We’d like to understand how you use our websites in order to improve them. Register your interest.

# Solutions of the third order Cauchy difference equation on groups

## Abstract

Let $f:G→H$ be a function, where $(G,⋅)$ is a group and $(H,+)$ is an abelian group. In this paper, the following third order Cauchy difference of $f: C ( 3 ) f( x 1 , x 2 , x 3 , x 4 )$ = $f( x 1 x 2 x 3 x 4 )$$f( x 1 x 2 x 3 )$$f( x 1 x 2 x 4 )$$f( x 1 x 3 x 4 )$$f( x 2 x 3 x 4 )$ + $f( x 1 x 2 )$ + $f( x 1 x 3 )$ + $f( x 1 x 4 )$ + $f( x 2 x 3 )$ + $f( x 2 x 4 )$ + $f( x 3 x 4 )$$f( x 1 )$$f( x 2 )$$f( x 3 )$$f( x 4 )$ ($∀ x 1 , x 2 , x 3 , x 4 ∈G$), is studied. We first give some special solutions of $C ( 3 ) f=0$ on free groups. Then sufficient and necessary conditions on finite cyclic groups and symmetric groups are also obtained.

MSC:39B52, 39A70.

## 1 Introduction

It is well known from  that Jensen’s functional equation

$f(x+y)+f(x−y)=2f(x),$
(1.1)

with the additional condition $f(0)=0$, is equivalent to Cauchy’s equation

$f(x+y)=f(x)+f(y)$

on the real line. Let $(G,⋅)$ be a group, $(H,+)$ be an abelian group. Let $e∈G$ and $0∈H$ denote the identity elements. The study of (1.1) was extended to groups for f maps G into H in , where the general solution for a free group H with two generators and $G=G L 2 (Z)$ was given, respectively. Later, the results were generalized to all free groups and $G=G L n (Z)$, $n≥3$ (see ). Since functional equations involve Cauchy difference, which made it become much more interesting . For a function $f:G→H$, its Cauchy difference, $C ( m ) f$, is defined by

$C ( 0 ) f=f,$
(1.2)
$C ( 1 ) f( x 1 , x 2 )=f( x 1 x 2 )−f( x 1 )−f( x 2 ),$
(1.3)
$C ( m + 1 ) f ( x 1 , x 2 , … , x m + 2 ) = C ( m ) f ( x 1 x 2 , x 3 , … , x m + 2 ) − C ( m ) f ( x 1 , x 3 , … , x m + 2 ) − C ( m ) f ( x 2 , x 3 , … , x m + 2 ) .$
(1.4)

The first order Cauchy difference $C ( 1 ) f$ will be abbreviated as Cf. In , by using the reduction formulas and relations, as given in [2, 3], the general solution of the second order Cauchy difference equation was provided on free groups. Particularly, the authors also gave the expression of general solutions on symmetric group and finite cyclic group.

In this paper, we consider the following functional equation:

$f ( x 1 x 2 x 3 x 4 ) − f ( x 1 x 2 x 3 ) − f ( x 1 x 2 x 4 ) − f ( x 1 x 3 x 4 ) − f ( x 2 x 3 x 4 ) + f ( x 1 x 2 ) + f ( x 1 x 3 ) + f ( x 1 x 4 ) + f ( x 2 x 3 ) + f ( x 2 x 4 ) + f ( x 3 x 4 ) − f ( x 1 ) − f ( x 2 ) − f ( x 3 ) − f ( x 4 ) = 0 ( ∀ x 1 , x 2 , x 3 , x 4 ∈ G ) .$
(1.5)

It follows from (1.4) that (1.5) is equivalent to the vanishing third order Cauchy difference equation

$C ( 3 ) f=0.$

The purpose of this paper is to determine the solutions of (1.5) on some given groups. Clearly, the general solution of (1.5) will be denoted by

(1.6)

Remark 1 (1) $Ker C ( 3 ) (G,H)$ is an abelian group under the pointwise addition of functions; (2) $Hom(G,H)⊆Ker C ( 3 ) (G,H)$.

## 2 Properties of solution

Lemma 1 Suppose that $f∈Ker C ( 3 ) (G,H)$. Then

$f(e)=0,$
(2.1)
$Cf(x,e)=0,Cf(e,y)=0,$
(2.2)
$C ( 2 ) f(e,y,z)=0, C ( 2 ) f(x,e,z)=0, C ( 2 ) f(x,y,e)=0,$
(2.3)
(2.4)
$f ( x n ) =nf(x)+ n ( n − 1 ) 2 Cf(x,x)+ n ( n − 1 ) ( n − 2 ) 6 C ( 2 ) f(x,x,x),$
(2.5)

for all $x,y,z,∈G$ and $n∈Z$.

Proof Putting $x 1 =e$ in (1.5) we get (2.1). Then from (2.1) we obtain (2.2)-(2.3). Furthermore, by the definition of $C ( 2 ) f$, we have

$C ( 2 ) f(x,yw,z)=f(xywz)−f(xyw)−f(xz)−f(ywz)+f(x)+f(yw)+f(z),$

and

$C ( 2 ) f ( x , y , z ) + C ( 2 ) f ( x , w , z ) = f ( x y z ) − f ( x y ) − f ( x z ) − f ( y z ) + f ( x ) + f ( y ) + f ( z ) + f ( x w z ) − f ( x w ) − f ( x z ) − f ( w z ) + f ( x ) + f ( w ) + f ( z ) .$

One can easily check that

$C ( 2 ) f(x,yw,z)− ( C ( 2 ) f ( x , y , z ) + C ( 2 ) f ( x , w , z ) ) = C ( 3 ) f(x,y,w,z)=0.$

Hence, the above relations imply that $Cf(x,⋅,z)$ is a homomorphism. Similarly, the fact is also true for both $Cf(⋅,y,z)$ and $Cf(x,y,⋅)$. This proves (2.4).

We now consider (2.5). Actually, it is trivial for $n=0,1,2$ by (2.1) and the definition of Cf. Suppose that (2.5) holds for all natural numbers smaller than $n≥4$, then

$f ( x n ) = f ( x n − 2 ⋅ x ⋅ x ) = 2 f ( x n − 1 ) + f ( x 2 ) − f ( x n − 2 ) − 2 f ( x ) + C ( 2 ) f ( x n − 2 , x , x ) = 2 [ ( n − 1 ) f ( x ) + ( n − 1 ) ( n − 2 ) 2 C f ( x , x ) + ( n − 1 ) ( n − 2 ) ( n − 3 ) 6 C ( 2 ) f ( x , x , x ) ] + 2 f ( x ) + C f ( x , x ) − [ ( n − 2 ) f ( x ) + ( n − 2 ) ( n − 3 ) 2 C f ( x , x ) + ( n − 2 ) ( n − 3 ) ( n − 4 ) 6 C ( 2 ) f ( x , x , x ) ] − 2 f ( x ) + ( n − 2 ) C ( 2 ) f ( x , x , x ) = n f ( x ) + n ( n − 1 ) 2 C f ( x , x ) + n ( n − 1 ) ( n − 2 ) 6 C ( 2 ) f ( x , x , x ) ,$

where the definition of $C ( 2 ) f$ and (2.4) are used in the second equation. This gives (2.5) for all $n≥0$. On the other hand, for any fixed integer $n>0$, by (1.4) and (2.1), we have

$C ( 2 ) f ( x n , x − n , x n ) = f ( x n ) − f ( e ) − f ( x 2 n ) − f ( e ) + f ( x n ) + f ( x − n ) + f ( x n ) = 3 f ( x n ) + f ( x − n ) − f ( x 2 n ) ,$

from which it follows that

$f ( x − n ) = f ( x 2 n ) − 3 f ( x n ) + C ( 2 ) f ( x n , x − n , x n ) = 2 n f ( x ) + 2 n ( 2 n − 1 ) 2 C f ( x , x ) + 2 n ( 2 n − 1 ) ( 2 n − 2 ) 6 C ( 2 ) f ( x , x , x ) − 3 [ n f ( x ) + n ( n − 1 ) 2 C f ( x , x ) + n ( n − 1 ) ( n − 2 ) 6 C ( 2 ) f ( x , x , x ) ] − n 3 C ( 2 ) f ( x , x , x ) = − n f ( x ) + − n ( − n − 1 ) 2 C f ( x , x ) + − n ( − n − 1 ) ( − n − 2 ) 6 C ( 2 ) f ( x , x , x ) ,$

from (2.4) and the above claim for $n>0$. This confirms (2.5) for $n<0$. □

Remark 2 For any function $f:G→H$, the following statements are pairwise equivalent:

1. (i)

The function $f∈Ker C ( 3 ) (G,H)$;

2. (ii)

$C ( 2 ) f(⋅,y,z)$ is a homomorphism;

3. (iii)

$C ( 2 ) f(x,⋅,z)$ is a homomorphism;

4. (iv)

$C ( 2 ) f(x,y,⋅)$ is a homomorphism.

Before presenting Proposition 1, we first introduce the following useful lemma, which was given in .

Lemma 2 (Lemma 2.4 in )

The following identity is valid for any function $f:G→H$ and $l∈N$:

$f( x 1 x 2 ⋯ x l )= ∑ m ≤ l ∑ 1 ≤ i 1 < i 2 < ⋯ < i m ≤ l C ( m − 1 ) f( x i 1 , x i 2 ,…, x i m ).$
(2.6)

Proposition 1 Suppose that $f∈Ker C ( 3 ) (G,H)$. Then

$f ( x 1 n 1 x 2 n 2 ⋯ x l n l ) = ∑ 1 ≤ i ≤ l [ n i f ( x i ) + n i ( n i − 1 ) 2 C f ( x i , x i ) + n i ( n i − 1 ) ( n i − 2 ) 6 C ( 2 ) f ( x i , x i , x i ) ] + ∑ 1 ≤ i 1 < i 2 ≤ l C f ( x i 1 n i 1 , x i 2 n i 2 ) + ∑ 1 ≤ i 1 < i 2 < i 3 ≤ l n i 1 n i 2 n i 3 C ( 2 ) f ( x i 1 , x i 2 , x i 3 ) ,$
(2.7)

for $n i ∈Z$ and all $x i ∈G$, $i=1,2,…,l$ such that $x j ≠ x j + 1$, $j=1,2,…,l−1$.

Proof Replacing $x i$ in (2.6) by $x i n i$ we have

$f ( x 1 n 1 x 2 n 2 ⋯ x l n l ) = ∑ m ≤ l ∑ 1 ≤ i 1 < i 2 < ⋯ < i m ≤ l C ( m − 1 ) f ( x i 1 n i 1 , x i 2 n i 2 , … , x i m n i m ) .$

The vanishing of $C ( m − 1 ) f$ for $m≥4$ yields

$f ( x 1 n 1 x 2 n 2 ⋯ x l n l ) = ∑ 1 ≤ i ≤ l f ( x i n i ) + ∑ 1 ≤ i 1 < i 2 ≤ l C f ( x i 1 n i 1 , x i 2 n i 2 ) + ∑ 1 ≤ i 1 < i 2 < i 3 ≤ l C ( 2 ) f ( x i 1 n i 1 , x i 2 n i 2 , x i 3 n i 3 ) .$

Therefore, by (2.5) and (2.4), we have

$f ( x i n i ) = n i f ( x i ) + n i ( n i − 1 ) 2 C f ( x i , x i ) + n i ( n i − 1 ) ( n i − 2 ) 6 C ( 2 ) f ( x i , x i , x i ) , C ( 2 ) f ( x i 1 n i 1 , x i 2 n i 2 , x i 3 n i 3 ) = n i 1 n i 2 n i 3 C ( 2 ) f ( x i 1 , x i 2 , x i 3 ) ,$

which is (2.7). This completes the proof. □

Remark 3 In particular, if $l=1$, then Proposition 1 holds.

## 3 Solution on a free group

In this section, we study the solutions on a free group. We first solve (1.5) for the free group G on a single letter a.

Theorem 1 Let G be the free group on one letter a. Then $f∈Ker C ( 3 ) (G,H)$ if and only if it is given by

$f ( a n ) =nf(a)+ n ( n − 1 ) 2 Cf(a,a)+ n ( n − 1 ) ( n − 2 ) 6 C ( 2 ) f(a,a,a),∀n∈Z.$
(3.1)

Proof Necessity. It can be obtained from (2.5) in Lemma 1.

Sufficiency. Taking (3.1) as the definition of f on $G=〈a〉$. By Remark 2, we only need to verify that $C ( 2 ) f$ is a homomorphism with respect to each variable and thus f belongs to $Ker C ( 3 ) (G,H)$. Let

$x= a m ,y= a n ,z= a p$

be any three elements of G. Then it follows from (1.4) and (3.1) that

$C ( 2 ) f ( x , y , z ) = C ( 2 ) f ( a m , a n , a p ) = f ( a m + n + p ) − f ( a m + n ) − f ( a m + p ) − f ( a n + p ) + f ( a m ) + f ( a n ) + f ( a p ) = [ ( m + n + p ) f ( a ) + ( m + n + p ) ( m + n + p − 1 ) 2 C f ( a , a ) + ( m + n + p ) ( m + n + p − 1 ) ( m + n + p − 2 ) 6 C ( 2 ) f ( a , a , a ) ] − [ ( m + n ) f ( a ) + ( m + n ) ( m + n − 1 ) 2 C f ( a , a ) + ( m + n ) ( m + n − 1 ) ( m + n − 2 ) 6 C ( 2 ) f ( a , a , a ) ] − [ ( m + p ) f ( a ) + ( m + p ) ( m + p − 1 ) 2 C f ( a , a ) + ( m + p ) ( m + p − 1 ) ( m + p − 2 ) 6 C ( 2 ) f ( a , a , a ) ] − [ ( n + p ) f ( a ) + ( n + p ) ( n + p − 1 ) 2 C f ( a , a ) + ( n + p ) ( n + p − 1 ) ( n + p − 2 ) 6 C ( 2 ) f ( a , a , a ) ] + [ m f ( a ) + m ( m − 1 ) 2 C f ( a , a ) + m ( m − 1 ) ( m − 2 ) 6 C ( 2 ) f ( a , a , a ) ] + [ n f ( a ) + n ( n − 1 ) 2 C f ( a , a ) + n ( n − 1 ) ( n − 2 ) 6 C ( 2 ) f ( a , a , a ) ] + [ p f ( a ) + p ( p − 1 ) 2 C f ( a , a ) + p ( p − 1 ) ( p − 2 ) 6 C ( 2 ) f ( a , a , a ) ] .$

By a tedious calculation, we have

$C ( 2 ) f ( a m , a n , a p ) =mnp C ( 2 ) f(a,a,a),$

which leads to the result that $C ( 2 ) f$ is a homomorphism with respect to each variable. □

At the end of this section, for the free group on an alphabet $〈A〉$ with $|A|≥2$, we discuss some special solutions of (1.5).

An element $x∈A$ can be written in the form

(3.2)

For each fixed $a∈A$ and fixed pair of distinct $a,b∈A$, define the functions W, $W 2$, $W 3$:

$W(x;a)= ∑ a i = a n i ,$
(3.3)
$W 2 (x;a,b)= ∑ i < j , a i = a , a j = b n i n j ,$
(3.4)
$W 3 (x;a,b)= ∑ i > j , a i = a , a j = b n i n j ,$
(3.5)

along with (3.2). Referring to [2, 3], the above functions are well defined. Furthermore, they satisfy the following relations:

$W(xy;a)=W(x;a)+W(y;a),$
(3.6)
$W 2 (x;a,b)= W 3 (x;b,a).$
(3.7)

Proposition 2 For any fixed $a∈A$ and fixed pair of distinct a, b in $A$, the following assertions hold:

1. (i)

$W(⋅;a)$ belongs to $Ker C ( 3 ) (〈A〉,Z)$;

2. (ii)

$W 2 (⋅;a,b)$ belongs to $Ker C ( 3 ) (〈A〉,Z)$;

3. (iii)

$W 3 (⋅;a,b)$ belongs to $Ker C ( 3 ) (〈A〉,Z)$.

Proof Claim (i) follows from the fact that $x↦W(x;a)$ is a morphism from $〈A〉$ to by (3.6).

Now we consider assertion (ii). Let x, y, z, w in the free group be written as

$x = a 1 r 1 a 2 r 2 ⋯ a l r l , y = b 1 s 1 b 2 s 2 ⋯ b p s p , z = c 1 t 1 c 2 t 2 ⋯ c q t q , w = d 1 m 1 d 2 m 2 ⋯ d v m v .$

Then

$W 2 ( x y z w ; a , b ) = ∑ i < j , a i = a , a j = b r i r j + ∑ i < j , b i = a , b j = b s i s j + ∑ i < j , c i = a , c j = b t i t j + ∑ i < j , d i = a , d j = b m i m j + ∑ a i = a , b j = b r i s j + ∑ a i = a , c j = b r i t j + ∑ a i = a , d j = b r i m j + ∑ b i = a , c j = b s i t j + ∑ b i = a , d j = b s i m j + ∑ c i = a , d j = b t i m j , W 2 ( x y z ; a , b ) = ∑ i < j , a i = a , a j = b r i r j + ∑ i < j , b i = a , b j = b s i s j + ∑ i < j , c i = a , c j = b t i t j + ∑ a i = a , b j = b r i s j + ∑ a i = a , c j = b r i t j + ∑ b i = a , c j = b s i t j , W 2 ( x y w ; a , b ) = ∑ i < j , a i = a , a j = b r i r j + ∑ i < j , b i = a , b j = b s i s j + ∑ i < j , d i = a , d j = b m i m j + ∑ a i = a , b j = b r i s j + ∑ a i = a , d j = b r i m j + ∑ b i = a , d j = b s i m j , W 2 ( x z w ; a , b ) = ∑ i < j , a i = a , a j = b r i r j + ∑ i < j , c i = a , c j = b t i t j + ∑ i < j , d i = a , d j = b m i m j + ∑ a i = a , c j = b r i t j + ∑ a i = a , d j = b r i m j + ∑ c i = a , d j = b t i m j , W 2 ( y z w ; a , b ) = ∑ i < j , b i = a , b j = b s i s j + ∑ i < j , c i = a , c j = b t i t j + ∑ i < j , d i = a , d j = b m i m j + ∑ b i = a , c j = b s i t j + ∑ b i = a , d j = b s i m j + ∑ c i = a , d j = b t i m j , W 2 ( x y ; a , b ) = ∑ i < j , a i = a , a j = b r i r j + ∑ i < j , b i = a , b j = b s i s j + ∑ a i = a , b j = b r i s j , W 2 ( x z ; a , b ) = ∑ i < j , a i = a , a j = b r i r j + ∑ i < j , c i = a , c j = b t i t j + ∑ a i = a , c j = b r i t j , W 2 ( x w ; a , b ) = ∑ i < j , a i = a , a j = b r i r j + ∑ i < j , d i = a , d j = b m i m j + ∑ a i = a , d j = b r i m j , W 2 ( y z ; a , b ) = ∑ i < j , b i = a , b j = b s i s j + ∑ i < j , c i = a , c j = b t i t j + ∑ b i = a , c j = b s i t j , W 2 ( y w ; a , b ) = ∑ i < j , b i = a , b j = b s i s j + ∑ i < j , d i = a , d j = b m i m j + ∑ b i = a , d j = b s i m j , W 2 ( z w ; a , b ) = ∑ i < j , c i = a , c j = b t i t j + ∑ i < j , d i = a , d j = b m i m j + ∑ c i = a , d j = b t i m j , W 2 ( x ; a , b ) = ∑ i < j , a i = a , a j = b r i r j , W 2 ( y ; a , b ) = ∑ i < j , b i = a , b j = b s i s j , W 2 ( z ; a , b ) = ∑ i < j , c i = a , c j = b t i t j , W 2 ( w ; a , b ) = ∑ i < j , d i = a , d j = b m i m j .$

Hence, we have

$W 2 ( x y z w ; a , b ) − W 2 ( x y z ; a , b ) − W 2 ( x y w ; a , b ) − W 2 ( x z w ; a , b ) − W 2 ( y z w ; a , b ) + W 2 ( x y ; a , b ) + W 2 ( x z ; a , b ) + W 2 ( x w ; a , b ) + W 2 ( y z ; a , b ) + W 2 ( y w ; a , b ) + W 2 ( z w ; a , b ) − W 2 ( x ; a , b ) − W 2 ( y ; a , b ) − W 2 ( z ; a , b ) − W 2 ( w ; a , b ) = 0 .$

This concludes assertion (ii). Claim (iii) follows from (3.7) directly. □

## 4 Solution on symmetric group $S n$

The symmetric group on a finite set X is the group whose elements are all bijective functions from X to X and whose group operation is that of function composition. If $X={1,2,…,n}$, then it is called the symmetric group of degree n and denoted $S n$.

In this section, we consider (1.5) for $G= S n$.

Lemma 3 If $f∈Ker C ( 3 ) ( S n ,H)$, then

$C ( 2 ) f( x 1 x 2 ⋯ x m ,y,z)= C ( 2 ) f( x π ( 1 ) x π ( 2 ) ⋯ x π ( m ) ,y,z),$
(4.1)
$C ( 2 ) f(x, y 1 y 2 ⋯ y m ,z)= C ( 2 ) f(x, y π ( 1 ) y π ( 2 ) ⋯ y π ( m ) ,z),$
(4.2)
$C ( 2 ) f(x,y, z 1 z 2 ⋯ z m )= C ( 2 ) f(x,y, z π ( 1 ) z π ( 2 ) ⋯ z π ( m ) ),$
(4.3)

for all $x,y,z, x i , y i , z i ∈ S n$, $i=1,2,…,m$, and all rearrangements π.

Proof Note that $C ( 2 ) f(⋅,y,z)$ is a homomorphism and H is an abelian group, which yields

$C ( 2 ) f ( x 1 x 2 ⋯ x n , y , z ) = C ( 2 ) f ( x 1 , y , z ) + C ( 2 ) f ( x 2 , y , z ) + ⋯ + C ( 2 ) f ( x n , y , z ) = C ( 2 ) f ( x π ( 1 ) , y , z ) + C ( 2 ) f ( x π ( 2 ) , y , z ) + ⋯ + C ( 2 ) f ( x π ( n ) , y , z ) = C ( 2 ) f ( x π ( 1 ) x π ( 2 ) ⋯ x π ( n ) , y , z ) .$

This proves (4.1). By a similar procedure, we can also verify (4.2)-(4.3). □

Lemma 4 Let τ be an arbitrary 2-cycle in $S n$ and $f∈Ker C ( 3 ) ( S n ,H)$, then

$f ( τ 2 ) =0,$
(4.4)
$Cf(τ,τ)=−2f(τ),$
(4.5)
$C ( 2 ) f(τ,τ,τ)=4f(τ),$
(4.6)
$C ( 3 ) f(τ,τ,τ,τ)=−8f(τ),8f(τ)=0.$
(4.7)

Proof It suffices to prove (4.7). The proofs for the rest of the statements are straightforward. Using $τ 2 =e$, $f(e)=0$ and (1.4) we get

$C ( 3 ) f ( τ , τ , τ , τ ) = f ( τ 4 ) − 4 f ( τ 3 ) + 6 f ( τ 2 ) − 4 f ( τ ) = − 8 f ( τ ) ,$

which implies that $8f(τ)=0$ since $C ( 3 ) f=0$. This completes the proof. □

Lemma 5 For any 2-cycle σ, τ, υ, and $f∈Ker C ( 3 ) ( S n ,H)$, we have

$C ( 2 ) f(σ,τ,υ)= C ( 2 ) f ( ( 12 ) , ( 12 ) , ( 12 ) ) .$
(4.8)

Proof For any 2-cycle σ, there exists $z∈ S n$ such that $σ=z(12) z − 1$. Hence, for any $x,y∈ S n$, by (4.3) we have

$C ( 2 ) f ( x , y , σ ) = C ( 2 ) f ( x , y , z ( 12 ) z − 1 ) = C ( 2 ) f ( x , y , ( 12 ) z z − 1 ) = C ( 2 ) f ( x , y , ( 12 ) ) .$
(4.9)

Similarly,

$C ( 2 ) f(σ,x,y)= C ( 2 ) f ( ( 12 ) , x , y ) ,$
(4.10)
$C ( 2 ) f(x,σ,y)= C ( 2 ) f ( x , ( 12 ) , y ) .$
(4.11)

In particular, (4.8) follows from (4.9)-(4.11). □

Lemma 6 Assume that $Cf(σ,τ)=Cf((12),(12))$ for every 2-cycle $σ,τ∈ S n$. Then for any $x,y,β, σ i ∈ S n$, $i=1,2,…,n$, rearrangement π, where $σ i$, β are 2-cycles, we have

$f( σ 1 σ 2 ⋯ σ l )=f( σ π ( 1 ) σ π ( 2 ) ⋯ σ π ( l ) ),$
(4.12)
$f(xβy)=f ( x ( 12 ) y ) ,$
(4.13)
$f(β)=f ( ( 12 ) )$
(4.14)

for every $f∈Ker C ( 3 ) ( S n ,H)$.

Proof Firstly, for any 2-cycle $σ i ∈ S n$, $i=1,2,…,l$ and rearrangement π, it follows from the assumption $Cf(σ,τ)=Cf((12),(12))$, Proposition 1, and (4.8) that

$f ( σ 1 σ 2 ⋯ σ l ) = ∑ i = 1 l f ( σ i ) + ∑ 1 ≤ i < j ≤ l C f ( σ i , σ j ) + ∑ 1 ≤ i < j < k ≤ l C ( 2 ) f ( σ i , σ j , σ k ) = ∑ i = 1 l f ( σ i ) + l ( l − 1 ) 2 C f ( ( 12 ) , ( 12 ) ) + l ( l − 1 ) ( l − 2 ) 6 C ( 2 ) f ( ( 12 ) , ( 12 ) , ( 12 ) ) = ∑ i = 1 l f ( σ π ( i ) ) + l ( l − 1 ) 2 C f ( ( 12 ) , ( 12 ) ) + l ( l − 1 ) ( l − 2 ) 6 C ( 2 ) f ( ( 12 ) , ( 12 ) , ( 12 ) ) = f ( σ π ( 1 ) σ π ( 2 ) ⋯ σ π ( l ) ) ,$

which gives (4.12).

On the other hand, for every $x,y,β∈ S n$ there exist 2-cycles $σ i , τ j ,z∈ S n$, $i=1,…,p$, $j=1,…,q$, such that $x= σ 1 σ 2 ⋯ σ p$, $y= τ 1 τ 2 ⋯ τ q$ and $β=z(12) z − 1$. Note that $z= δ 1 δ 2 ⋯ δ r$ for some 2-cycles $δ 1 ,…, δ r ∈ S n$, we have

$f ( x β y ) = f ( σ 1 σ 2 ⋯ σ p δ 1 δ 2 ⋯ δ r ( 12 ) δ r − 1 δ r − 1 − 1 ⋯ δ 1 − 1 τ 1 τ 2 ⋯ τ q ) = f ( σ 1 σ 2 ⋯ σ p ( 12 ) δ 1 δ 2 ⋯ δ r δ r − 1 δ r − 1 − 1 τ 1 τ 2 ⋯ τ q ) = f ( x ( 12 ) y )$

by (4.12). In particular, taking $x=y=e$ in (4.13), we obtain (4.14). This completes the proof. □

According to Lemma 6, we give the following main result in this section.

Theorem 2 Assume that $Cf(σ,τ)=Cf((12),(12))$ for every 2-cycle $σ,τ∈ S n$. Then $f∈Ker C ( 3 ) ( S n ,H)$ if and only if there is an $h 0 ∈H$ such that $8 h 0 =0$ and

(4.15)

Proof Necessity. Let $f∈Ker C ( 3 ) ( S n ,H)$. Then for any $x∈ S n$, there exist 2-cycles $α i ∈ S n$, $i=1,2,…,p$, such that $x= α 1 α 2 ⋯ α p$. In view of (4.8), (4.14), (4.5)-(4.6), and Proposition 1, we get

$f ( x ) = f ( α 1 α 2 ⋯ α p ) = ∑ i = 1 p f ( α i ) + ∑ 1 ≤ i < j ≤ p C f ( α i , α j ) + ∑ 1 ≤ i < j < k ≤ p C ( 2 ) f ( α i , α j , α k ) = p f ( ( 12 ) ) + p ( p − 1 ) 2 C f ( ( 12 ) , ( 12 ) ) + p ( p − 1 ) ( p − 2 ) 6 C ( 2 ) f ( ( 12 ) , ( 12 ) , ( 12 ) ) = p f ( ( 12 ) ) + p 2 − p 2 ( − 2 f ( ( 12 ) ) ) + p 3 − 3 p 2 + 2 p 6 ⋅ 4 f ( ( 12 ) ) = ( 2 3 p 3 − 3 p 2 + 10 3 p ) f ( 12 ) .$
(4.16)

Let $g(p)= 2 3 p 3 −3 p 2 + 10 3 p$, we claim that

(4.17)

We first prove the even case. Obviously, (4.17) is true for $p=2$ since $g(2)=0$. For an inductive proof, suppose that (4.17) also holds for $p=2n$, $n∈Z$. Then we compute that

$g ( 2 n + 2 ) = 2 3 ( 2 n + 2 ) 3 − 3 ( 2 n + 2 ) 2 + 10 3 ( 2 n + 2 ) = 2 3 ( ( 2 n ) 3 + 6 ( 2 n ) 2 + 24 n + 8 ) − 3 ( ( 2 n ) 2 + 8 n + 4 ) + 10 3 ( 2 n + 2 ) = 2 3 ( 2 n ) 3 − 3 ( 2 n ) 2 + 10 3 ⋅ 2 n + 16 n 2 − 8 n = g ( 2 n ) + 16 n 2 − 8 n ,$

which yields $g(2n+2)∈8N$. This confirms the even case of (4.17). When p is odd, (4.17) is true for $p=1$ because of $g(1)=1$. Suppose that (4.17) holds for $p=2n−1$, and then we get

$g ( 2 n + 1 ) = 2 3 ( 2 n + 1 ) 3 − 3 ( 2 n + 1 ) 2 + 10 3 ( 2 n + 1 ) = 2 3 ( ( 2 n − 1 ) 3 + 6 ( 2 n − 1 ) 2 + 12 ( 2 n − 1 ) + 8 ) − 3 ( ( 2 n − 1 ) 2 + 4 ( 2 n − 1 ) + 4 ) + 10 3 ( 2 n − 1 + 2 ) = 2 3 ( 2 n − 1 ) 3 − 3 ( 2 n − 1 ) 2 + 10 3 ( 2 n − 1 ) + 4 ( 2 n − 1 ) 2 + 8 ( 2 n − 1 ) + 16 3 − 12 ( 2 n − 1 ) − 12 + 20 3 = 2 3 ( 2 n − 1 ) 3 − 3 ( 2 n − 1 ) 2 + 10 3 ( 2 n − 1 ) + 16 n 2 − 24 n + 8 = g ( 2 n − 1 ) + 16 n 2 − 24 n + 8 .$

This completes the proof of (4.17). According to (4.17), (4.16) becomes

This proves that f must have the form (4.15) with $h 0 =f((12))$.

Sufficiency. Let $f: S n →H$ be defined by (4.15), where $h 0$ is a constant with $8 h 0 =0$. In order to prove the identity of (1.5), by the symmetry of $x 1$, $x 2$, $x 3$, $x 4$ it suffices to verify the following four cases: case (i) $x 1$ is odd, and $x 2$, $x 3$, $x 4$ are even; case (ii) $x 1$, $x 2$ are odd, and $x 3$, $x 4$ are even; case (iii) $x 1$, $x 2$, $x 3$ are odd, and $x 4$ is even; case (iv) $x 1$, $x 2$, $x 3$ and $x 4$ are odd.

In fact, for case (i) it is easy to see that $x 1$, $x 1 x 2$, $x 1 x 3$, $x 1 x 4$, $x 1 x 2 x 3$, $x 1 x 3 x 4$, $x 1 x 2 x 4$, $x 1 x 2 x 3 x 4$ are odd, and $x 2$, $x 3$, $x 4$, $x 2 x 3$, $x 2 x 4$, $x 3 x 4$, $x 2 x 3 x 4$ are even, which leads to the equality of (1.5). The proofs of the other cases are similar. □

## 5 Solution on the finite cyclic group $C n$

Let $C n =〈a∣ a n =e〉$ be a cyclic group of order n with generator a. In this section, we study the general solution on the finite cyclic group $C n$.

Theorem 3 Assume that n is odd and $nCf(a,a)=0$. Then $f∈Ker C ( 3 ) ( C n ,H)$ if and only if it is given by

$f ( a p ) =pf(a)+ p ( p − 1 ) 2 Cf(a,a)+ p ( p − 1 ) ( p − 2 ) 6 C ( 2 ) f(a,a,a),∀p∈Z,$
(5.1)

where $f(a)$ and $C ( 2 ) f(a,a,a)$ satisfy

$n C ( 2 ) f(a,a,a)=0,$
(5.2)
$nf(a)+ n ( n − 1 ) ( n − 2 ) 6 C ( 2 ) f(a,a,a)=0.$
(5.3)

Proof Necessity. Let $f: C n →H$ be a function satisfying (1.5). Then by (2.5), we see that f also satisfies (5.1), i.e.,

$f ( a p ) =pf(a)+ p ( p − 1 ) 2 Cf(a,a)+ p ( p − 1 ) ( p − 2 ) 6 C ( 2 ) f(a,a,a),∀p∈Z.$

Let $p=n$ in (5.1), since $nCf(a,a)=0$ and $n ( n − 1 ) 2$ is an integer, the summand $n ( n − 1 ) 2 Cf(a,a)=0$ and by the fact that $a n =e$, $f(e)=0$, we obtain

$nf(a)+ n ( n − 1 ) ( n − 2 ) 6 C ( 2 ) f(a,a,a)=0.$

Furthermore, by using (2.4), (2.3), and $a n =e$, we get

$n C ( 2 ) f(a,a,a)= C ( 2 ) f ( a n , a , a ) = C ( 2 ) f(e,a,a)=0.$

This proves (5.2)-(5.3).

Sufficiency. We claim that (5.1)-(5.3) defines a function on $C n$. Indeed, for each $p∈Z$, by (5.1) and (5.3) we have

$f ( a p + n ) − f ( a ) = [ ( p + n ) f ( a ) + ( p + n ) ( p + n − 1 ) 2 C f ( a , a ) + ( p + n ) ( p + n − 1 ) ( p + n − 2 ) 6 C ( 2 ) f ( a , a , a ) ] − [ p f ( a ) + p ( p − 1 ) 2 C f ( a , a ) + p ( p − 1 ) ( p − 2 ) 6 C ( 2 ) f ( a , a , a ) ] = [ n f ( a ) + n ( n − 1 ) 2 C f ( a , a ) + n ( n − 1 ) ( n − 2 ) 6 C ( 2 ) f ( a , a , a ) ] + p n C f ( a , a ) + ( p ( p + n ) 2 − p ) n C ( 2 ) f ( a , a , a ) = p n C f ( a , a ) + ( p ( p + n ) 2 − p ) n C ( 2 ) f ( a , a , a ) = 0 ,$

where the last identity is obtained because $nCf(a,a)=0$, (5.2), and n is odd.

Finally, for any $x= a m$, $y= a p$, $z= a q$, and $w= a l ∈ C n$, we have

$f ( x y z w ) − f ( x y z ) − f ( x y w ) − f ( x z w ) − f ( y z w ) + f ( x y ) + f ( x z ) + f ( x w ) + f ( y z ) + f ( y w ) + f ( z w ) − f ( x ) − f ( y ) − f ( z ) − f ( w ) = f ( a m + p + q + l ) − f ( a m + p + q ) − f ( a m + p + l ) − f ( a m + q + l ) − f ( a p + q + l ) + f ( a m + p ) + f ( a m + q ) + f ( a m + l ) + f ( a p + q ) + f ( a p + l ) + f ( a q + l ) − f ( a m ) − f ( a p ) − f ( a q ) − f ( a l ) = ( m + p + q + l ) f ( a ) + ( m + p + q + l ) ( m + p + q + l − 1 ) 2 C f ( a , a ) + ( m + p + q + l ) ( m + p + q + l − 1 ) ( m + p + q + l − 2 ) 6 C ( 2 ) f ( a , a , a ) − [ ( m + p + q ) f ( a ) + ( m + p + q ) ( m + p + q − 1 ) 2 C f ( a , a ) + ( m + p + q ) ( m + p + q − 1 ) ( m + p + q − 2 ) 6 C ( 2 ) f ( a , a , a ) ] − [ ( m + p + l ) f ( a ) + ( m + p + l ) ( m + p + l − 1 ) 2 C f ( a , a ) + ( m + p + l ) ( m + p + l − 1 ) ( m + p + l − 2 ) 6 C ( 2 ) f ( a , a , a ) ] − [ ( m + q + l ) f ( a ) + ( m + q + l ) ( m + q + l − 1 ) 2 C f ( a , a ) + ( m + q + l ) ( m + q + l − 1 ) ( m + q + l − 2 ) 6 C ( 2 ) f ( a , a , a ) ] − [ ( p + q + l ) f ( a ) + ( p + q + l ) ( p + q + l − 1 ) 2 C f ( a , a ) + ( p + q + l ) ( p + q + l − 1 ) ( p + q + l − 2 ) 6 C ( 2 ) f ( a , a , a ) ] + [ ( m + p ) f ( a ) + ( m + p ) ( m + p − 1 ) 2 C f ( a , a ) + ( m + p ) ( m + p − 1 ) ( m + p − 2 ) 6 C ( 2 ) f ( a , a , a ) ] + [ ( m + q ) f ( a ) + ( m + q ) ( m + q − 1 ) 2 C f ( a , a ) + ( m + q ) ( m + q − 1 ) ( m + q − 2 ) 6 C ( 2 ) f ( a , a , a ) ] + [ ( m + l ) f ( a ) + ( m + l ) ( m + l − 1 ) 2 C f ( a , a ) + ( m + l ) ( m + l − 1 ) ( m + l − 2 ) 6 C ( 2 ) f ( a , a , a ) ] + [ ( p + q ) f ( a ) + ( p + q ) ( p + q − 1 ) 2 C f ( a , a ) + ( p + q ) ( p + q − 1 ) ( p + q − 2 ) 6 C ( 2 ) f ( a , a , a ) ] + [ ( p + l ) f ( a ) + ( p + l ) ( p + l − 1 ) 2 C f ( a , a ) + ( p + l ) ( p + l − 1 ) ( p + l − 2 ) 6 C ( 2 ) f ( a , a , a ) ] + [ ( q + l ) f ( a ) + ( q + l ) ( p + l − 1 ) 2 C f ( a , a ) + ( q + l ) ( q + l − 1 ) ( q + l − 2 ) 6 C ( 2 ) f ( a , a , a ) ] − m f ( a ) + m ( m − 1 ) 2 C f ( a , a ) − m ( m − 1 ) ( m − 2 ) 6 C ( 2 ) f ( a , a , a ) − p f ( a ) + p ( p − 1 ) 2 C f ( a , a ) − p ( p − 1 ) ( p − 2 ) 6 C ( 2 ) f ( a , a , a ) − q f ( a ) + q ( q − 1 ) 2 C f ( a , a ) − q ( q − 1 ) ( q − 2 ) 6 C ( 2 ) f ( a , a , a ) − l f ( a ) + l ( l − 1 ) 2 C f ( a , a ) − l ( l − 1 ) ( l − 2 ) 6 C ( 2 ) f ( a , a , a ) ,$

which, after a long and tedious computation, gives 0. Consequently, $f∈Ker C ( 3 ) ( C n ,H)$. This completes the proof. □

## References

1. 1.

Aczél J: Lectures on Functional Equations and Their Applications. Academic Press, New York; 1966.

2. 2.

Ng CT: Jensen’s functional equation on groups. Aequ. Math. 1990, 39: 85-99. 10.1007/BF01833945

3. 3.

Ng CT: Jensen’s functional equation on groups II. Aequ. Math. 1999, 58: 311-320. 10.1007/s000100050118

4. 4.

Baron K, Kannappan P: On the Cauchy difference. Aequ. Math. 1993, 46: 112-118. 10.1007/BF01834002

5. 5.

Brzdȩk J: On the Cauchy difference on normed spaces. Abh. Math. Semin. Univ. Hamb. 1996, 66: 143-150. 10.1007/BF02940800

6. 6.

Ebanks B: Generalized Cauchy difference functional equations. Aequ. Math. 2005, 70: 154-176. 10.1007/s00010-004-2739-5

7. 7.

Ebanks B: Generalized Cauchy difference equations, II. Proc. Am. Math. Soc. 2008, 136: 3911-3919. 10.1090/S0002-9939-08-09379-9

8. 8.

Ng CT, Zhao H: Kernel of the second order Cauchy difference on groups. Aequ. Math. 2013, 86: 155-170. 10.1007/s00010-012-0174-6

## Acknowledgements

The authors are grateful to the editor and the referees for their valuable comments and suggestions, especially for the improvement of Theorem 2. This paper is supported by the National Science Foundation of China (11301226, 61202109), Zhejiang Provincial Natural Science Foundation of China under Grant No. LQ13A010017.

## Author information

Authors

### Corresponding author

Correspondence to Lin Li.

### Competing interests

The authors declare that they have no competing interests.

### Authors’ contributions

The authors contributed equally to the writing of this paper. All authors read and approved the final manuscript.

## Rights and permissions 