# Stability of a functional equation connected with Reynolds operator

## Abstract

Let G be a commutative semigroup, $K=R$ or and $F:G→ K n$. Generalizing the stability of the functional equation $F(x∘g(y))−F(x)F(y)=0$ 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 $F=( f 1 , f 2 ,…, f n )$.

MSC:39B82.

## 1 Main results

Throughout this paper, $〈G,∘〉$ is a commutative semigroup with an identity e, the set of real numbers, the set of complex numbers, $K=R$ or , $ϵ≥0$, and $g:G→G$ and $ϕ:G→[0,∞)$ are given functions. For $( a 1 , a 2 ,…, a n ),( b 1 , b 2 ,…, b n )∈ K n$, we define $( a 1 , a 2 ,…, a n )( b 1 , b 2 ,…, b n )=( a 1 b 1 , a 2 b 2 ,…, a n b n )$. A function $σ:G→G$ is said to be an involution if $σ(x∘y)=σ(x)∘σ(y)$ and $σ(σ(x))=x$ for all $x,y∈G$. A function $m:G→ K n$ is called an exponential function provided that $m(x∘y)=m(x)m(y)$ for all $x,y∈G$.

Generalizing the result of Ger and Šemrl , Najdecki  proved the stability of the functional equation

$F ( x ∘ g ( y ) ) −F(x)F(y)=0$
(1.1)

in the class of functions $F:G→ K n$. The particular cases of (1.1) are the exponential equation $f(xy)=f(x)f(y)$ (see Aczél and Dhombres  and Baker ) and the equation

$f ( x f ( y ) ) =f(x)f(y)$
(1.2)

for all $x,y∈K∖{0}$, where $f:K∖{0}→K∖{0}$ (see Brzdȩk , Brzdȩk, Najdecki and Xu  and Chudziak and Tabor  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 ), the averaging operator and the multiplicatively symmetric operator (see ). Moreover, the equation (1.2) is connected with a description of some associative operations, i.e., the binary operation $∘:(K∖{0})×(K∖{0})→K∖{0}$ defined by $x∘f(y)=xf(y)$ is associative if and only if f satisfies (1.2) (see  for more details). We also refer the reader to Belluot, Brzdȩk and Ciepliński  and Brzdȩk and Ciepliński  for some recent developments on the issues of stability and superstability for functional equations.

The main result of Najdecki  is the following.

Theorem 1.1 Let $F:G→ K n$, $F=( f 1 , f 2 ,…, f n )$ satisfy

$∥ F ( x ∘ g ( y ) ) − F ( x ) F ( y ) ∥ ≤ϵ$
(1.3)

for all $x,y∈G$ with any norm $∥⋅∥$ in $K n$. Then there exist ideals $I,J⊂ K n$ such that $K n =I⊕J$, PF is bounded and QF satisfies (1.1), where $P: K n →I$, $Q: K n →J$ are the natural projections.

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

$∥ F ( x ∘ g ( y ) ) − F ( x ) F ( y ) ∥ ≤ϕ(y),$
(1.4)
$∥ F ( x ∘ g ( y ) ) − F ( x ) F ( y ) ∥ ≤ϕ(x)$
(1.5)

for all $x,y∈G$ with any norm $∥⋅∥$ in $K n$ (see  for related results).

Throughout this paper we denote

where $F=( f 1 , f 2 ,…, f n )$.

Theorem 1.2 Let $F:G→ K n$, $F=( f 1 , f 2 ,…, f n )$ satisfy (1.4) for all $x,y∈G$ with any norm $∥⋅∥$ in $K n$. Assume that one of the following two conditions is fulfilled.

1. (i)

g is an involution,

2. (ii)

for each $j∈K$, there exists a sequence $x n$, $n=1,2,3,…$ (possibly depending on j) such that

(1.6)

Then there exist ideals $I,J⊂ K n$ such that $K n =I⊕J$, PF is bounded and QF satisfies (1.1), where $P: K n →I$, $Q: K n →J$ are the natural projections. Moreover, $Q(F∘ g − 1 )$ is exponential provided g is bijective.

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

Theorem 1.3 Let $F:G→ K n$, $F=( f 1 , f 2 ,…, f n )$ satisfy (1.5) for all $x,y∈G$ with any norm $∥⋅∥$ in $K n$. Assume that g is an involution. Then there exist ideals $I,J⊂ K n$ such that $K n =I⊕J$, PF is bounded, QF satisfies (1.1), where $P: K n →I$, $Q: K n →J$ are the natural projections.

If we replace $∥⋅∥$ by the usual norm $∥ ⋅ ∥ u$ on $K n$ defined by

$∥ ( a 1 , a 2 , … , a n ) ∥ u = | a 1 | 2 + | a 2 | 2 + ⋯ + | a n | 2 ,$

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

Theorem 1.4 The following two statements are valid.

1. (a)

If $F:G→ K n$, $F=( f 1 , f 2 ,…, f n )$ satisfies (1.4), then PF satisfies

$∥ P F ( y ) ∥ u ≤ | L | 2 ( 1 + 1 + 4 ϕ ( y ) )$
(1.7)

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

$1 2 ( 1 + 1 − 4 ϕ ( y ) ) ≤ ∥ P F ( y ) ∥ u ≤ 1 2 ( 1 + 1 + 4 ϕ ( y ) )$
(1.8)

for all $y∈B:={y∈G:ϕ(y)< 1 4 }$, or

$∥ P F ( y ) ∥ u ≤ 1 2 ( 1 − 1 − 4 ϕ ( y ) )$
(1.9)

for all $y∈B$.

1. (b)

If $F:G→ K n$, $F=( f 1 , f 2 ,…, f n )$ 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 $g:G→G$ and $ϕ:G→[0,∞)$ be given. We first consider the stability of the functional equation

$f ( x ∘ g ( y ) ) −f(x)f(y)=0$
(2.1)

in the class of functions $f:G→K$, i.e., we investigate both bounded and unbounded functions $f:G→K$ satisfying the functional inequalities

$|f ( x ∘ g ( y ) ) −f(x)f(y)|≤ϕ(y),$
(2.2)
$| f ( x ∘ g ( y ) ) − f ( x ) f ( y ) | ≤ϕ(x)$
(2.3)

for all $x,y∈G$.

Lemma 2.1 Assume that $g=σ$ is an involution and $f:G→K$ 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

$f(x)=m ( x ∘ σ ( x ) 2 )$
(2.4)

for all $x∈G$, where $m:G→K$ is an exponential function.

Proof Choose a sequence $x n ∈G$, $n=1,2,3,…$ , such that $|f( x n )|→∞$ as $n→∞$. Putting $x= x n$, $n=1,2,3,…$ , in (2.2), dividing the result by $|f( x n )|$ and letting $n→∞$ we have

$f(y)= lim n → ∞ f ( x n ∘ σ ( y ) ) f ( x n )$
(2.5)

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

$f ( y ) f ( x ) = lim n → ∞ f ( x n ∘ σ ( y ) ) f ( x ) f ( x n ) = lim n → ∞ f ( x n ∘ σ ( y ) ∘ σ ( x ) ) f ( x n ) = lim n → ∞ f ( x n ∘ σ ( y ∘ x ) ) f ( x n ) = f ( y ∘ x )$
(2.6)

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

$| f ( x ) | | f ( σ ( y ) ) − f ( y ) | ≤ϕ(y)$
(2.7)

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

$f ( σ ( y ) ) =f(y)$
(2.8)

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

$f ( x ) = f ( x 2 ∘ x 2 ) = f ( x 2 ∘ σ ( x 2 ) ) = f ( x 2 ∘ σ ( x ) 2 ) = m ( x ∘ σ ( x ) 2 )$
(2.9)

for all $x∈G$. This completes the proof. □

Lemma 2.2 Let $f:G→K$ be an unbounded function satisfying (2.2). Assume that there exists a sequence $x n$, $n=1,2,3,…$ , satisfying

$lim n → ∞ | f ( x n ) | 1 + ϕ ( x n ) =∞.$
(2.10)

Then f satisfies (2.1).

Proof Note that (2.10) implies

$lim n → ∞ 1 | f ( x n ) | =0and lim n → ∞ ϕ ( x n ) | f ( x n ) | =0.$

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

$| f ( x ) − f ( x ∘ g ( x n ) ) f ( x n ) | ≤ ϕ ( x n ) | f ( x n ) |$
(2.11)

for all $x∈G$, $n=1,2,3,…$ . Letting $n→∞$ in (2.11) we have

$f(x)= lim n → ∞ f ( x ∘ g ( x n ) ) f ( x n )$
(2.12)

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

$f ( x ) f ( y ) = lim n → ∞ f ( x ∘ g ( x n ) ) f ( y ) f ( x n ) = lim n → ∞ f ( x ∘ g ( x n ) ∘ g ( y ) ) f ( x n ) = lim n → ∞ f ( x ∘ g ( y ) ∘ g ( x n ) ) f ( x n ) = f ( x ∘ g ( y ) )$
(2.13)

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

Lemma 2.3 Assume that g is bijective and $f:G→K$ is an unbounded function satisfying the inequality (2.2). Then $f∘ g − 1$ is an exponential function.

Proof Choose a sequence $x n ∈G$, $n=1,2,3,…$ , such that $|f( x n )|→∞$ as $n→∞$. Putting $x= x n$, $n=1,2,3,…$ , in (2.2), dividing the result by $|f( x n )|$, replacing y by $g − 1 (y)$ and letting $n→∞$ we have

$f ( g − 1 ( y ) ) = lim n → ∞ f ( x n ∘ y ) f ( x n )$
(2.14)

for all $y∈G$. Multiplying both sides of (2.14) by $f( g − 1 (x))$ and using (2.2) and (2.14) we have

$f ( g − 1 ( y ) ) f ( g − 1 ( x ) ) = lim n → ∞ f ( x n ∘ y ) f ( g − 1 ( x ) ) f ( x n ) = lim n → ∞ f ( x n ∘ y ∘ x ) f ( x n ) = f ( g − 1 ( y ∘ x ) )$
(2.15)

for all $x,y∈G$. Thus, $f∘ g − 1$ 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 $α>0$ such that

$| f j ( x ∘ g ( y ) ) − f j ( x ) f j ( y ) | ≤ ∥ F ( x ∘ g ( y ) ) − F ( x ) F ( y ) ∥ u ≤ α ∥ F ( x ∘ g ( y ) ) − F ( x ) F ( y ) ∥ ≤ α ϕ ( y )$
(2.16)

for all $x,y∈G$ and all $j∈{1,2,…,n}$. For the case (i), by Lemma 2.1, $f j$ satisfies (2.1) for all $j∈K$. For the case (ii), by Lemma 2.2, $f j$ satisfies (2.1) for all $j∈K$. Let , . Then it follows that $K n =I⊕J$, PF is bounded and QF satisfies (1.1). If g is bijective, then by Lemma 2.3, $f j ∘ g − 1$ are exponential function for all $j∈K$, which implies $Q(F∘ g − 1 )$ is an exponential function. This completes the proof. □

Lemma 2.4 Assume that $g=σ$ is an involution and $f:G→K$ 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

$f(x)=m ( x ∘ σ ( x ) 2 )$
(2.17)

for all $x∈G$, where $m:G→K$ is an exponential function.

Proof Choose a sequence $y n ∈G$, $n=1,2,3,…$ , such that $|f( y n )|→∞$ as $n→∞$. Putting $y= y n$, $n=1,2,3,…$ , in (2.3), dividing the result by $|f( y n )|$ and letting $n→∞$ we have

$f(x)= lim n → ∞ f ( x ∘ σ ( y n ) ) f ( y n ) .$
(2.18)

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

$| f ( y ) − f ( e ) f ( σ ( y ) ) | ≤ϕ(e)$
(2.19)

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

$f ( y ) f ( x ) = lim n → ∞ f ( y ) f ( x ∘ σ ( y n ) ) f ( y n ) = lim n → ∞ f ( y ∘ σ ( x ∘ σ ( y n ) ) ) f ( y n ) = lim n → ∞ f ( e ) f ( σ ( y ) ∘ x ∘ σ ( y n ) ) f ( y n ) = f ( e ) f ( σ ( y ) ∘ x )$
(2.20)

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

$f(y)=f ( σ ( y ) )$
(2.21)

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

$| f ( y ) | | 1 − f ( e ) | ≤ϕ(e)$
(2.22)

for all $y∈G$. 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 $α>0$ such that

$| f j ( x ∘ g ( y ) ) − f j ( x ) f j ( y ) | ≤αϕ(x)$
(2.23)

for all $x,y∈G$, $j∈{1,2,…,n}$. Applying Lemma 2.4 to (2.23) for each $j∈K$ we find that $f j$ satisfies (2.1) for all $j∈K$, 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, 1113] for bounded solutions of an exponential functional equation).

Lemma 2.5 Let $f:G→K$ be a bounded function satisfying (2.2). Then f satisfies

$| f ( y ) | ≤ 1 2 ( 1 + 1 + 4 ϕ ( y ) )$
(2.24)

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

$1 2 ( 1 + 1 − 4 ϕ ( y ) ) ≤ | f ( y ) | ≤ 1 2 ( 1 + 1 + 4 ϕ ( y ) )$
(2.25)

for all $y∈B$, or

$| f ( y ) | ≤ 1 2 ( 1 − 1 − 4 ϕ ( y ) )$
(2.26)

for all $y∈B$.

Proof Let $M f = sup x ∈ G |f(x)|$. Using the triangle inequality with (2.2) we have

$| f ( x ) f ( y ) | ≤ | f ( x ∘ g ( y ) ) | +ϕ(y)≤ M f +ϕ(y)$
(2.27)

for all $x,y∈G$. Taking the supremum of the left hand side of (2.27) with respect to $x∈G$ we get $M f |f(y)|≤ M f +ϕ(y)$ for all $y∈G$. Thus, we have

$M f ( | f ( y ) | − 1 ) ≤ϕ(y)$
(2.28)

for all $y∈G$. From (2.28) we have

$| f ( y ) | ( | f ( y ) | − 1 ) ≤ϕ(y)$
(2.29)

for all $y∈G$. Solving the inequality (2.29) we get (2.24). Now, we assume that G is a group. Replacing x by $x∘g ( y ) − 1$ in (2.2) and using the triangle inequality we have

$| f ( x ) | ≤ | f ( x ∘ g ( y ) − 1 ) f ( y ) | +ϕ(y)≤ M f | f ( y ) | +ϕ(y)$
(2.30)

for all $x,y∈G$. Taking the supremum of the left hand side of (2.30) with respect to $x∈G$ we get $M f ≤ M f |f(y)|+ϕ(y)$ for all $y∈G$. Thus, we have

$M f ( 1 − | f ( y ) | ) ≤ϕ(y)$
(2.31)

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

$| f ( y ) | | 1 − | f ( y ) | | ≤ M f | 1 − | f ( y ) | | ≤ϕ(y)$
(2.32)

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

$1 2 ( 1 + 1 − 4 ϕ ( y ) ) ≤ | f ( y ) | ≤ 1 2 ( 1 + 1 + 4 ϕ ( y ) ) ,$
(2.33)

or

$| f ( y ) | ≤ 1 2 ( 1 − 1 − 4 ϕ ( y ) ) .$
(2.34)

Now, assume that there exist a bounded function f and $y 1 , y 2 ∈B$ such that

$| f ( y 1 ) | ≤ 1 2 ( 1 − 1 − 4 ϕ ( y 1 ) ) , | f ( y 2 ) | ≥ 1 2 ( 1 + 1 − 4 ϕ ( y 2 ) ) .$
(2.35)

Then from (2.31) we have

$| f ( y 2 ) | ( 1 − | f ( y 1 ) | ) ≤ M f ( 1 − | f ( y 1 ) | ) ≤ϕ( y 1 ).$
(2.36)

On the other hand, from (2.35) we have

$| f ( y 2 ) | ( 1 − | f ( y 1 ) | ) ≥ 1 2 ( 1 + 1 − 4 ϕ ( y 2 ) ) ( 1 − 1 2 ( 1 − 1 − 4 ϕ ( y 1 ) ) ) > 1 2 ( 1 − 1 − 4 ϕ ( y 1 ) ) ( 1 − 1 2 ( 1 − 1 − 4 ϕ ( y 1 ) ) ) = ϕ ( y 1 ) ,$

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

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

Proof Using the triangle inequality with (2.3) we have

$| f ( x ) f ( y ) | ≤ | f ( x ∘ g ( y ) ) | +ϕ(x)≤ M f +ϕ(x)$
(2.37)

for all $x,y∈G$. Taking the supremum of the left hand side of (2.37) with respect to $y∈G$ we get $M f |f(x)|≤ M f +ϕ(x)$ for all $x∈G$. Thus, we have

$M f ( | f ( x ) | − 1 ) ≤ϕ(x)$
(2.38)

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

$| f ( z ) | ≤ | f ( x ) f ( w ) | +ϕ(x)≤ | f ( x ) | M f +ϕ(x)$
(2.39)

for all $x,z∈G$. Taking the supremum of the left hand side of (2.39) we get $M f ≤ M f |f(x)|+ϕ(x)$ for all $x∈G$. Thus, we have

$M f ( 1 − | f ( x ) | ) ≤ϕ(x)$
(2.40)

for all $x∈G$. 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 $j∈L$ we have

$| f j ( y ) | ≤ 1 2 ( 1 + 1 + 4 ϕ ( y ) )$
(2.41)

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

$∥ P F ( y ) ∥ u = ∑ j ∈ L | f j ( y ) | 2 ≤ | L | 2 ( 1 + 1 + 4 ϕ ( y ) )$

for all $y∈G$, which gives (1.7). Now, if $|L|=1$, say $L={ j 0 }$ we have

$∥ P F ( y ) ∥ u = | f j 0 ( y ) |$

for all $y∈G$. 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.

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## Acknowledgements

The author is very thankful to the referees for valuable suggestions that improved the presentation of the paper. This work was supported by Basic Science Research Program through the National Foundation of Korea (NRF) funded by the Korea Government (no. 2012R1A1A008507).

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