1 Introduction

It is well known that any Hamel basis H (algebraic basis of a linear space X over the field \({\mathbb Q}\) of rationals) has the property that any function \(f_0\) mapping H into another linear space over \({\mathbb Q}\) may uniquely be extended to a solution \(f: X \rightarrow Y\) of the fundamental Cauchy functional equation

$$\begin{aligned} f(x+y) = f(x) + f(y) \,. \end{aligned}$$
(A)

In connection with this, the following general problem arises (S. Goła̧b’s question, see [10] and [14]): given a functional equation

$$\begin{aligned} E_1(\varphi ) = E_2(\varphi ) \end{aligned}$$
(*)

with the unknown function \(\varphi : X \longrightarrow Y\), what must the set \(\emptyset \ne Z \subset X\) be like in order to ensure that an arbitrary function \(\varphi _0: Z \longrightarrow Y\) admits a unique function \(\varphi : X \longrightarrow Y\) solving equation (\(*\)) and such that \(\varphi _{|_Z} = \varphi _0\); in the sequel, if such a set does exist it will be named a basic set. The requirement of uniqueness guarantees here that a basic set is the largest possible in the sense of inclusion.

Remark 1

The statement of this definition (following the one existing in the literature) requires a clarification. It is clear that the answer to the question whether a given equation admits the existence of basic sets depends (at least formally) on the choice of the spaces X and Y that are involved. In fact, this dependence is actual: in his paper [12] Sablik shows that d’Alembert’s (cosine) functional equation

$$\begin{aligned} f(x+y) + f(x-y) = 2f(x)f(y) \end{aligned}$$
(C)

has no basic set in the class of complex functions (\(Y = {\mathbb C}\)—the set of all complex numbers) defined on a linear space X over the field of rationals. In what follows, we shall show that basic sets do exist provided that we deal with real solutions of (C), i.e. in the case where \(Y = {\mathbb R}\)—the set of all real numbers.

Analyzing this definition in a more detailed way, observe further that there is no problem in the case where a given equation has a solution that is unique in a sense. If this is not the case, we should speak about a basic set for a concrete solution rather than for the equation itself. To illustrate the question spoken of, let us consider the following functional equation of conditional (alternative) character (see e.g. the article by Dhombres and Ger [5]):

$$\begin{aligned} \left( f(x+y) - f(x) - f(y) \right) \left( f(x+y) - f(x)f(y) \right) = 0. \end{aligned}$$
(AE)

Obviously it has at least two solutions f mapping \({\mathbb R}\) into itself: \(f = a\)—an additive function and \(f = m\)—an exponential one. As we have seen already, additivity has a basic set (any Hamel basis) whereas exponentiality fails to have any basic set since for any nonempty set \(Z \subset {\mathbb R}\) no function \(m_0: Z \rightarrow (- \infty , 0)\) admits an extension to a solution \(m: {\mathbb R}\longrightarrow {\mathbb R}\) of the equation

$$\begin{aligned} m(x+y) = m(x)m(y) \qquad (\mathrm{of\,\, exponential\,\, functions}) \end{aligned}$$
(E)

because (E) forces m to be nonnegative.

Nevertheless, any Hamel base of \({\mathbb R}\) (considered as a vector space over the field of rationals) yields a basic set for equation (AE) because it is basic for equation (A).

Remark 2

In general, constant solutions of equation (\(*\)) are excluded as potential candidates for proving that (\(*\)) has a basic set. Indeed, constant functions are uniquely determined by their values at a single point and, therefore, the corresponding possible basic sets should be of cardinality 1. More precisely, if for some \(x_0 \in X\), the set \(Z:= \{x_0\}\) were basic with a corresponding constant solution \(\varphi (x) \equiv c\), then for every function \(\varphi _0: Z \longrightarrow Y\) one has \(\varphi _{|_Z} = \varphi _0\); in other words,

$$\begin{aligned} \mathrm{for\,\, every} \,\,y \in Y \,\, \mathrm{one\,\, has}\,\, c = \varphi (x_0) = y, \end{aligned}$$

which is impossible except for the case where every constant function would satisfy equation (\(*\)) or the range Y has cardinality 1.

The existence of a basic set regarding the Fréchet functional equation defining polynomial functions was proved by the present author in [8] (see also Gajda [7]). An extensive study of the problem of the existence of basic sets for various functional equations may be found in M. Sablik’s habilitation thesis [14]. In an earlier paper (see [13]) M. Sablik states that in the class of functions mapping an Abelian group \((X,+)\) into a divisible Abelian group \((Y, \cdot )\) containing elements of all orders a basic set for (E) does exist if and only if \((X,+)\) is a free group.

In an elementary way we have already shown above, that even in the case of real valued functions equation (E) fails to admit any basic set. Beggars can’t be choosers. Whilst it is very much a second best, we offer the following

Proposition 1

Let X be a linear space over the field \({\mathbb Q}\) and let H stand for its arbitrary Hamel basis. Then any function \(m_0: H \rightarrow (0,\infty )\) admits a unique extension to a real functional m on X solving equation \((\mathrm{E)}\).

Proof

Fix arbitrarily a function \(m_0: H \rightarrow (0,\infty )\) and define a function \(A_0: H \rightarrow {\mathbb R}\) by the formula \(A_0(h):= \log m_0(h), h \in H.\) Since each real function on H admits a (unique) extension to an additive function on X, denote by A the corresponding extension of \(A_0\). Then setting

$$\begin{aligned} m(x): = e^{A(x)},\, x \in X, \end{aligned}$$

we obtain a solution \(m: X \rightarrow {\mathbb R}\) of equation (E) such that \(m|_H = m_0\) because for every \(h \in H\) one has

$$\begin{aligned} m(h) = e^{A(h)} = e^{A_0(h)} = e^{\log m_0(h)} = m_0(h). \end{aligned}$$

To prove the uniqueness, note that in view of the fact that each real valued nonzero solution of equation (E) is of the form \(X \ni x \longmapsto M(x):= e^{B(x)}\) with an additive function \(B: X \rightarrow {\mathbb R}\) (see e.g. Kuczma [11]), we get

$$\begin{aligned} e^{A_0(h)} = m_0(h) = M(h) = e^{B(h)}, \, h \in H, \quad \mathrm{we \,\, get} \quad A_0(h) = B(h), \,h \in H, \end{aligned}$$

forcing the equality \(A = B\) and, consequently, \(m = M\). This completes the proof. \(\square \)

2 Equations with solutions of the form: a given function of an additive one

In general, when looking for possible basic sets, analytic formula of solutions of a given equation helps notably. In particular, the following simple result holds true.

Proposition 2

Given a linear space X over the field \({\mathbb Q}\) and a functional equation such that for some nonconstant function \(\varphi : {\mathbb R}\rightarrow {\mathbb R}\) and any real additive functional on X the composition \(\varphi \circ a\) solves the equation spoken of, any Hamel basis of X yields a basic set for the equation in question.

Proof

Let H stand for a fixed Hamel basis of the space X and let \(f_0: H \rightarrow {\mathbb R}\) be a given function. Let further \(a: X \rightarrow {\mathbb R}\) be a (unique) extension of \(f_0\) to an additive map. Then, by assumption, for some nonconstant function \(\varphi : {\mathbb R}\rightarrow {\mathbb R}\) the composition \( \varphi \circ a\) yields a solution of a functional equation spoken of. Obviously such an extension is unique which completes the proof. \(\square \)

In what follows we will use Proposition 2 to show that d’Alembert’s equation (C), the sine equation

$$\begin{aligned} \qquad \qquad f(x+y) f(x-y) = f(x)^2 - f(y)^2 , \end{aligned}$$
(S)

as well as the equation

$$\begin{aligned} \left( f(x+y) - f(x)f(y)\right) ^2= \left( 1 - f(x)^2\right) \left( 1 - f(y)^2\right) , \end{aligned}$$
(Cuc)

admit basic sets.

Theorem 1

Let X be a linear space over the field \({\mathbb Q}\) and let H stand for its arbitrary Hamel basis. Then any function \(f_0: H \rightarrow {\mathbb R}\) admits a unique extension to a nonzero real functional f on X solving equation \((\mathrm{C)}\) or equation \((\mathrm{S)}\) or equation (Cuc).

Proof

You may treat is as folklore or derive it single-handedly from the monograph [2] by Aczél and Dhombres (Theorem 16 and Theorem 14 in Chapter 13) that functions \(f: X \rightarrow {\mathbb R}\) are given by

$$\begin{aligned} f = 0, \quad f = \cosh \circ \, a, \quad f = \cos \circ \, a, \end{aligned}$$

where \(a: X \rightarrow {\mathbb R}\) is an additive map, solve d’Alembert’s equation (C)

(resp. the sine functional equation (S) has solutions of the form)

$$\begin{aligned} f = \alpha \sinh \circ \,a, \quad f = \alpha \sin \circ \,a, \quad f = a). \end{aligned}$$

The equation (Cuc), written alternatively in the form

$$\begin{aligned} f(x)^2 + f(y)^2 + f(x+y)^2 - 2f(x)f(y)f(x+y) = 1, \end{aligned}$$

is connected to a question asked by a French mathematician Roger Cuculière in [4] (see also the author’s paper [9]). It is not hard to check that for any additive map \(a: X \longrightarrow {\mathbb R}\) the functions

$$\begin{aligned} f = - \frac{1}{2}, \quad \cosh \circ \, a, \quad f = \cos \circ \, a, \end{aligned}$$
(**)

solve the equation (Cuc).

Now, it remains to apply Proposition 2 to finish the proof. \(\square \)

Remark 3

It seems to me that functions occurring in \( (**)\) describe the general solution of the equation (Cuc) (up to some functions with finite ranges). Obviously, the fact that it is still an open problem does not disprove the latter part of the assertion of Theorem 1.

Hamel base viewed as basic sets for equations (C), (S) and (Cuc) may happen to be quite large subsets of the linear space considered (nonmeasurable sets, second Baire category sets in \({\mathbb R}\), for instance; see e.g Kuczma [11], Chapter 11]). It turns out that in some cases basic sets for other trigonometric equations might occur to be extremely small (sets with cardinality 2, for instance). To illustrate it, let us consider the following rather special trigonometric functional equation:

$$\begin{aligned} \ f(x+y) + f(x-y) = 2f(x)\cos y\,. \end{aligned}$$
(cos)

Without any regularity condition whatsoever, one may prove (see J. Aczél [1]) that the general solution \(f: {\mathbb R}\rightarrow {\mathbb R}\) of this equation is of the form

$$\begin{aligned} f(x) = \alpha \cos x + \beta \sin x, \quad x \in {\mathbb R}, \end{aligned}$$

where \(\alpha \) and \(\beta \) are arbitrary real constants. Therefore, the totality of all possible basic sets for equation (cos) coincides with the set family

$$\begin{aligned} \left\{ \{2k\pi , \frac{\pi }{2} +2\ell \pi \}: \, k, \ell \in {\mathbb Z}\,\, \mathrm{(the\,\, set\,\, of \,\, all\,\, integers)} \right\} \end{aligned}$$

of 2-point subsets of the real line. In fact, fix arbitrarily elements \(k, \ell \in {\mathbb Z}\); then any function \(f_0: \{2k\pi , \frac{\pi }{2} +2\ell \pi \} \rightarrow {\mathbb R}\) is uniquely extendable to a solution \(f: {\mathbb R}\rightarrow {\mathbb R}\) of equation (cos), given by the formula

$$\begin{aligned} f(x) = f_0(2k\pi ) \cos x + f_0\left( \frac{\pi }{2} +2\ell \pi \right) \sin x, \, x \in {\mathbb R}, \end{aligned}$$

which states nothing else but the fact that the set \(\{2k\pi , \frac{\pi }{2} +2\ell \pi \}\) is basic for the equation (cos).

Basic sets of cardinality 2 exist also in the case of derivations, i.e. additive mappings (solutions of (A)) such that

$$\begin{aligned} f(xy) = xf(y) + yf(x) \quad \mathrm{for\,\, all} \,\, x,y \,\, \mathrm{from\,\, the\,\, underlying\,\, domain}. \end{aligned}$$

This results from the following two facts:

  • there exist nonzero derivations of \({\mathbb R}\) (see Theorem 14.2.2 in Kuczma’s monograph [11]) and

  • a special case (\({\mathbb K}= {\mathbb C}\),  \({\mathbb {F}}= {\mathbb R}\),  \(S = \{1,i\}\))  of the following result (see Kuczma [11, Theorem 14.2.1]): Let \(({\mathbb K}; +, \cdot )\) be a field of characteristic zero, let \(({\mathbb {F}}; +, \cdot )\) be a subfield of \(({\mathbb K}; +, \cdot )\), let S be an algebraic base of \({\mathbb K}\) over \({\mathbb {F}}\), if it exists, and let \(S = \emptyset \) otherwise. Let \(f: {\mathbb {F}}\rightarrow {\mathbb K}\) be a derivation. Then any function \(f_0: S \rightarrow {\mathbb K}\) is uniquely extendable to a derivation \(g: {\mathbb K}\rightarrow {\mathbb K}\)such that \(g\vert _{{\mathbb {F}}} = f\) and \(g\vert _S = f_0.\)

3 Another trigonometric equation

The following functional equation

$$\begin{aligned} f(x+ y) = \frac{f(x) + f(y)}{1 + f(x)f(y)} \quad (\mathrm{of\,\,the\,\, tanh\,\, function})\, \end{aligned}$$
(1)

valid for suitable \(x,y \in {\mathbb R},\, \) is an example of the so called addition formulas i.e. equations of the form

$$\begin{aligned} f(x+y) = F(f(x),f(y)). \end{aligned}$$

In this particular case the given binary operation

$$\begin{aligned} F(u,v) = \frac{u+v}{1+uv} \end{aligned}$$

is associative but its domain of definition is disconnected (admits “singularities”). One of the papers in which that functional equation (in a much more general setting) has been discussed in details is [6] written jointly by Domańska and the present author. The methods applied over there might serve also while solving addition formulas with the corresponding binary operations

$$\begin{aligned} F(u,v):= \frac{u +v}{1-uv} , \quad u,v \in {\mathbb R}, \, uv \ne 1, \quad (\mathrm{of\,\,the\,\, tan\,\, function}), \\ F(u,v):= \frac{uv-1}{u+v} , \quad u,v \in {\mathbb R}, \, u+v \ne 0, \quad (\mathrm{of\,\,the\,\, cot\,\, function}), \end{aligned}$$

and others. When singularities are omitted the following results (very special cases of Theorems 1  and 2 in [6]) hold true.

Theorem DG1

A function \(f: {\mathbb R}\longrightarrow [-1,1]\) yields a solution to the functional Eq. (1) if and only if there exists an additive map \(a: {\mathbb R}\longrightarrow {\mathbb R}\) such that

$$\begin{aligned} f(x) = \textrm{tanh}\, a(x), \quad x \in {\mathbb R}, \end{aligned}$$

provided that \(f({\mathbb R}) \subset (-1, 1).\) Moreover, if there existed an \(x_0 \in {\mathbb R}\) such that \(f(x_0) = 1\) (resp. \(f(x_0) = -1\)), then \(f(x) \equiv 1\) provided that \(f(x) \ne -1\) (resp. \(f(x) \equiv -1\) provided that \(f(x) \ne 1\)).

Theorem DG2

Equation (1) has no solutions \(f: {\mathbb R}\longrightarrow {\mathbb R}\) with the range \(f({\mathbb R})\) contained in the set \((- \infty , -1) \cup (1, \infty ).\)

When singularities are admissible, then we have (cf. [6, Theorem 3] and [6, Corollary] again):

Theorem DG3

Let \((G,+)\) be a group (not necessarily commutative) and let \(f: G \rightarrow {\mathbb R}\) be a solution of Eq. (1) such that \(f(0) = 0\) and

$$\begin{aligned} 1 \not \in f(G) \not \subset (-1,1). \end{aligned}$$
(2)

Then there exists a subgroup \((\Gamma ,+)\) of the group \((G,+)\) of index 2, and an additive function \(a: G \longrightarrow {\mathbb R}\)  such that \(\ker a \subset \Gamma \)  and

$$\begin{aligned} f(x)= & {} \left\{ \begin{array}{ll} \textrm{tanh} \circ a (x) &{}\quad \hbox { for}\ x\in \Gamma \\ \textrm{coth} \circ a (x) &{}\quad \hbox { for}\ x\in G \setminus \Gamma . \end{array} \right. \end{aligned}$$

Conversely, any function f of that form yields a solution to Eq. (1) and satisfies condition (2).

If the group \((G,+)\) does not admit subgroups of index 2, then the only real solutions of Eq. (1) having 1 off their ranges are the canonical ones: tanh \(\circ \,a\) where \(a: G \rightarrow {\mathbb R}\) is an arbitrary additive function.

Looking for solutions with ranges in the interval (\(-1,1\)), by means of Theorem DG1 and Proposition 2, we infer that

Theorem 2

Any Hamel basis of a linear space X over the field \({\mathbb Q}\) yields a basic set for the family of all solutions of Eq. (1) mapping X into the interval \((-1,1)\). There are no basic sets for the family of all solutions of Eq. (1) mapping X into the union \((- \infty , -1) \cup (1, \infty ).\)

The latter assertion of Theorem 2 is trivially satisfied because, by Theorem DG2, in this case we have no solutions at all.

A simple, but not quite trivial, consequence of Theorem DG3, is the following result involving the additive group \(({\mathbb Z}, +)\) of all integers.

Theorem 3

The family of all solutions \(f: {\mathbb Z}\longrightarrow {\mathbb R}\) of Eq. (1), vanishing at zero and such that \(1 \not \in f({\mathbb Z}) \not \subset (-1,1)\) admits basic sets. More precisely, if \(f: {\mathbb Z}\longrightarrow {\mathbb R}\) stands for a noncanonical solution of (1) with 1 off its range then,

$$\begin{aligned} f(n)= & {} \left\{ \begin{array}{ll} \textrm{tanh}\, cn &{}\quad \hbox { for even}\ n\\ \textrm{coth}\, cn &{}\quad \hbox { for odd}\ n \end{array} \right. , \end{aligned}$$
(3)

(see Theorem DG3  applied for \((G,+) = ({\mathbb Z},+)\), \(\Gamma = 2{\mathbb Z}\)) and for every \(k \in Z\) the singleton {k} yields a basic set.

Proof

Suppose that we deal with a noncanonical solution of (1). Observe that every additive function \(a: {\mathbb Z}\rightarrow {\mathbb R}\) is of the form \(f(n) = cn, n \in {\mathbb Z},\) where c is a real number. Clearly, in our case, \(c \ne 0\) because, otherwise, formula (3) would not make sense. In view of the fact that both hyperbolic tangent as well as hyperpolic cotangent are injective,  singletons are the only candidates for possible basic sets with the hope that the unrestricted parameter c in formula (3) might do the job. That is really the case. In fact, to prove that a singleton \(Z:= \{k\}\) is a basic set for a noncanonical solution \(f: {\mathbb Z}\longrightarrow {\mathbb R}\) of (1), we have to show that for an arbitrary number \(\beta \in {\mathbb R}\setminus \{0,1\}\) there exists a real number \(c \ne 0\) such that

$$\begin{aligned} \beta = f(k)\!\!=\!\! & {} \left\{ \begin{array}{ll} \textrm{tanh}\, ck &{}\quad \text{ whenever } \text{ k } \text{ is } \text{ even }\\ \textrm{coth}\, ck &{}\quad \text{ provided } \text{ that } \text{ k } \text{ is } \text{ odd } \end{array} \right. . \end{aligned}$$

So, fix arbitrarily a \(\beta \in {\mathbb R}\setminus \{0,1\}\) and distinguish two cases:

  • \(\beta \in (-1,1)\). Then k is forced to be even and to have the equality \(\beta = f(k) = \textrm{tanh}\, ck\) it suffices to take

    $$\begin{aligned} c: = \frac{1}{2k} \log \frac{1 + \beta }{1 - \beta }. \end{aligned}$$

  • \(\beta \in (- \infty , -1) \cup (1, \infty ).\) Then k is forced to be odd and to have the equality \(\beta = f(k) = \textrm{coth}\, ck\) it suffices to take

    $$\begin{aligned} c: = \frac{1}{2k} \log \frac{\beta +1}{\beta - 1}. \end{aligned}$$

Thus the proof has been completed. \(\square \)

4 Concluding remarks

We terminate this paper with several comments.

  1. 1.

    The existence of a basic set is a rather rare phenomenon. However, if a basic set does exist and is sufficiently large (infinite, at least) it allows us to realize how potentially irregular solutions might occur.

  2. 2.

    Once more we want to emphasize that, dealing with scalar functions, the question of existence of basic sets, happens to have an affirmative answer more frequently in the real case than in the world of complex solutions.

  3. 3.

    From the set-theoretical point of view basic sets may happen to be pretty large as well as considerably small, even when they are concerned with the same equation. We have seen that an alternative functional equation

    $$\begin{aligned} \left( f(x+y) + f(x-y) - 2f(x)\cos y \right) (f(x+y) + f(x-y) - 2f(x)f(y)) = 0 \end{aligned}$$

    on the real line admits both quite “thick” basic sets (nonmeasurable or second category Hamel bases) as well as 2-point ones.

  4. 4.

    Since we have realized already that it seems more reasonable to speak about a basic set for a concrete solution rather than for the equation itself, we should (at least in some cases) look for basic sets in a prescribed class of solutions (measurable, continuous, etc.). For instance, in the case of Lebesgue measurable solutions of the equation (Cuc) the knowing their analytic form (see the author’s paper [9]) might prove to be helpful.

  5. 5.

    As we know (see the paper [3] written by Aczél and Erdős) no analogue of a Hamel base exists for nonnegative real numbers. Nevertheless, reasonable subsets of the natural domain in question should also be considered when dealing with the problem of finding potential basic sets for various important functional equations and their solutions.