A Pexiderized Cosine Functional Equation

Bruce Ebanks

doi:10.2478/amsil-2026-0004

Full Article

1.

Introduction

In a recent paper [3] we introduced a new cosine functional equation (1.1) $\begin{array}{l} g (x y z) - [g (x) g (y z) + g (y) g (x z) + g (z) g (x y)] \\ + 2 g (x) g (y) g (z) = 0, x, y, z \in S, \end{array}$ \eqalign{ & g\left( {xyz} \right) - \left[ {g\left( x \right)g\left( {yz} \right) + g\left( y \right)g\left( {xz} \right) + g\left( z \right)g\left( {xy} \right)} \right] \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, + 2g\left( x \right)g\left( y \right)g\left( z \right) = 0,\;x,\;y,\;z \in S, \cr} for unknown g : S → K, where S is a semigroup and K is a field. (We called it a “new” cosine equation since d’Alembert’s equation was previously called a cosine functional equation.) One of the main results of that paper showed that if g is a solution of (1.1) then there exists a function f : S → K such that the pair (f, g) is a solution of the sine addition formula (1.2) $f (x y) = f (x) g (y) + g (x) f (y), x, y \in S .$ f\left( {xy} \right) = f\left( x \right)g\left( y \right) + g\left( x \right)f\left( y \right)\;,\;x,\;y \in S.

Conversely if the pair (f, g) satisfies (1.2) and f ≠ 0, then g satisfies (1.1). (A simpler proof of this result has been found by Stetkaer [5].)

The other main result in [3] gave the general solution of (1.1) for the case K = ℂ, and that forms part of the foundation for the present paper. If X is a topological space, let C(X) denote the algebra of continuous functions mapping X into ℂ. A function m: S → ℂ is multiplicative if m(xy) = m(x)m(y) for all x, y ∈ S. The following is [3, Corollary 3.3].

Proposition 1.1

Let S be a topological semigroup. A function g ∈ C(S) is a solution of (1.1) if and only if there exist multiplicative functions m₁, m₂ ∈ C(S) such that $g = \frac{m_{1} + m_{2}}{2} .$ g = {{{m_1} + {m_2}} \over 2}.

Here we consider the more general functional equation (1.3) $\begin{array}{l} f (x y z) + g (x) g (y z) + g (y) g (x z) + g (z) g (x y) \\ + h (x) h (y) h (z) = 0, x, y, z \in S, \end{array}$ \eqalign{ & f\left( {xyz} \right) + g\left( x \right)g\left( {yz} \right) + g\left( y \right)g\left( {xz} \right) + g\left( z \right)g\left( {xy} \right) \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, + h\left( x \right)h\left( y \right)h\left( z \right) = 0,\;x,\;y,\;z \in S, \cr} for three unknown functions f, g, h: S → ℂ, where S is a monoid.

The next section contains some terminology, notation, and other preliminaries. There we describe the solution form of a special case of (1.2) that arises in the preparations for solving (1.3). In the third section we prove some preparatory results about a simple Levi-Civita equation and an extension of Proposition 1.1. The main result is Theorem 4.1, which describes the complex-valued solutions of (1.3) on a monoid. It shows that all such solutions are specified linear combinations of at most two multiplicative functions.

All results are stated for topological semigroups, but the discrete topology is allowed.

2.

Preliminaries

Let ℂ^* = ℂ \ {0}.

A function A: S → ℂ is said to be additive if A(xy) = A(x) + A(y) for all x, y ∈ S.

If m: S → ℂ is multiplicative and m ≠ 0 then we call m an exponential.

For any multiplicative m: S → ℂ we define the sets $I_{m} : = {x \in S | m (x) = 0}$ {I_m}\;: = \{ x \in S|m\left( x \right) = 0\} and $P_{m} : = {p \in I_{m} \ I_{m} I_{m} | u p, p v, u p v \in I_{m} \ I_{m} I_{m} for all u, v \in S \ I_{m}} .$ P_m : = \left\{ {p \in I_m \backslash I_m I_m |up,pv,upv \in I_m \backslash I_m I_m \,\,{\text{for}}\,{\text{all}}\,u,v \in S\backslash I_m } \right\}.

It follows that if p ∈ P_m and u, v ∈ S \ I_m, then up, pv, upv ∈ P_m also.

A monoid is a semigroup with an identity element that we denote e. If S is a monoid and m: S → K is multiplicative, then m(e) = m(e²) = m(e)² so m(e) ∈ {0, 1}. If m(e) = 0 then m(x) = m(xe) = m(x)m(e) = 0 for all x ∈ S, so m = 0. Therefore if m ≠ 0 then m(e) = 1.

The general solution of (1.2) for f, g : S → ℂ is given in [1, Theorem 3.1], but here we need only the following corollary describing the solutions in the special case that g is an exponential and f is not the zero function.

Proposition 2.1

Let S be a topological semigroup. Suppose f, m ∈ C(S) satisfy (2.1) $f (x y) = f (x) m (y) + m (x) f (y), x, y \in S,$ f\left( {xy} \right) = f\left( x \right)m\left( y \right) + m\left( x \right)f\left( y \right)\;,\;x,\;y \in S, where f ≠ 0 and m is an exponential. Then there exists an additive function A ∈ C(S \ I_m) and a function f_P ∈ C(P_m) such that

(a)
$f (x) = {\begin{array}{l} A (x) m (x) & f o r x \in S \ I_{m}, \\ f_{P} (x) & f o r x \in P_{m}, \\ 0 & f o r x \in I_{m} \ P_{m}, \end{array}$ f\left( x \right) = \left\{ {\matrix{ {A\left( x \right)m\left( x \right)} \hfill & {for\;x \in S\backslash {I_m},} \hfill \cr {{f_P}\left( x \right)} \hfill & {for\;x \in {P_m},} \hfill \cr 0 \hfill & {for\;x \in {I_m}\backslash {P_m},} \hfill \cr } } \right.
(b)
f(qt) = f(tq) = 0 for all q ∈ I_m \ P_m and t ∈ S \ I_m, and
(c)
for p ∈ P_m and u, v ∈ S \ I_m we have f_P (up) = f_P (p)m(u), f_P (pv) = f_P (p)m(v), and f_P (upv) = f_P (p)m(uv).

Conversely, for any exponential m the function f described above satisfies (2.1).

Proof

This is an immediate consequence of [1, Theorem 3.1].

Definition 2.2

For f, m as described in Proposition 2.1 we refer to the pair (f, m) as a sine-exponential pair.

A function γ on S is said to be central if γ(xy) = γ(yx) for all x, y ∈ S.

3.

Preparatory results

We will use the following, which is [2, Theorem 4.4].

Proposition 3.1

Let S be a topological monoid, and suppose f, g₁, g₂, h₁, h₂ ∈ C(S) satisfy the Levi-Civita equation $f (x y) = g_{1} (x) h_{1} (y) + g_{2} (x) h_{2} (y), x, y \in S,$ f\left( {xy} \right) = {g_1}\left( x \right){h_1}\left( y \right) + {g_2}\left( x \right){h_2}\left( y \right)\;,\;x,\;y \in S, with f central, and with {g₁, g₂} and {h₁, h₂} linearly independent. Then the solutions are given by the following families, where m, m₁, m₂ ∈ C(S) are exponentials with m₁ ≠ m₂, and (ϕ, m) is a continuous sine-exponential pair.

(a)
There exist constants a_i, b_i, c_i, d_i, e_i ∈ ℂ satisfying $(\begin{matrix} b_{1} & d_{1} \\ b_{2} & d_{2} \end{matrix}) (\begin{matrix} c_{1} & c_{2} \\ e_{1} & e_{2} \end{matrix}) = (\begin{matrix} a_{1} & 0 \\ 0 & a_{2} \end{matrix})$ \left( {\matrix{ {{b_1}} & {{d_1}} \cr {{b_2}} & {{d_2}} \cr } } \right)\left( {\matrix{ {{c_1}} & {{c_2}} \cr {{e_1}} & {{e_2}} \cr } } \right) = \left( {\matrix{ {{a_1}} & 0 \cr 0 & {{a_2}} \cr } } \right) such that $\begin{matrix} f = a_{1} m_{1} + a_{2} m_{2}, g_{1} = b_{1} m_{1} + b_{2} m_{2}, g_{2} = d_{1} m_{1} + d_{2} m_{2}, \\ h_{1} = c_{1} m_{1} + c_{2} m_{2}, h_{2} = e_{1} m_{1} + e_{2} m_{2} . \end{matrix}$ \matrix{ {f = {a_1}{m_1} + {a_2}{m_2},\;{g_1} = {b_1}{m_1} + {b_2}{m_2},\;{g_2} = {d_1}{m_1} + {d_2}{m_2},} \cr {{h_1} = {c_1}{m_1} + {c_2}{m_2},\;{h_2} = {e_1}{m_1} + {e_2}{m_2}.} \cr }
(b)
There exist constants a_i, b_i, c_i, d_i, e_i ∈ ℂ satisfying $(\begin{matrix} b_{1} & d_{1} \\ b_{2} & d_{2} \end{matrix}) (\begin{matrix} c_{1} & c_{2} \\ e_{1} & e_{2} \end{matrix}) = (\begin{matrix} a_{1} & a_{2} \\ a_{2} & 0 \end{matrix})$ \left( {\matrix{ {{b_1}} & {{d_1}} \cr {{b_2}} & {{d_2}} \cr } } \right)\left( {\matrix{ {{c_1}} & {{c_2}} \cr {{e_1}} & {{e_2}} \cr } } \right) = \left( {\matrix{ {{a_1}} & {{a_2}} \cr {{a_2}} & 0 \cr } } \right) such that $\begin{matrix} f = a_{1} m + a_{2} φ, g_{1} = b_{1} m + b_{2} φ, g_{2} = d_{1} m + d_{2} φ, \\ h_{1} = c_{1} m + c_{2} φ, h_{2} = e_{1} m + e_{2} φ . \end{matrix}$ \matrix{ {f = {a_1}m + {a_2}\varphi ,\;{g_1} = {b_1}m + {b_2}\varphi ,\;{g_2} = {d_1}m + {d_2}\varphi ,} \cr {{h_1} = {c_1}m + {c_2}\varphi ,\;{h_2} = {e_1}m + {e_2}\varphi .} \cr }
Moreover all matrices above are invertible.

We also need some simple linear independence results.

Lemma 3.2

Let S be a semigroup.

(i)
Any set of distinct exponentials on S into ℂ is linearly independent.
(ii)
If (ϕ, m) is a sine-exponential pair on S into ℂ, then {m, ϕ} is linearly independent.

Proof

Part (i) is [4, Theorem 3.18(b)]. Part (ii) is [1, Lemma 5.1(b)].

The next step is to prove the following consequence of Proposition 3.1.

Corollary 3.3

Let S be a topological monoid. The functions F, G, H ∈ C(S) satisfy (3.1) $F (x y) = G (x) G (y) + H (x) H (y), x, y \in S,$ F\left( {xy} \right) = G\left( x \right)G\left( y \right) + H\left( x \right)H\left( y \right)\;,\;x,\;y \in S, if and only if they belong to one of the following families, where m ∈ C(S) is multiplicative, m₁, m₂ ∈ C(S) are exponentials with m₁ ≠ m₂, (ϕ, m) is a continuous sine-exponential pair, and a, b, c ∈ ℂ.

(i)
F = c²m, G = cm, and H = 0.
(ii)
F = 0 and G = ±iH, where H is an arbitrary nonzero function.
(iii)
For c ≠ 0, b ≠ ±i, and m ≠ 0 we have $\begin{matrix} F = (b^{2} + 1) c^{2} m, & G = b c m, & H = c m . \end{matrix}$ \;\matrix{ {F = \left( {{b^2} + 1} \right){c^2}m,} & {G = bcm,} & {H = cm.} \cr }
(iv)
For bc(a² + c²) ≠ 0 we have $\begin{matrix} F = (a^{2} + c^{2}) m_{1} + b^{2} (1 + \frac{a^{2}}{c^{2}}) m_{2}, & G = a m_{1} + b m_{2}, & H = c m_{1} - \frac{a b}{c} m_{2} . \end{matrix}$ \matrix{ {F = \left( {{a^2} + {c^2}} \right){m_1} + {b^2}\left( {1 + {{{a^2}} \over {{c^2}}}} \right){m_2},} & {G = a{m_1} + b{m_2},} & {H = c{m_1} - {{ab} \over c}{m_2}.} \cr }
(v)
For b(a + ic) ≠ 0 we have $\begin{matrix} F = (a^{2} + c^{2}) m + b (a + i c) φ, & G = a m + b φ, & H = \pm (c m + i b φ) . \end{matrix}$ \matrix{ {F = \left( {{a^2} + {c^2}} \right)m + b\left( {a + ic} \right)\varphi ,} & {G = am + b\varphi ,} & {H = \pm \left( {cm + ib\varphi } \right).} \cr }

Proof

It is easily checked that the triples (F, G, H) in each case (i)–(v) satisfy (3.1), using the identity ϕ(xy) = ϕ(x)m(y) + m(x)ϕ(y) in case (v).

For the converse suppose F, G, H ∈ C(S) satisfy (3.1). First we consider the case that {G, H} is linearly dependent. If H = 0 then (3.1) reduces to $\begin{matrix} F (x y) = G (x) G (y), & x, y \in S, \end{matrix}$ \matrix{ {F\left( {xy} \right) = G\left( x \right)G\left( y \right),} & {x,\;y \in S,} \cr } which with x = e yields F = G(e)G. Thus we have G(e)G(xy) = G(x)G(y) for all x, y ∈ S. If G(e) = 0 it follows that G = 0, so F = 0 and we are in solution family (i) with c = 0. If G(e) ≠ 0, then m := G(e)⁻¹G is multiplicative. Defining c := G(e) ∈ ℂ* we are in family (i) with c ≠ 0.

If H ≠ 0 then by linear dependence we have G = bH for some b ∈ ℂ, and (3.1) reduces to $\begin{matrix} F (x y) = (b^{2} + 1) H (x) H (y), & x, y \in S . \end{matrix}$ \matrix{ {F\left( {xy} \right) = \left( {{b^2} + 1} \right)H\left( x \right)H\left( y \right),\;} & {x,\;y \in S.} \cr }

With x = e we get F = (b² + 1)H(e)H, so (3.2) $\begin{matrix} (b^{2} + 1) H (e) H (x y) = (b^{2} + 1) H (x) H (y), & x, y \in S . \end{matrix}$ \matrix{ {\left( {{b^2} + 1} \right)H\left( e \right)H\left( {xy} \right) = \left( {{b^2} + 1} \right)H\left( x \right)H\left( y \right),\;} & {x,\;y \in S.} \cr }

If b = ±i then F = 0 and we are in family (ii). If b ≠ ±i then we get H(e) ≠ 0 from (3.2) since H ≠ 0. Defining c := H(e) and proceeding as in the previous paragraph, we are in family (iii).

Henceforth we assume that {G, H} is linearly independent. Since the right hand side of (3.1) is symmetric in x and y, the function F is central. Therefore we may apply Proposition 3.1 with f := F , g₁ = h₁ := G, and g₂ = h₂ := H.

The result in case (a) is $\begin{matrix} F = a_{1} m_{1} + a_{2} m_{2}, & G = b_{1} m_{1} + b_{2} m_{2}, & H = d_{1} m_{1} + d_{2} m_{2}, \end{matrix}$ \matrix{ {F = {a_1}{m_1} + {a_2}{m_2},\;\;} & {G = {b_1}{m_1} + {b_2}{m_2},} & {H = {d_1}{m_1} + {d_2}{m_2},} \cr } where m₁, m₂ ∈ C(S) are exponentials with m₁ ≠ m₂, and the coefficients satisfy $(\begin{matrix} b_{1} & d_{1} \\ b_{2} & d_{2} \end{matrix}) (\begin{matrix} b_{1} & b_{2} \\ d_{1} & d_{2} \end{matrix}) = (\begin{matrix} a_{1} & 0 \\ 0 & a_{2} \end{matrix})$ \left( {\matrix{ {{b_1}} & {{d_1}} \cr {{b_2}} & {{d_2}} \cr } } \right)\left( {\matrix{ {{b_1}} & {{b_2}} \cr {{d_1}} & {{d_2}} \cr } } \right) = \left( {\matrix{ {{a_1}} & 0 \cr 0 & {{a_2}} \cr } } \right)

Since the matrices are non-singular, we have a₁a₂ ≠ 0, b₁d₂ − b₂d₁ ≠ 0, $b_{j}^{2} + d_{j}^{2} = a_{j}$ b_j^2 + d_j^2 = {a_j} for j ∈ {1, 2}, and b₁b₂ + d₁d₂ = 0. There are several sub-cases to consider. If b₁ = 0, then b₂d₁ ≠ 0 and d₂ = 0. Defining c := d₁ ∈ ℂ^* and b := b₂ ∈ ℂ^*, we are in solution family (iv) with a = 0. Similarly, if b₂ = 0 (or d₁ = 0 or d₂ = 0) we are again in solution family (iv) (switching the roles of m₁ and m₂ if needed) with a = 0. Finally, suppose b₁b₂d₁d₂ ≠ 0. Defining a := b₁, b := b₂, and c := d₁ we have d₂ = −ab/c, and we arrive at solution family (iv) with a ≠ 0.

From case (b) of Proposition 3.1 we get $\begin{matrix} F = a_{1} m + a_{2} φ, & G = b_{1} m + b_{2} φ, & H = d_{1} m + d_{2} φ, \end{matrix}$ \matrix{ {F = {a_1}m + {a_2}\varphi ,\;\;} & {G = {b_1}m + {b_2}\varphi ,} & {H = {d_1}m + {d_2}\varphi ,} \cr } where m ∈ C(S) is an exponential, (ϕ, m) is a continuous sine-exponential pair, and the coefficients satisfy $(\begin{matrix} b_{1} & d_{1} \\ b_{2} & d_{2} \end{matrix}) (\begin{matrix} b_{1} & b_{2} \\ d_{1} & d_{2} \end{matrix}) = (\begin{matrix} a_{1} & a_{2} \\ a_{2} & 0 \end{matrix})$ \left( {\matrix{ {{b_1}} & {{d_1}} \cr {{b_2}} & {{d_2}} \cr } } \right)\left( {\matrix{ {{b_1}} & {{b_2}} \cr {{d_1}} & {{d_2}} \cr } } \right) = \left( {\matrix{ {{a_1}} & {{a_2}} \cr {{a_2}} & 0 \cr } } \right) where the matrices are invertible. Thus we have a₂ ≠ 0, b₁d₂ − b₂d₁ ≠ 0, $b_{1}^{2} + d_{1}^{2} = a_{1}$ b_1^2 + d_1^2 = {a_1} , b₁b₂ + d₁d₂ = a₂, and $b_{2}^{2} + d_{2}^{2} = 0$ b_2^2 + d_2^2 = 0 . Defining a := b₁, b := b₂, and c := d₁, we have d₂ = ±ib, a₁ = a² + c², and a₂ = (a ± ic)b. If d₂ = ib then a₂ = (a + ic)b ≠ 0 implies that a ≠ −ic and we are in solution family (v) with the plus sign in the formula for H. If d₂ = −ib we arrive by similar calculations at $\begin{matrix} F = (a^{2} + c^{2}) m + b (a - i c) φ, & G = a m + b φ, & H = c m - i b φ . \end{matrix}$ \matrix{ {F = \left( {{a^2} + {c^2}} \right)m + b\left( {a - ic} \right)\varphi ,} & {G = am + b\varphi ,} & {H = cm - ib\varphi .} \cr }

Replacing c by −c we arrive at family (v) with the minus sign in the formula for H.

We also need the following small extension of Proposition 1.1.

Corollary 3.4

Let S be a topological monoid. A function g ∈ C(S) is a solution of (3.3) $\begin{array}{l} g {(e)}^{2} g (x y z) - g (e) [g (x) g (y z) + g (y) g (x z) + g (z) g (x y)] \\ + 2 g (x) g (y) g (z) = 0, x, y, z \in S, \end{array}$ \matrix{ {g{{(e)}^2}g\left( {xyz} \right) - g\left( e \right)\left[ {g\left( x \right)g\left( {yz} \right) + g\left( y \right)g\left( {xz} \right) + g\left( z \right)g\left( {xy} \right)} \right]} \hfill \cr {\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, + 2g\left( x \right)g\left( y \right)g\left( z \right) = 0,\;x,\;y,\;z \in S,} \hfill \cr } if and only if there exist multiplicative functions m₁, m₂ ∈ C(S) such that (3.4) $g = g (e) \frac{m_{1} + m_{2}}{2} .$ g = g\left( e \right){{{m_1} + {m_2}} \over 2}.

Proof

Suppose g ∈ C(S) satisfies (3.3). If g(e) = 0, then (3.3) reduces to g(x)g(y)g(z) = 0 for all x, y, z ∈ S, so g = 0 and we have (3.4).

If g(e) ≠ 0, then multiplying (3.3) by g(e)⁻³ and defining g^′ : S → ℂ by $g^{'} : = g {(e)}^{- 1} g$ g': = g{(e)^{ - 1}}g we find that g^′ is a solution of (1.1). Since g^′ ∈ C(S) we get (3.4) by Proposition 1.1.

The converse is straightforward.

4.

The main result

Recall that (1.3) is the functional equation $\begin{array}{l} f (x y z) + g (x) g (y z) + g (y) g (x z) + g (z) g (x y) + h (x) h (y) h (z) = 0, & x, y, z \in S . \end{array}$ \matrix{ {f\left( {xyz} \right) + g\left( x \right)g\left( {yz} \right) + g\left( y \right)g\left( {xz} \right) + g\left( z \right)g\left( {xy} \right) + h\left( x \right)h\left( y \right)h\left( z \right) = 0,} \hfill & {x,y,z \in S.} \hfill \cr }

Theorem 4.1

Let S be a topological monoid. The functions f, g, h ∈ C(S) satisfy (1.3) if and only if they belong to one of the following families, where m ∈ C(S) is multiplicative, m₁, m₂ ∈ C(S) are exponentials with m₁ ≠ m₂, and a, b, c ∈ ℂ.

(a)
f = −(3b² + c³)m, g = bm, and h = cm.
(b)
For some a ∈ ℂ^* we have $\begin{array}{l} f = \frac{a^{6}}{4} (m_{1} + m_{2}), & g = \pm \frac{i a^{3}}{2 \sqrt{2}} (m_{1} + m_{2}), & h = \frac{a^{2}}{2} (m_{1} + m_{2}) \end{array}$ \matrix{ {f = {{{a^6}} \over 4}\left( {{m_1} + {m_2}} \right),} \hfill & {g = \pm {{i{a^3}} \over {2\sqrt 2 }}\left( {{m_1} + {m_2}} \right),} \hfill & {h = {{{a^2}} \over 2}\left( {{m_1} + {m_2}} \right)} \hfill \cr }
(c)
For b, c ∈ ℂ^* we have $f = - 3 b^{2} m_{1} - c^{3} m_{2}, g = b m_{1}, h = c m_{2} .$ f = - 3{b^2}{m_1} - {c^3}{m_2},\;g = b{m_1},\;h = c{m_2}.
(d)
For b, c ∈ ℂ^* we have $\begin{array}{l} f = - (3 b^{2} + c^{3}) m_{1} - c^{3} (\frac{3 c^{3}}{b^{2}} + 1) m_{2}, & g = b m_{1} - \frac{c^{3}}{b} m_{2}, & h = c (m_{1} + m_{2}) . \end{array}$ \matrix{ {f = - \left( {3{b^2} + {c^3}} \right){m_1} - {c^3}\left( {{{3{c^3}} \over {{b^2}}} + 1} \right){m_2},} \hfill & {g = b{m_1} - {{{c^3}} \over b}{m_2},} \hfill & {h = c\left( {{m_1} + {m_2}} \right).} \hfill \cr }

Proof

It is easy to check that each family (a)–(d) is a solution of (1.3). Conversely, suppose (f, g, h) is a (continuous) solution of (1.3). Putting z = e we find that (4.1) $\begin{array}{l} - f (x y) - g (e) g (x y) = 2 g (x) g (y) + h (e) h (x) h (y), & x, y \in S . \end{array}$ \matrix{ { - f\left( {xy} \right) - g\left( e \right)g\left( {xy} \right) = 2g\left( x \right)g\left( y \right) + h\left( e \right)h\left( x \right)h\left( y \right),\;} \hfill & {x,\;y \in S.} \hfill \cr }

Choose a ∈ ℂ such that a² = h(e), and define F, G, H ∈ C(S) by (4.2) $\begin{array}{l} F : = - f - g (e) g, & G : = \sqrt{2} g, & H : = a h . \end{array}$ \matrix{ {F: = - f - g\left( e \right)g,} \hfill & {G: = \sqrt 2 g,} \hfill & {H: = ah.} \hfill \cr }

Then (4.1) becomes (3.1): $\begin{array}{l} F (x y) = G (x) G (y) + H (x) H (y), & x, y \in S, \end{array}$ \matrix{ {F\left( {xy} \right) = G\left( x \right)G\left( y \right) + H\left( x \right)H\left( y \right),} \hfill & {x,\;y \in S,} \hfill \cr } so by Corollary 3.3 we have solution families (i)–(v) for the triple (F, G, H).

Case 1: Suppose (F, G, H) are given by (i) or (iii). Then F, G, H ∈ span{m} for some multiplicative m ∈ C(S). By (4.2) this means f = dm and g = bm for some constants b, d ∈ ℂ. Putting these forms into (1.3) we get (4.3) $\begin{array}{l} (d + 3 b^{2}) m (x) m (y) m (z) + h (x) h (y) h (z) = 0, & x, y, z \in S . \end{array}$ \matrix{ {\left( {d + 3{b^2}} \right)m\left( x \right)m\left( y \right)m\left( z \right) + h\left( x \right)h\left( y \right)h\left( z \right) = 0,} \hfill & {x,\;y,\;z \in S.} \hfill \cr } \;

If m = 0 then we have f = g = h = 0, which is included in family (a) by taking m = 0. Now suppose m ≠ 0, so m(e) = 1. With y = z = e in (4.3) we see that $\begin{array}{l} (d + 3 b^{2}) m (x) + h {(e)}^{2} h (x) = 0, & x \in S . \end{array}$ \matrix{ {\left( {d + 3{b^2}} \right)m\left( x \right) + h{{(e)}^2}h\left( x \right) = 0,} \hfill & {x \in S.} \hfill \cr } \;

If h(e) = 0 then d + 3b² = 0, so by (4.3) we have h = 0. This solution is the case c = 0 and d = −3b² of family (a). On the other hand if h(e) ≠ 0 then we have h = cm, where c = h(e) ≠ 0. Now (4.3) yields $\begin{array}{l} (d + 3 b^{2} + c^{3}) m (x) m (y) m (z) = 0, & x, y, z \in S, \end{array}$ \matrix{ {\left( {d + 3{b^2} + {c^3}} \right)m\left( x \right)m\left( y \right)m\left( z \right) = 0,} \hfill & {x,\;y,\;z \in S,} \hfill \cr } and since m ≠ 0 we have d = −3b² − c³. That is again in family (a), and that concludes Case 1.

Case 2: Suppose (F, G, H) are given by (ii), that is F = 0 and G = ±iH for arbitrary H ≠ 0. By (4.2) we have h ≠ 0 and $\begin{array}{l} g = \pm \frac{i a h}{\sqrt{2}}, & f = \frac{h {(e)}^{2} h}{2}, \end{array}$ \matrix{ {g = \pm {{iah} \over {\sqrt 2 }},} \hfill & {f = {{h{{(e)}^2}h} \over 2},} \hfill \cr } where a² = h(e) ≠ 0. Inserting these forms into (1.3) we find that $\frac{h {(e)}^{2}}{2} h (x y z) - \frac{h (e)}{2} [h (x) h (y z) + h (y) h (x z) + h (z) h (x y)] + h (x) h (y) h (z) = 0$ {{h{{(e)}^2}} \over 2}h\left( {xyz} \right) - {{h\left( e \right)} \over 2}\left[ {h\left( x \right)h\left( {yz} \right) + h\left( y \right)h\left( {xz} \right) + h\left( z \right)h\left( {xy} \right)} \right] + h\left( x \right)h\left( y \right)h\left( z \right) = 0 for all x, y, z ∈ S. That is, h is a solution of (3.3). Therefore by Corollary 3.4 there exist multiplicative functions m₁, m₂ ∈ C(S) such that $h = h (e) \frac{m_{1} + m_{2}}{2} .$ h = h\left( e \right){{{m_1} + {m_2}} \over 2}.

Since h ≠ 0, at least one of m₁ or m₂ must be an exponential.

If either m₁ = m₂ ≠ 0, or m₂ = 0 ≠ m₁, or m₁ = 0 ≠ m₂, then we have (F, G, H) ∈ span{m₁} or (F, G, H) ∈ span {m₂}. Hence in these cases we are back in Case 1, where m := m₁ or m := m₂.

If m₁ and m₂ are distinct exponentials, then the solution is in family (b) (see Case 3).

For the rest of the proof (Cases 3 and 4) we have h(e) ≠ 0, since by (4.2) we have ah = H ≠ 0 and h(e) = a².

Case 3: Suppose (F, G, H) are given by (iv). Then F, G, H ∈ span{m₁, m₂} for distinct exponentials m₁, m₂ ∈ C(S). By (4.2) the same is true for f, g, h, since h(e) ≠ 0. Let $\begin{array}{l} f = a_{1} m_{1} + a_{2} m_{2}, & g = b_{1} m_{1} + b_{2} m_{2}, & h = c_{1} m_{1} + c_{2} m_{2} \end{array}$ \matrix{ {f = {a_1}{m_1} + {a_2}{m_2},} \hfill & {g = {b_1}{m_1} + {b_2}{m_2},} \hfill & {h = {c_1}{m_1} + {c_2}{m_2}} \hfill \cr } for some a_j, b_j, c_j ∈ ℂ. Inserting these forms into (1.3) we have $\begin{array}{l} 0 = & a_{1} m_{1} (x y z) + a_{2} m_{2} (x y z) + [b_{1} m_{1} (x) + b_{2} m_{2} (x)] [b_{1} m_{1} (y z) + b_{2} m_{2} (y z)] \\ + [b_{1} m_{1} (y) + b_{2} m_{2} (y)] [b_{1} m_{1} (x z) + b_{2} m_{2} (x z)] \\ + [b_{1} m_{1} (z) + b_{2} m_{2} (z)] [b_{1} m_{1} (x y) + b_{2} m_{2} (x y)] \\ + [c_{1} m_{1} (x) + c_{2} m_{2} (x)] [c_{1} m_{1} (y) + c_{2} m_{2} (y)] [c_{1} m_{1} (z) + c_{2} m_{2} (z)] \\ = (a_{1} + 3 b_{1}^{2} + c_{1}^{3}) m_{1} (x) m_{1} (y) m_{1} (z) + (a_{2} + 3 b_{2}^{2} + c_{2}^{3}) m_{2} (x) m_{2} (y) m_{2} (z) \\ + (b_{1} b_{2} + c_{1} c_{2}^{2}) [m_{1} (x) m_{2} (y) m_{2} (z) + m_{2} (x) m_{1} (y) m_{2} (z) \\ + m_{2} (x) m_{2} (y) m_{1} (z)] + (b_{1} b_{2} + c_{1}^{2} c_{2}) [m_{2} (x) m_{1} (y) m_{1} (z) \\ + m_{1} (x) m_{2} (y) m_{1} (z) + m_{1} (x) m_{1} (y) m_{2} (z)] \end{array}$ \matrix{ {0 = } \hfill & {{a_1}{m_1}\left( {xyz} \right) + {a_2}{m_2}\left( {xyz} \right) + \left[ {{b_1}{m_1}\left( x \right) + {b_2}{m_2}\left( x \right)} \right]\left[ {{b_1}{m_1}\left( {yz} \right) + {b_2}{m_2}\left( {yz} \right)} \right]} \hfill \cr {} \hfill & { + \left[ {{b_1}{m_1}\left( y \right) + {b_2}{m_2}\left( y \right)} \right]\left[ {{b_1}{m_1}\left( {xz} \right) + {b_2}{m_2}\left( {xz} \right)} \right]} \hfill \cr {} \hfill & { + \left[ {{b_1}{m_1}\left( z \right) + {b_2}{m_2}\left( z \right)} \right]\left[ {{b_1}{m_1}\left( {xy} \right) + {b_2}{m_2}\left( {xy} \right)} \right]} \hfill \cr {} \hfill & { + \left[ {{c_1}{m_1}\left( x \right) + {c_2}{m_2}\left( x \right)} \right]\left[ {{c_1}{m_1}\left( y \right) + {c_2}{m_2}\left( y \right)} \right]\left[ {{c_1}{m_1}\left( z \right) + {c_2}{m_2}\left( z \right)} \right]} \hfill \cr {} \hfill & { = \left( {{a_1} + 3b_1^2 + c_1^3} \right){m_1}\left( x \right){m_1}\left( y \right){m_1}\left( z \right) + \left( {{a_2} + 3b_2^2 + c_2^3} \right){m_2}\left( x \right){m_2}\left( y \right){m_2}\left( z \right)} \hfill \cr {} \hfill & { + \left( {{b_1}{b_2} + {c_1}c_2^2} \right)[{m_1}\left( x \right){m_2}\left( y \right){m_2}\left( z \right) + {m_2}\left( x \right){m_1}\left( y \right){m_2}\left( z \right)} \hfill \cr {} \hfill & { + {m_2}\left( x \right){m_2}\left( y \right){m_1}\left( z \right)\left] { + \left( {{b_1}{b_2} + c_1^2{c_2}} \right)} \right[{m_2}\left( x \right){m_1}\left( y \right){m_1}\left( z \right)} \hfill \cr {} \hfill & { + {m_1}\left( x \right){m_2}\left( y \right){m_1}\left( z \right) + {m_1}\left( x \right){m_1}\left( y \right){m_2}\left( z \right)]} \hfill \cr } for all x, y, z ∈ S. By Lemma 3.2 all of the coefficients must vanish, that is $\begin{array}{l} a_{j} + 3 b_{j}^{2} + c_{j}^{3} = 0, & j \in {1, 2}, & and & b_{1} b_{2} + c_{1} c_{2}^{2} = 0 = b_{1} b_{2} + c_{1}^{2} c_{2} . \end{array}$ \matrix{ {{a_j} + 3b_j^2 + c_j^3 = 0,} \hfill & {j \in \left\{ {1,2} \right\},} \hfill & {{\rm{and}}} \hfill & {{b_1}{b_2} + {c_1}c_2^2 = 0 = {b_1}{b_2} + c_1^2{c_2}.} \hfill \cr }

It follows that c₁c₂(c₂ − c₁) = 0.

If c₁ = 0 then b₁b₂ = 0. In this case if b₁ = 0 then a₁ = 0, so we have f, g, h ∈ span{m₂} and are back Case 1. If b₁ ≠ 0 then b₂ = 0 so we have $a_{1} = - 3 b_{1}^{2}$ {a_1} = - 3b_1^2 and $a_{2} = - c_{2}^{3}$ {a_2} = - c_2^3 , thus $\begin{array}{l} f = - 3 b_{1}^{2} m_{1} - c_{2}^{3} m_{2}, & g = b_{1} m_{1}, & h = c_{2} m_{2} . \end{array}$ \matrix{ {f = - 3b_1^2{m_1} - c_2^3{m_2},} \hfill & {g = {b_1}{m_1},} \hfill & {h = {c_2}{m_2}.} \hfill \cr }

If c₂ = 0 too, then again we are back to family (a). If not, then we are in solution family (c) with b := b₁ and c := c₂.

The case c₁ ≠ 0 and c₂ = 0 is similar to the preceding one and leads to the same results by interchanging m₁ and m₂.

Lastly, if c₁c₂ ≠ 0 then we have c₂ = c₁ =: c ∈ ℂ^*. Then b₁b₂ = −c³ ≠ 0 and $a_{j} = - 3 b_{j}^{2} - c^{3}$ {a_j} = - 3b_j^2 - {c^3} for j = 1, 2. Defining b := b₁ ∈ ℂ^* we are in solution family (d).

Case 4: Suppose (F, G, H) are given by (v). Then F, G, H ∈ span{M, ϕ} for a continuous sine-exponential pair (ϕ, M). As in the preceding case, the same is true for f, g, h. It is helpful to recall that (ϕ, M) satisfy the sine addition formula ϕ(xy) = ϕ(x)M(y) + M(x)ϕ(y). So for x = y = e we have ϕ(e) = 2ϕ(e) since M(e) = 1, therefore ϕ(e) = 0. Let $\begin{array}{l} f = a_{1} M + a_{2} φ, & g = b_{1} M + b_{2} φ, & h = c_{1} M + c_{2} φ, \end{array}$ \matrix{ {f = {a_1}M + {a_2}\varphi ,} \hfill & {g = {b_1}M + {b_2}\varphi ,} \hfill & {h = {c_1}M + {c_2}\varphi ,} \hfill \cr } where a_j, b_j, c_j ∈ ℂ. Inserting these forms into (1.3) we have $\begin{array}{l} 0 = & a_{1} M (x y z) + a_{2} (φ (x) M (y z) + M (x) φ (y) M (z) + M (x) M (y) φ (z)) \\ + [b_{1} M (x) + b_{2} φ (x)] [b_{1} M (y z) + b_{2} (φ (y) M (z) + M (y) φ (z))] \\ + [b_{1} M (y) + b_{2} φ (y)] [b_{1} M (x z) + b_{2} (φ (x) M (z) + M (x) φ (z))] \\ + [b_{1} M (z) + b_{2} φ (z)] [b_{1} M (x y) + b_{2} (φ (x) M (y) + M (x) φ (y))] \\ + [c_{1} M (x) + c_{2} φ (x)] [c_{1} M (y) + c_{2} φ (y)] [c_{1} M (z) + c_{2} φ (z)] \\ = (a_{1} + 3 b_{1}^{2} + c_{1}^{3}) M (x) M (y) M (z) + c_{2}^{3} φ (x) φ (y) φ (z) \\ + (a_{2} + 3 b_{1} b_{2} + c_{2} c_{1}^{2}) [φ (x) M (y) M (z) + M (x) φ (y) M (z) + M (x) M (y) φ (z)] \\ + (2 b_{2}^{2} + c_{1} c_{2}^{2}) [φ (x) φ (y) M (z) + φ (x) M (y) φ (z) + M (x) φ (y) φ (z)] \end{array}$ \matrix{ {0 = } \hfill & {{a_1}M\left( {xyz} \right) + {a_2}\left( {\varphi \left( x \right)M\left( {yz} \right) + M\left( x \right)\varphi \left( y \right)M\left( z \right) + M\left( x \right)M\left( y \right)\varphi \left( z \right)} \right)} \hfill \cr {} \hfill & { + \left[ {{b_1}M\left( x \right) + {b_2}\varphi \left( x \right)} \right]\left[ {{b_1}M\left( {yz} \right) + {b_2}\left( {\varphi \left( y \right)M\left( z \right) + M\left( y \right)\varphi \left( z \right)} \right)} \right]} \hfill \cr {} \hfill & { + \left[ {{b_1}M\left( y \right) + {b_2}\varphi \left( y \right)} \right]\left[ {{b_1}M\left( {xz} \right) + {b_2}\left( {\varphi \left( x \right)M\left( z \right) + M\left( x \right)\varphi \left( z \right)} \right)} \right]} \hfill \cr {} \hfill & { + \left[ {{b_1}M\left( z \right) + {b_2}\varphi \left( z \right)} \right]\left[ {{b_1}M\left( {xy} \right) + {b_2}\left( {\varphi \left( x \right)M\left( y \right) + M\left( x \right)\varphi \left( y \right)} \right)} \right]} \hfill \cr {} \hfill & { + \left[ {{c_1}M\left( x \right) + {c_2}\varphi \left( x \right)} \right]\left[ {{c_1}M\left( y \right) + {c_2}\varphi \left( y \right)} \right]\left[ {{c_1}M\left( z \right) + {c_2}\varphi \left( z \right)} \right]} \hfill \cr {} \hfill & { = \left( {{a_1} + 3b_1^2 + c_1^3} \right)M\left( x \right)M\left( y \right)M\left( z \right) + c_2^3\varphi \left( x \right)\varphi \left( y \right)\varphi \left( z \right)} \hfill \cr {} \hfill & { + \left( {{a_2} + 3{b_1}{b_2} + {c_2}c_1^2} \right)\left[ {\varphi \left( x \right)M\left( y \right)M\left( z \right) + M\left( x \right)\varphi \left( y \right)M\left( z \right) + M\left( x \right)M\left( y \right)\varphi \left( z \right)} \right]} \hfill \cr {} \hfill & { + \left( {2b_2^2 + {c_1}c_2^2} \right)\left[ {\varphi \left( x \right)\varphi \left( y \right)M\left( z \right) + \varphi \left( x \right)M\left( y \right)\varphi \left( z \right) + M\left( x \right)\varphi \left( y \right)\varphi \left( z \right)} \right]} \hfill \cr } for all x, y, z ∈ S. By Lemma 3.2 we see that all coefficients vanish, so in particular c₂ = 0, and for the rest we have $a_{1} + 3 b_{1}^{2} + c_{1}^{3} = a_{2} + 3 b_{1} b_{2} = 2 b_{2}^{2} = 0$ {a_1} + 3b_1^2 + c_1^3 = {a_2} + 3{b_1}{b_2} = 2b_2^2 = 0 . Thus a₂ = b₂ = 0 also. Now we have f, g, h ∈ span{M} and revert to family (a).

This completes the proof.

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