Extensions of the Cosine-Sine functional equation

TL;DR

This paper explores new extensions of the cosine-sine functional equation, aiming to broaden its applicability and deepen understanding of its mathematical properties.

Contribution

It introduces novel extensions of the cosine-sine functional equation, expanding the theoretical framework and potential applications.

Findings

Extended the functional equation to new forms

Provided proofs for the generalized equations

Discussed implications for mathematical analysis

Abstract

The aim of the present paper is to give extensions of the cosine-sine functional equation.

Equations

$$f(xy)=g_1(x)h_1(y)+g(x)h_2(y)+\chi (x)h(y),x,y\in G,$$

$$f(xy)=g(x)h(y)+\sum _{j=1}^N{\chi }_j(x)h_j(y),x,y\in G,$$

$$f(xy)=g_1(x)h_1(y)+g(x)h_2(y)+\chi (x)A(x)h(y),x,y\in G,$$

$$f(xy)=g_1(x)h_1(y)+\mu (x)h_2(y)+\chi (x)A(x)h(y),x,y\in G,$$

$$f(xy)=f(x)h_1(y)+g(x)h_2(y)+\chi (x)h(y),x,y\in G,$$

$$f(xy)=f(x)h(y)+\sum _{j=1}^N{\chi }_j(x)h_j(y),x,y\in G,$$

$$f(xy)=f(x)h_1(y)+g(x)h_2(y)+\chi (x)A(x)h(y),x,y\in G,$$

$$f(xy)=f(x)h_1(y)+\mu (x)h_2(y)+\chi (x)A(x)h(y),x,y\in G.$$

$$f(xy)=\sum _{j=1}^N{g}_j(x)h_j(y),x,y\in G,$$

$$f(x+y)=\sum _{j=1}^N{g}_j(x)h_j(y),x,y\in G,$$

$$f(xy)=g_1(x)h_1(y)+g(x)h_2(y),$$

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

$$f(xy)=f(x)g(y)+f(y)g(x)+h(x)h(y),x,y\in G,$$

$$f(xy)=f(x){\chi }_1(y)+{\chi }_2(x)f(y),$$

$$f(xy)=k(x)M(y)+g(x)h(y),x,y\in S,$$

$$f(xy)=f(x){\chi }_1(y)+{\chi }_2(x)f(y),x,y\in G$$

$$f(xy)=g_1(x)h_1(y)+{\chi }_2(x)h_2(y),x,y\in G$$

$$f(xy)=g_1(x)h_1(y)+g(x)h_2(y),$$

$$f(xy)=g(x)h(y)+\sum _{j=1}^n{g}_j(x)h_j(y),x,y\in M$$

$$f(xy)=f(x)g(y)+g(x)f(y)+k(x)k(y)$$

$$f(xy)=f(x)g(y)+\frac{\mu (x)+\chi (x)}{2}f(y)+\chi (x)A(x)h(y)$$

$$f(xy)=g_1(x)h_1(y)+\frac{\mu (x)+\chi (x)}{2}{h}_2(y)+\chi (x)A(x)h(y)$$

$$f(xy)=g_1(x)h_1(y)+\mu (x)h_2(y)+\chi (x)A(x)h(y)$$

$$f(xy)=f(x){\chi }_1(y)+{\chi }_2(x)f(y),x,y\in G.$$

$$f(x)=\alpha ({\chi }_1(x)-{\chi }_2(x))+A([y_0,x]){\chi }_1(x),x\in G,$$

$$A(xc{x}^{-1})=\frac{{\chi }_2(x)}{{\chi }_1(x)}A(c)\text{for all}x\in G\text{and}c\in [G,G].$$

$$\begin{array}{rl}& a{\mu }_1(y){\mu }_1(x)+a{\mu }_2(y){\mu }_2(x)+b\chi (y)\chi (x)\\ & =\frac{h_2(y)}{2}{\mu }_1(x)+\frac{h_2(y)}{2}{\mu }_2(x)+h(y)\chi (x),\end{array}$$

$$f(xy)={\alpha }_1{\mu }_1(x){\mu }_1(y)+{\alpha }_2{\mu }_2(x){\mu }_2(y)+{\alpha }_3\chi (x)\chi (y)$$

$$\begin{array}{rl}& g_1(x)h_1(y)=[{\alpha }_1{\mu }_1(y)-\frac{1}{2}{h}_2(y)]{\mu }_1(x)+[{\alpha }_2{\mu }_2(y)-\frac{1}{2}{h}_2(y)]{\mu }_2(x)\\ & +[{\alpha }_3\chi (y)-h(y)]\chi (x)\end{array}$$

Full text

2010 Mathematics Subject Classification. Primary 39B52; Secondary 39B32

1. Introduction

1.1. Connection to the literature

In the present paper we solve the following functional equations:

[TABLE]

on a group $G$, where $g$ is the average of two different characters $\mu_{1}$ and $\mu_{2}$ of $G$ and $\chi$ is a character of $G$ such that $\mu_{1}\neq\chi$ and $\mu_{2}\neq\chi$;

[TABLE]

where $\chi_{1},\cdot\cdot\cdot,\chi_{N}$ are $N$ distinct characters of a group $G$;

[TABLE]

where $g$ is the average of two given nonzero multiplicative functions $\mu$ and $\chi$ on a monoid $G$ and $A$ is an additive function on $G$ such that $\chi A\neq 0$;

[TABLE]

where $\mu$ and $\chi$ are two different characters of a group $G$ and $A:G\rightarrow\mathbb{C}$ is a nonzero additive function on $G$. To solve the functional equations (1.1), (1.2), (1.3) and (1.4) we will need to solve respectively the corresponding functional equations

[TABLE]

[TABLE]

[TABLE]

and

[TABLE]

We notice that (1.1), (1.2), (1.3), (1.4), (1.5), (1.6), (1.7) and (1.8) are examples of Levi-Civita’s functional equation

[TABLE]

on which lots of work has been done. Székelyhidi [12, Section 10] computes for any abelian group $(G,+)$ the solution of

[TABLE]

under the assumption that each of the sets $\{g_{1},g_{2},\cdot\cdot\cdot,g_{N}\}$ and $\{h_{1},h_{2},\cdot\cdot\cdot,h_{N}\}$ is linearly independent. The structure of any solution of Levi-Civita’s functional equation on monoids was described by using matrix-coefficients of the right regular representations [10, Theorem 5.2]. In [3, pp . 25-27] Vincze’s equation, which is a particular case of our functional equations, was solved. We refer also to [1, 8, 9, 13].

If $h=0$ then the functional equations (1.1) and (1.3) become

[TABLE]

which was recently solved by Ebanks [6], Stetkær [11], and Ebanks and Stetkær [7].

If $h=0$, $f=g_{1}=h_{2}$ and $h_{1}=g$ then Eq. (1.1) and (1.3) reduce to the sine addition law

[TABLE]

for which solutions are known on semigroups, see for example [10, Theorem 4.1].

Chung, Kannappan and Ng [5] solved the functional equation

[TABLE]

where $G$ is a group. Recently, the results of [5] were extended by the authors [2] to semigroups generated by their squares.

We relate the solutions of the functional equations (1.1), (1.5), (1.2), (1.6), (1.4) and (1.8) to those of a functional equation of type

[TABLE]

where $\chi_{1}$ and $\chi_{2}$ are two characters of $G$, which was solved by Stetkær [11].

1.2. Motivation

In [6] Ebanks solved the functional equation

[TABLE]

for four unknown central functions $f,g,h,k$ on certain semigroups $S$, where $M$ is a fixed multiplicative function on $S$.

In [11] Stetkær solved, by removing the restriction of the solutions, the functional equation

[TABLE]

and its Pexiderized version

[TABLE]

on a group $G$, where $\chi_{1}$ and $\chi_{2}$ are fixed characters of $G$.

In [7] Ebanks and Stetkær studied the more general variant of (1.10)

[TABLE]

for four unknown $f,g_{1},h_{1},h_{2}$ on a monoid, where $g$ is a linear combination of $n\geq 2$ distinct multiplicative functions.

Recently Belfakih and Elqorachi [4, Theorem 2.7] solved the functional equation

[TABLE]

on a monoid $M$, where $g_{j}$ are linear combinations of at least $2$ distinct nonzero multiplicative functions on $M$, and $f,g,h,h_{j}$ are the unknown functions.

Let $G$ be a group. Let $\mu$ and $\chi$ be two different characters of $G$ and let $A:G\rightarrow\mathbb{C}$ be a non zero additive function. Let $c_{1},c_{2}\in\mathbb{C}$ be two constants satisfying $1+c_{1}c_{2}^{2}=0$. We check by elementary computations that the functions $f,g,k:G\rightarrow\mathbb{C}$ defined by $f:=-c_{1}\mu+c_{1}\chi-c_{1}c_{2}\chi\,A$, $g:=\dfrac{\mu+\chi}{2}-\dfrac{c_{2}}{2}\chi\,A$ and $k:=\chi\,A$ satisfy the functional equation

[TABLE]

on the group $G$, which implies, by simple computations, that the triple $(f,g,h)$ is a solution of the functional equation

[TABLE]

for all $x,y\in G$, where $h:=-\dfrac{1}{2}c_{2}f+\chi\,A$ and $h\neq\chi\,A$ because $c_{1},c_{2}\in\mathbb{C}\setminus\{0\}$. Moreover, for a given character $m$ of $G$, we check easily that the quadruple $(m,\dfrac{m+\mu+\chi}{2}+\chi\,A,2m,-2m,-2m)$ satisfies the functional equation

[TABLE]

while the functions $f:=\mu$, $g_{1}:=-\mu+\chi\,A$, $h_{1}:=\chi$, $h_{2}:=\mu+\chi$ and $h:=-\chi$ satisfy the functional equation

[TABLE]

with $g_{1}\neq f$, $h_{2}\neq f$, $h\neq 0$ and $h\neq\chi\,A$.

The papers cited above motivate us to deal with the functional equations (1.1), (1.2), (1.3), (1.4), (1.5), (1.6), (1.7) and (1.8), which are not covered by the cited works.

1.3. About our results

In the present paper we solve the functional equations (1.1), (1.2), (1.3), (1.4), (1.5), (1.6), (1.7) and (1.8), which are not in the literature. We express the solutions of the functional equations (1.1), (1.2), (1.3) and (1.4) respectively in term of the solutions of (1.5), (1.6), (1.7) and (1.8). We relate some categories of solutions to the solutions of the functional equation $f(xy)=f(x)\chi_{1}(y)+\chi_{2}(x)f(y)$ where $\chi_{1}$ and $\chi_{2}$ are two characters of the group $G$, which was recently solved by Stetkær [11]. Our results are extensions of those of [11] and [7].

There is no connection between the functional equations (1.1), (1.2), (1.3) and (1.4) nor between (1.5), (1.6), (1.7) and (1.8), but the reasonings for the functional equations with additive function , i.e. (1.3), (1.4), (1.7) and (1.8) are similar to those for the functional equations that not contain the additive function, i.e. (1.1), (1.2), (1.5), (1.6). This is due to fact that the resolution of this functional equations

(1) make use the linear independence between some unknown functions and the given arguments in equations, namely multiplicative functions, additive function or characters.

(2) relate some of them to the functional equation $f(xy)=f(x)\chi_{1}(y)+\chi_{2}(x)f(y)$, where $\chi_{1}$ and $\chi_{2}$ are two characters of the group $G$.

The organization of the paper is as follows. In the next section we give some definitions and notations. In the third section we give preliminary results that we need in the paper. In sections 4, 5, 6 and 7 we prove our main results and give some examples.

Our methods are elementary algebraic manipulations (no analysis or geometry come into play).

2. Set Up and Terminology

Definition 2.1**.**

Let $f:S\to\mathbb{C}$ be a function on a semigroup $S$. We say that

$f$ is central if $f(xy)=f(yx)$ for all $x,y\in G$.

$f$ is additive if $f(xy)=f(x)+f(y)$ for all $x,y\in S$.

$f$ is multiplicative if $f(xy)=f(x)f(y)$ for all $x,y\in S$.

If $S=G$ is a group, then we say that $f$ is a character of $G$ if $f$ is a homomorphism of $G$ into $\mathbb{C}^{*}$, where $\mathbb{C}^{*}$ denotes the multiplicative group of nonzero complex numbers.

If $G$ is a group, then $[G,G]$ denotes the commutator of $G$, i.e., the subgroup of $G$ generated by the set of commutators $[x,y]:=xyx^{-1}y^{-1},\,x,y\in G$.

If $\chi_{1}$ and $\chi_{2}$ are two characters of a group $G$, we denote by $\varphi_{\chi_{1},\chi_{2}}$ a solution of the functional equation

[TABLE]

Blanket assumption: Throughout this paper $G$ denotes a group or a monoid with identity element $e$.

3. Preliminaries

In this section we give the essential tools that we will need in the paper.

Proposition 3.1**.**

[10*, Theorem 3.18(a)]** Let $S$ be a semigroup and $n\in\mathbb{N}$. Let $\chi,\chi_{2},\cdot\cdot\cdot,\chi_{n}:S\to\mathbb{C}$ be $n$ different multiplicative functions and let $a_{1},a_{2},\cdot\cdot\cdot,a_{n}\in\mathbb{C}$.

If $a_{1}\chi+a_{2}\chi_{2}+\cdot\cdot\cdot+a_{n}\chi_{n}=0$, then $a_{1}\chi=a_{2}\chi_{2}=\cdot\cdot\cdot=a_{n}\chi_{n}=0$, in particular, any set of non zero multiplicative functions is linearly independent.*

Lemma 3.2**.**

[2, pp. 1122-1123]** Let $a:S\rightarrow\mathbb{C}$ be an additive function and $m:S\rightarrow\mathbb{C}$ be a multiplicative function on a semigroup $S$.

If $ma=\Sigma_{j=1}^{N}c_{j}\mu_{j}$, where $c_{j}\in\mathbb{C}$ and $\mu_{j}:S\rightarrow\mathbb{C}$ is multiplicative for each $j=1,2,\cdot\cdot\cdot{},N$, then $ma=0$.

Let $\chi_{1}$ and $\chi_{2}$ be characters of $G$. The following result, which is taken from [11, Proposition 4, Theorem 11], solves the functional equation

[TABLE]

Theorem 3.3**.**

[11*, p.5, pp .8-9]** Let $G$ be a group.

(1) If $\chi_{1}=\chi_{2}$ then the solutions of (3.1) are the functions $f=\chi_{1}\,A$ where $A:G\to\mathbb{C}$ is an additive function on $G$.

(2) If $\chi_{1}\neq\chi_{2}$ then the solutions of (3.1) are the functions*

[TABLE]

where $y_{0}\in G$ such that $\chi_{1}(y_{0})\neq\chi_{2}(y_{0})$, $\alpha$ ranges over $\mathbb{C}$ and $A:[G,G]\to\mathbb{C}$ ranges over the additive functions on $[G,G]$ with the transformation property

[TABLE]

Remark 3.4**.**

According to [11, Proposition 4], if $\chi_{1}\neq\chi_{2}$ the central solutions of (3.1) are the functions $f=\alpha(\chi_{1}-\chi_{2})$ where $\alpha$ ranges over $\mathbb{C}$.

Lemma 3.5**.**

*Let $G$ be a group. Let $(f,g_{1},h_{1},h_{2},h)$ be a solution of the functional equation (1.1).

(1) If $h_{1}=0$ then $f=h=b\chi$, $h_{2}=0$ and $g_{1}$ is arbitrary, where $b\in\mathbb{C}$ is a constant.

(2) If $h_{1}\neq 0$ and the set $\{f,\mu_{1},\mu_{2},\chi\}$ is linearly dependent then the set $\{g_{1},\mu_{1},\mu_{2},\chi\}$ is also linearly dependent.*

Proof.

(1) Assume that $h_{1}=0$. Then $g_{1}$ is arbitrary. By putting $y=e$ in (1.1), we get that $f=a\mu_{1}+a\mu_{2}+b\chi$ where $a:=\dfrac{h_{2}(e)}{2}$ and $b:=h(e)$. Substituting this back into (1.1) we obtain

[TABLE]

for all $x,y\in G$. Since the characters $\mu_{1}$, $\mu_{2}$ and $\chi$ are different the set $\{\mu_{1},\mu_{2},\chi\}$ is linearly independent. Then we get, from the identity above, that $\dfrac{h_{2}}{2}=a\mu_{1}=a\mu_{2}$ and $h=b\chi$. So that $a=0$ because $\mu_{1}\neq\mu_{2}$. Hence, $f=h=b\chi$ and $h_{2}=0$.

(2) Assume that $h_{1}\neq 0$ and $\{f,\mu_{1},\mu_{2},\chi\}$ is linearly dependent. Since the set $\{\mu_{1},\mu_{2},\chi\}$ is linearly independent there exist $\alpha_{1},\alpha_{2},\alpha_{3}\in\mathbb{C}$ such that $f=\alpha_{1}\mu_{1}+\alpha_{2}\mu_{2}+\alpha_{3}\chi$. So that

[TABLE]

for all $x,y\in G$, from which we get, taking (1.1) into account, that

[TABLE]

for all $x,y\in G$. As $h_{1}\neq 0$ we deduce from the identity above, by choosing $y_{0}\in G$ such that $h_{1}(y_{0})\neq 0$, that the set $\{g_{1},\mu_{1},\mu_{2},\chi\}$ is linearly dependent. ∎

Proposition 3.6**.**

*Let $G$ be a monoid. For $f=0$ the solutions of the functional equation (1.3) can be listed as follows

(1) $g_{1}$ arbitrary and $h_{1}=h_{2}=h=0$;

(2) $g_{1}=a\dfrac{\mu+\chi}{2}+b\,\chi\,A$, $h_{1}\neq 0$ arbitrary, $h_{2}=-ah_{1}$ and $h=-bh_{1}$, where $a$ and $b$ range over $\mathbb{C}$.*

Proof.

(1) Assume that $h_{1}=0$. Then $g_{1}$ is arbitrary. Let $y\in G$ be arbitrary. Since $f=0$ the functional equation (1.3) implies

[TABLE]

for all $x\in G$. So, the set $\{\mu,\chi,\chi\,A\}$ being linearly independent, $h_{2}(y)=h(y)=0$ and $y$ arbitrary, we deduce that $h_{2}=h=0$.

(2) Assume that $h_{1}\neq 0$. Taking $x=e$ in (1.3) and seeing that $\mu(e)=\chi(e)=1$ and $A(e)=0$ we get that $g_{1}(e)h_{1}(y)+h_{2}(y)=0$ for all $y\in G$. So that

[TABLE]

where $a:=g_{1}(e)$. Substituting (3.3) back into (1.3) we get that

[TABLE]

for all $x,y\in G$. Choosing $y_{0}\in G$ such that $h_{1}(y_{0})\neq 0$ we deduce from the last functional equation that

[TABLE]

where $b:=-\dfrac{h(y_{0})}{h_{1}(y_{0})}$.

Substituting (3.3) and (3.3) back into (1.3) we get that

[TABLE]

for all $x,y\in G$. Since $\chi\,A\neq 0$ we derive that

[TABLE]

Conversely, it is easy to check that the formulas of (1) and (2) define solutions of (1.3) for $f=0$. ∎

Proposition 3.7**.**

*Let $G$ be a group and let $\alpha\in\mathbb{C}$ be a constant. For $f=\alpha\mu$ the solutions of the functional equation (1.4) can be listed as follows

(1) $g_{1}$ arbitrary, $h_{1}=0$, $h_{2}=\alpha\mu$ and $h=0$;

(2) $g_{1}=a\mu+b\chi\,A$, $h_{1}\neq 0$ arbitrary, $h_{2}=-ah_{1}+\alpha\mu$ and $h=-bh_{1}$, where $a$ and $b$ range over $\mathbb{C}$.*

Proof.

By proceeding as in the proof of Proposition 3.6. ∎

4. Solutions of Eq. (1.5) and Eq. (1.1)

In this section $\mu_{1},\,\mu_{2}$ and $\chi$ are different characters of a group $G$, and $g:=\dfrac{\mu_{1}+\mu_{2}}{2}$.

4.1. Solutions of Eq. (1.5)

In this subsection we solve the functional equation (1.5), i.e.,

[TABLE]

where $f,h_{1},h_{2},h:G\to\mathbb{C}$ are the unknown functions to be determined.

Proposition 4.1 solves the functional equation (1.5) when the set $\{f,\mu_{1},\mu_{2},\chi\}$ is linearly independent.

Proposition 4.1**.**

Let $(f,h_{1},h_{2},h)$ be a solution of (1.5) such that the set $\{f,\mu_{1},\mu_{2},\chi\}$ is linearly independent. Then

[TABLE]

where $\mu$ is a character of $G$ and $c\in\mathbb{C}$ is a constant.

Proof.

Let $x,y,z\in G$ be arbitrary. We compute $f(xyz)$ as $f(x(yz))$ and then as $f((xy)z)$. Using Eq. (1.5) and that $g=\dfrac{\mu_{1}+\mu_{2}}{2}$ we obtain

[TABLE]

On the other hand, by applying Eq. (1.5) to the pair $(xy,z)$ we get that

[TABLE]

So that

[TABLE]

As the set $\{f,\mu_{1},\mu_{2},\chi\}$ is linearly independent and $x,y,z$ are arbitrary, we derive from (4.1) and (4.2) that $(h_{1},h_{2},h)$ is a solution of the following functional equations as a system

[TABLE]

[TABLE]

[TABLE]

and

[TABLE]

for all $x,y\in G$. The functional equation (4.3) implies that $h_{1}=\mu$ where $\mu:G\to\mathbb{C}$ is a multiplicative function. By subtracting (4.4) from (4.5) we get that

[TABLE]

for all $x,y\in G$. Since $\mu_{1}\neq\mu_{2}$ we have $h_{2}(y)=0$. So, $y$ being arbitrary, we deduce that $h_{2}=0$.

If $\mu=0$, then $h_{1}=0$ and the functional equation (4.6) reduces to

[TABLE]

for all $x,y\in G$, which yields, by putting $y=e$, that $h=\alpha\chi\,\,\text{where}\,\,\alpha:=h(e)$. Thus the functional equation (1.5) becomes $f(xy)=\alpha\chi(xy)$ for all $x,y\in G$. This implies, by putting $y=e$, that $f=\alpha\chi$, which contradicts the linear independence of the set $\{f,\mu_{1},\mu_{2},\chi\}$.

Hence $\mu\neq 0$ and then $\mu$ is a character of $G$. So, the functional equation (4.6) implies that

[TABLE]

So that the functional equation (1.5) reduces to

[TABLE]

for all $x,y\in G$, from which we get, by putting $x=e$ and $c=f(e)$, that $f=c\mu+\varphi_{\mu,\chi}$. ∎

Theorem 4.2**.**

*The solutions $(f,h_{1},h_{2},h)$ of (1.5) are:

(1) $f=c\mu+\varphi_{\mu,\chi},\,h_{1}=\mu,\,h_{2}=0$ and $h=\varphi_{\mu,\chi}$, where $\mu$ is a character of $G$ and $c\in\mathbb{C}$ is a constant;

(2) $f=c\chi,\,h_{1}\,\,\text{is arbitrary},\,h_{2}=0\,\,\text{and}\,\,h=c(\chi-h_{1})$, where $c\in\mathbb{C}$ is a constant;

(3)*

[TABLE]

[TABLE]

where $a,b,c\in\mathbb{C}$ are constants such that $a\neq b$.

Proof.

It is easy to check that the formulas in parts (1)-(3) define solutions of the functional equation (1.5), so left is to show that any solution $(f,h_{1},h_{2},h)$ fits into (1)-(3).

If the set $\{f,\mu_{1},\mu_{2},\chi\}$ is linearly independent we get, by Proposition 4.1, the part (1) of Theorem 6.2.

In the remainder of the proof we assume that $\{f,\mu_{1},\mu_{2},\chi\}$ is linearly dependent. Recall that the set $\{\mu_{1},\mu_{2},\chi\}$ is linearly independent because $\mu_{1}$, $\mu_{2}$ and $\chi$ are different characters of $G$. Then there exists a triple $(a,b,c)\in\mathbb{C}^{3}$ such that

[TABLE]

Substituting this back into (1.5) we get by a small computation that

[TABLE]

for all $x,y\in G$. Since $\mu_{1}$, $\mu_{2}$ and $\chi$ are different characters of $G$, we derive by Proposition 3.1 that $(h_{1},h_{2},h)$ is a solution of the following identities as a system

[TABLE]

[TABLE]

and

[TABLE]

When we subtract (4.9) from (4.8) we obtain

[TABLE]

If $a=b$ the identity (4.11) reduces to $a(\mu_{1}-\mu_{2})=0$. Hence $a=b=0$ because $\mu_{1}\neq\mu_{2}$. So, taking (4.7) and (4.10) into account, we get that

[TABLE]

which is solution (2).

If $a\neq b$ the identity (4.11) implies that

[TABLE]

Substituting this back into (4.8) and (4.10) we derive that

[TABLE]

and

[TABLE]

The solution occurs in part (3). This completes the proof. ∎

4.2. Solutions of Eq. (1.1)

In Theorem 4.3 we solve the functional equation (1.1), i.e.,

[TABLE]

where $f,g_{1},h_{1},h_{2},h:G\to\mathbb{C}$ are the unknown functions to be determined.

Theorem 4.3**.**

*The solutions $(f,g_{1},h_{1},h_{2},h)$ of (1.1) can be listed as follows:

(1) $f=b\chi$, $g_{1}$ is arbitrary, $h_{1}=h_{2}=0$ and $h=b\chi$, where $b\in\mathbb{C}$ is a constant;

(2) $f=b\chi$, $g_{1}=\alpha\mu_{1}+\alpha\mu_{2}+\gamma\chi$, $h_{1}\neq 0$ is arbitrary, $h_{2}=-2\alpha h_{1}$ and $h=b\chi-\gamma h_{1}$, where $b\in\mathbb{C}\setminus\{0\}$ and $\alpha,\gamma\in\mathbb{C}$ are constants;

(3)*

[TABLE]

[TABLE]

*where $a_{1},a_{2},a_{3},d_{1},d_{2},d_{3}\in\mathbb{C}$ are constants such that $d_{1}\neq d_{2}$;

(4) $f=a_{1}b\mu+b\varphi_{\mu,\chi}$, $g_{1}=-\frac{1}{2}a_{2}\mu_{1}-\frac{1}{2}a_{2}\mu_{2}-a_{3}\chi+a_{1}\mu+\varphi_{\mu,\chi}$, $h_{1}=b\mu$, $h_{2}=a_{2}b\mu$ and $h=a_{3}b\mu+b\varphi_{\mu,\chi}$, where $b\in\mathbb{C}\setminus\{0\},a_{1},a_{2},a_{3}\in\mathbb{C}$ are constants and $\mu$ is a character of $G$;

(5)*

[TABLE]

[TABLE]

where $b\in\mathbb{C}\setminus\{0\}$ and $a_{1},\,a_{2},\,a_{3}\in\mathbb{C}$ are constants.

Proof.

We check by elementary computations that if $f,g_{1},h_{1},h_{2}$ and $h$ are of the forms (1)-(5) then $(f,g_{1},h_{1},h_{2},h)$ is a solution of (1.1), so left is that any solution $(f,g_{1},h_{1},h_{2},h)$ of (1.1) fits into (1)-(5).

By putting $y=e$ in (1.1) we get that

[TABLE]

We split the discussion into the subcases $h_{1}(e)=0$ and $h_{1}(e)\neq 0$.

Case 1: Suppose $h_{1}(e)=0$. Then (4.12) gives

[TABLE]

where $a:=\dfrac{h_{2}(e)}{2}$ and $b:=h(e)$. Hence, the set $\{f,\mu_{1},\mu_{2},\chi\}$ is linearly dependent.

If $h_{1}=0$ then, according to Lemma 3.5(1), we get that $g_{1}$ is arbitrary, $h_{2}=0$ and $f=h=b\chi$, which is solution (1).

If $h_{1}\neq 0$. So, by Lemma 3.5(2), the set $\{g_{1},\mu_{1},\mu_{2},\chi\}$ is linearly dependent. Since the set $\{\mu_{1},\mu_{2},\chi\}$ is linearly independent there exists a triple $(\alpha,\beta,\gamma)\in\mathbb{C}^{3}$ such that

[TABLE]

By substituting (4.13) and (4.14) in (1.1) we derive by simple computations that

[TABLE]

for all $x,y\in G$. So, the set $\{\mu_{1},\mu_{2},\chi\}$ being linearly independent, we derive from the identity above that $(h_{1},h_{2},h)$ satisfies the following identities as a system

[TABLE]

[TABLE]

and

[TABLE]

By subtracting (4.16) from (4.15) we get that

[TABLE]

If $\alpha=\beta$ then (4.18) implies that $a=0$ because $\mu_{1}\neq\mu_{2}$. Hence, from (4.13), (4.14), (4.15) and (4.18) we get that

[TABLE]

This is solution (2).

If $\alpha\neq\beta$ then we obtain, from (4.18), that

[TABLE]

By substituting (4.19) in (4.15) and in (4.17) we get that

[TABLE]

and

[TABLE]

Let

[TABLE]

[TABLE]

Notice that $d_{1}\neq d_{2}$ because $\alpha-\beta\neq 0$. By using (4.13), (4.14), (4.19), (4.20) and (4.21) we obtain by small computations

[TABLE]

[TABLE]

[TABLE]

and

[TABLE]

which is solution (3).

Case 2: Suppose $h_{1}(e)\neq 0$. Then from (4.12) we get that

[TABLE]

Substituting this back into (1.1) we obtain

[TABLE]

for all $x,y\in G$. So that

[TABLE]

for all $x,y\in G$, i.e., the quadruple $(f,h_{1}/h_{1}(e),h_{2}-\dfrac{h_{2}(e)}{h_{1}(e)}h_{1},h-\dfrac{h(e)}{h_{1}(e)}h_{1})$ is a solution of the functional equation (1.5), so, according to Theorem 6.2, that quadruple falls into three categories:

(i) $f=c\mu+\varphi_{\mu,\chi},\,h_{1}/h_{1}(e)=\mu,\,h_{2}-\dfrac{h_{2}(e)}{h_{1}(e)}h_{1}=0,\,h-\dfrac{h(e)}{h_{1}(e)}h_{1}=\varphi_{\mu,\chi}$,

where $\mu$ is a character of $G$ and $c\in\mathbb{C}$ is a constant. We define complex constants by $b:=h_{1}(e)\in\mathbb{C}\setminus\{0\}$, $a_{1}:=\dfrac{c}{h_{1}(e)}$, $a_{2}:=\dfrac{h_{2}(e)}{h_{1}(e)}$ and $a_{3}:=\dfrac{h(e)}{h_{1}(e)}$. So, the identities above and (4.22) yield, by writing $\varphi_{\mu,\chi}$ instead of $\dfrac{1}{b}\varphi_{\mu,\chi}$, that

[TABLE]

[TABLE]

which is solution (4).

(ii)

[TABLE]

where $c\in\mathbb{C}$ is a constant.

Defining complex constants $b:=h_{1}(e)\in\mathbb{C}\setminus\{0\}$, $a_{1}:=\dfrac{c}{h_{1}(e)}\in\mathbb{C}$, $a_{2}:=\dfrac{h_{2}(e)}{h_{1}(e)}$ and $a_{3}:=\dfrac{h(e)}{h_{1}(e)}$ we obtain, by a small computation, from the identities above and (4.22) that

[TABLE]

[TABLE]

which is solution (5).

(iii)

[TABLE]

[TABLE]

where $a,b,c\in\mathbb{C}$ such that $a\neq b$. Using (4.22) and the identities above we get that

[TABLE]

Defining

[TABLE]

[TABLE]

we have $a=\frac{1}{2}a_{1}(d_{1}-d_{2})$, $b=-\frac{1}{2}(d_{1}-d_{2})$ and $c=-\frac{1}{2}a_{3}(d_{1}-d_{3})$. Sine $a\neq b$ we have $d_{1}\neq d_{2}$. Moreover the identities $f=a\mu_{1}+b\mu_{2}+c\chi$ and (4.23) read

[TABLE]

[TABLE]

and

[TABLE]

which is solution (3). This completes the proof of Theorem 4.3. ∎

From Case 1 of the proof of Theorem 4.3 we derive that if $f,g_{1},h_{1},h_{2},h:G\to\mathbb{C}$ satisfy the functional equation (1.1) and $h_{1}(e)=0$, then $f$ and $g_{1}$ are abelian, or $f,\,g_{1},\,h_{1},\,h_{2}$ and $h$ are abelian.

According to [11, Proposition 5], if $\mu\neq\chi$ and $G$ is an abelian group then the function $\varphi_{\mu,\chi}$, in the formulas of solutions of the form (4) in Theorem 4.3, take the form $\varphi_{\mu,\chi}=c(\mu-\chi)$, where $c$ ranges over $\mathbb{C}$. On groups that need not be abelian, the expression of $\varphi_{\mu,\chi}$ is given by Theorem 3.3.

Example 4.4**.**

If we choice $b=\gamma=0$ in Theorem 4.3(2), or $a_{3}=d_{3}=0$ in Theorem 4.3(3), or $a_{3}=0$ and $\varphi_{\mu,\chi}=0$ in Theorem 4.3(4), we get that $h=0$, then the functional equation (1.1) reduces to

[TABLE]

*which was solved in [7, Theorem 8].

(1) for $b=\gamma=0$ in Theorem 4.3(2), we get, by putting $c=2\alpha$, that*

[TABLE]

*which is the solution obtained in [7, Theorem 8(d)].

(2) for $a_{3}=d_{3}=0$ in Theorem 4.3(3) we obtain*

[TABLE]

and

[TABLE]

*where $a_{1},\,a_{2},\,d_{1},\,d_{2}\in\mathbb{C}$ are constants. Furthermore for $f\neq 0$ we have $d_{1}\neq d_{2}$ and $(a_{1},a_{2})\neq(0,0)$. The solution was obtained in [7, Proposition 6(b), Theorem 8(b)].

(3) for $a_{3}=0$ and $\varphi_{\mu,\chi}=0$ in Theorem 4.3(4),*

[TABLE]

where $a_{1},\,a_{2},\,\in\mathbb{C}$ are constants. Furthermore for $f\neq 0$ we have $a_{1},\,b\in\mathbb{C}\setminus\{0\}$. The solution was obtained in [7, Theorem 8(a)].

Example 4.5**.**

By taking $d_{1}=-d_{2}$, $a_{1}=a_{2}=\dfrac{1}{2}$ and $a_{3}=d_{3}=0$ in Theorem 4.3(3) we get that

[TABLE]

which is the solution of the classic sine addition law obtained in [10, Theorem 4.1(c)].

5. Solutions of Eq. (1.6) and Eq. (1.2)

In this section $\chi_{1},\cdot\cdot\cdot,\chi_{N}$ are $N$ distinct characters of a group $G$.

5.1. Solutions of Eq. (1.6)

In Theorem 5.1 we solve the functional equation (1.6), i.e.,

[TABLE]

where $f,\,h,\,h_{1},\cdot\cdot\cdot,\,h_{N}:G\to\mathbb{C}$ are the unknown functions to be determined.

Theorem 5.1**.**

*The solutions $(f,\,h,\,h_{1},\cdot\cdot\cdot,\,h_{N})$ of (1.6) are:

(1)*

[TABLE]

*for all $j=1,\cdot\cdot\cdot,N$, where $\alpha_{1},\cdot\cdot\cdot,\alpha_{N}\in\mathbb{C}$ are constants ;

(2)*

[TABLE]

for all $j=1,\cdot\cdot\cdot,N$, where $\chi$ is a character of $G$ and $a\in\mathbb{C}$ is a constant.

Proof.

Elementary computations show that if $f,h,h_{1},\cdot\cdot\cdot,h_{N}$ are of the forms (1)-(2) then $(f,h,h_{1},\cdot\cdot\cdot,h_{N})$ is a solution of (1.6), so left is that any solution $(f,h,h_{1},\cdot\cdot\cdot,h_{N})$ of (1.6) fits into (1)-(2). We consider two cases.

Case 1: Suppose that the set $\{f,\chi_{1},\cdot\cdot\cdot,\chi_{N}\}$ is linearly dependent. Since $\chi_{1},\cdot\cdot\cdot,\chi_{N}$ are different characters then, according to Proposition 3.1, the set $\{\chi_{1},\cdot\cdot\cdot,\chi_{N}\}$ is linearly independent. So there exist $\alpha_{1},\cdot\cdot\cdot,\alpha_{N}\in\mathbb{C}$ such that

[TABLE]

Substituting this in (1.6) we get that

[TABLE]

for all $x,y\in G$. So, by using the linear independence of $\{\chi_{1},\cdot\cdot\cdot,\chi_{N}\}$, we obtain

[TABLE]

for all $y\in G$ and all $j=1,\cdot\cdot\cdot,N$. Hence

[TABLE]

for all $j=1,\cdot\cdot\cdot,N$, with $h$ arbitrary. The result occurs in part (1).

Case 2: Suppose that the set $\{f,\chi_{1},\cdot\cdot\cdot,\chi_{N}\}$ is linearly independent. Let $x,y,z\in G$ be arbitrary. We compute $f(zxy)$ by using the associativity of the operation of $G$. We have

[TABLE]

On the other hand, by using (1.6), and that $\chi_{1},\cdot\cdot\cdot,\chi_{N}$ are multiplicative, we obtain

[TABLE]

Since the set $\{f,\chi_{1},\cdot\cdot\cdot,\chi_{N}\}$ is linearly independent and $x,y,z$ are arbitrary we derive from (5.1) and (5.2) that

[TABLE]

and

[TABLE]

for all $x,y\in G$ and all $j=1,\cdot\cdot\cdot,N$. The functional equation (5.3) says that $\chi:=h$ is a multiplicative function of $G$. If there exists $x_{0}\in G$ such that $\chi(x_{0})=0$ then $h=\chi=0$. So, by putting $y=e$ in (1.6), we get that $f=\sum_{j=1}^{N}h_{j}(e)\chi$, which contradicts that the set $\{f,\chi_{1},\cdot\cdot\cdot,\chi_{N}\}$ is linearly independent. Thus $\chi$ is a character of $G$ and the functional equations (5.5) becomes

[TABLE]

for all $x,y\in G$ and all $j=1,\cdot\cdot\cdot,N$, i.e., $h_{j}=\varphi_{\chi,\chi_{j}}$ for all $j=1,\cdot\cdot\cdot,N$.

To find $f$ we put $x=e$ in (1.6) which yields that

[TABLE]

where $a:=f(e)$. The result occurs in part (2). This completes the proof of Theorem 5.1. ∎

5.2. Solutions of Eq. (1.2)

In Theorem 5.2 we solve the functional equation (1.2), i.e.,

[TABLE]

where $f,\,g,\,h,\,h_{1},\cdot\cdot\cdot,\,h_{N}:G\to\mathbb{C}$ are the unknown functions to be determined.

Theorem 5.2**.**

*The solutions $(f,\,g,\,h,\,h_{1},\cdot\cdot\cdot,\,h_{N})$ of (1.2) are:

(1)*

[TABLE]

*for all $j=1,\cdot\cdot\cdot,N$, where $a_{1},\cdot\cdot\cdot,a_{N}\in\mathbb{C}$ are constants;

(2)*

[TABLE]

*for all $j=1,\cdot\cdot\cdot,N$, where $a_{1},\cdot\cdot\cdot,a_{N},\beta_{1},\cdot\cdot\cdot,\beta_{N}\in\mathbb{C}$ are constants;

(3)*

[TABLE]

for all $j=1,\cdot\cdot\cdot,N$, where $\chi$ is a character of $G$, and $\alpha,a_{1},\cdot\cdot\cdot,a_{N}\in\mathbb{C}$ and $b\in\mathbb{C}\setminus\{0\}$ are constants.

Proof.

We check by elementary computations that if $f,\,g,\,h,\,h_{1},\cdot\cdot\cdot,\,h_{N}$ are of the forms (1)-(3) then $(f,\,g,\,h,\,h_{1},\cdot\cdot\cdot,\,h_{N})$ is a solution of (1.2), so left is that any solution $(f,\,g,\,h,\,h_{1},\cdot\cdot\cdot,\,h_{N})$ of (1.2) fits into (1)-(3).

We will discuss two cases according to whether the set $\{f,\chi_{1},\cdot\cdot\cdot,\chi_{N}\}$ is linearly independent or not.

Case 1: Suppose that the set $\{f,\chi_{1},\cdot\cdot\cdot,\chi_{N}\}$ is linearly dependent. As seen earlier the set $\{\chi_{1},\cdot\cdot\cdot,\chi_{N}\}$ is linearly independent. Then there exist $a_{1},\cdot\cdot\cdot,a_{N}\in\mathbb{C}$ such that

[TABLE]

There are two subcases to consider.

Subcase 1.1: Suppose that $h=0$. Then $g$ is arbitrary, and taking (5.6) into account, the functional equation (1.2) implies that

[TABLE]

for all $x,y\in G$ and all $j=1,\cdot\cdot\cdot,N$. So, the set $\{\chi_{1},\cdot\cdot\cdot,\chi_{N}\}$ being linearly independent, and $x$ and $y$ arbitrary, we get, according to Proposition 3.1, that $h_{j}=a_{j}\chi_{j}$ for all $j=1,\cdot\cdot\cdot,N$. The result occurs in part (1).

Subcase 1.2: Suppose that $h\neq 0$. Then, we derive from Eq. (1.2) that there exist constants $\beta_{1},\cdot\cdot\cdot,\beta_{N}\in\mathbb{C}$ such that

[TABLE]

By substituting (5.7) in (1.2) we obtain by a small computation, that

[TABLE]

for all $x,y\in G$. Since the set $\{\chi_{1},\cdot\cdot\cdot,\chi_{N}\}$ is linearly independent, and $x$ and $y$ arbitrary, we get, according to Proposition 3.1, that

[TABLE]

for all $j=1,\cdot\cdot\cdot,N$. The identities (5.6), (5.7) and (5.8), with $h\neq 0$ arbitrary, constitute the result (2).

Case 2: Suppose that the set $\{f,\chi_{1},\cdot\cdot\cdot,\chi_{N}\}$ is linearly independent. By putting $y=e$ in the functional equation (1.2) we get that

[TABLE]

Since the set $\{f,\chi_{1},\cdot\cdot\cdot,\chi_{N}\}$ is linearly independent the identity (5.9) impose that $h(e)\neq 0$, which implies that

[TABLE]

Substituting this back into (1.2) we get, by elementary computation, that

[TABLE]

for all $x,y\in G$. Then the functions $f,h/h(e),h_{1}-\dfrac{h_{1}(e)}{h(e)}h,\cdot\cdot\cdot,h_{N}-\dfrac{h_{N}(e)}{h(e)}h$ satisfy the functional equation (1.6). Hence, according to Theorem 5.1 and seeing that the set $\{f,\chi_{1},\cdot\cdot\cdot,\chi_{N}\}$ is linearly independent, we derive that:

[TABLE]

for all $j=1,\cdot\cdot\cdot,N$, where $\chi$ is a character of $G$ and $a\in\mathbb{C}$ is a constant. Using this and the identity (5.10) we get that

[TABLE]

[TABLE]

[TABLE]

and

[TABLE]

for all $j=1,\cdot\cdot\cdot,N$. Defining complex constants

[TABLE]

and written $\varphi_{\chi,\chi_{j}}$ instead of $\dfrac{1}{b}\varphi_{\chi,\chi_{j}}$ for all $j=1,\cdot\cdot\cdot,N$, the formulas above read

[TABLE]

for all $j=1,\cdot\cdot\cdot,N$. The result occurs in part (3). This completes the proof. ∎

We close this section with relating our results to two functional equations in the literature. The proof of Theorem 5.2 reveals that if the set $\{f,\chi_{1},\cdot\cdot\cdot,\chi_{N}\}$ is linearly independent, then the solutions $(f,\,g,\,h,\,h_{1},\cdot\cdot\cdot,\,h_{N})$ of the functional equation (1.2) are of the form (3) in Theorem 5.2. This is an extension of [11, Theorem 14] as example 5.3 says.

Example 5.4 relate Theorem 5.2(3) to the simple generalization of Prexider’s functional equation

[TABLE]

where $\chi$ is a character of $G$ (see [11, Proposition 15]).

Example 5.3**.**

If we choice $a_{j}=0$ and $\varphi_{\chi,\chi_{j}}=0$ for all $j=2,\cdot\cdot\cdot,N$ in part (3) of Theorem 5.2 we get that $h_{j}=0$ for all $j=2,\cdot\cdot\cdot,N$. With the notations $c_{1}:=\alpha b,\,c_{2}:=a_{1}b-c_{1}$ and $f_{\chi,\chi_{1}}:=c_{1}(\chi-\chi_{1})+b\varphi_{\chi,\chi_{1}}$ we get that

[TABLE]

where $b\in\mathbb{C}\setminus\{0\}$ and $c_{1},c_{2}\in\mathbb{C}$ are constants, and $f_{\chi,\chi_{1}}:G\to\mathbb{C}$ satisfies the functional equation

[TABLE]

which is the solution obtained in [11, Theorem 14].

Example 5.4**.**

If we take $b=1$, $\chi=\chi_{1}$, $a_{j}=0$ and $\varphi_{\chi,\chi_{j}}=0$ for all $j=2,\cdot\cdot\cdot,N$ in part (3) of Theorem 5.2, we get that $h_{j}=0$ for all $j=2,\cdot\cdot\cdot,N$ and $\varphi_{\chi,\chi_{1}}=\varphi_{\chi,\chi}=\chi A$ where $A$ is an additive function on $G$. With the notations $c_{1}:=\alpha-a_{1}\,\text{and}\,\,c_{2}:=a_{1}$ we obtain

[TABLE]

where $c_{1},c_{2}\in\mathbb{C}$ are constants, which is the solution obtained in [11, Proposition 15].

6. Solutions of Eq. (1.7) and Eq. (1.3)

Throughout this section $G$ denotes a monoid with identity element $e$, $\mu,\chi:G\to\mathbb{C}$ different nonzero multiplicative functions, $A:G\to\mathbb{C}$ an additive function such that $\chi A\neq 0$ and $g:=\dfrac{\mu+\chi}{2}$.

6.1. Solutions of Eq. (1.7)

Proposition 6.1 solves the functional equation (1.7), i.e.,

[TABLE]

under the assumption that the set $\{f,\mu,\chi,\chi\,A\}$ is linearly independent.

Proposition 6.1**.**

*The solutions $(f,h_{1},h_{2},h)$ of (1.7) such that the set $\{f,\mu,\chi,\chi\,A\}$ is linearly independent are:

[TABLE]

where $m$ is a non zero multiplicative function and $c\in\mathbb{C}\setminus\{0\}$ is a constant such that $m\neq\mu$ and $m\neq\chi$.

Proof.

It is easy to check that the formulas in Proposition 6.1 define solutions of Eq. (1.7), so left is that any solution $(f,h_{1},h_{2},h)$ is of that form.

Let $x,y,z\in G$ be arbitrary. First we compute $f(xyz)$ as $f(x(yz))$ and then as $f((xy)z)$. Using Eq. (1.7) and that $g=\dfrac{\mu+\chi}{2}$ we obtain

[TABLE]

On the other hand, by making use that $\mu$ and $\chi$ are multiplicative and that $A$ is additive, we get from Eq. (1.7) that

[TABLE]

So that

[TABLE]

As the set $\{f,\mu,\chi,\chi\,A\}$ is linearly independent and $x,y,z$ are arbitrary, we derive from (6.1) and (6.2) that $(h_{1},h_{2},h)$ is a solution of the following functional equations as a system

[TABLE]

[TABLE]

and

[TABLE]

for all $x,y\in G$. The functional equation (6.3) implies that $h_{1}=m$ where $m:G\rightarrow\mathbb{C}$ is a multiplicative function. From (6.4) and (6.5) we get that

[TABLE]

for all $x,y\in G$, from which we deduce that $h(y)=h_{2}(y)=0$ for all $y\in G$, and then $h_{2}=h=0$. Hence, the functional equation (1.7) reduces to

[TABLE]

for all $x,y\in G$, which yields, by putting $x=e$, that

[TABLE]

If $c=0$ or $m\in\{0,\mu,\chi\}$, then the set $\{f,\mu,\chi,\chi\,A\}$ is linearly dependent, contradicting the hypothesis. Thus $c\in\mathbb{C}\setminus\{0\}$, $m\neq 0$, $m\neq\mu$ and $m\neq\chi$. ∎

Theorem 6.2**.**

*The solutions $(f,h_{1},h_{2},h)$ of (1.7) are:

(1) $f=cm,\,h_{1}=m,\,h_{2}=h=0$, where $m$ is a non zero multiplicative function and $c\in\mathbb{C}\setminus\{0\}$ is a constant such that $m\neq\mu$ and $m\neq\chi$;

(2) $f=0,\,h_{1}\,\,\text{is arbitrary}\,\,\text{and}\,\,h_{2}=h=0$;

(3)*

[TABLE]

[TABLE]

where $a,b,c\in\mathbb{C}$ are constants such that $(a,b,c)\neq(0,0,0)$ and $a\neq b$.

Proof.

(1) If the set $\{f,\mu,\chi,\chi\,A\}$ is linearly independent we get, by Proposition 6.1, the part (1) of Theorem 6.2.

In the remainder of the proof we assume that $\{f,\mu,\chi,\chi\,A\}$ is linearly dependent.

(2) If $f=0$ then $h_{1}$ is arbitrary and the functional equation (1.7) implies that

[TABLE]

for all $x,y\in G$. So $h_{2}=h=0$. In what follows we assume that $f\neq 0$.

(3) By using Lemma 3.2 we get that the set $\{\mu,\chi,\chi\,A\}$ is linearly independent. So, seeing that $f\neq 0$ there exists a triple $(a,b,c)\in\mathbb{C}^{3}\setminus\{(0,0,0)\}$ such that

[TABLE]

Substituting this back into (1.7) we get by a small computation that

[TABLE]

for all $x,y\in G$. We derive that $(h_{1},h_{2},h)$ is a solution of the following identities as a system

[TABLE]

[TABLE]

and

[TABLE]

When we subtract (6.7) from (6.6) we obtain

[TABLE]

We have $a\neq b$. Indeed, if $a=b$ the identity (6.9) reduces to

[TABLE]

Hence $a=b=c=0$, which contradicts the assumption on $f$. Hence, the identity (6.9) implies that

[TABLE]

Substituting this back into (6.6) and (6.8) we derive that

[TABLE]

and

[TABLE]

Conversely, simple computations show that the formulas in (1)-(3) define solutions of (1.7). ∎

6.2. Solutions of Eq. (1.3)

In Theorem 6.3 we solve the functional equation (1.3), i.e.,

[TABLE]

where $f,g_{1},h_{1},h_{2},h:G\rightarrow\mathbb{C}$ are the unknown functions to be determined.

Theorem 6.3**.**

*The solutions $(f,g_{1},h_{1},h_{2},h)$ of (1.3) can be listed as follows:

(1) $f=0$, $g_{1}$ is arbitrary and $h_{1}=h_{2}=h=0$;

(2) $f=0$, $g_{1}=a\dfrac{\mu+\chi}{2}+b\,\chi\,A$, $h_{1}\neq 0$ is arbitrary, $h_{2}=-ah_{1}$ and $h=-bh_{1}$, where $a,b\in\mathbb{C}$ are constants.

(3)*

[TABLE]

[TABLE]

and

[TABLE]

*where $a_{1},a_{2},a_{3},d_{1},d_{2},d_{3}\in\mathbb{C}$ are constants such that $d_{1}\neq d_{2}$ and $(a_{1},a_{2},a_{3})\neq(0,0,0)$.

(4) $f=a_{1}\alpha m$, $g_{1}=\alpha m-a_{2}\dfrac{\mu+\chi}{2}-a_{3}\chi\,A$, $h_{1}=a_{1}m$, $h_{2}=a_{1}a_{2}m$ and $h=a_{1}a_{3}m$, where $\alpha,a_{1}\in\mathbb{C}\setminus\{0\}$ and $a_{2},a_{3}\in\mathbb{C}$ are constants, and $m:G\rightarrow\mathbb{C}$ is a non zero multiplicative function.*

Proof.

We check by elementary computations that if $f,g_{1},h_{1},h_{2}$ and $h$ are of the forms (1)-(4) then $(f,g_{1},h_{1},h_{2},h)$ is a solution of (1.3), so left is that any solution $(f,g_{1},h_{1},h_{2},h)$ of (1.3) fits into (1)-(4).

There are two cases to consider.

Case 1. Suppose $f=0$. By Proposition 3.6, we obtain solutions (1) and (2).

Case 2. Suppose $f\neq 0$. Taking $y=e$ in (1.3) we get that

[TABLE]

We split the discussion into the subcases $h_{1}(e)=0$ and $h_{1}(e)\neq 0$.

Subcases 2.1. Suppose $h_{1}(e)=0$. Then (6.10) gives

[TABLE]

where $a:=\dfrac{h_{2}(e)}{2}$, $b:=h(e)$ and $(a,b)\neq(0,0)$. Hence, the set $\{f,\mu,\chi,\chi\,A\}$ is linearly dependent. As $f\neq 0$ we get, according to Lemma 3.5(1), that $h_{1}\neq 0$. So, by Lemma 3.5(2), the set $\{g_{1},\mu,\chi,\chi\,A\}$ is linearly dependent. Since the set $\{\mu,\chi,\chi\,A\}$ is linearly independent there exists a triple $(\alpha,\beta,\gamma)\in\mathbb{C}^{3}$ such that

[TABLE]

By substituting (6.11) and (6.12) in (1.3) we derive by simple computations that

[TABLE]

for all $x,y\in G$. So, the set $\{\mu,\chi,\chi\,A\}$ being linearly independent, we derive from the identity above that $(h_{1},h_{2},h)$ satisfies the following identities as a system

[TABLE]

[TABLE]

and

[TABLE]

The first and the second identities imply that $(\alpha-\beta)h_{1}=a\mu-a\chi-b\chi\,A$. Since $(a,b)\neq(0,0)$ and $h_{1}\neq 0$ we have $\alpha-\beta\neq 0$, so that

[TABLE]

Substituting this in the first identity we get that

[TABLE]

hence,

[TABLE]

The third identity implies that

[TABLE]

so that

[TABLE]

Let

[TABLE]

[TABLE]

Notice that $d_{1}\neq d_{2}$ and $(a_{1},a_{2},a_{3})\neq(0,0,0)$ because $\alpha-\beta\neq 0$ and $(a,b)\neq(0,0)$. From (6.11), (6.12), (6.13) and (6.14) we get that

[TABLE]

[TABLE]

and

[TABLE]

From (6.15) we obtain

[TABLE]

which is solution (3).

Subcases 2.2. Suppose $h_{1}(e)\neq 0$. Then from (6.10) we get that

[TABLE]

Substituting this back into (1.3) we obtain

[TABLE]

for all $x,y\in G$. So that

[TABLE]

for all $x,y\in G$, i.e., the quadruple $(f,h_{1}/h_{1}(e),h_{2}-\dfrac{h_{2}(e)}{h_{1}(e)}h_{1},h-\dfrac{h(e)}{h_{1}(e)}h_{1})$ is a solution of the functional equation (1.7), so, according to Theorem 6.2, it falls into two categories:

(i) $f=cm,\,h_{1}/h_{1}(e)=m,\,h_{2}-\dfrac{h_{2}(e)}{h_{1}(e)}h_{1}=0,\,h-\dfrac{h(e)}{h_{1}(e)}h_{1}=0$,

where $m$ is a non zero multiplicative function and $c\in\mathbb{C}\setminus\{0\}$ is a constant such that $m\neq\mu$ and $m\neq\chi$. We define complex constants by $\alpha:=\dfrac{c}{h_{1}(e)}\in\mathbb{C}\setminus\{0\}$, $a_{1}:=h_{1}(e)\in\mathbb{C}\setminus\{0\}$, $a_{2}:=\dfrac{h_{2}(e)}{h_{1}(e)}$ and $a_{3}:=\dfrac{h(e)}{h_{1}(e)}$. So, the identities above and (6.16) yield that

$f=a_{1}\alpha m$, $g_{1}=\alpha m-a_{2}\dfrac{\mu+\chi}{2}-a_{3}\chi\,A$, $h_{1}=a_{1}m$, $h_{2}=a_{1}a_{2}m$ and $h=a_{1}a_{3}m$, which is solution (4).

(ii)

[TABLE]

[TABLE]

where $a,b,c\in\mathbb{C}$ such that $(a,b,c)\neq(0,0,0)$ and $a\neq b$. Taking (6.16) into account and using the identities above we get, by simple computations, that

[TABLE]

Defining

[TABLE]

[TABLE]

we have $d_{1}\neq d_{2}$ because $a\neq b$, $(a_{1},a_{2},a_{3})\neq(0,0,0)$ because $(a,b,c)\neq(0,0,0)$, $c=-\dfrac{1}{2}a_{3}d_{1}+\dfrac{1}{2}a_{3}d_{2}$, then (6.17) reads

[TABLE]

[TABLE]

and

[TABLE]

which is solution (3). ∎

Remark 6.4**.**

If $f,g_{1},h_{1},h_{2},h:G\rightarrow\mathbb{C}$ satisfy the functional equation (1.3), the formulas in parts (3) and (4) of Theorem 6.3 reveal that if $f\neq 0$ the functions $f,\,g_{1},\,h_{1},\,h_{2}$ and $h$ are abelian.

Example 6.5**.**

If we choice $a_{3}=d_{3}=0$ in Theorem 6.3(3) or $a_{3}=0$ in Theorem 6.3(4) we get that $h=0$, then the functional equation (1.3) reduces to

[TABLE]

*which was solved in [7, Theorem 8], and we obtain:

(a) for $a_{3}=d_{3}=0$ in Theorem 6.3(3),*

[TABLE]

and

[TABLE]

*which is the solution obtained in [7, Theorem 8(b)].

(b) for $a_{3}=0$ in Theorem 6.3(4),*

[TABLE]

where $\alpha,a_{1},a_{2}\in\mathbb{C}$ are constants and $m:G\rightarrow\mathbb{C}$ is a non zero multiplicative function such that $\alpha\neq 0$ and $a_{1}\neq 0$, which is the solution obtained in [7, Theorem 8(a)].

Example 6.6**.**

By taking $d_{1}=-d_{2}$, $a_{1}=a_{2}=\dfrac{1}{2}$ and $a_{3}=d_{3}=0$ in Theorem 6.3(3) we get that

[TABLE]

which is the solution of the classic sine addition law in case $\mu\neq\chi$ obtained in [10, Theorem 4.1(c)].

7. Solutions of Eq. (1.8) and Eq. (1.4)

Throughout this section $G$ denotes a group with identity element $e$, $\mu,\chi:G\to\mathbb{C}^{*}$ different characters and $A:G\to\mathbb{C}$ a nonzero additive function. We deal with the functional equations (1.8) and (1.4)

7.1. Solutions of Eq. (1.8)

In Theorem 7.1 we solve the functional equation (1.8), i.e.,

[TABLE]

where $f,h_{1},h_{2},h$ are the unknown functions to be determined.

Theorem 7.1**.**

*The solutions $(f,h_{1},h_{2},h)$ of (1.8) are:

(1) $f=a\mu$, $h_{1}$ arbitrary, $h_{2}=a\mu-ah_{1}$ and $h=0$, where $a$ ranges over $\mathbb{C}$;

(2) $f=a\mu_{1}+\varphi_{\mu_{1},\mu}$, $h_{1}=\mu_{1}$, $h_{2}=\varphi_{\mu_{1},\mu}$ and $h=0$, where $\mu_{1}$ is a character of $G$ and $a$ ranges over $\mathbb{C}$;

(3) $f=a\mu+b\chi+b\alpha\chi\,A$, $h_{1}=\chi+\alpha\chi\,A$, $h_{2}=a\mu-a\chi-a\alpha\chi\,A$ and $h=-\alpha^{2}b\chi\,A$, where $a$ and $\alpha$ range over $\mathbb{C}$, and $b$ over $\mathbb{C}\setminus\{0\}$.*

Proof.

Elementary computations show that if $f,h_{1},h_{2}$ and $h$ are of the forms (1)-(3) then $(f,h_{1},h_{2},h)$ is a solution of (1.8), so left is that any solution $(f,h_{1},h_{2},h)$ of (1.8) fits into (1)-(3).

If $f=0$ then $h_{1}$ is arbitrary and the functional equation (1.8) becomes $\mu(x)h_{2}(y)+\chi(x)A(x)h(y)=0$ for all $x,y\in G$. Since the set $\{\mu,\chi\,A\}$ is linearly independent we get that $h_{2}=h=0$, which occurs in (1) for $a=0$. In the rest of the proof we assume that $f\neq 0$. We consider two cases.

Case 1. Suppose that the set $\{f,\mu,\chi,\chi\,A\}$ is linearly independent. Let $x,y,z\in G$ be arbitrary. We compute $f(zxy)$ by using the associativity of the operation of $G$. We have

[TABLE]

On the other hand, by using (1.8), that $\mu$ and $\chi$ are multiplicative and $A$ is additive, we get that

[TABLE]

Since the set $\{f,\mu,\chi,\chi\,A\}$ is linearly independent and $x,y,z$ are arbitrary we derive from (7.1) and (7.2) that

[TABLE]

[TABLE]

[TABLE]

and

[TABLE]

for all $x,y\in G$. The functional equation (7.3) says that $\mu_{1}:=h_{1}$ is a multiplicative function on $G$. If there exists $x_{0}\in G$ such that $\mu_{1}(x_{0})=0$ then $\mu_{1}=0$. So, by putting $y=e$ in (1.8), we get that $f=h_{2}(e)\mu+h(e)\chi\,A$, which contradicts that the set $\{f,\mu,\chi,\chi\,A\}$ is linearly independent. Thus $\mu_{1}$ is a character of $G$.

Since $\chi$ is a character of $G$ and $A$ is an additive function on $G$ we derive from (7.5) that $h=0$.

Taking into account that $h_{1}=\mu_{1}$ the functional equation (7.4) becomes

[TABLE]

for all $x,y\in G$, i.e., $h_{2}=\varphi_{\mu_{1},\mu}$.

To find $f$ we put $x=e$ in (1.8) which yields that

[TABLE]

where $a:=f(e)$. The result occurs in part (2).

Case 2. Suppose that the set $\{f,\mu,\chi,\chi\,A\}$ is linearly dependent. Since $\{\mu,\chi,\chi\,A\}$ is linearly independent and $f\neq 0$ there exists a triple $(a,b,c)\in\mathbb{C}^{3}\setminus\{(0,0,0)\}$ such that

[TABLE]

Combining (1.8) and (7.7) we get by a small computation that

[TABLE]

for all $x,y\in G.$

On the other hand (7.7) gives

[TABLE]

for all $x,y\in G$. Since the set $\{\mu,\chi,\chi\,A\}$ is linearly independent we get from (7.8) and (7.9) that

[TABLE]

[TABLE]

and

[TABLE]

If $b=0$ then we get from (7.11) that $c=0$. So, (7.7) and (7.12) imply that $f=a\mu$ and $h=0$. From (7.10) we obtain $h_{2}=a\mu-ah_{1}$ with $h_{1}$ arbitrary. The result occurs in part (1).

If $b\neq 0$ then we derive from (7.10), (7.11) and (7.12) that

[TABLE]

In (7.7) and the identities above we put $\alpha=\dfrac{c}{b}$, which yields that

[TABLE]

The result occurs in part (3). ∎

7.2. Solutions of Eq. (1.4)

In Theorem 7.2 we solve the functional equation (1.4), i.e.,

[TABLE]

where $f,g_{1},h_{1},h_{2},h$ are the unknown functions to be determined.

Theorem 7.2**.**

*The solutions $(f,g_{1},h_{1},h_{2},h)$ of (1.4) are:

(1) $f=a\mu$, $g_{1}$ arbitrary, $h_{1}=0$, $h_{2}=a\mu$ and $h=0$, where $a$ ranges over $\mathbb{C}$;

(2) $f=a\mu$, $g_{1}=-b\mu+c\chi\,A$, $h_{1}\neq 0$ arbitrary, $h_{2}=a\mu+bh_{1}$ and $h=-ch_{1}$, where $a$, $b$ and $c$ range over $\mathbb{C}$;

(3) $f=a\mu+b\chi\,A,\,g_{1}=\alpha\mu+\beta\chi+\gamma\chi\,A,\,h_{1}=c\chi\,A,\,h_{2}=a\mu-\alpha c\chi\,A$ and $h=b\chi-\gamma c\chi\,A$, where $a,\,\alpha,\,\beta,\,\gamma,\,b$ and $c$ are constants such that $\beta bc\neq 0$ and $b=\beta c$;

(4) $f=a_{1}\mu+\varphi_{\mu_{1},\mu},\,g_{1}=(\varphi_{\mu_{1},\mu}-a_{2}\mu)/a+a_{3}\chi\,A,\,h_{1}=a\mu_{1},\,h_{2}=a_{1}\mu+a_{2}\mu_{1}+\varphi_{\mu_{1},\mu}$

$\quad\text{and}\quad h=-aa_{3}\mu_{1}$, where $\mu_{1}$ is a character of $G$, $a_{1},\,a_{2}$ and $a_{3}$ range over $\mathbb{C}$ and $a$ over $\mathbb{C}\setminus\{0\}$;

(5) $f=a\mu+ca_{2}\chi+\alpha ca_{2}\chi\,A,\,g_{1}=a_{1}\mu+a_{2}\chi+a_{3}\chi\,A,\,h_{1}=c\chi+\alpha c\chi\,A,\\ h_{2}=a\mu-ca_{1}\chi-\alpha ca_{1}\chi\,A\quad\text{and}\quad h=c(\alpha a_{2}-a_{3})\chi-\alpha ca_{3}\chi\,A$, where $\alpha,\,a,\,a_{1}$ and $a_{3}$ range over $\mathbb{C}$, and $c$ and $a_{2}$ over $\mathbb{C}\setminus\{0\}$.*

Proof.

We check by elementary computations that if $f,g_{1},h_{1},h_{2}$ and $h$ are of the forms (1)-(5) then $(f,g_{1},h_{1},h_{2},h)$ is a solution of (1.4), so left is that any solution $(f,g_{1},h_{1},h_{2},h)$ of (1.4) fits into (1)-(5).

There are two cases to consider.

Case 1. $f$ is proportional to $\mu$. By Proposition 3.7 we get solutions (1) and (2).

Case 2. $f$ is not proportional to $\mu$. By putting $y=e$ in the functional equation (1.4) we get that

[TABLE]

We consider two subcases according to whether $h_{1}(e)=0$ or not.

Subcase 2.1. $h_{1}(e)=0$. Then (7.13) becomes

[TABLE]

where $a:=h_{2}(e)$ and $b:=h(e)$. So the set $\{f,\mu,\chi,\chi\,A\}$ is linearly dependent. As $f$ is not proportional to $\mu$ we derive, according to Lemma 3.5(1), that $h_{1}\neq 0$. So, by Lemma 3.5(2), the set $\{g_{1},\mu,\chi,\chi\,A\}$ is linearly dependent. As the set $\{\mu,\chi,\chi\,A\}$ is linearly independent we deduce that there exists a triple $(\alpha,\beta,\gamma)\in\mathbb{C}^{3}$ such that

[TABLE]

Let $y\in G$ be arbitrary. By substituting (7.14) and (7.15) in (1.4) we obtain

[TABLE]

for all $x\in G$. Since the set $\{\mu,\chi,\chi\,A\}$ is linearly independent we derive from the identity above that $\alpha h_{1}(y)+h_{2}(y)=a\mu(y)$, $\beta h_{1}(y)=b\chi(y)A(y)$ and $\gamma h_{1}(y)+h(y)=b\chi(y)$. So, $y$ being arbitrary, we deduce that

[TABLE]

Since $f$ is not proportional to $\mu$ we get from (7.14) that $b\neq 0$. Hence, the second identity in (7.16) implies that $\beta\neq 0$ because $\chi\,A\neq 0$. So, by putting $c=\dfrac{b}{\beta}$ we derive from (7.16) that

[TABLE]

The result occurs in part (3).

Subcase 2.2. $h_{1}(e)\neq 0$. Here we get from (7.13) that

[TABLE]

Substituting this back into (1.4) we get, by elementary computation, that

[TABLE]

for all $x,y\in G$. Then the quadruple $(f,h_{1}/h_{1}(e),h_{2}-\dfrac{h_{2}(e)}{h_{1}(e)}h_{1},h-\dfrac{h(e)}{h_{1}(e)}h_{1})$ satisfies the functional equation (1.8), hence, according to Theorem 7.1 and seeing that $f$ is not proportional to $\mu$, that quadruple falls into two categories:

(i) $f=a_{1}\mu_{1}+\varphi_{\mu_{1},\mu}$, $h_{1}/h_{1}(e)=\mu_{1}$, $h_{2}-\dfrac{h_{2}(e)}{h_{1}(e)}h_{1}=\varphi_{\mu_{1},\mu}$ and $h-\dfrac{h(e)}{h_{1}(e)}h_{1}=0$, where $\mu_{1}$ is a character of $G$ and $a_{1}$ ranges over $\mathbb{C}$. Then

[TABLE]

So, (7.17) gives

[TABLE]

Defining

[TABLE]

and written $\varphi_{\mu_{1},\mu}$ instead of $a_{1}(\mu_{1}-\mu)+\varphi_{\mu_{1},\mu}$ the formulas above read

[TABLE]

Moreover since $f$ is not proportional to $\mu$, $\varphi_{\mu_{1},\mu}$ is non zero. The result obtained occurs in part (4).

(ii) $f=a\mu+b\chi+\alpha b\chi\,A$, $h_{1}/h_{1}(e)=\chi+\alpha\chi\,A$, $h_{2}-\dfrac{h_{2}(e)}{h_{1}(e)}h_{1}=a\mu-a\chi-\alpha a\chi\,A$ and $h-\dfrac{h(e)}{h_{1}(e)}h_{1}=-\alpha^{2}b\chi\,A$, where $a$ and $\alpha$ range over $\mathbb{C}$, and $b$ over $\mathbb{C}\setminus\{0\}$. Then

[TABLE]

So, by using (7.17), we get that

[TABLE]

Defining $c:=h_{1}(e)\neq 0$, $a_{1}:=\dfrac{a-h_{2}(e)}{h_{1}(e)}$, $a_{2}:=\dfrac{b}{h_{1}(e)}\neq 0$ and $a_{3}:=\dfrac{\alpha b-h(e)}{h_{1}(e)}$, we get, by using the formula of $f$ above, (7.18) and (7.19), that

[TABLE]

The result occurs in part (5). ∎

Example 7.3**.**

If we choice $a_{3}=0$ in part (4) of Theorem 7.2 we get that $h=0$ and

[TABLE]

with the same constraints on $\mu_{1},\,\varphi_{\mu_{1},\mu},\,a,\,a_{1}$ and $a_{2}$, which is the solution obtained in [11, Theorem 14].

Example 7.4**.**

If we choice $\alpha=a_{3}=0$ in part (5) we get that $h=0$ and

[TABLE]

With the notations $c_{1}:=a+cc_{2},\,c_{2}:=-ca_{1}-ca_{2},\,\lambda:=ca_{2}\neq 0$ and $\widetilde{f}:=\lambda(\chi-\mu)$ the formulas above read

[TABLE]

which is a solution obtained in [11, Theorem 14] corresponding to the non zero central solutions $\widetilde{f}=\lambda(\chi-\mu)$ of the functional equation

[TABLE]

We close with non-examples. On finite monoids or groups the only additive function is $A=0$, so the functional equations (1.3) and (1.4) become

[TABLE]

where $G$ is a monoid, $g$ is the average of two distinct nonzero multiplicative functions $\mu_{1}$ and $\mu_{2}$ on $G$, and

[TABLE]

where $G$ is a group and $\mu$ is a character of $G$, which was recently solved in [7] and [10].

on the other hand, on perfect groups, i.e. groups $G$ such that $G=[G,G]$, the only character is $\chi=1$ so that the function $g=\dfrac{\mu_{1}+\mu_{2}}{2}$ cannot exist. As examples of perfect groups we cite connected groups and semi-simple Lie groups like $SO(n)$, $SL(n,\mathbb{R})$ and $SL(n,\mathbb{C})$ for $n\geq 2$.