On the Alienation of Multiplicative and Additive Functions

The solutions f, g : S → ℂ of the functional equation fxy+gxy=fx+fy+gxgy, x, y∈S, f\left( {xy} \right) + g\left( {xy} \right) = f\left( x \right) + f\left( y \right) + g\left( x \right)g\left( y \right), \;\;\; x,\;y \in S, are the following pairs:

• (a)

  g=χ+α2α2+1 g = {{\chi + {\alpha ^2}} \over {{\alpha ^2} + 1}} , f=A−α2α2+12χ−1 f = A - {{{\alpha ^2}} \over {{{\left( {{\alpha ^2} + 1} \right)}^2}}}\left( {\chi - 1} \right) , where α ∈ ℂ\ {i,−i} and A is additive and χ is multiplicative;

• (b)

  g = A + 1, f=B+12A2 f = B + {1 \over 2}{A^2} , where A, B are additive.

Let the pair (f, g) be a solution of equation (2.3). If g = 1 or g = 0, then f is additive. This occurs in case (a) with χ = 1 or with χ = 0 and α = 0. From now on we assume that g ≠ 1 and g ≠ 0 and we follow the proof given by Ger [8] in the commutative case. It is well known that the Cauchy difference C_f (x, y) := f(xy) − f(x) − f(y), x, y ∈ S satisfies the cocycle equation (3.1) Cfxy, z+Cfx, y=Cfx, yz+Cfy, z, x, y, z∈S. {C_f}\left( {xy,\;z} \right) + {C_f}\left( {x,\;y} \right) = {C_f}\left( {x,\;yz} \right) + {C_f}\left( {y,\;z} \right),\,\,\,\,\,\,x,\;y,\;z \in S.

From (2.3) we find that Cfx, y=gxgy−gxy, x, y∈S. {C_f}\left( {x,\;y} \right) = g\left( x \right)g\left( y \right) - g\left( {xy} \right), \;\;\; x,\;y \in S.

Inserting the last identity into (3.1), we obtain the following functional equation (3.2) gxygz−gygz+gyz =gxgyz−gxgy+gxy, x, y, z∈S. \eqalign{ & g\left( {xy} \right)g\left( z \right) - g\left( y \right)g\left( z \right) + g\left( {yz} \right) \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = g\left( x \right)g\left( {yz} \right) - g\left( x \right)g\left( y \right) + g\left( {xy} \right),\,\,\,\,\,\,\;x,\;y,\;z \in S. \cr}

Now, the rest of the proof takes another way, we simply notice that g satisfies the cosine addition law. Indeed, by subtracting g(x)g(y)g(z) from both sides of (3.2), we get gz−1gxy−gxgy=gx−1gyz−gygz, x,y,z∈S. \left( {g\left( z \right) - 1} \right)\left[ {g\left( {xy} \right) - g\left( x \right)g\left( y \right)} \right] = \left( {g\left( x \right) - 1} \right)\left[ {g\left( {yz} \right) - g\left( y \right)g\left( z \right)} \right], \;\;\; x,y,z \in S.

Since g ≠ 1, there exists x₀ ∈ S such that (3.3) gxy−gxgy =(gx0−1)−1 gx−1gyx0−gygx0, y∈S. \eqalign{ & g\left( {xy} \right) - g\left( x \right)g\left( y \right) \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = {(g\left( {{x_0}} \right) - 1)^{ - 1}}\left( {g\left( x \right) - 1} \right)\left[ {g\left( {y{x_0}} \right) - g\left( y \right)g\left( {{x_0}} \right)} \right],\,\,\,\,y \in S. \cr} <math display='block'></math>

Using (3.3), the term g(yx₀) − g(y)g(x₀) can be written as gyx0−gygx0=gx0−1−1gy−1gx02−gx02, x,y∈S. g\left( {y{x_0}} \right) - g\left( y \right)g\left( {{x_0}} \right) = {\left( {g\left( {{x_0}} \right) - 1} \right)^{ - 1}}\left( {g\left( y \right) - 1} \right)g\left( {x_0^2} \right) - \left[ {\left( {g{{\left( {{x_0}} \right)}^2}} \right)} \right], \;\;\; x,y \in S.

It follows from (3.3) that g satisfies the cosine functional equation (3.4) gxy=gxgy+α2gx−1gy−1, x, y∈S, g\left( {xy} \right) = g\left( x \right)g\left( y \right) + {\alpha ^2}\left( {g\left( x \right) - 1} \right)\left( {g\left( y \right) - 1} \right),\,\,\,\,\,\,\,x,\;y \in S, where α2=gx0−1−2gx02−gx02 {\alpha ^2} = {\left( {g\left( {{x_0}} \right)\left. { - 1} \right)} \right)^{ - 2}}\left[ {{{\left( {g\left( {x_0^2} \right) - g\left( {{x_0}} \right)} \right)}^2}} \right] .

If α = 0, then g is multiplicative and, consequently, f is additive. This occurs in case (a). Otherwise we have to deal with a particular case of the cosine addition law (3.4) with α ≠ 0. According to [3, Theorem 3.2] (see [1, Theorem 2.2] for simpler formulas), the pair (g, α(g − 1)) has one of the forms included in the following list:

• (i)

  g = 0 on S², where S² = {xy | x, y ∈ S} and α(g − 1) = ±ig.

• (ii)

  g=11+c2m g = {1 \over {1 + {c^2}}}m and αg−1=c1+c2m \alpha \left( {g - 1} \right) = {c \over {1 + {c^2}}}m , where c ∈ ℂ \ {0, i,−i} and m is a nonzero multiplicative function.

• (iii)

  g=δ−1χ1+δχ2δ−1+δ g = {{{\delta ^{ - 1}}{\chi _1} + \delta {\chi _2}} \over {{\delta ^{ - 1}} + \delta }} and αg−1=χ2−χ1δ−1+δ \alpha \left( {g - 1} \right) = {{{\chi _2} - {\chi _1}} \over {{\delta ^{ - 1}} + \delta }} , where χ₁ and χ₂ are two multiplicative functions such that χ₁ ≠ χ₂ and δ ∈ ℂ \ {0, i,−i}.

• (iv)

  g = χ ± ϕ and α(g − 1) = −iϕ, where χ is a nonzero multiplicative function and ϕ is a solution of the special sine addition law

(3.5) ϕxy=ϕxχy+ϕyχx, x, y∈S. \phi \left( {xy} \right) = \phi \left( x \right)\chi \left( y \right) + \phi \left( y \right)\chi \left( x \right),\,\,\,\,x,\;y \in S.

First case: the pair (g, α(g − 1)) has the form (i). This case is omitted because the identities α(g − 1) = ±ig and g = 0 on S² imply that α = 0.

Second case: the pair (g, α(g − 1)) has the form (ii). In this case we have α ≠ c and g=αα−c g = {\alpha \over {\alpha - c}} . Since g is a solution of (3.4), we check by elementary computations that αc = −1 and hence g=α2α2+1 g = {{{\alpha ^2}} \over {{\alpha ^2} + 1}} . Using this formulas in (2.3), we get that f−α2(α2+1)2 f - {{{\alpha ^2}} \over {{{({\alpha ^2} + 1)}^2}}} is an additive function. That is f=A+α2(α2+1)2 f = A + {{{\alpha ^2}} \over {{{({\alpha ^2} + 1)}^2}}} , where A is additive. This occurs in case (a) with χ = 0.

Third case: the pair (g, α(g − 1)) has the form (iii). In this case we deduce that (3.6) δ−1+α−1δ−1+δχ1+δ−α−1δ−1+δχ2−1=0. \left[ {{{{\delta ^{ - 1}} + {\alpha ^{ - 1}}} \over {{\delta ^{ - 1}} + \delta }}} \right]{\chi _1} + \left[ {{{\delta - {\alpha ^{ - 1}}} \over {{\delta ^{ - 1}} + \delta }}} \right]{\chi _2} - 1 = 0.

Using [11, Theorem 3.18], equation (3.6) shows that χ₁ = 1 and δ = α⁻¹ or χ₂ = 1 and δ = α since χ₁ ≠ χ₂, then g=χ+α2α2+1 g = {{\chi + {\alpha ^2}} \over {{\alpha ^2} + 1}} , where χ is a multiplicative function. Using this form of g in equation (2.3) we get after some rearrangements that f=A−α2(α2+1)2χ−1 f = A - {{{\alpha ^2}} \over {{{({\alpha ^2} + 1)}^2}}}\left( {\chi - 1} \right) , where A is an additive function. This occurs in case (a).

Fourth case: the pair (g, α(g − 1)) has the form (iv). In this case we have (3.7) χ+iα−1±1ϕ−1=0. \chi + \left( {i{\alpha ^{ - 1}} \pm 1} \right)\phi - 1 = 0.

Since χ ≠ 0, we deduce that ϕ ≠ χ because otherwise, we will have χ(xy) = 2χ(x)χ(y) for all x, y ∈ S which implies χ=0. In view of [4, Lemma 4.2], the identity (3.7) shows that ϕ = 1 or χ = 1.

The case ϕ = 1 does not occur because, otherwise, we will have χ = 1, which contradicts the fact that ϕ ≠ χ. So we have to take χ = 1. It follows that ϕ is additive, consequently g = A + 1, where A is an additive function. Then, from (2.3) we deduce that f and g satisfy f(xy) = f(x)+f(y)−g(xy)+g(x)g(y), x, y ∈ S, or equivalently (3.8) fxy=fx+fy+AxAy, x, y∈S. f\left( {xy} \right) = f\left( x \right) + f\left( y \right) + A\left( x \right)A\left( y \right),\,\,\,\,\,x,\;y \in S.

Consider the particular Levi–Civita equation (3.9) ψxy=ψxχy+χxψy+ϕxϕy, x, y∈S, \psi \left( {xy} \right) = \psi \left( x \right)\chi \left( y \right) + \chi \left( x \right)\psi \left( y \right) + \phi \left( x \right)\phi \left( y \right),\,\,\,\,\,\,\,\,x,\;y \in S, for unknown ψ: S → ℂ, where χ is a nonzero multiplicative function and ϕ is a nonzero solution of the special sine addition law (3.5). Taking χ = 1 and ϕ = A, one can observe that the functional equation (3.8) is a particular instance of (3.9). According to [4, Theorem 4.1] with χ = 1, we have f=B+12A2 f = B + {1 \over 2}{A^2} , where B is an additive function. This is case (b), and this completes the proof.

Let S be a semigroup. If functions f : S → ℂ and g : S → ℂ are not quadratically equivalent then they satisfy the functional equation fxy+gxy=fx+fy+gxgy, x, y∈S, f\left( {xy} \right) + g\left( {xy} \right) = f\left( x \right) + f\left( y \right) + g\left( x \right)g\left( y \right), \;\;\; x,\;y \in S, if and only if g is multiplicative and f is additive.

The “if” part of the proof is obvious.

Let the pair (f, g) be a solution of (2.3). If the pair (f, g) has the form as in Theorem 3.1(b), then Δy3f+g=0 \Delta _y^3\left( {f + g} \right) = 0 . This is a contradiction. If (f, g) has the form as in Theorem 3.1(a) such that α ≠ 0, then we find that Δy3f−c−1g=Δy3A−c−1=0 \Delta _y^3\left[ {f - \left( {c - 1} \right)g} \right] = \Delta _y^3\left[ {A - \left( {c - 1} \right)} \right] = 0 , where c=1α2+1 c = {1 \over {{\alpha ^2} + 1}} , which contradicts the fact that f and g are not quadratically equivalent. So, necessarily we have α = 0. Then g is multiplicative and f is additive. This completes the proof.

By replacing x by xy and y by ks in (2.7) and integrating the result obtained with respect to k and s we get (4.1) ∫S∫S∫Sfxykstdμkdμsdμt =fxy∫Sdμt2+∫S∫Sfksdμkdμs . \eqalign{ & \int_S {\int_S {\int_S {f\left( {xykst} \right)d\mu \left( k \right)d\mu \left( s \right)d\mu \left( t \right)} } } \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = f\left( {xy} \right){\left( {\int_S {d\mu \left( t \right)} } \right)^2} + \int_S {\int_S {f\left( {ks} \right)d\mu \left( k \right)d\mu \left( s \right)} } . \cr}

By replacing x by xyk and y by s in (2.7) and integrating the result obtained with respect to k and s we obtain (4.2) ∫S∫S∫Sfxykstdμkdμsdμt =∫Sfxykdμk ∫Sdμt+ ∫Sfsdμs ∫Sdμt =∫Sdμt fx+fy+∫Sfsdμs . \eqalign{ & \int_S {\int_S {\int_S {f\left( {xykst} \right)d\mu \left( k \right)d\mu \left( s \right)d\mu \left( t \right)} } } \cr & \,\,\,\,\,\, = \int_S {f\left( {xyk} \right)d\mu \left( k \right)} \int_S {d\mu \left( t \right) + } \int_S {f\left( s \right)d\mu \left( s \right)} \int_S {d\mu \left( t \right)} \cr & \,\,\,\,\,\, = \int_S {d\mu \left( t \right)} \left[ {f\left( x \right) + f\left( y \right) + \int_S {f\left( s \right)d\mu \left( s \right)} } \right]. \cr} <math display='block'></math>

By comparing (4.1) and (4.2) and taking into account that ∫_S dμ(t) = 1 we deduce that f + ∫_S f(s)dμ(s) −∫_S ∫_S f(ks)dμ(k)dμ(s) is additive. Consequently f = A −∫_S f(s)dμ(s) + ∫_S ∫_S f(ks)dμ(k)dμ(s), where A is an additive function. Putting this back into (2.7) we find that −∫_S f(s)dμ(s) + ∫_S ∫_S f(ks)dμ(k)dμ(s) = ∫_S A(s)dμ(s). This completes the proof.

The solutions f, g : S → ℂ of the functional equation ∫Sfxytdμt+∫Sgxytdμt=fx+fy+gxgy, x,y∈S, \int_S {f\left( {xyt} \right)d\mu \left( t \right) + } \int_S {g\left( {xyt} \right)d\mu \left( t \right) = f\left( x \right) + f\left( y \right) + g\left( x \right)g\left( y \right), }\;\;\; x,y \in S, are the following pairs:

• (a)

  g = c, f = A + ∫_S A(t)dμ(t) + c − c², where c ∈ ℂ and A is additive;

• (b)

  g = A, f=B+∫SBtdμt+12A2+∫SA2tdμt−1 f = B + \int_S {B\left( t \right)d\mu \left( t \right) + {1 \over 2}\left[ {{A^2} + \int_S {{A^2}\left( t \right)d\mu \left( t \right)} } \right] - 1} , where A and B are additive, and ∫_S A(t)dμ(t) = −1;

• (c)

  g = c (φ − 1), f = A+ ∫_S A(t)dμ(t)+c² (φ − 1)−c, where c ∈ ℂ\ {−1, 0}, φ is a multiplicative function such that φ ≠ 1, ∫Sφtdμt=cc+1 \int_S {\varphi \left( t \right)d\mu \left( t \right) = {c \over {c + 1}}} and A is additive;

• (d)

  g = χ ∫_S χ(t)dμ(t), f = A + ∫_S A(t)dμ(t), where χ is an nonzero multiplicative function such that χ ≠ 1 and A is additive;

• (e)

  g=rχ+α2α2+1 g = r{{\chi + {\alpha ^2}} \over {{\alpha ^2} + 1}} , f=r2A+∫SAtdμt−α2(α2+1)2χ−1+r1−rα2α2+1 f = {r^2}\left[ {A + \int_S {A\left( t \right)d\mu \left( t \right) - {{{\alpha ^2}} \over {{{({\alpha ^2} + 1)}^2}}}\left( {\chi - 1} \right)} } \right] + r\left( {1 - r} \right){{{\alpha ^2}} \over {{\alpha ^2} + 1}} , where α ∈ ℂ\ {0, i,−i}, r ∈ ℂ\ {0} and r≠α2+1α2 r \ne {{{\alpha ^2} + 1} \over {{\alpha ^2}}} , A is additive and χ is a multiplicative function such that χ ≠ 1 and ∫Sχtdμt=rα21−r+1 \int_S {\chi \left( t \right)d\mu \left( t \right) = {r \over {{\alpha ^2}\left( {1 - r} \right) + 1}}} ;

• (f)

  g = r(A+1), f=r2B+∫SBtdμt+12A2+12∫SA2tdμt −(r−1)2 f = {r^2}\left[ {B + \int_S {B\left( t \right)d\mu \left( t \right) + {1 \over 2}{A^2} + {1 \over 2}\int_S {{A^2}\left( t \right)d\mu \left( t \right)} } } \right] - {(r - 1)^2} , where r ∈ ℂ\ {0} and A, B are additive such that A ≠ 0 and ∫SAtdμt=r−1r \int_S {A\left( t \right)d\mu \left( t \right) = {{r - 1} \over r}} .

Let the pair (f, g) be a solution of (2.6). If g is constant, that is g = c, where c ∈ ℂ, we see from (2.6) that f − c + c² is a solution of equation (2.7). Then, using Lemma 4.1, we conclude that f − c + c² = A + ∫_S A(t)dμ(t), where A is additive. This proves (a). From now on we assume that g is not constant.

Using the associative property of the semigroup operation, we compute the term ∫_S f(xyzt)dμ(t) + ∫_S g(xyzt)dμ(t) first as ∫_S f((xy)zt)dμ(t) + ∫_S g((xy)zt)dμ(t), then as ∫_S f(x(yz)t)dμ(t) + ∫_S g(x(yz)t)dμ(t). Comparing the results, we obtain the following (4.3) fxy+fz+gxygz=fx+fyz+gxgyz, x,y,z∈S. f\left( {xy} \right) + f\left( z \right) + g\left( {xy} \right)g\left( z \right) = f\left( x \right) + f\left( {yz} \right) + g\left( x \right)g\left( {yz} \right), \;\;\; x,y,z \in S.

Subtracting g(x)g(y)g(z) + f(y) from both sides of (4.3), we get (4.4) fxy−fx−fy+gzgxy−gxgy =fyz−fy−fz+gxgyz−gygz, x, y, z∈S. \eqalign{ & f\left( {xy} \right) - f\left( x \right) - f\left( y \right) + g\left( z \right)\left[ {g\left( {xy} \right) - g\left( x \right)g\left( y \right)} \right] \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\, = f\left( {yz} \right) - f\left( y \right) - f\left( z \right) + g\left( x \right)\left[ {g\left( {yz} \right) - g\left( y \right)g\left( z \right)} \right], & x,\;y,\;z \in S. \cr}

Since g is not constant, there exist z₁, z₂ ∈ S such that g(z₁) ≠ g(z₂). So, from (4.4) we get fxy−fx−fy+gz1gxy−gxgy=k1y+gxk2y, x,y∈S, f\left( {xy} \right) - f\left( x \right) - f\left( y \right) + g\left( {{z_1}} \right)\left[ {g\left( {xy} \right) - g\left( x \right)g\left( y \right)} \right] = {k_1}\left( y \right) + g\left( x \right){k_2}\left( y \right), \;\;\; x,y \in S, and fxy−fx−fy+gz2gxy−gxgy=k3y+gxk4y, x,y∈S, f\left( {xy} \right) - f\left( x \right) - f\left( y \right) + g\left( {{z_2}} \right)\left[ {g\left( {xy} \right) - g\left( x \right)g\left( y \right)} \right] = {k_3}\left( y \right) + g\left( x \right){k_4}\left( y \right), \;\;\; x,y \in S, for some functions k_j, j = 1, 2, 3, 4, or equivalently 1gz11gz2 fxy−fx−fygxy−gxgy =k1yk2yk3yk4y 1gx , x,y∈S. \left[ {\matrix{ 1 \hfill & {g\left( {{z_1}} \right)} \hfill \cr 1 \hfill & {g\left( {{z_2}} \right)} \hfill \cr } } \right]\left[ {\matrix{ {f\left( {xy} \right) - f\left( x \right) - f\left( y \right)} \cr {g\left( {xy} \right) - g\left( x \right)g\left( y \right)} \cr } } \right] = \left[ {\matrix{ {{k_1}\left( y \right)} \hfill & {{k_2}\left( y \right)} \hfill \cr {{k_3}\left( y \right)} \hfill & {{k_4}\left( y \right)} \hfill \cr } } \right]\left[ {\matrix{ 1 \cr {g\left( x \right)} \cr } } \right], \;\;\; x,y \in S.

Since g(z₁) ≠ g(z₂) we have (4.5) fxy−fx−fy=ρ1y+gxρ2y, x, y∈S, f\left( {xy} \right) - f\left( x \right) - f\left( y \right) = {\rho _1}\left( y \right) + g\left( x \right){\rho _2}\left( y \right),\,\,\,\,\,\,\,x,\;y \in S, and (4.6) gxy−gxgy=ρ3y+gxρ4y, x, y∈S, g\left( {xy} \right) - g\left( x \right)g\left( y \right) = {\rho _3}\left( y \right) + g\left( x \right){\rho _4}\left( y \right),\,\,\,\,\,\,\,\;x,\;y \in S, for some functions ρ_j, j = 1, 2, 3, 4.

Putting this back into (4.4) we get after some rearrangements that (4.7) gxρ2y+gzρ4y−ρ3z−gyρ4z =ρ1z+gyρ2z−ρ1y−gzρ3y, x, y, z∈S. \eqalign{ & g\left( x \right)\left[ {{\rho _2}\left( y \right) + g\left( z \right){\rho _4}\left( y \right) - {\rho _3}\left( z \right) - g\left( y \right){\rho _4}\left( z \right)} \right] \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = {\rho _1}\left( z \right) + g\left( y \right){\rho _2}\left( z \right) - {\rho _1}\left( y \right) - g\left( z \right){\rho _3}\left( y \right),\,\,\,\,\,\,\,\,\;x,\;y,\;z \in S. \cr}

Taking into account that g is not constant, (4.7) shows that (4.8) ρ1z+gyρ2z=ρ1y+gzρ3y, y, z∈S, {\rho _1}\left( z \right) + g\left( y \right){\rho _2}\left( z \right) = {\rho _1}\left( y \right) + g\left( z \right){\rho _3}\left( y \right),\,\,\,\,\,\,\,\,\;y,\;z \in S, and (4.9) ρ2y+gzρ4y=ρ3z+gyρ4z, y, z∈S. {\rho _2}\left( y \right) + g\left( z \right){\rho _4}\left( y \right) = {\rho _3}\left( z \right) + g\left( y \right){\rho _4}\left( z \right),\,\,\,\,\,\,\,y,\;z \in S.

Consequently, for the same reason, (4.8) and (4.9) show that (4.10) ρj=aj+bjg, j=1,2,3,4, {\rho _j} = {a_j} + {b_j}g, \;\;\; j = \;1,2,3,4, where a_j, b_j are constants.

Substituting (4.10) into (4.8) and (4.9) we infer that b1=a2=a3 and b3=b2=a4. {b_1} = {a_2} = {a_3}\; {\rm{and}} \; {b_3} = {b_2} = {a_4}.

Then, from (4.5) and (4.6) we deduce that f and g satisfy the functional equations (4.11) fxy=fx+fy+b1gx+b1gy+b2gxgy+a1, x,y∈S, f\left( {xy} \right) = f\left( x \right) + f\left( y \right) + {b_1}g\left( x \right) + {b_1}g\left( y \right) + {b_2}g\left( x \right)g\left( y \right) + {a_1}, \;\;\; x,y \in S, <math display='block'></math> and (4.12) gxy=b4+1gygx+b2gx+b2gy+b1, x, y∈S. g\left( {xy} \right) = \left( {{b_4} + 1} \right)g\left( y \right)g\left( x \right) + {b_2}g\left( x \right) + {b_2}g\left( y \right) + {b_1},\;\,\,\,\,\,\,x,\;y \in S.

By replacing y by t in (4.11) and (4.12) and integrating the results obtained with respect to t we get (4.13) ∫Sfxtdμt=fx+a0gx+b0, x∈S, \int_S {f\left( {xt} \right)d\mu \left( t \right) = f\left( x \right) + {a_0}g\left( x \right) + {b_0},\,\,\,\,\,\,\;x \in S,} and (4.14) ∫Sgxtdμt=c0gx+d0, x∈S, \int_S {g\left( {xt} \right)d\mu \left( t \right) = {c_0}g\left( x \right) + {d_0},\,\,\,\,\,\;x \in S,} where a₀ := b₁ + b₂ ∫_S g(t)dμ(t), b₀ := ∫_S f(t)dμ(t) + b₁ ∫_S g(t)dμ(t) + a₁, c₀ := (b₄ + 1) ∫_S g(t)dμ(t) + b₂ and d₀ := b₂ ∫_S g(t)dμ(t) + b₁. Now, by using the relations (4.13) and (4.14) in (2.6) we get (4.15) fxy+rgxy+q=fx+fy+gxgy, x, y∈S, f\left( {xy} \right) + rg\left( {xy} \right) + q = f\left( x \right) + f\left( y \right) + g\left( x \right)g\left( y \right),\,\,\,\,\,\,\,\,\;x,\;y \in S, where r := a₀ + c₀ and q = b₀ + d₀.

First case: r = 0. In this case (4.15) becomes (4.16) fxy+q=fx+fy+gxgy, x, y∈S. f\left( {xy} \right) + q = f\left( x \right) + f\left( y \right) + g\left( x \right)g\left( y \right),\,\,\,\,\,\,\,\;x,\;y \in S.

Equation (4.16) shows that the Cauchy difference C_f has the form Cfx, y=gxgy−q, x, y∈S. {C_f}\left( {x,\;y} \right) = g\left( x \right)g\left( y \right) - q, \;\;\; x,\;y \in S.

Using again the cocycle equation (3.1) leads to gzgxy−gy=gxgyz−gy, x, y, z∈S. g\left( z \right)\left[ {g\left( {xy} \right) - g\left( y \right)} \right] = g\left( x \right)\left[ {g\left( {yz} \right) - g\left( y \right)} \right], \;\;\; x,\;y,\;z \in S.

Since g ≠ 0, we deduce that (4.17) gxy=gxφy+gy, x, y∈S, g\left( {xy} \right) = g\left( x \right)\varphi \left( y \right) + g\left( y \right),\,\,\,\,\,\,\;x,\;y \in S, where φ(y) := (g(s⁰))⁻¹ [g(ys₀) − g(y)], for some s₀ ∈ S such that g(s₀) ≠ 0.

Now, from [12, Proposition 3], the identity (4.17) shows that φ is multiplicative. In view of (4.12) g is central: g(xy) = g(yx), x, y ∈ S. Using the centrality of g, (4.17) shows that gxφy−1=gyφx−1, x, y∈S. g\left( x \right)\left[ {\varphi \left( y \right) - 1\left] { = g\left( y \right)} \right[\varphi \left( x \right) - 1} \right], \;\;\; x,\;y \in S.

We may distinguish two subcases here.

If φ = 1, then (4.17) shows that g = A, where A is additive, and since f and g satisfy equation (4.16), we conclude that fxy−q=fx−q+fy−q+AxAy, x, y∈S. f\left( {xy} \right) - q = \left[ {f\left( x \right) - q} \right] + \left[ {f\left( y \right) - q} \right] + A\left( x \right)A\left( y \right), \;\;\; x,\;y \in S.

Then from [4, Theorem 4.1] we have f=B+12A2+q f = B + {1 \over 2}{A^2} + q , where B is an additive function. Since the pair (f, g) satisfies (2.6), by taking x = y we have the following relation (4.18) 2∫SAtdμt+1 Ax =q−∫SAtdμt−12 ∫SA2tdμt− ∫SBtdμt, x∈S. \eqalign{ & 2\left[ {\int_S {A\left( t \right)d\mu \left( t \right) + 1} } \right]A\left( x \right) \cr & \,\,\,\,\,\,\,\, = q - \int_S {A\left( t \right)d\mu \left( t \right) - {1 \over 2}} \int_S {{A^2}\left( t \right)d\mu \left( t \right) - } \int_S {B\left( t \right)d\mu \left( t \right),\,\,\,\,\,\;x \in S.} \cr}

Recall that g is a nonconstant function. So, from (4.18) we deduce that ∫_S A(t)dμ(t) = −1 and q=∫SBtdμt+12∫SA2tdμt−1 q = \int_S {B\left( t \right)d\mu \left( t \right) + {1 \over 2}\int_S {{A^2}\left( t \right)d\mu \left( t \right) - 1} } . This proves (b).

If φ ≠ 1, then (4.19) gx=cφx−1, x∈S, g\left( x \right) = c\left( {\varphi \left( x \right) - 1} \right),\,\,\,\,\,\,\;x \in S, where c ∈ ℂ \ {0}. Inserting (4.19) into (4.16), we deduce that f−c² (φ − 1)−q is an additive function, that is f = A + c² (φ − 1) + q, where A is additive. Putting back the new forms of f and g into (2.6) yields (4.20) c∫Sφtdμt+c∫Sφtdμt−c φxφy =q−∫SAtdμt+c, x, y∈S. \eqalign{ & c\left[ {\int_S {\varphi \left( t \right)d\mu \left( t \right) + c\int_S {\varphi \left( t \right)d\mu \left( t \right) - c} } } \right]\varphi \left( x \right)\varphi \left( y \right) \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = q - \int_S {A\left( t \right)d\mu \left( t \right) + c,\,\,\,\,\,\,\,\,\,x,\;y \in S.} \cr}

From (4.19) we deduce that φ can not be constant. So (4.20) shows that ∫_S φ(t)dμ(t) + c ∫_S φ(t)dμ(t) − c = 0 and q −∫_S A(t)dμ(t) + c = 0. Then f(x) = A(x)+ ∫_S A(t)dμ(t)+c² (φ − 1)−c, x ∈ S and ∫Sφtdμt=cc+1 \int_S {\varphi \left( t \right)d\mu \left( t \right) = {c \over {c + 1}}} . This proves (c).

Second case: r ≠ 0. From equation (4.15) one can easily verify that F := r⁻² (f − q) and G := r⁻¹g solve the functional equation (2.3). So, we know from Theorem 3.1 that there are only the following two possibilities:

• (i)

  g=rχ+α2α2+1 g = r{{\chi + {\alpha ^2}} \over {{\alpha ^2} + 1}} and f=r2A−α2(α2+1)2χ−1+q f = {r^2}\left[ {A - {{{\alpha ^2}} \over {{{({\alpha ^2} + 1)}^2}}}\left( {\chi - 1} \right)} \right] + q , where α ∈ ℂ \ {i,−i}, A is additive and χ is multiplicative.

• (ii)

  g = r (A + 1) and f=r2B+12A2+q f = {r^2}\left[ {B + {1 \over 2}{A^2}} \right] + q , where A, B are additive functions.

Subcase 1: If the pair (f, g) has the form (i). If α = 0, then f and g satisfy the following equations (4.21) fxy−q=fx−q+fy−q, x, y∈S, f\left( {xy} \right) - q = \left[ {f\left( x \right) - q} \right] + \left[ {f\left( y \right) - q} \right],\,\,\,\,\,\,\,\,x,\;y \in S, and (4.22) r−1gxy=r−1gxr−1gy, x, y∈S, {r^{ - 1}}g\left( {xy} \right) = \left[ {{r^{ - 1}}g\left( x \right)} \right]\left[ {{r^{ - 1}}g\left( y \right)} \right],\,\,\,\,\,\,\,x,\;y \in S, respectively.

From (4.21) and (4.22) we deduce that (4.23) ∫Sfxytdμt=fx+fy+∫Sftdμt−2q, x, y∈S \int_S {f\left( {xyt} \right)d\mu \left( t \right) = f\left( x \right) + f\left( y \right) + } \int_S {f\left( t \right)d\mu \left( t \right) - 2q,\,\,\,\,\,x,\;y \in S} and (4.24) ∫Sgxytdμt=r−2gxgy∫Sgtdμt, x, y∈S. \int_S {g\left( {xyt} \right)d\mu \left( t \right) = {r^{ - 2}}g\left( x \right)g\left( y \right)} \int_S {g\left( t \right)d\mu \left( t \right),\,\,\,\,\,\,x,\;y \in S.}

Now, by inserting (4.23) and (4.24) into (2.6), we see that r−2∫Sgtdμt−1 gxgy=2q−∫Sftdμt, x, y∈S. \left[ {{r^{ - 2}}\int_S {g\left( t \right)d\mu \left( t \right) - 1} } \right]g\left( x \right)g\left( y \right) = 2q - \int_S {f\left( t \right)d\mu \left( t \right), \;\;\; x,\;y \in S.}

Since g is not constant, the last identity implies that r⁻² ∫_S g(t)dμ(t) = 1 and 2q −∫_S f(t)dμ(t) = 0. It follows that f and g are solutions of (2.7) and (2.8) respectively. Then, from Lemma 4.1 and [5, Corollary 2.5], we deduce the desired forms of f and g. This proves (d).

From now on we assume that α ≠ 0. Using the notation c:=1α2+1 c: = {1 \over {{\alpha ^2} + 1}} , f and g can be written as f = r² [A − c(1 − c)(χ − 1)] + q and g = r [cχ + (1 − c)] respectively. Putting this back into (2.6), one gets (4.25) rc1−r1−c∫Sχtdμt−rc χxχy =q−r2∫SAtdμt−r1−r1−c, x, y∈S. \eqalign{ & rc\left[ {\left( {1 - r\left( {1 - c} \right)} \right)\int_S {\chi \left( t \right)d\mu \left( t \right) - rc} } \right]\chi \left( x \right)\chi \left( y \right) \cr & = q - {r^2}\int_S {A\left( t \right)d\mu \left( t \right) - r\left( {1 - r} \right)\left( {1 - c} \right), \;\;\;\;\;\;\;\;\;\;\;\;\;\;\; \;x,\;y \in S.} \cr}

As χ is not constant, (4.25) implies that 1 − r(1 − c) ≠ 0 and (1 − r(1 − c)) ∫_S χ(t)dμ(t)−rc = 0. Consequently we have r≠11−c r \ne {1 \over {1 - c}} , ∫Sχtdμt=rc1−r1−c \int_S {\chi \left( t \right)d\mu \left( t \right) = {{rc} \over {1 - r\left( {1 - c} \right)}}} and q − r² ∫_S A(t)dμ(t) − r(1 − c)(1 − r) = 0. This proves (e).

Subcase 2: If the pair (f, g) has the form (ii). Then, taking x = y, (2.6) implies that (4.26) 2rr∫SAtdμt+1−r Ax=q−r+r2 −r22∫SA2tdμt−r ∫SAtdμt−r2 ∫SBtdμt, x∈S. \eqalign{ & 2r\left[ {r\int_S {A\left( t \right)d\mu \left( t \right) + 1 - r} } \right]A\left( x \right) = q - r + {r^2} \cr & \,\,\,\,\,\,\,\, - {{{r^2}} \over 2}\int_S {{A^2}\left( t \right)d\mu \left( t \right) - r} \int_S {A\left( t \right)d\mu \left( t \right) - {r^2}} \int_S {B\left( t \right)d\mu \left( t \right), x \in S.} \cr}

Since A is not constant, we deduce from (4.26) that r ∫_SA(t)dμ(t)+1−r = 0. Consequently we have ∫SAtdμt=r−1r \int_S {A\left( t \right)d\mu \left( t \right) = {{r - 1} \over r}} and q=r22∫SA2tdμt+r2∫SBtdμt−(r−1)2 q = {{{r^2}} \over 2}\int_S {{A^2}\left( t \right)d\mu \left( t \right) + {r^2}} \int_S {B\left( t \right)d\mu \left( t \right) - {{(r - 1)}^2}} . This proves case (f).

Let S be a semigroup and μ is a linear combination of Dirac measures (δ_zi)_i∈I for some fixed elements (z_i)_i∈I contained in S such that ∫_S dμ = 1. If functions f : S → ℂ and g : S → ℂ are not quadratically equivalent then they satisfy the equation ∫Sfxytdμt+∫Sgxytdμt=fx+fy+gxgy, x,y∈S, \int_S {f\left( {xyt} \right)d\mu \left( t \right) + } \int_S {g\left( {xyt} \right)d\mu \left( t \right) = f\left( x \right) + f\left( y \right) + g\left( x \right)g\left( y \right), \;\;\; x,y \in S,} if and only if f and g solve the system ∫Sfxytdμt=fx+fy , ∫Sgxytdμt=gxgy, x, y∈S. \left\{ {\matrix{ {\int_S {f\left( {xyt} \right)d\mu \left( t \right) = f\left( x \right) + f\left( y \right)\;,} } \hfill \cr {\int_S {g\left( {xyt} \right)d\mu \left( t \right) = g\left( x \right)g\left( y \right), \;\;\; x,\;y \in S.} } \hfill \cr } } \right.

The “if”" part of the proof is obvious.

Let the pair (f, g) be a solution of (2.6). We can easily check that if (f, g) has one of the forms (a), (b) and (f) of Theorem 4.2, then we have Δy3f+g=0 \Delta _y^3\left( {f + g} \right) = 0 , which contradicts our hypothesis about f and g. In the case when (f, g) has the form (c) we obtain Δy3f−cg=0 \Delta _y^3\left( {f - cg} \right) = 0 and when (f, g) has the form (e) we have Δy3f+r0α2α2+1g=0 \Delta _y^3\left( {f + {r_0}{{{\alpha ^2}} \over {{\alpha ^2} + 1}}g} \right) = 0 . For the same reason, the two last cases does not occur. It follows that the only possible case is when (f, g) has the form (d). This completes the proof.