le \varphi (y),$$

(2.2)

$$|f(x\circ g(y))-f(x)f(y)|\le \varphi (x)$$

(2.3)

for all $x,y\in G$.

Lemma 2.1 Assume that $g=\sigma$ is an involution and $f:G\to \mathbb{K}$ is an unbounded function satisfying the inequality (2.2). Then f is exponential and satisfies (2.1). In particular, if G is 2-divisible, then f has the form

$$f(x)=m\left(\frac{x\circ \sigma (x)}{2}\right)$$

(2.4)

for all $x\in G$, where $m:G\to \mathbb{K}$ is an exponential function.

Proof Choose a sequence $x_n\in G$, $n=1,2,3,\dots$ , such that $|f(x_n)|\to \mathrm{\infty }$ as $n\to \mathrm{\infty }$. Putting $x=x_n$, $n=1,2,3,\dots$ , in (2.2), dividing the result by $|f(x_n)|$ and letting $n\to \mathrm{\infty }$ we have

$$f(y)=\underset{n\to \mathrm{\infty }}{lim}\frac{f(x_n\circ \sigma (y))}{f(x_n)}$$

(2.5)

for all $y\in G$. Multiplying both sides of (2.5) by $f(x)$ and using (2.2) and (2.5) we have

$$\begin{array}{rl}f(y)f(x)& =\underset{n\to \mathrm{\infty }}{lim}\frac{f(x_n\circ \sigma (y))f(x)}{f(x_n)}=\underset{n\to \mathrm{\infty }}{lim}\frac{f(x_n\circ \sigma (y)\circ \sigma (x))}{f(x_n)}\\ =\underset{n\to \mathrm{\infty }}{lim}\frac{f(x_n\circ \sigma (y\circ x))}{f(x_n)}=f(y\circ x)\end{array}$$

(2.6)

for all $x,y\in G$. Thus, f is an exponential function, say $f=m$. From (2.2) and (2.6) we have

$$|f(x)||f(\sigma (y))-f(y)|\le \varphi (y)$$

(2.7)

for all $x,y\in G$. Since f is unbounded, from (2.7) we have

$$f(\sigma (y))=f(y)$$

(2.8)

for all $y\in G$. Replacing y by $\sigma (y)$ in (2.6) and using (2.8) we get the equation (2.1). In particular, if G is 2-divisible, then we can write

$$\begin{array}{rl}f(x)& =f(\frac{x}{2}\circ \frac{x}{2})=f(\frac{x}{2}\circ \sigma \left(\frac{x}{2}\right))\\ =f(\frac{x}{2}\circ \frac{\sigma (x)}{2})=m\left(\frac{x\circ \sigma (x)}{2}\right)\end{array}$$

(2.9)

for all $x\in G$. This completes the proof. □

Lemma 2.2 Let $f:G\to \mathbb{K}$ be an unbounded function satisfying (2.2). Assume that there exists a sequence $x_n$, $n=1,2,3,\dots$ , satisfying

$$\underset{n\to \mathrm{\infty }}{lim}\frac{|f(x_n)|}{1+\varphi (x_n)}=\mathrm{\infty }.$$

(2.10)

Then f satisfies (2.1).

Proof Note that (2.10) implies

$$\underset{n\to \mathrm{\infty }}{lim}\frac{1}{|f(x_n)|}=0\text{and}\underset{n\to \mathrm{\infty }}{lim}\frac{\varphi (x_n)}{|f(x_n)|}=0.$$

Putting $y=x_n$, $n=1,2,3,\dots$ , in (2.2) and dividing the result by $|f(x_n)|$ we have

$$|f(x)-\frac{f(x\circ g(x_n))}{f(x_n)}|\le \frac{\varphi (x_n)}{|f(x_n)|}$$

(2.11)

for all $x\in G$, $n=1,2,3,\dots$ . Letting $n\to \mathrm{\infty }$ in (2.11) we have

$$f(x)=\underset{n\to \mathrm{\infty }}{lim}\frac{f(x\circ g(x_n))}{f(x_n)}$$

(2.12)

for all $x\in G$. Multiplying both sides of (2.12) by $f(y)$ and using (2.2) and (2.12) we have

$$\begin{array}{rl}f(x)f(y)& =\underset{n\to \mathrm{\infty }}{lim}\frac{f(x\circ g(x_n))f(y)}{f(x_n)}=\underset{n\to \mathrm{\infty }}{lim}\frac{f(x\circ g(x_n)\circ g(y))}{f(x_n)}\\ =\underset{n\to \mathrm{\infty }}{lim}\frac{f(x\circ g(y)\circ g(x_n))}{f(x_n)}=f(x\circ g(y))\end{array}$$

(2.13)

for all $x,y\in G$. This completes the proof. □

Lemma 2.3 Assume that g is bijective and $f:G\to \mathbb{K}$ is an unbounded function satisfying the inequality (2.2). Then $f\circ g^{-1}$ is an exponential function.

Proof Choose a sequence $x_n\in G$, $n=1,2,3,\dots$ , such that $|f(x_n)|\to \mathrm{\infty }$ as $n\to \mathrm{\infty }$. Putting $x=x_n$, $n=1,2,3,\dots$ , in (2.2), dividing the result by $|f(x_n)|$, replacing y by $g^{-1}(y)$ and letting $n\to \mathrm{\infty }$ we have

$$f(g^{-1}(y))=\underset{n\to \mathrm{\infty }}{lim}\frac{f(x_n\circ y)}{f(x_n)}$$

(2.14)

for all $y\in G$. Multiplying both sides of (2.14) by $f(g^{-1}(x))$ and using (2.2) and (2.14) we have

$$\begin{array}{rl}f(g^{-1}(y))f(g^{-1}(x))& =\underset{n\to \mathrm{\infty }}{lim}\frac{f(x_n\circ y)f(g^{-1}(x))}{f(x_n)}\\ =\underset{n\to \mathrm{\infty }}{lim}\frac{f(x_n\circ y\circ x)}{f(x_n)}=f(g^{-1}(y\circ x))\end{array}$$

(2.15)

for all $x,y\in G$. Thus, $f\circ g^{-1}$ is an exponential function. This completes the proof. □

Proof of Theorem 1.2 Since every two norms in $K^n$ are equivalent, from (1.4) there exists $\alpha >0$ such that

$$\begin{array}{rl}|f_j(x\circ g(y))-f_j(x)f_j(y)|& \le {\parallel F(x\circ g(y))-F(x)F(y)\parallel }_u\\ \le \alpha \parallel F(x\circ g(y))-F(x)F(y)\parallel \le \alpha \varphi (y)\end{array}$$

(2.16)

for all $x,y\in G$ and all $j\in \{1,2,\dots ,n\}$. For the case (i), by Lemma 2.1, $f_j$ satisfies (2.1) for all $j\in K$. For the case (ii), by Lemma 2.2, $f_j$ satisfies (2.1) for all $j\in K$. Let $I=\{(a_1,a_2,\dots ,a_n):a_i=0\text{ for }i\in K\}$, $J=\{(a_1,a_2,\dots ,a_n):a_i=0\text{ for }i\in L\}$. Then it follows that ${\mathbb{K}}^n=I\oplus J$, PF is bounded and QF satisfies (1.1). If g is bijective, then by Lemma 2.3, $f_j\circ g^{-1}$ are exponential function for all $j\in K$, which implies $Q(F\circ g^{-1})$ is an exponential function. This completes the proof. □

Lemma 2.4 Assume that $g=\sigma$ is an involution and $f:G\to \mathbb{K}$ is an unbounded function satisfying the inequality (2.3). Then f satisfies (2.1). In particular, if G is 2-divisible, then f has the form

$$f(x)=m\left(\frac{x\circ \sigma (x)}{2}\right)$$

(2.17)

for all $x\in G$, where $m:G\to \mathbb{K}$ is an exponential function.

Proof Choose a sequence $y_n\in G$, $n=1,2,3,\dots$ , such that $|f(y_n)|\to \mathrm{\infty }$ as $n\to \mathrm{\infty }$. Putting $y=y_n$, $n=1,2,3,\dots$ , in (2.3), dividing the result by $|f(y_n)|$ and letting $n\to \mathrm{\infty }$ we have

$$f(x)=\underset{n\to \mathrm{\infty }}{lim}\frac{f(x\circ \sigma (y_n))}{f(y_n)}.$$

(2.18)

Putting $x=e$ in (2.3) and replacing y by $\sigma (y)$ in the result we have

$$|f(y)-f(e)f(\sigma (y))|\le \varphi (e)$$

(2.19)

for all $x,y\in G$. Multiplying both sides of (2.18) by $f(y)$ and using (2.3), (2.18), and (2.19) we have

$$\begin{array}{rl}f(y)f(x)& =\underset{n\to \mathrm{\infty }}{lim}\frac{f(y)f(x\circ \sigma (y_n))}{f(y_n)}=\underset{n\to \mathrm{\infty }}{lim}\frac{f(y\circ \sigma (x\circ \sigma (y_n)))}{f(y_n)}\\ =\underset{n\to \mathrm{\infty }}{lim}\frac{f(e)f(\sigma (y)\circ x\circ \sigma (y_n))}{f(y_n)}=f(e)f(\sigma (y)\circ x)\end{array}$$

(2.20)

for all $x,y\in G$. Putting $x=e$ in (2.20) we have

$$f(y)=f(\sigma (y))$$

(2.21)

for all $y\in G$. From (2.19) and (2.21) we have

$$|f(y)||1-f(e)|\le \varphi (e)$$

(2.22)

for all $y\in G$. Since f is unbounded, from (2.22) we have $f(e)=1$. Thus, f satisfies (2.1). This completes the proof. □

Proof of Theorem 1.3 From (1.5), as in (2.16) there exists $\alpha >0$ such that

$$|f_j(x\circ g(y))-f_j(x)f_j(y)|\le \alpha \varphi (x)$$

(2.23)

for all $x,y\in G$, $j\in \{1,2,\dots ,n\}$. Applying Lemma 2.4 to (2.23) for each $j\in K$ we find that $f_j$ satisfies (2.1) for all $j\in K$, which implies that QF satisfies (1.1). This completes the proof. □

Now, we investigate bounded functions satisfying each of (2.2) and (2.3) (see [4, 11–13] for bounded solutions of an exponential functional equation).

Lemma 2.5 Let $f:G\to \mathbb{K}$ be a bounded function satisfying (2.2). Then f satisfies

$$|f(y)|\le \frac{1}{2}(1+\sqrt{1+4\varphi (y)})$$

(2.24)

for all $y\in G$. In particular, G is a group and let $B=\{y\in G:\varphi (y)<\frac{1}{4}\}$, then f satisfies either

$$\frac{1}{2}(1+\sqrt{1-4\varphi (y)})\le |f(y)|\le \frac{1}{2}(1+\sqrt{1+4\varphi (y)})$$

(2.25)

for all $y\in B$, or

$$|f(y)|\le \frac{1}{2}(1-\sqrt{1-4\varphi (y)})$$

(2.26)

for all $y\in B$.

Proof Let $M_f={sup}_{x\in G}|f(x)|$. Using the triangle inequality with (2.2) we have

$$|f(x)f(y)|\le \left|f(x\circ g(y))\right|+\varphi (y)\le M_f+\varphi (y)$$

(2.27)

for all $x,y\in G$. Taking the supremum of the left hand side of (2.27) with respect to $x\in G$ we get $M_f|f(y)|\le M_f+\varphi (y)$ for all $y\in G$. Thus, we have

$$M_f(|f(y)|-1)\le \varphi (y)$$

(2.28)

for all $y\in G$. From (2.28) we have

$$|f(y)|(|f(y)|-1)\le \varphi (y)$$

(2.29)

for all $y\in G$. Solving the inequality (2.29) we get (2.24). Now, we assume that G is a group. Replacing x by $x\circ g{(y)}^{-1}$ in (2.2) and using the triangle inequality we have

$$|f(x)|\le |f(x\circ g{(y)}^{-1})f(y)|+\varphi (y)\le M_f|f(y)|+\varphi (y)$$

(2.30)

for all $x,y\in G$. Taking the supremum of the left hand side of (2.30) with respect to $x\in G$ we get $M_f\le M_f|f(y)|+\varphi (y)$ for all $y\in G$. Thus, we have

$$M_f(1-|f(y)|)\le \varphi (y)$$

(2.31)

for all $y\in G$. From (2.28) and (2.31) we have

$$|f(y)||1-|f(y)||\le M_f|1-|f(y)||\le \varphi (y)$$

(2.32)

for all $y\in G$. For each fixed $y\in B$, solving the inequality (2.32) we get

$$\frac{1}{2}(1+\sqrt{1-4\varphi (y)})\le |f(y)|\le \frac{1}{2}(1+\sqrt{1+4\varphi (y)}),$$

(2.33)

or

$$|f(y)|\le \frac{1}{2}(1-\sqrt{1-4\varphi (y)}).$$

(2.34)

Now, assume that there exist a bounded function f and $y_1,y_2\in B$ such that

$$|f(y_1)|\le \frac{1}{2}(1-\sqrt{1-4\varphi (y_1)}),|f(y_2)|\ge \frac{1}{2}(1+\sqrt{1-4\varphi (y_2)}).$$

(2.35)

Then from (2.31) we have

$$|f(y_2)|(1-|f(y_1)|)\le M_f(1-|f(y_1)|)\le \varphi (y_1).$$

(2.36)

On the other hand, from (2.35) we have

$$\begin{array}{rl}|f(y_2)|(1-|f(y_1)|)& \ge \frac{1}{2}(1+\sqrt{1-4\varphi (y_2)})(1-\frac{1}{2}(1-\sqrt{1-4\varphi (y_1)}))\\ >\frac{1}{2}(1-\sqrt{1-4\varphi (y_1)})(1-\frac{1}{2}(1-\sqrt{1-4\varphi (y_1)}))=\varphi (y_1),\end{array}$$

which contradicts (2.36). Thus, f satisfies (2.25) for all $y\in B$, or it satisfies (2.26) for all $y\in B$. This completes the proof. □

Lemma 2.6 Let $f:G\to \mathbb{K}$ be a bounded function satisfying (2.3). Then f satisfies (2.24) for all $y\in G$. In particular, if G is a group and g is surjective, then f satisfies (2.25) for all $y\in B:=\{y\in G:\varphi (y)<\frac{1}{4}\}$, or satisfies (2.26) for all $y\in B$.

Proof Using the triangle inequality with (2.3) we have

$$|f(x)f(y)|\le \left|f(x\circ g(y))\right|+\varphi (x)\le M_f+\varphi (x)$$

(2.37)

for all $x,y\in G$. Taking the supremum of the left hand side of (2.37) with respect to $y\in G$ we get $M_f|f(x)|\le M_f+\varphi (x)$ for all $x\in G$. Thus, we have

$$M_f(|f(x)|-1)\le \varphi (x)$$

(2.38)

for all $x\in G$. From (2.38) we get (2.24) as in the proof of Lemma 2.5. We assume that G is a group. For given $x,z\in G$, choosing $w\in G$ such that $g(w)=x^{-1}\circ z$, putting $y=w$ in (2.3) and using the triangle inequality we have

$$|f(z)|\le |f(x)f(w)|+\varphi (x)\le |f(x)|M_f+\varphi (x)$$

(2.39)

for all $x,z\in G$. Taking the supremum of the left hand side of (2.39) we get $M_f\le M_f|f(x)|+\varphi (x)$ for all $x\in G$. Thus, we have

$$M_f(1-|f(x)|)\le \varphi (x)$$

(2.40)

for all $x\in G$. Now, the remaining parts of the proof are the same as those of Lemma 2.5. □

Proof of Theorem 1.4 From Lemma 2.5 and Lemma 2.6, for each $j\in L$ we have

$$|f_j(y)|\le \frac{1}{2}(1+\sqrt{1+4\varphi (y)})$$

(2.41)

for all $y\in G$. Thus, from (2.41) we have

$${\parallel PF(y)\parallel }_u=\sqrt{\sum _{j\in L}{|f_j(y)|}^2}\le \frac{\sqrt{|L|}}{2}(1+\sqrt{1+4\varphi (y)})$$

for all $y\in G$, which gives (1.7). Now, if $|L|=1$, say $L=\{j_0\}$ we have

$${\parallel PF(y)\parallel }_u=|f_{j_0}(y)|$$

for all $y\in G$. Thus, the inequalities (1.8) and (1.9) follow immediately from (2.25) and (2.26). This completes the proof. □

Author’s contributions

The author is the only person who is responsible to this work.