On a Generalized Conjecture by Alzer and Matkowski

Every solution f : ℝ → ℝ of (1.1) has a zero.

Assume that ϕ and f solve (1.2) and (2.1) ∃z0∈X ϕz0,z0≠0. {\exists_{{z_0} \in X}}\,\,\,\phi \left({{z_0},{z_0}} \right) \ne 0. Then there exists a unique constant a ∈ 𝕂 \ {0} such that (2.2) fx=aϕx,z0+1, x∈X, f\left(x \right) = a\phi \left({x,{z_0}} \right) + 1,\,\,\,\,\,\,x \in X, and moreover (2.3) a2ϕ(x,z0)2=ϕx,x, x∈X. {a^2}\phi {(x,{z_0})^2} = \phi \left({x,x} \right),\,\,\,\,\,\,x \in X.

Substituting y = z₀ and then y = −z₀ in (1.2) we obtain fx+z0=fxfz0−ϕx,z0, x∈X f\left({x + {z_0}} \right) = f\left(x \right)f\left({{z_0}} \right) - \phi \left({x,{z_0}} \right),\,\,\,\,\,x \in X and fx−z0=fxf−z0+ϕx, z0, x∈X. f\left({x - {z_0}} \right) = f\left(x \right)f\left({- {z_0}} \right) + \phi \left({x,\,{z_0}} \right),\,\,\,\,\,x \in X. Replace x by x + z₀ in the latter formula and join it with the former one to arrive at fx=fx+z0f−z0+ϕx+z0,z0=fxfz0−ϕx,z0f−z0+ϕx,z0+ϕz0,z0, x∈X. \matrix{{f\left(x \right)} \hfill & {= f\left({x + {z_0}} \right)f\left({- {z_0}} \right) + \phi \left({x + {z_0},{z_0}} \right)} \hfill \cr {} \hfill & {= \left[ {f\left(x \right)f\left({{z_0}} \right) - \phi \left({x,{z_0}} \right)} \right]f\left({- {z_0}} \right) + \phi \left({x,{z_0}} \right) + \phi \left({{z_0},{z_0}} \right),\,\,\,\,x \in X.} \hfill \cr}

Denote c := f(z₀), d := f(−z₀) and β = ϕ(z₀, z₀) ≠ 0. We get 1−cdfx=1−dϕx,z0+β, x∈X. \left({1 - cd} \right)f\left(x \right) = \left({1 - d} \right)\phi \left({x,{z_0}} \right) + \beta,\,\,\,\,\,x \in X. As stated in the proof of Theorem 1 in [1], it follows that f(0) = 1. The argument works in our case, as well. Indeed, substitution x = y = 0 in (1.2) gives us f(0)² = f(0), so f(0) = 0 or f(0) = 1. But f(0) = 0 would imply β = 0, which is a contradiction with the definition of β.

Therefore, from (1.2) applied for x = z₀ and y = −z₀ we deduce 1=fz0−z0=fz0f−z0+β, 1 = f\left({{z_0} - {z_0}} \right) = f\left({{z_0}} \right)f\left({- {z_0}} \right) + \beta, thus 1 − cd = β. Since β ≠ 0, then denoting a := (1 − d)/β we get (2.2). The case a = 0 is impossible, since it leads to a contradiction with (2.1).

To prove equality (2.3) apply (1.2) with substitution y = −x to obtain fxf−x=1−ϕx,x, x∈X. f\left(x \right)f\left({- x} \right) = 1 - \phi \left({x,x} \right),\,\,\,\,\,x \in X. Now, use the already proven formula (2.2) to derive (2.3) after some reductions.

From Theorem 1, we see that under assumption (2.1) and with a fixed functional ϕ there are always either no solutions or exactly two solutions f of (1.2). Indeed, if f is a solution, then it must be of the form (2.2) with some constant a ∈ 𝕂 \ {0}. Substituting this into (1.2) leads us to the equality: a2ϕx,z0ϕy,z0=ϕx,y, x,y∈X, {a^2}\phi \left({x,{z_0}} \right)\phi \left({y,{z_0}} \right) = \phi \left({x,y} \right),\,\,\,\,\,\,x,y \in X, which is true for two different values of a ≠ 0. Therefore, in case there do exist solutions, functional ϕ is of the form ϕx,y=a2Fx⋅Fy, x,y∈X, \phi \left({x,y} \right) = {a^2}F\left(x \right) \cdot F\left(y \right),\,\,\,\,\,\,x,y \in X, with an additive, nonzero functional F : X → 𝕂, and the two possible functions f are given by (2.2).

Assume that 𝕂 = ℝ and (2.4) ∃z0∈X ϕz0,z0<0. {\exists_{{z_0} \in X}}\,\,\,\phi \left({{z_0},{z_0}} \right) < 0. Then equation (1.2) has no solutions.

Inequality (2.4) implies that condition (2.1) holds true. Then, from Theorem 1 we obtain formula (2.3). However, in the real case formula (2.3) implies that ϕ(x, x) ≥ 0 for all x ∈ X, which leads to a contradiction with (2.1).

Assume that 𝕂 = ℂ, ϕ and f solve (1.2), ϕ satisfies (2.1) and w: X → ℂ is a map such that w2x=ϕx,x, x∈X. {w^2}\left(x \right) = \phi \left({x,x} \right),\,\,\,\,\,x \in X. Then fx=wx+1, x∈X, f\left(x \right) = w\left(x \right) + 1,\,\,\,\,\,x \in X, or fx=−wx+1, x∈X. f\left(x \right) = - w\left(x \right) + 1,\,\,\,\,x \in X.

Assume that ϕ and f solve (1.2) and ∀z∈X ϕz,z=0. {\forall_{z \in X}}\,\,\,\phi \left({z,z} \right) = 0. Then ϕ = 0 on X × X and fx+y=fxfy, x,y∈X. f\left({x + y} \right) = f\left(x \right)f\left(y \right),\,\,\,\,\,x,y \in X. Consequently, either f = 0 or there exists an additive functional A: X → 𝕂 such that f = exp ∘ A.

It suffices to apply a well-known result, which states that if a multiadditive function vanishes on a diagonal, then it vanishes everywhere, cf. [3, Corollary 15.9.1, p. 448]. The final part follows from the form of solutions of the exponential Cauchy equation, cf. [3, Theorem 13.1.1, p. 343].

Assume that α ∈ ℝ and f : ℝ → ℝ solves (1.1). Then α ≥ 0 and moreover, in case α > 0 either fx=1−αx f\left(x \right) = 1 - \sqrt \alpha x , or fx=1+αx f\left(x \right) = 1 + \sqrt \alpha x for x ∈ ℝ. Conversely, both mappings solve (1.1).

Firstly, substituting α = 0 in (1.1) we obtain the exponential Cauchy’s equation, for which the solutions are known. Now assume that α ≠ 0, 𝕂 = ℝ, X = ℝ and ϕ(x, y) := αxy. Let z₀ = 1. Then β = ϕ(1, 1) = α ≠ 0. From (2.2) we have fx=aϕx,1+1=aαx+1, x∈ℝ. f\left(x \right) = a\phi \left({x,1} \right) + 1 = a\alpha x + 1,\,\,\,\,\,x \in {\mathbb R}. From (2.3) we obtain a2(αx)2=ϕx,x=αx2, x∈ℝ. {a^2}{(\alpha x)^2} = \phi \left({x,x} \right) = \alpha {x^2},\,\,\,\,\,x \in {\mathbb R}. We get a² = 1/α, so α > 0 and a=±1/α a = \pm 1/\sqrt \alpha . After substitution to the equation for f we arrive at fx=±αx+1 f\left(x \right) = \pm \sqrt \alpha x + 1 . Conversely, it is easy to check that both such mappings solve (1.1).

Assume that α ∈ ℂ \ {0} and f : ℂ → ℂ solves (2.5) fx+y=fxfy−αxy, x,y∈ℂ. f\left({x + y} \right) = f\left(x \right)f\left(y \right) - \alpha xy,\,\,\,\,\,\,x,y \in {\mathbb C}. Then either f(x) = 1 + w₁x, or f(x) = 1 + w₂x for x ∈ ℂ, where w₁,w₂ are two complex roots of the second order of α. Conversely, both mappings solve (2.5).

Assume that 𝕂 = ℂ, X = ℂ and ϕ(x, y) := αxy. By repeating steps from the previous proof (this time without assuming α > 0), we obtain demanded results.

Let X be an inner product space of dimension at least 2 and define ϕ := 〈·, ·〉. Then, Theorem 1 implies that the potential solutions f : X → 𝕂 of (1.2) are of the form f(x)=x,ξ+1, x∈X, f(x) = \left\langle {x,\xi} \right\rangle + 1,\,\,\,\,\,x \in X, with some ξ ∈ X. One can check that such mapping solves (1.2) if and only if x,ξy,ξ=x,y, x,y∈X, \left\langle {x,\xi} \right\rangle \left\langle {y,\xi} \right\rangle = \left\langle {x,y} \right\rangle,\,\,\,\,\,\,x,y \in X, which is impossible if dim X ≥ 2.

Let X be a complex linear space and A: X → ℂ an additive nonzero functional. Define ϕ(x, y) := −A(x) · A(y) for x, y ∈ X. Then, according to Theorem 1 every solution f : X → ℂ of (1.2) is of the form: fx=γAx+1, x∈X f\left(x \right) = \gamma A\left(x \right) + 1,\,\,\,\,\,\,x \in X with some constant γ ∈ ℂ. A direct calculation shows that f is indeed a solution if and only if γ = ±i.

We can choose A in such a way that A(x) ≠ 0 whenever x ≠ 0, or such that A has a bigger set of zeros. Therefore, for every complex linear space X there is an abundance of nontrivial solutions (f, ϕ) to (1.2).

This example also illustrates that the assertion of Corollary 1 does not hold in the case of complex spaces, even when the values of ϕ are real (since A may only attain real values, as it does not necessarily have to be linear).

Let X be a linear space over the field 𝕂 and A: X → 𝕂 an additive nonzero functional. Define ϕ(x, y) := A(x) · A(y) for x, y ∈ X. Then, similarly every solution f : X → 𝕂 of (1.2) is of the form: fx=δAx+1, x∈X f\left(x \right) = \delta A\left(x \right) + 1,\,\,\,\,\,x \in X with some constant δ ∈ 𝕂. Further, f is indeed a solution if and only if δ = ±1.