D’Alembert’s and Wilson’s equations on semigroups

Abstract

In this paper we consider a generalization of d’Alembert’s equation and Wilson’s equation on commutative semigroups using only the semigroup operation, ie. we consider the functional equation

$$\begin{aligned}&h(x+2y)+h(x)=2f(y)h(x+y),\ x,y\in S, \end{aligned}$$

where $f,h:S\rightarrow \mathbb {K}$, $(S,+)$ is a commutative semigroup, $\mathbb {K}$ is a quadratically closed field, $\text {char}\,\mathbb {K}\ne 2$.

1 Introduction

If we look at d’Alembert’s functional equation

$$\begin{aligned} f(x+y)+f(x-y)=2f(x)f(y),\ x,y\in S, \end{aligned}$$

(1.1)

on a group S, then we have two possible ways of generalizing it to semigroups.

The first one is the functional equation

$$\begin{aligned} f(x+y)+f(x+\sigma (y))=2f(x)f(y),\ x,y\in S, \end{aligned}$$

(1.2)

where $(S,+)$ is an abelian semigroup, $\sigma \in \text {Aut}(S)$ is an involution, which has been studied by many mathematicians, (see [3, 4, 7, 12,13,14, 16, 18, 19]). For non-abelian groups the solutions of d’Alembert’s functional equation may be different from those of the abelian case (see [8, 10, 11, 20]).

The above equation has a generalization of the form

$$\begin{aligned} \int \limits _{K} f(x+\lambda y) d\mu (\lambda ) = f(x)f(y) ,\ x,y\in G, \end{aligned}$$

where $(G,+)$ is a locally compact group, K is a compact subgroup of the automorphism group of G with the normalized Haar measure $\mu$, $f:G\rightarrow \mathbb {C}$. It is a generalization of the cosine equation and it is studied in the theory of group representations, being the relation defining K-spherical functions (for the terminology see [5, p. 88]). This equation has been studied by many mathematicians (for example see [6, 15, 17, 21, 22]).

The second way is the functional equation

$$\begin{aligned} f(x+2y)+f(x)=2f(x+y)f(y),\ x,y\in S, \end{aligned}$$

(1.3)

which we obtain by the substitution $x\mapsto x+y$. This equation is equivalent to d’Alembert’s functional equation on groups and we will show that its solutions are the same as those of d’Alembert’s functional equation.

2 Preliminaries

Throughout the present paper, we assume that $(S,+)$ is an abelian semigroup and the relation $\sim \subseteq S\times S$ is given by

$$\begin{aligned} \forall _{x,y\in S} \left( x\sim y \Leftrightarrow \exists _{z\in S}\ (x+z=y+z )\right) , \end{aligned}$$

(2.1)

$\mathbb {K}$ is a quadratically closed field, $\mathrm{char}\,\mathbb {K}\ne 2$.

Lemma 2.1

([15, Lemma 2.2]) The relation $\sim$ given by (2.1) is an equivalence relation, $S/_\sim$ with the operation $+:{S/_\sim }^2 \rightarrow S/_\sim$ defined by

$$\begin{aligned}{}[x]_\sim + [y]_\sim := [x+y]_\sim ,\ x,y\in S, \end{aligned}$$

(2.2)

is a cancellative abelian semigroup and the function $\varkappa :S\rightarrow S/_\sim$ given by

$$\begin{aligned} \varkappa (x)=[x]_\sim ,\ x\in S, \end{aligned}$$

(2.3)

is a semigroup epimorphism.

Theorem 2.2

([18, Theorem 1]) Let $\sigma :S\rightarrow S$ be an involution, $f:S\rightarrow \mathbb {K}$. Then f satisfies Eq. (1.2) iff there exists an exponential function $m:S\rightarrow \mathbb {K}$ such that

$$\begin{aligned} f(x) =\frac{m(x)+m(\sigma x)}{2},\ x\in S. \end{aligned}$$

(2.4)

The exponential function m from the above theorem is on groups either zero everywhere or non-zero everywhere (e.g. $m:S\rightarrow \mathbb {K}^*$ is a homomorphism). Similarly on semigroups, m has the same zero behavior as in groups iff f satisfies

$$\begin{aligned} f(x+\sigma x)\ne 0,\ x\in S, \end{aligned}$$

(2.5)

(see [15, Theorem 2.8]). But generally on semigroups there may exist exponential functions which have zeros on some non-trivial subset of S.

Example

Let $c\in \mathbb {K}\setminus \{0\}$, $\mathbb {N}_0=\mathbb {N}\cup \{0\}$, $S=\mathbb {N}_0\times \mathbb {N}_0$, $\sigma :S\rightarrow S$ be a function given by $\sigma (n,k)=(k,n)$ for $n,k\in \mathbb {N}_0$. We define functions $f,m:S\rightarrow \mathbb {K}$ by the formulas

$$\begin{aligned} f(n,k)={\left\{ \begin{array}{ll} 1,&{} n=k=0\\ 0 &{} n,k\ge 1\\ 2^{n-1}c^n,&{} k=0,\ n\ge 1\\ 2^{k-1}c^k,&{} n=0,\ k\ge 1 \end{array}\right. }, \end{aligned}$$

$$\begin{aligned} m(n,k)={\left\{ \begin{array}{ll} 0 &{} k\ge 1\\ (2c)^n,&{} k=0 \end{array}\right. }. \end{aligned}$$

Let further $n,k,p,q\in \mathbb {N}_0$. We observe that:

• if $k\ge 1$ or $q\ge 1$, then

  $$\begin{aligned}&m(n+p,k+q)=0=m(n,k)\cdot m(p,q). \end{aligned}$$

• if $k=q=0$, then

  $$\begin{aligned} m(n+p,k+q)&=m(n+p,0)=(2c)^{n+p}=(2c)^n \cdot (2c)^p\\&=m(n,0)\cdot m(p,0)=m(n,k)\cdot m(p,q). \end{aligned}$$

Hence m is an exponential function. We observe also that

• if $k=n=0$, then

  $$\begin{aligned}&m(n,k)+m(k,n)=2=2f(n,k). \end{aligned}$$

• if $k,n\ge 1$, then

  $$\begin{aligned}&m(n,k)+m(k,n)=0=f(n,k). \end{aligned}$$

• if $k\ge 1$ and $n=0$, then

  $$\begin{aligned}&m(n,k)+m(k,n)=0+(2c)^k=2\cdot 2^{k-1}c^k=f(n,k). \end{aligned}$$

• if $k=0$ and $n\ge 1$, then

  $$\begin{aligned}&m(n,k)+m(k,n)=(2c)^n+0=2\cdot 2^{n-1}c^n=f(n,k). \end{aligned}$$

Hence f satisfies Eq. (2.4) and in view of Theorem 2.2 it satisfies Eq. (1.2). We have also

$$\begin{aligned}&f((n,m)+(m,n))=f(n+m,m+n)=0,\ n\ne 0 \vee m\ne 0. \end{aligned}$$

3 Main results

Lemma 3.1

Let $f:S\rightarrow \mathbb {K}$ satisfy Eq. (1.3). Then the function $\widetilde{f}:S/_{\sim } \rightarrow \mathbb {K}$ given by the formula $\widetilde{f}(\varkappa (x))=f(x)$ for $x\in S$ is well-defined and $\widetilde{f}$ satisfies Eq. (1.3).

Proof

Let $x,y,z\in S$ be such that $x+z=y+z$. Then

$$\begin{aligned} f(x)=2f(x+z)f(z)-f(x+2z)=2f(y+z)f(z)-f(y+2z)=f(y), \end{aligned}$$

so $\widetilde{f}$ is well-defined. We have also

$$\begin{aligned}&\widetilde{f} (\varkappa (x)) +\widetilde{f} (\varkappa (x)+2\varkappa (y))=f(x)+f(x+2y)\\&\quad =2f(x+y)f(y)=2\widetilde{f} (\varkappa (x)+\varkappa (y))\widetilde{f} (\varkappa (y)),\ x,y\in S. \end{aligned}$$

$\square$

Lemma 3.2

Assume that S is abelian and cancellative, G is an abelian group such that $G=S-S$. Let $f:S\rightarrow \mathbb {K}$ satisfy Eq. (1.3). Then the function $F:G\rightarrow \mathbb {K}$ given by

$$\begin{aligned} F(x-y)=2f(x)f(y)-f(x+y),\ x,y\in S, \end{aligned}$$

(3.1)

is well-defined, $F|_S =f$ and F satisfies the equation

$$\begin{aligned} F(x+y)+F(x-y)=2F(x)F(y),\ x,y\in G. \end{aligned}$$

Proof

Let $x,y,u,v\in S$, $x-y=u-v$. Then $x+v=y+u$ and

$$\begin{aligned}&f(x+y)+2f(u)f(v)=2f(x+y+v)f(v)-f(x+y+2v)+2f(u)f(v)\\&\quad =2f(u+2y)f(v)-f(u+v+2y)+2f(u)f(v)\\&\quad =2\big ( f(u+2y)+f(u)\big ) f(v)-f(u+v+2y)\\&\quad =4f(u+y)f(y)f(v) -f(u+v+2y)=4f(x+v)f(v)f(y)-f(u+v+2y)\\&\quad =2f(x+2v)f(y)+2f(x)f(y)-f(u+v+2y)\\&\quad =2f(u+v+y)f(y)+2f(x)f(y)-f(u+v+2y)\\&\quad =f(u+v)+2f(x)f(y), \end{aligned}$$

which means that

$$\begin{aligned}&2f(x)f(y)-f(x+y)=2f(u)f(v)-f(u+v), \end{aligned}$$

so F is well-defined. We observe also that

$$\begin{aligned}&F(x)=F(2x-x)=2f(2x)f(x)-f(3x)=f(x),\ x\in S. \end{aligned}$$

Now we show that

$$\begin{aligned} F(x-y-z)+F(x-y+z)=2F(x-y)f(z),\ x,y,z\in S. \end{aligned}$$

(3.2)

Indeed, for $x,y,z\in S$ we have

$$\begin{aligned}&2F(x-y)f(z)=2F(x+z-z-y)f(z)\\&\quad =4f(x+z)f(y+z)f(z)-2f(x+z+y+z)f(z)\\&\quad =2\bigg ( f(x)+f(x+2z) \bigg )f(y+z)-f(x+z+y)-f(x+z+y+2z)\\&\quad =2f(x)f(y+z)-f(x+y+z)+2f(x+2z)f(y+z)-f(x+2z+y+z)\\&\quad =F(x-y-z)+F(x+2z-y-z)=F(x-y-z)+F(x-y+z). \end{aligned}$$

Hence, for $x,y,u,v\in S$ we get

$$\begin{aligned}&2F(x-y)F(u-v)=4F(x-y)f(u)f(v)-2F(x-y)f(u+v)\\&\quad =2F(x-y+u)f(v) +2F(x-y-u)f(v) -F(x-y+u+v)\\&\qquad -F(x-y-u-v) =F(x-y+u+v) +F(x-y+u-v)\\&\qquad +F(x-y-u-v) +F(x-y-u+v) -F(x-y+u+v)\\&\qquad -F(x-y-u-v)=F(x-y+u-v)+F(x-y-u+v), \end{aligned}$$

which ends the proof.$\square$

Using Lemmas 2.1, 3.1, 3.2 and the fact that $\varkappa$ is a homomorphism we easily obtain the following result.

Corollary 3.3

Let G be an abelian group such that $G=S/_\sim -S/_\sim$.

• 1. Let $f:S\rightarrow \mathbb {K}$ satisfy Eq. (1.3). Then the function $F:G\rightarrow \mathbb {K}$ given by

  $$\begin{aligned} F(\varkappa (x)-\varkappa (y))=2f(x)f(y)-f(x+y),\ x,y\in S, \end{aligned}$$

  (3.3)

  is well-defined, $F\circ \varkappa =f$ and F satisfies d’Alembert’s functional equation.

• 2. Let $F:G\rightarrow \mathbb {K}$ satisfy d’Alembert’s functional equation. Then $f=F\circ \varkappa :S\rightarrow \mathbb {K}$ satisfies Eq. (1.3).

Theorem 3.4

Let $f:S\rightarrow \mathbb {K}$ be a non-zero function. Then f satisfies Eq. (1.3) iff there exists a homomorphism $m:S\rightarrow \mathbb {K}^*$ such that

$$\begin{aligned} f(x)=\frac{m(x)+m(x)^{-1}}{2},\ x\in S. \end{aligned}$$

(3.4)

Proof

It is easy to check that the function given by (3.4) satisfies Eq. (1.3).

Assume that f satisfies Eq. (1.3). In view of Corollary 3.3 there exists a function $F:G\rightarrow \mathbb {K}$ such that $F\circ \varkappa =f$, F satisfies d’Alembert’s functional equation, where G is an abelian group such that $G=S/_\sim - S/_\sim$. Hence there exists a homomorphism $M:G\rightarrow \mathbb {K}^*$ such that

$$\begin{aligned} F(x)=\frac{M(x)+M(-x)}{2}=\frac{M(x)+M(x)^{-1}}{2},\ x\in G. \end{aligned}$$

We define $m:S\rightarrow \mathbb {K}^*$ by the formula

$$\begin{aligned} m(x)=M(\varkappa (x)),\ x\in S. \end{aligned}$$

Since M and $\varkappa$ are homomorphisms, m is a homomorphism. We have also

$$\begin{aligned}&f(x)=F(\varkappa (x))=\frac{M(\varkappa (x))+M(\varkappa (x))^{-1}}{2}=\frac{m(x)+m(x)^{-1}}{2}, \end{aligned}$$

which ends the proof.$\square$

A well-known generalization of d’Alembert’s functional equation is Wilson’s functional equation (see e.g. [1] for references).

Solutions of this equation can be found as a special case of some more general functional equation in [2], but we use a more readable result from the paper [9, Theorem 8].

Theorem 3.5

Let G be an abelian group, $F,H:G\rightarrow \mathbb {K}$. The ordered pair (F, H) satisfies Wilson’s functional equation

$$\begin{aligned} H(x+y)+H(x-y)=2F(y)H(x),\ x,y\in G, \end{aligned}$$

iff F, H have one of the following forms:

• 1. $H=0$ and F is arbitrary;

• 2. $F(x)=\frac{M(x)+M(x)^{-1}}{2}$, $H(x)=c\frac{M(x)+M(x)^{-1}}{2}+d\frac{M(x)-M(x)^{-1}}{2}$ for $x\in G$;

• 3. $F(x)=M(x)$, $H(x)=M(x)\Big ( A(x)+c \Big )$, $M(x)\in \{1,-1\}$ for $x\in G$;

where $M:G\rightarrow \mathbb {K}^*$ is a homomorphism, $A:G\rightarrow \mathbb {K}$ is additive, $c,d\in \mathbb {K}$.

We can equivalently write Wilson’s functional equation in the form

$$\begin{aligned} h(x+2y)+h(x)=2f(y)h(x+y),\ x,y\in S, \end{aligned}$$

(3.5)

and now we can consider it on semigroups.

Lemma 3.6

Let $f,h:S\rightarrow \mathbb {K}$, $h\ne 0$, (f, h) satisfies (3.5) and. Then f is a non-zero function which satisfies (1.3).

Proof

We observe that

$$\begin{aligned}&2\Big ( f(y+2z)+f(y)-2f(y+z)f(z) \Big ) h(x+y+2z)\\&\quad =2f(y+2z)h(x+y+2z)+2f(y)h(x+y+2z)\\&\qquad -4f(y+z)f(z)h(z+y+2z)=h(x+2y+4z)+h(x)+h(x+2y+2z)\\&\qquad +h(x+2z)-2f(y+z)h(x+y+3z)-2f(y+z)h(x+y+z)\\&\quad =h(x+2y+4z)+h(x)+h(x+2y+2z)+h(x+2z)-h(x+2y+2z)\\&\qquad -h(x)-h(x+2y+4z)-h(x+2z)=0,\ x,y,z\in S. \end{aligned}$$

Suppose that there exist $y,z\in S$ such that $h(x+y+2z)=0$ for all $x\in S$. Then

$$\begin{aligned}&h(x+y+z)=h(x+y+z)+h(x+y+3z)\\&\quad =2f(z)h(x+y+2z)=0,\ x\in S, \end{aligned}$$

so

$$\begin{aligned}&h(x+y)=h(x+y)+h(x+y+2z)=2f(z)h(x+y+z)=0,\ x\in S, \end{aligned}$$

and

$$\begin{aligned}&h(x)=h(x)+h(x+2y)=2f(y)h(x+y)=0,\ x\in S, \end{aligned}$$

which gives us a contradiction. Hence f satisfies (1.3).

Suppose that $f=0$. Then

$$\begin{aligned} h(2x)+h(2x+2y)=0=h(2y)+h(2y+2x),\ x,y\in S, \end{aligned}$$

which means that $h(2x)=0$ for all $x\in S$. We have also

$$\begin{aligned} h(x+2y)=-h(x),\ x,y\in S, \end{aligned}$$

so

$$\begin{aligned}&2h(x)=h(x)+h(x)=-h(x+2y)-h(x+4y)=0,\ x,y\in S, \end{aligned}$$

which gives us a contradiction.$\square$

Lemma 3.7

Let $f,h:S\rightarrow \mathbb {K}$, $h\ne 0$, (f, h) satisfies Eq. (3.5). Then functions $\widetilde{f},\widetilde{h}:S/_{\sim } \rightarrow \mathbb {K}$ given by the formulas $\widetilde{f}(\varkappa (x))=f(x)$, $\widetilde{h}(\varkappa (x))=h(x)$ for $x\in S$ are well-defined and $(\widetilde{f},\widetilde{h})$ satisfies Eq. (3.5).

Proof

In view of Lemmas 3.1 and 3.6 the map $\widetilde{f}$ is well-defined.

Let $x,y,z\in S$ be such that $x+z=y+z$. Then

$$\begin{aligned} h(x)=2h(x+z)f(z)-h(x+2z)=2h(y+z)f(z)-h(y+2z)=h(y), \end{aligned}$$

so $\widetilde{h}$ is well-defined. We have also

$$\begin{aligned}&\widetilde{h} (\varkappa (x)) +\widetilde{h} (\varkappa (x)+2\varkappa (y))=h(x)+h(x+2y)\\&\quad =2h(x+y)f(y)=2\widetilde{h} (\varkappa (x)+\varkappa (y))\widetilde{f} (\varkappa (y)),\ x,y\in S. \end{aligned}$$

$\square$

Lemma 3.8

Assume that S is abelian and cancellative, G is an abelian group such that $G=S-S$. Let $f,h:S\rightarrow \mathbb {K}$, $h\ne 0$, (f, h) satisfies Eq. (3.5). Then functions $F,H:G\rightarrow \mathbb {K}$ given by

$$\begin{aligned}&F(x-y)=2f(x)f(y)-f(x+y),\ x,y\in S,\end{aligned}$$

(3.6)

$$\begin{aligned}&H(x-y)=2h(x)f(y)-h(x+y),\ x,y\in S, \end{aligned}$$

(3.7)

are well-defined, $F|_S =f$, $H|_S =h$ and (F, H) satisfies the equation

$$\begin{aligned} H(x+y)+H(x-y)=2H(x)F(y),\ x,y\in G. \end{aligned}$$

Proof

In view of Lemmas 3.2 and 3.6 the map F is well-defined.

Let $x,y,u,v\in S$, $x-y=u-v$. Then $x+v=y+u$ and

$$\begin{aligned}&h(x+y)+2h(u)f(v)=2h(x+y+v)f(v)-h(x+y+2v)+2h(u)f(v)\\&\quad =2h(u+2y)f(v)-h(u+v+2y)+2h(u)f(v)\\&\quad =2\big ( h(u+2y)+h(u)\big ) f(v)-h(u+v+2y)\\&\quad =4h(u+y)f(y)f(v) -h(u+v+2y)=4h(x+v)f(v)f(y)-h(u+v+2y)\\&\quad =2h(x+2v)f(y)+2h(x)f(y)-h(u+v+2y)\\&\quad =2h(u+v+y)f(y)+2h(x)f(y)-h(u+v+2y)\\&\quad =h(u+v)+2h(x)f(y), \end{aligned}$$

which means that

$$\begin{aligned}&2h(x)f(y)-h(x+y)=2h(u)f(v)-h(u+v), \end{aligned}$$

so H is well-defined. We observe also that

$$\begin{aligned}&H(x)=H(2x-x)=2h(2x)f(x)-h(3x)=h(x),\ x\in S. \end{aligned}$$

Now we show that

$$\begin{aligned} H(x-y-z)+H(x-y+z)=2f(z)H(x-y),\ x,y,z\in S. \end{aligned}$$

(3.8)

Indeed, we have

$$\begin{aligned} 2&H(x-y)f(z)=2H(x+z-z-y)f(z)\\&\quad =4h(x+z)f(y+z)f(z)-2h(x+z+y+z)f(z)\\&\quad =2\bigg ( h(x)+h(x+2z) \bigg )f(y+z)-h(x+z+y)-h(x+z+y+2z)\\&\quad =2h(x)f(y+z)-h(x+y+z)+2h(x+2z)f(y+z)-h(x+2z+y+z)\\&\quad =H(x-y-z)+H(x+2z-y-z)\\&\quad =H(x-y-z)+H(x-y+z),\ x,y,z\in S. \end{aligned}$$

Hence, for $x,y,u,v\in S$ we get

$$\begin{aligned}&2F(u-v)H(x-y)=4H(x-y)f(u)f(v)-2H(x-y)f(u+v)\\&\quad =2H(x-y+u)f(v) +2H(x-y-u)f(v) -H(x-y+u+v)\\&\qquad -H(x-y-u-v) =H(x-y+u+v) +H(x-y+u-v)\\&\qquad +H(x-y-u-v) +H(x-y-u+v)-H(x-y+u+v) \\&\qquad -H(x-y-u-v)=H(x-y+u-v)+H(x-y-u+v), \end{aligned}$$

which ends the proof.$\square$

Using Lemmas 2.1, 3.7, 3.8 and the fact that $\varkappa$ is a homomorphism we easily obtain the following result.

Corollary 3.9

Let G be an abelian group such that $G=S/_\sim -S/_\sim$.

• 1. Let $f,h:S\rightarrow \mathbb {K}$, $h\ne 0$, (f, h) satisfies Eq. (3.5). Then functions $F,H: G\rightarrow \mathbb {K}$ given by

  $$\begin{aligned}&F(\varkappa (x)-\varkappa (y))=2f(x)f(y)-f(x+y),\ x,y\in S,\end{aligned}$$

  (3.9)
  $$\begin{aligned}&H(\varkappa (x)-\varkappa (y))=2h(x)f(y)-h(x+y),\ x,y\in S, \end{aligned}$$

  (3.10)

  are well-defined, $F\circ \varkappa =f$, $H\circ \varkappa =h$ and (F, H) satisfies Wilson’s functional equation.

• 2. Let $F,H:G\rightarrow \mathbb {K}$, $f=F\circ \varkappa , h=H\circ \varkappa :S\rightarrow \mathbb {K}$, (F, H) satisfies Wilson’s functional equation. Then (f, h) satisfies Eq. (3.5).

Theorem 3.10

Let $f,h:S\rightarrow \mathbb {K}$. Then (f, h) satisfies (3.5) iff f, h have one of the following forms:

• 1. $h=0$ and f is arbitrary;

• 2. $f(x)=\frac{m(x)+m(x)^{-1}}{2}$, $h(x)=c\frac{m(x)+m(x)^{-1}}{2}+d\frac{m(x)-m(x)^{-1}}{2}$ for $x\in S$;

• 3. $f(x)=m(x)$, $h(x)=m(x)\Big ( a(x)+c \Big )$, $m(x)\in \{1,-1\}$ for $x\in S$;

where $m:S\rightarrow \mathbb {K}^*$ is a homomorphism, $a:S\rightarrow \mathbb {K}$ is additive, $c,d\in \mathbb {K}$.

Proof

It is easy to check that for functions f, h given by the forms 1–3 the pair (f, h) satisfies Eq. (3.5).

Assume that (f, h) satisfies Eq. (3.5). In view of Corollary 3.9 there exist functions $F,H:G\rightarrow \mathbb {K}$ such that $F\circ \varkappa =f$, $H\circ \varkappa =h$, (F, H) satisfies Wilson’s functional equation, where G is an abelian group such that $G=S/_\sim - S/_\sim$. Hence we get that F, H have one of the following forms:

• 1. $H=0$ and F is arbitrary;

• 2. $F(x)=\frac{M(x)+M(x)^{-1}}{2}$, $H(x)=c\frac{M(x)+M(x)^{-1}}{2}+d\frac{M(x)-M(x)^{-1}}{2}$ for $x\in G$;

• 3. $F(x)=M(x)$, $H(x)=M(x)\Big ( A(x)+c \Big )$, $M(x)\in \{1,-1\}$ for $x\in G$;

where $M:G\rightarrow \mathbb {K}^*$ is a homomorphism, $A:G\rightarrow \mathbb {K}$ is additive, $c,d\in \mathbb {K}$.

We define $m:S\rightarrow \mathbb {K}^*$, $a:S\rightarrow \mathbb {K}$ by the formulas

$$\begin{aligned}&m(x)=M(\varkappa (x)),\ x\in S,\\&a(x)=A(\varkappa (x)),\ x\in S. \end{aligned}$$

Since M and $\varkappa$ are homomorphisms, m is a homomorphism. Since A and $\varkappa$ are additive, a is additive. We have also:

• In case 2

  $$\begin{aligned} f(x)&=F(\varkappa (x))=\frac{M(\varkappa (x))+M(\varkappa (x))^{-1}}{2}=\frac{m(x)+m(x)^{-1}}{2},\ x\in S,\\ h(x)&=H(\varkappa (x))= c\frac{M(\varkappa (x))+M(\varkappa (x))^{-1}}{2}+d\frac{M(\varkappa (x))-M(\varkappa (x))^{-1}}{2}\\&=c\frac{m(x)+m(x)^{-1}}{2}+d\frac{m(x)-m(x)^{-1}}{2},\ x\in S, \end{aligned}$$

• In case 3

  $$\begin{aligned}&f(x)=F(\varkappa (x))=M(\varkappa (x))=m(x),\ x\in S,\\&h(x)=H(\varkappa (x))=M(\varkappa (x))\Big ( A(\varkappa (x))+c \Big )=m(x)\Big ( a(x)+c \Big ),\ x\in S, \end{aligned}$$

which ends the proof.$\square$