On a Generalized Conjecture by Alzer and Matkowski

Włodzimierz Fechner; Marta Pierzchałka; Gabriela Smejda

doi:10.2478/amsil-2025-0007

1. Introduction

Alzer and Matkowski [1] recently studied the following functional equation:

(1.1)

f (x + y) = f (x) f (y) - α xy, x, y \in ℝ,

where α ∈ ℝ is a non-zero parameter and f : ℝ → ℝ is an unknown function. They proved two theorems on equation (1.1). The first result with a short proof [1, Theorem 1] completely describes solutions of (1.1) in case f has a zero. More precisely, they showed that if f solves (1.1) and it has a zero, then α > 0 and either

f (x) = 1 - \sqrt{α} x

, or

f (x) = 1 + \sqrt{α} x

for x ∈ ℝ. The second theorem with a longer proof [1, Theorem 2] provides the solutions to equation (1.1) under the assumption that f : ℝ → ℝ is differentiable at least at one point. In this case, there are the same two solutions (clearly, both are differentiable and have a zero). In [1] the authors formulated the following conjecture:

Conjecture (Alzer and Matkowski)

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

This conjecture has been answered affirmatively by T. Małolepszy, see [4]. In the present note, we will determine the solutions of a more general equation, namely

(1.2)

f (x + y) = f (x) f (y) - ϕ (x, y), x, y \in X,

where X is a linear space over the field 𝕂 ∈{ℝ,ℂ}, ϕ: X × X → 𝕂 is a biadditive functional and f : X → 𝕂 is a function. The motivation for such a generalization comes from an article by K. Baron and Z. Kominek [2], in which the authors, in connection with a problem proposed by S. Rolewicz [5], studied mappings defined on a real linear space with the additive Cauchy difference bounded from below by a bilinear functional.

2. Main results

In this section, it is assumed that X is a linear space over 𝕂 ∈ {ℝ,ℂ}, ϕ: X × X → 𝕂 is a biadditive functional and f : X → 𝕂. We will consider two situations, depending on the behavior of the biadditive functional ϕ on the diagonal.

Theorem 1

Assume that ϕ and f solve (1.2) and

(2.1)

\exists_{z_{0} \in X} ϕ (z_{0}, z_{0}) \neq 0 .

Then there exists a unique constant a ∈ 𝕂 \ {0} such that

(2.2)

f (x) = a ϕ (x, z_{0}) + 1, x \in X,

and moreover

(2.3)

a^{2} ϕ {(x, z_{0})}^{2} = ϕ (x, x), x \in X .

Proof

Substituting y = z₀ and then y = −z₀ in (1.2) we obtain

f (x + z_{0}) = f (x) f (z_{0}) - ϕ (x, z_{0}), x \in X

and

f (x - z_{0}) = f (x) f (- z_{0}) + ϕ (x, z_{0}), x \in X .

Replace x by x + z₀ in the latter formula and join it with the former one to arrive at

\begin{array}{l} f (x) & = f (x + z_{0}) f (- z_{0}) + ϕ (x + z_{0}, z_{0}) \\ = f (x) f (z_{0}) - ϕ (x, z_{0})] f (- z_{0}) + ϕ (x, z_{0}) + ϕ (z_{0}, z_{0}), x \in X . \end{array}

Denote c := f(z₀), d := f(−z₀) and β = ϕ(z₀, z₀) ≠ 0. We get

(1 - cd) f (x) = (1 - d) ϕ (x, z_{0}) + β, 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 = f (z_{0} - z_{0}) = f (z_{0}) f (- z_{0}) + β,

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

f (x) f (- x) = 1 - ϕ (x, x), x \in X .

Now, use the already proven formula (2.2) to derive (2.3) after some reductions.

Remark 1

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:

a^{2} ϕ (x, z_{0}) ϕ (y, z_{0}) = ϕ (x, y), 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

ϕ (x, y) = a^{2} F (x) \cdot F (y), x, y \in X,

with an additive, nonzero functional F : X → 𝕂, and the two possible functions f are given by (2.2).

We have the following corollary in the real case.

Corollary 1

Assume that 𝕂 = ℝ and

(2.4)

\exists_{z_{0} \in X} ϕ (z_{0}, z_{0}) < 0 .

Then equation (1.2) has no solutions.

Proof

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).

In the complex case, every element of the field has a complex root of second order. Therefore, we can state the next corollary.

Corollary 2

Assume that 𝕂 = ℂ, ϕ and f solve (1.2), ϕ satisfies (2.1) and w: X → ℂ is a map such that

w^{2} (x) = ϕ (x, x), x \in X .

Then

f (x) = w (x) + 1, x \in X,

f (x) = - w (x) + 1, x \in X .

The next theorem deals with the remaining case for ϕ and is easy to prove.

Theorem 2

Assume that ϕ and f solve (1.2) and

\forall_{z \in X} ϕ (z, z) = 0 .

Then ϕ = 0 on X × X and

f (x + y) = f (x) f (y), x, y \in X .

Consequently, either f = 0 or there exists an additive functional A: X → 𝕂 such that f = exp ∘ A.

Proof

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].

The following corollary is immediate and offers an alternative proof of the conjecture by Alzer and Matkowski.

Corollary 3 (T. Małolepszy)

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

Proof

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

f (x) = a ϕ (x, 1) + 1 = a α x + 1, x \in ℝ .

From (2.3) we obtain

a^{2} {(α x)}^{2} = ϕ (x, x) = α x^{2}, x \in ℝ .

We get a² = 1/α, so α > 0 and

a = \pm 1 / \sqrt{α}

. After substitution to the equation for f we arrive at

f (x) = \pm \sqrt{α} x + 1

. Conversely, it is easy to check that both such mappings solve (1.1).

Our last corollary is a complex counterpart of Corollary 3.

Corollary 4

Assume that α ∈ ℂ \ {0} and f : ℂ → ℂ solves

(2.5)

f (x + y) = f (x) f (y) - α xy, x, y \in 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).

Proof

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.

3. Examples and final remarks

We observe that Theorem 1 generally works only in one direction, that is, the converse implications do not necessarily hold.

Example 1

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 \in X,

with some ξ ∈ X. One can check that such mapping solves (1.2) if and only if

x, ξ〉 y, ξ〉 = x, y〉, x, y \in X,

which is impossible if dim X ≥ 2.

It may be suspected that in higher dimensions, there are no solutions to (1.2). However, the following example demonstrates that this is not the case.

Example 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:

f (x) = γ A (x) + 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).

A counterpart of the above example that works in both cases, real and complex, is also possible.

Example 3

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:

f (x) = δ A (x) + 1, x \in X

with some constant δ ∈ 𝕂. Further, f is indeed a solution if and only if δ = ±1.

On a Generalized Conjecture by Alzer and Matkowski

Full Article

1. Introduction

(1.1)

Conjecture (Alzer and Matkowski)

(1.2)

2. Main results

Theorem 1

(2.1)

(2.2)

(2.3)

Proof

Remark 1

Corollary 1

(2.4)

Proof

Corollary 2

Theorem 2

Proof

Corollary 3 (T. Małolepszy)

Proof

Corollary 4

(2.5)

Proof

3. Examples and final remarks

Example 1

Example 2

Example 3

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