COMMENTATIONES MATHEMATICAE Vol. 50, No. 1 (2010), 69-72
Eliza Jabłońska
Baire measurable solutions of a generalized Gołąb–Schinzel equation
Abstract. J. Brzdęk [1] characterized Baire measurable solutions f : X → 𝕂 of the functional equation
f (x + f(x)
ny) = f(x)f(y)
under the assumption that X is a Fr´echet space over the field 𝕂 of real or complex numbers and n is a positive integer. We prove that his result holds even if X is a linear topological space over 𝕂; i.e. completeness and metrizability are not necessary.
2000 Mathematics Subject Classification: 39B52.
Key words and phrases: generalized Gołąb–Schinzel equation, net, finer net, Baire measurability.
For the first time, in connection with examining of subgroups of the centroaffine group of the plane, the functional equation
(1) f (x + f (x)y) = f (x)f (y)
has been studied by S. Gołąb and A. Schinzel [3] in the class of continuous real functions. In 1965 C.G. Popa [7] proved that every Lebesgue measurable real solu- tion of (1) is continuous or equal to zero almost everywhere. Following this idea J.
Brzdęk [2] showed that the same is true for each Christensen measurable solution of the generalized Gołąb–Schinzel equation
(2) f (x + f (x) n y) = f (x)f (y)
mapping a real or complex separable Fr´echet space into the field of real or complex numbers, respectively, where n is a positive integer.
J. Brzdęk [1] proved also an analogous result for Baire measurable solutions of the equation
(3) f (x + f (x) n y) = tf (x)f (y)
70 Baire measurable solutions of a generalized Gołąb–Schinzel equation
mapping a Fr´echet space over the field 𝕂 of real or complex numbers into 𝕂, where n is a positive integer and t ∈ 𝕂 \ {0} . To prove this fact he used the open mapping theorem (see, for example, [5, 11.4]). We will show that it is enough to assume that X is a linear topological space over the field of real or complex numbers; i.e. X need not be complete metrizable. Thereby we must ”go around”the open mapping theorem.
Throughout the paper ℕ, ℝ and ℂ stand for the set of all positive integers, reals and complex numbers, respectively. Moreover, we use some basic facts concerning nets, which can be found in [4, Chapter I (3.10), p.14]. Here, we recall only that a net {x σ
0: σ 0 ∈ Σ 0 } in a topological space X (where Σ 0 is directed by the relation
¬ Σ
0) is finer than a net {x σ : σ ∈ Σ} in X (where Σ is directed by ¬ Σ ), if there exists a function ϕ : Σ 0 → Σ fulfilling the following two conditions:
✓ for every σ 0 ∈ Σ there is σ 0 0 ∈ Σ 0 such that σ 0 Σ
0σ 0 0 implies ϕ(σ 0 ) Σ σ 0 ;
✓ for every σ 0 ∈ Σ 0 we have x ϕ(σ
0) = x σ
0.
To generalize the result of J. Brzdęk [1] we need the following
Proposition 1 (cf. [1, Lemma 5]) Let X be a linear–topological space over 𝕂 ∈ {ℝ, ℂ} and V ⊂ 𝕂. If g : X → 𝕂 is a nontrivial continuous linear functional and g −1 (V ) has a subset of second category and with the Baire property, then the set V has a subset of second category and with the Baire property.
Proof Let Y := ker g. Then Y is a closed linear subspace of X and there exists a point x 0 ∈ X \ Y such that g −1 (V ) = V x 0 + Y. Since Y is a linear–topological space (with the induced topology), Y × 𝕂 is a linear–topological space with the product topology.
Define a mapping
h : Y × 𝕂 3 (y, r) → y + rx 0 ∈ Y + 𝕂x 0 .
It is easy to see that h is a continuous linear bijection. We prove that H = h −1 is continuous; i.e. for every net {r σ : σ ∈ Σ} ⊂ 𝕂 and {y σ : σ ∈ Σ} ⊂ Y such that lim σ∈Σ (r σ x 0 + y σ ) = 0 we have lim σ∈Σ (r σ , y σ ) = (0, 0).
To this end we prove that for each net S = {r σ : σ ∈ Σ}, there is a net S 0 = {r σ
0: σ 0 ∈ Σ 0 } finer than S, such that
(4) either S 0 is convergent or lim
σ
0∈Σ
01 r σ
0= 0.
First consider the case, when there exists a net S 0 finer than S such that r σ
06= 0 for each σ 0 ∈ Σ 0 . If S 0 is bounded, then S 0 ⊂ {z ∈ 𝕂 : |z| ¬ M} for some M > 0 and {z ∈ 𝕂 : |z| ¬ M} is a compact topological space (with the induced topology).
Hence there is a convergent net {r σ
00: σ 00 ∈ Σ 00 } finer than S 0 . So assume that S 0 is unbounded. Then, for each M > 0, there is σ 0 ∈ Σ 0 such that |r 1
σ0
| < M 1 . Denote
E. Jabłońska 71
ε = M 1 and a σ
0= r 1
σ0