On Solutions of Some Functional Equations of Radical Type

Abstract

Many authors have studied various functional equations of forms patterned on the equation expressed as $g\left(\sqrt{a^2+b^2}\right)=g\left(a\right)+g\left(b\right)$, which can be considered, e.g., for real functions. Such equations are usually referred to as radical functional equations or of the radical type. Authors mainly study the so-called Ulam stability of such equations, i.e., they investigate how much the mappings satisfying the equations approximately (in a sense) differ from the exact solutions of these equations. Quite often, information about the solutions of these equations is also provided, but unfortunately, sometimes, such information is given in a misleading or incomplete way. It seems, therefore, that there is a need for a publication containing simple descriptions of such solutions (with appropriate examples), which would help in easy correction of such information and avoidance of similar problems for future authors. This is the main motivation for this expository paper. We present a general approach to the topic and consider two general forms of such equations. Moreover, the results presented in this paper show significant symmetry between the solutions of numerous functional equations and the solutions of equations of the radical type that correspond to them. To make this publication accessible to a wider audience, we omit various related information, avoid advanced generalizations, and present several simple examples.

1. Introduction

In this paper, $\mathbb{N}$, $\mathbb{Z}$, $\mathbb{Q}$, $\mathbb{R}$, and $\mathbb{C}$ denote, as usual, sets of positive integers, integers, rational numbers, and real and complex numbers, respectively. Moreover ${\mathbb{N}}_0:=\mathbb{N}\cup \left\{0\right\}$, $\mathbb{K}\in \{\mathbb{R},\mathbb{C}\}$, and S and T are nonempty sets.

At the 16th International Conference on Functional Equations and Inequalities (Będlewo, Poland, 17–23 May 2015), W. Sintunavarat gave a talk ([1]) on the Ulam stability of the following functional equation:

$$\begin{array}{c}\hfill g\left(\sqrt{a^2+b^2}\right)=g\left(a\right)+g\left(b\right)\end{array}$$

(1)

in the class of all functions $g:\mathbb{R}\to \mathbb{R}$.

For more information on Ulam stability, we refer to [2,3,4]. Here, let us only mention that this stability (often also called Hyers–Ulam stability) concerns the flowing problem: how much the mappings satisfying an equation approximately (in a sense) differ from the exact solutions of the equation. This problem is closely related to the issues considered in the theories of approximation, optimization, perturbation, and shadowing.

Motivated by the talk of W. Sintunavarat, J. Schwaiger raised the problem to determine the general solution of the equation, which was answered by the author of this paper (see [1], p. 196), who showed that $g:\mathbb{R}\to \mathbb{R}$ is a solution to (1) if and only if

$$g\left(x\right)=A\left(x^2\right),x\in \mathbb{R},$$

with some $A:\mathbb{R}\to \mathbb{R}$ that is additive (i.e., fulfills the equality $A(a+b)=A\left(a\right)+A\left(b\right)$ for every $a,b\in \mathbb{R}$).

It is worth mentioning that, earlier, in [5,6,7,8,9], Ulam stability was studied for other analogous equations, which are usually called radical functional equations. For examples of some later similar results, we refer to [10,11,12,13,14,15,16,17,18,19]. The authors of these papers sometimes provide some descriptions of the solutions to such equations, but in some cases, these descriptions are incomplete, unclear, or even misleading (see Remarks 3 and 5 and the comments after Equation (78)). This is the reason why this paper was written. However, the main purpose of this paper is not to identify and discuss all such imperfections, but to present a general, easy method that allows for the establishment of sensible, simple, and complete descriptions of solutions to numerous equations of the radical type.

Now, let us present some examples of these equations of the radical type. First, note that (1) is a special case ($k=2$) of the following functional equation that has been considered (in various situations) in [5,9,10,11,13,17] for $k=2,3,4$:

$$\begin{array}{c}\hfill g\left(\sqrt[k]{a^k+b^k}\right)=g\left(a\right)+g\left(b\right).\end{array}$$

(2)

Next, let us mention the following equation that was studied in [16] for functions g mapping $(0,\infty )$ into $\mathbb{R}\setminus \left\{0\right\}$ and has been called Pythagorean mean functional equation:

$$\begin{array}{c}\hfill g\left(\sqrt{a^2+b^2}\right)={\displaystyle \frac{g\left(a\right)g\left(b\right)}{g\left(a\right)+g\left(b\right)}}.\end{array}$$

(3)

Of course, we must exclude in (3) all $a,b\in (0,\infty )$ such that $g\left(a\right)+g\left(b\right)=0$ (this was not clearly mentioned in [16]).

Solutions to and the stability of the following functional equation were investigated in [9] for functions from $\mathbb{R}$ in a real linear space:

$$\begin{array}{c}\hfill g\left(\sqrt{\alpha a^2+\beta b^2}\right)=\alpha g\left(a\right)+\beta g\left(b\right),\end{array}$$

(4)

with positive $\alpha ,\beta \in \mathbb{R}$ such that $\alpha +\beta \ne 1$.

Moreover, the equation

$$\begin{array}{c}\hfill g\left(\sqrt{a^2+b^2}\right)+g\left(\sqrt{|a^2-b^2|}\right)=2g\left(a\right)+2g\left(b\right)\end{array}$$

(5)

and its generalized form

$$\begin{array}{c}\hfill g\left(\sqrt{\alpha a^2+\beta b^2}\right)+g\left(\sqrt{|\alpha a^2-\beta b^2|}\right)=2{\alpha }^{2}g\left(a\right)+2{\beta }^{2}g\left(b\right),\end{array}$$

(6)

were considered in [5,8,17] for functions from $\mathbb{R}$ in a real linear space, with positive $\alpha ,\beta \in \mathbb{R}$, $\alpha +\beta \ne 1$.

The following functional equations

$$\begin{array}{c}\hfill \varphi \left(\sqrt[k]{a^k+b^k}\right)+\varphi \left(\sqrt[k]{a^k-b^k}\right)=h\left(a\right)g\left(b\right),\end{array}$$

(7)

$$\begin{array}{c}\hfill \varphi \left(\sqrt[k]{a^k+b^k}\right)-\varphi \left(\sqrt[k]{a^k-b^k}\right)=h\left(a\right)g\left(b\right),\end{array}$$

(8)

$$\begin{array}{c}\hfill \varphi {\left(\sqrt[k]{{\displaystyle \frac{1}{2}}\left(a^k+b^k\right)}\right)}^2-\varphi {\left(\sqrt[k]{{\displaystyle \frac{1}{2}}(a^k-b^k)}\right)}^2=h\left(a\right)g\left(b\right)\end{array}$$

(9)

and some particular cases of them (e.g., with $\varphi =g$, $\varphi =h$, or $\varphi =g=h$) have been studied for $\varphi ,g,h:\mathbb{R}\to \mathbb{R}$, e.g., in [15,18,19].

Clearly, all these functional equations are particular cases of the general functional equation, of the form

$$\begin{array}{cc}\hfill H(g_1\left(\sqrt[n]{F_1(x_1^n,\dots ,x_m^n)}\right)& ,\dots ,g_k\left(\sqrt[n]{F_k(x_1^n,\dots ,x_m^n)}\right))\hfill \\ \hfill & =G\left(g_{k+1}\left(x_1\right),\dots ,g_{k+m}\left(x_m\right)\right),\hfill \end{array}$$

(10)

considered for unknown functions $g_1,\dots ,g_{k+m}:\mathbb{R}\to T$ with given functions $H:T^k\to S$, $G:T^m\to S$, $F_1,\dots ,F_k:P^m\to P$, where n, k, and m are fixed positive integers with $n>1$, T and S are nonempty sets, and $P:=\{a^n:a\in \mathbb{R}\}$.

A description of solutions to (10) was provided in [20], but only when $g_1=\dots =g_{k+m}$, i.e., for the following equation:

$$\begin{array}{c}\hfill H\left(f\left(\sqrt[n]{F_1(x_1^n,\dots ,x_m^n)}\right),\dots ,f\left(\sqrt[n]{F_k(x_1^n,\dots ,x_m^n)}\right)\right)=G\left(f\left(x_1\right),\dots ,f\left(x_m\right)\right).\end{array}$$

In the next section, we provide an analogous result for the more general Equation (10).

A somewhat different generalization of (1) and (2) was investigated in [21]. In particular, the conditional equation

$$g\left(p\right(\mathrm{\Pi }\left(a\right)★\mathrm{\Pi }\left(b\right)\left)\right)=g\left(a\right)\diamond g\left(b\right),\mathrm{\Pi }\left(a\right)\diamond \mathrm{\Pi }\left(b\right)\in P,$$

(11)

was considered for functions $g:S\to W$, where S is a nonempty set, $(Y,★)$ and $(W,\diamond )$ are groupoids (i.e., Y and W are nonempty sets endowed with binary operations $★:Y^2\to Y$ and $\diamond :W^2\to W$), $\mathrm{\Pi }:S\to Y$, $P_0:=\mathrm{\Pi }\left(S\right)$, $P\subset P_0$ is nonempty, and $p:P_0\to S$ is a selection of $\mathrm{\Pi }$, i.e.,

$$\mathrm{\Pi }\left(p\left(x\right)\right)=x,x\in P_0.$$

Only some very weak additional assumptions concerning the neutral elements in $(Y,★)$ or the cancellation property in $(W,\diamond )$ were used.

Remark 1. 

To avoid any ambiguity, let us explain that we understand conditional Equation (11) for a given function $g:S\to W$ in the following way: $g\left(p\right(\mathrm{\Pi }\left(a\right)★\mathrm{\Pi }\left(b\right)\left)\right)=g\left(a\right)\diamond g\left(b\right)$ for every $a,b\in S$ such that $\mathrm{\Pi }\left(a\right)\diamond \mathrm{\Pi }\left(b\right)\in P$. Any other conditional equation considered in this article should be understood in a similar way.

Clearly, if $S=\mathbb{R}$, $(Y,★)$ is the additive group of real numbers, $n\in \mathbb{N}$, $\mathrm{\Pi }\left(x\right):=x^n$ for $x\in S$, and $p\left(u\right):=\sqrt[n]{u}$ for $u\in P_0$, then (11) takes the following form:

$$f\left(\sqrt[n]{x^n+y^n}\right)=f\left(x\right)\diamond f\left(y\right).$$

It is easily seen that Equations (1), (2), and (3) are special cases of (11).

Another special case of (11) is

$$\begin{array}{c}\hfill g\left(x+y+2\sqrt{xy}\right)=g\left(x\right)+g\left(y\right),\end{array}$$

(12)

which was studied in [22] for $g:[0,\infty )\to \mathbb{R}$. Clearly, in this case, $\mathrm{\Pi }\left(x\right):=\sqrt{x}$ and $p\left(x\right):=x^2$ for $x\in [0,\infty )$.

In [21], two more simple examples of (11) were mentioned (for real-valued functions with real variables). The first one, for $S\subset \mathbb{R}$, is a conditional equation of the form

$$\begin{array}{c}\hfill g(\lfloor x\rfloor +\lfloor y\rfloor )=g\left(x\right)\diamond g\left(y\right),\lfloor x\rfloor +\lfloor y\rfloor \in \mathrm{\Pi }\left(S\right),\end{array}$$

(13)

where $\mathrm{\Pi }:S\to \mathbb{Z}$ is the floor function, i.e., $\mathrm{\Pi }\left(x\right):=\lfloor x\rfloor$ for $x\in S$ ($\lfloor x\rfloor$ denotes the largest integer not greater than a real number x), S is such that $\mathrm{\Pi }\left(S\right)\subset S$, and $p\left(n\right)=n$ for $n\in \mathrm{\Pi }\left(S\right)$.

The second one, also for $S\subset \mathbb{R}$, has the form

$$\begin{array}{c}\hfill g\left(\right\{x\}+\{y\left\}\right)=g\left(x\right)\diamond g\left(y\right),\left\{x\right\}+\left\{y\right\}\in \mathrm{\Pi }\left(S\right),\end{array}$$

(14)

when $\mathrm{\Pi }:S\to [0,1)$ is given by $\mathrm{\Pi }\left(x\right)=\left\{x\right\}:=x-\lfloor x\rfloor$ for $x\in S$, S is such that $\mathrm{\Pi }\left(S\right)\subset S$, and $p\left(x\right)=x$ for $x\in \mathrm{\Pi }\left(S\right)$.

Below, we provide three more examples of particular forms of (11) in the case when the ★ operation is just the usual addition + in $\mathbb{R}$ or $\mathbb{C}$ and $n\in \mathbb{N}$ is fixed.

In particular, if $\mathrm{\Pi }\left(x\right)=1/x^n$ for $x\in (0,\infty )$, then $p\left(x\right)=1/\sqrt[n]{x}$ and Equation (11) takes the following form:

$$g\left({\displaystyle \frac{xy}{\sqrt[n]{x^n+y^n}}}\right)=g\left(x\right)\diamond g\left(y\right).$$

If $\mathrm{\Pi }\left(x\right)=1/\sqrt[n]{x}$ for $x\in (0,\infty )$, then $p\left(x\right)=1/x^n$ and Equation (11) has the following form:

$$g\left({\displaystyle \frac{xy}{{\left(\sqrt[n]{x}+\sqrt[n]{y}\right)}^n}}\right)=g\left(x\right)\diamond g\left(y\right).$$

Finally, if $\mathrm{\Pi }\left(z\right)=\sqrt[n]{\left|z\right|}$ for $z\in \mathbb{C}$ ($\left|z\right|$ stands for the modulus of a complex number z) and $p\left(a\right)=a^n$, then (11) becomes

$$\begin{array}{c}\hfill g\left({\left(\sqrt[n]{\left|z\right|}+\sqrt[n]{\left|w\right|}\right)}^n\right)=g\left(z\right)\diamond g\left(w\right).\end{array}$$

(15)

The main purpose of this expository article is to provide simple general descriptions of solutions to various functional equations of the radical type and to present appropriate examples showing how to apply them. We believe that this should be helpful for authors considering similar topics in the future.

Moreover, the results presented in this paper show that there is a significant symmetry between solutions of numerous functional equations and solutions to the equations of the radical type that correspond to them.

2. First General Result on Solutions

Unless explicitly stated otherwise, throughout this paper, S and T are nonempty sets, $n,k,m\in \mathbb{N}$ and $n>1$, $H:T^k\to S$, $G:T^m\to S$, $P_0:=\{a^n:a\in \mathbb{R}\}$ and $F_1,\dots ,F_k:P_0^m\to P_0$.

In [20], the following general theorem was proven.

Theorem 1. 

Let $g:\mathbb{R}\to T$ and one of the following two conditions be valid.

(i)  

n is odd;

(ii) 

There exist $c_1,\dots ,c_{m-1}\in g\left(\mathbb{R}\right)$ with

$$\begin{array}{c}\hfill G(c_1,\dots ,c_{m-1},u)\ne G(c_1,\dots ,c_{m-1},v),u,v\in T,u\ne v.\end{array}$$

(16)

Then, g fulfills the functional equation

$$\begin{array}{c}\hfill H\left(g\left(\sqrt[n]{F_1(a_1^n,\dots ,a_m^n)}\right),\dots ,g\left(\sqrt[n]{F_k(a_1^n,\dots ,a_m^n)}\right)\right)\\ \hfill =G\left(g\left(a_1\right),\dots ,g\left(a_m\right)\right)\end{array}$$

(17)

if and only if there is $h:P_0\to T$ satisfying the equation

$$H\left(h\left(F_1(a_1,\dots ,a_m)\right),\dots ,h\left(F_k(a_1,\dots ,a_m)\right)\right)=G\left(h\left(a_1\right),\dots ,h\left(a_m\right)\right)$$

(18)

such that $g\left(a\right)=h\left(a^n\right)$ for each $a\in \mathbb{R}$.

Below, we show that the following generalization of Theorem 1 is possible.

Theorem 2. 

Let $U_1,\dots ,U_{k+m}\subset \mathbb{R}$ be nonempty sets with

$$\sqrt[n]{F_i(a_1^n,\dots ,a_m^n)}\in U_i,(a_1,\dots ,a_m)\in U_1\times \dots \times U_m,i=1,\dots ,k,$$

(19)

$$\sqrt[n]{a_i^n}\in U_i,a_i\in U_i,i=1,\dots ,k+m,$$

(20)

$g_i:U_i\to T$ and $V_i:=\{a^n:a\in U_i\}$ for $i=1,\dots ,k+m$, and let one of the following two conditions be valid:

(i)  

n is odd;

(ii)  

For every $j\in \{1,\dots ,m\}$, there exist $c_{ij}\in g_{k+i}\left(U_{k+i}\right)$ for $i\in \{1,\dots ,m\}\setminus \left\{j\right\}$ such that the mapping

$$G_j:T\ni a\to G(c_{1,j},\dots ,c_{j-1,j},a,c_{j+1,j},\dots ,c_{m,j})$$

is injective.

Then, functions $g_1,\dots ,g_{k+m}$ fulfill the functional equation

$$\begin{array}{c}\hfill H\left(g_1\left(\sqrt[n]{F_1(a_1^n,\dots ,a_m^n)}\right),\dots ,g_k\left(\sqrt[n]{F_k(a_1^n,\dots ,a_m^n)}\right)\right)\\ \hfill =G\left(g_{k+1}\left(a_1\right),\dots ,g_{k+m}\left(a_m\right)\right)\end{array}$$

(21)

if and only if there exist functions $h_i:V_i\to T$ for $i=1,\dots ,k+m$ satisfying the equation

$$H\left(h_1\left(F_1(a_1,\dots ,a_m)\right),\dots ,h_k\left(F_k(a_1,\dots ,a_m)\right)\right)=G\left(h_{k+1}\left(a_1\right),\dots ,h_{k+m}\left(a_m\right)\right),$$

(22)

such that $g_j\left(a_j\right)=h_j\left(a_j^n\right)$ for $a_j\in {\widehat{U}}_j$, $j=1,\dots ,k$, and $g_{k+i}\left(a_{k+i}\right)=h_{k+i}\left(a_{k+i}^n\right)$ for $a_{k+i}\in U_{k+i}$, $i=1,\dots ,m$, where

$${\widehat{U}}_j:=\left\{\begin{array}{cc}{U}_j\hfill & ifn is odd;\hfill \\ U_j\cap [0,\infty )\hfill & ifn is even,\hfill \end{array}\right.j=1,\dots ,k.$$

Proof. 

Assume that $g_1,\dots ,g_{k+m}$ satisfy (21). For $i=1,\dots ,k+m$, $h_i:V_i\to T$ is defined by

$$\begin{array}{c}\hfill h_i\left(a_i\right):=g_i\left(\sqrt[n]{a_i}\right),a_i\in V_i,i=1\dots ,k+m.\end{array}$$

(23)

In view of (20), this definition is correct. We prove that equality (22) is valid for all $(a_1,\dots ,a_m)\in V_1\times \dots \times V_m$.

Therefore, fix $a_i\in V_i$ for $i=1\dots ,k+m$, and let $b_i:=\sqrt[n]{a_i}\in U_i$ for $i=1,\dots ,k+m$ (see (20)). Clearly,

$$a_i:=b_i^n\in V_i,i=1,\dots ,k+m,$$

and according to (19),

$$F_i(a_1,\dots ,a_m)\in V_i,i=1,\dots ,k.$$

(24)

Hence (21) implies that

$$\begin{array}{cc}\hfill H\left(h_1\right(F_1(a_1,& \dots ,a_m\left)\right),\dots ,h_k\left(F_k(a_1,\dots ,a_m)\right))\hfill \\ \hfill & =H\left(g_1\left(\sqrt[n]{F_1(b_1^n,\dots ,b_m^n)}\right),\dots ,g_k\left(\sqrt[n]{F_k(b_1^n,\dots ,b_m^n)}\right)\right)\hfill \\ \hfill & =G\left(g_{k+1}\left(b_1\right),\dots ,g_{k+m}\left(b_m\right)\right)=G\left(g_{k+1}\left(\sqrt[n]{a_1}\right),\dots ,g_{k+m}\left(\sqrt[n]{a_m}\right)\right)\hfill \\ \hfill & =G\left(h_{k+1}\left(a_1\right),\dots ,h_{k+m}\left(a_m\right)\right),\hfill \end{array}$$

(25)

i.e., Equation (22) is fulfilled for each $(a_1,\dots ,a_m)\in V_1\times \dots \times V_m$.

If n is odd, then $\sqrt[n]{a^n}=a$ for each $a\in \mathbb{R}$, and according to (23),

$$g_i\left(a_i\right)=h_i\left(a_i^n\right),a_i\in U_i,i=1,\dots ,k+m.$$

Now, consider the case where n is even. Then, $V_i\subset [0,\infty )$ for $i=1,\dots ,k+m$, and according to (20) and (23),

$$\begin{array}{c}\hfill g_i\left(a_i\right)=h_i\left(a_i^n\right),a_i\in U_i\cap [0,\infty ),i=1,\dots ,k+m.\end{array}$$

(26)

This implies that $g_j\left(a_j\right)=h_j\left(a_j^n\right)$ for $a_j\in {\widehat{U}}_j$, $j=1,\dots ,k$.

Furthermore, in view of (ii), for every $j\in \{1,\dots ,m\}$, there exist $c_{ij}\in g_{k+i}\left(U_{k+i}\right)$ for $i\in \{1,\dots ,m\}\setminus \left\{j\right\}$ such that the mapping

$$\begin{array}{c}\hfill G_j:T\ni a\to G(c_{1,j},\dots ,c_{j-1,j},a,c_{j+1,j},\dots ,c_{m,j})\end{array}$$

(27)

is injective.

Fix $j\in \{1,\dots ,m\}$ and $v_i\in U_{k+i}$ for $i=1,\dots ,m$, $j\ne i$, with

$$c_{i,j}=g_{k+i}\left(v_i\right),i=1,\dots ,m,j\ne i.$$

Observe that for each $a\in U_{k+j}$ with $-a\in U_{k+j}$, we have

$$\begin{array}{cc}\hfill G(& c_{1,j},\dots ,c_{j-1,j},g_{k+j}(-a),c_{j+1,j},\dots ,c_{m,j})\hfill \\ \hfill & =G\left(g_{k+1}\left(v_1\right),\dots ,g_{k+j-1}\left(v_j\right),g_{k+j}(-a),g_{k+j+1}\left(v_{j+1}\right),\dots ,g_{k+m}\left(v_m\right)\right)\hfill \\ \hfill & =H\left(g_1\left(\sqrt[n]{F_1(v_1^n,\dots ,v_{j-1}^n,{(-a)}^n,v_{j+1}^n,\dots ,v_m^n}\right),\dots ,g_k\left(\sqrt[n]{F_k(v_1^n,\dots ,v_{j-1}^n,{(-a)}^n,v_{j+1}^n,\dots ,v_m^n}\right)\right)\hfill \\ \hfill & =H\left(g_1\left(\sqrt[n]{F_1(v_1^n,\dots ,v_{j-1}^n,a^n,v_{j+1}^n,\dots ,v_m^n}\right),\dots ,g_k\left(\sqrt[n]{F_k(v_1^n,\dots ,v_{j-1}^n,a^n,v_{j+1}^n,\dots ,v_m^n}\right)\right)\hfill \\ \hfill & =G\left(g_{k+1}\left(v_1\right),\dots ,g_{k+j-1}\left(v_j\right),g_{k+j}\left(a\right),g_{k+j+1}\left(v_{j+1}\right),\dots ,g_{k+m}\left(v_m\right)\right)\hfill \\ \hfill & =G\left(c_{1,j},\dots ,c_{j-1,j},g_{k+j}\left(a\right),c_{j+1,j},\dots ,c_{m,j}\right).\hfill \end{array}$$

Since the mapping (27) is injective, we have thus proven that

$$\begin{array}{c}\hfill g_{k+j}(-a)=g_{k+j}\left(a\right)\end{array}$$

(28)

for $a\in U_{k+j}$ with $-a\in U_{k+j}$, and (on account of (26))

$$g_{k+j}\left(a\right)=h_{k+j}\left(a^n\right),a\in U_{k+j}.$$

Now, we prove the converse implication. Therefore, let $g_i:U_i\to T$ for $i=1,\dots ,k+m$ be such that $g_i\left(a_i\right)=h_i\left(a_i^n\right)$ for every $a_i\in {\widehat{U}}_i$, $i=1,\dots ,k$, and $g_j\left(b_j\right)=h_j\left(b_j^n\right)$ for every $b_j\in U_j$, $j=k+1,\dots ,k+m$, with some $h_i:V_i\to T$ for $i=1,\dots ,k+m$, satisfying equation (22). Then, (23) holds. We show that $g_1,\dots ,g_{k+m}$ fulfill functional Equation (21).

Therefore, take $a_i\in U_i$ for $i=1,\dots ,m$. Then, $a_i^n\in V_i$ for $i=1,\dots ,m$ and, consequently,

$$\begin{array}{cc}\hfill H\left(g_1\right(& \sqrt[n]{F_1(a_1^n,\dots ,a_m^n)}),\dots ,g_k\left(\sqrt[n]{F_k(a_1^n,\dots ,a_m^n)}\right))\hfill \\ \hfill & =H\left(h_1\left(F_1\left(a_1^n,\dots ,a_m^n\right)\right),\dots ,h_k\left(F_k\left(a_1^n,\dots ,a_m^n\right)\right)\right)\hfill \\ \hfill & =G\left(h_{k+1}\left(a_1^n\right),\dots ,h_{k+m}\left(a_m^n\right)\right)=G\left(g_{k+1}\left(a_1\right),\dots ,g_{k+m}\left(a_m\right)\right).\hfill \end{array}$$

□

Remark 2. 

Note that condition (20) is always fulfilled when n is odd. When n is even, the condition simply means that $\left|a\right|\in U_i$ for $a\in U_i$, $i=1,\dots ,k+m$.

3. Applications of Theorem 2

Below, we provide two corollaries with simplified versions of Theorem 2 and show some applications of them. We start with the following one for $k=1$ and $m=2$ (as before, $P_0:=\{a^n:a\in \mathbb{R}\}$).

Corollary 1. 

Let $g_i:\mathbb{R}\to T$ for $i=1,2,3$ and one of the the following two conditions be fulfilled.

(a)  

n is odd;

(b)  

There exist $c_i\in g_{i+1}\left(\mathbb{R}\right)$ for $i=1,2$ such that the mappings expressed as

$$T\ni a\to G(c_1,a),T\ni a\to G(a,c_2)$$

are injective.

Then, functions $g_1$, $g_2$, and $g_3$ fulfill the functional equation

$$\begin{array}{c}\hfill g_1\left(\sqrt[n]{F_1(a_1^n,a_2^n)}\right)=G\left(g_2\left(a_1\right),g_3\left(a_2\right)\right)\end{array}$$

(29)

if and only if there exist functions $h_i:P_0\to T$ for $i=1,2,3$ satisfying the equation

$$h_1\left(F_1(a_1,a_2)\right))=G\left(h_2\left(a_1\right),h_3\left(a_2\right)\right),$$

such that $g_1\left(a_1\right)=h_1\left(a_1^n\right)$ for $a_1\in P_0$ and $g_i\left(a\right)=h_i\left(a^n\right)$ for $a\in \mathbb{R}$, $i=2,3$.

Now, we show applications of this corollary. Let $(X,+)$ be a group. We start with the functional equation

$$\begin{array}{c}\hfill g_1\left(\sqrt[n]{a^n+b^n}\right)=g_2\left(a\right)+g_3\left(b\right),\end{array}$$

(30)

for $g_1,g_2,g_3$ mapping $\mathbb{R}$ into X. Clearly, (30) is a radical version of the Pexider equation, i.e.,

$$\begin{array}{c}\hfill h_1(a+b)=h_2\left(a\right)+h_3\left(b\right).\end{array}$$

(31)

It is said quite often that the Pexider equation (31) is the so-called pexiderization of the well known Cauchy additive equation, i.e.,

$$\begin{array}{c}\hfill h(a+b)=h\left(a\right)+h\left(b\right).\end{array}$$

(32)

The terms Pexider equation and pexiderization refer to the name of J.V. Pexider, who studied Equation (31) in [23] (cf. [24], Ch. 4.3 and [25], Ch. 2.2).

Clearly, similar pexiderizations of other functional equations can be considered, as we do later. Now, let us focus on Equation (30) for functions mapping $\mathbb{R}$ into X.

First, let us recall (see, e.g., [24], Ch. 4.3, p. 43, Theorem 9) that $h_1,h_2,h_3:P_0\to X$ satisfy (31) if and only if they have the forms of

$$\begin{array}{c}\hfill h_i\left(a\right)=h\left(a\right)+c_i,a\in P_0,i=1,2,3,\end{array}$$

(33)

where $h:P_0\to X$ is a solution to (32) and $c_1,c_2,c_3\in X$ are constants with $c_1=c_2+c_3$. Moreover, if $(X,+)$ is the group $(\mathbb{R},+)$ and at least one of functions $h_1$, $h_2$, and $h_3$ is continuous at least at one point, then (see, e.g., [24], Ch. 2.1, pp. 15 and 18, Corollaries 5 and 9) there is a real constant $\alpha$ such that $h\left(a\right)=\alpha a$ for $a\in P_0$, and consequently,

$$\begin{array}{c}\hfill h_i\left(a\right)=\alpha a+c_i,a\in P_0,i=1,2,3.\end{array}$$

(34)

Observe that (30) is Equation (29) with $F_1(a_1,a_2)=a_1+a_2$ for $a_1,a_2\in \mathbb{R}$ and $G(x_1,x_2)=x_1+x_2$ for $x_1,x_2\in X$. Therefore, the assumption (b) of Corollary 2 is fulfilled, and functions $g_1,g_2,g_3:\mathbb{R}\to X$ satisfy (30) if and only if there exist $h_1,h_2,h_3:P_0\to X$ fulfilling (31) such that, for odd n,

$$\begin{array}{c}\hfill g_i\left(a\right)=h_i\left(a^n\right),a\in \mathbb{R},i=1,2,3\end{array}$$

(35)

and, for even n,

$$\begin{array}{c}\hfill g_1\left(a\right)=h_1\left(a^n\right),a\in [0,\infty ),g_i\left(a\right)=h_i\left(a^n\right),a\in \mathbb{R},i=2,3.\end{array}$$

(36)

Furthermore, in view of (33), condition (35) can be rewritten as

$$\begin{array}{c}\hfill g_i\left(a\right)=h\left(a^n\right)+c_i,a\in \mathbb{R},i=1,2,3,\end{array}$$

(37)

and condition (36) takes the following form

$$\begin{array}{c}\hfill g_1\left(a\right)=h\left(a^n\right)+c_1,a\in [0,\infty ),g_i\left(a\right)=h\left(a^n\right)+c_i,a\in \mathbb{R},i=2,3.\end{array}$$

(38)

Thus, we have proven the following result.

Corollary 2. 

Let $(X,+)$ be a group. Functions $g_1,g_2,g_3:\mathbb{R}\to X$ fulfill functional Equation (30) if and only if there exist $c_1,c_2,c_3\in X$ with $c_1=c_2+c_3$ and a solution $h:P_0\to X$ of Equation (32) such that

(i)  

In the case where n is odd, condition (37) is valid;

(ii)  

In the case where n is even, condition (38) holds.

Moreover, if $X=\mathbb{R}$ and there is $i\in \{1,2,3\}$ such that $g_i$ is continuous at least at one point, then $h\left(a\right)=\alpha a$ for $a\in P_0$, with some real constant α.

Proof. 

The necessary forms of $g_1$, $g_2$, and $g_3$ were determined by the reasoning presented before this corollary. It only remains to add that one can easily check that if one of conditions (i) and (ii) holds, then $g_1$, $g_2$, and $g_3$ satisfy functional Equation (30). □

The next example shows possible applications of Corollary 2 to functional equations with one unknown function.

Example 1. 

Let us consider a generalized form of Equation (4), namely

$$\begin{array}{c}\hfill f\left(\sqrt{{\alpha }_1{a}^2+{\beta }_1{b}^2}\right)={\alpha }_{2}f\left(a\right)+{\beta }_{2}f\left(b\right),\end{array}$$

(39)

for functions f mapping $\mathbb{R}$ into a real linear space X (as in [8,9]) and with fixed ${\alpha }_1,{\alpha }_2,{\beta }_1,{\beta }_2\in \mathbb{R}$, ${\alpha }_1>0$ and ${\beta }_1>0$. Let $f:\mathbb{R}\to X$ be a solution to (39) and

$$\begin{array}{c}\hfill g_2\left(a\right)={\alpha }_{2}f\left({\displaystyle \frac{a}{\sqrt{{\alpha }_1}}}\right),g_3\left(a\right)={\beta }_{2}f\left({\displaystyle \frac{a}{\sqrt{{\beta }_1}}}\right),a\in \mathbb{R}.\end{array}$$

(40)

Then,

$$\begin{array}{cc}\hfill f\left(\sqrt{a^2+b^2}\right)& =f\left(\sqrt{{\alpha }_1{\left({\displaystyle \frac{a}{\sqrt{{\alpha }_1}}}\right)}^2+{\beta }_1{\left({\displaystyle \frac{b}{\sqrt{{\beta }_1}}}\right)}^2}\right)\hfill \end{array}$$

(41)

$$\begin{array}{cc}\hfill & ={\alpha }_{2}f\left({\displaystyle \frac{a}{\sqrt{{\alpha }_1}}}\right)+{\beta }_{2}f\left({\displaystyle \frac{b}{\sqrt{{\beta }_1}}}\right)=g_2\left(a\right)+g_3\left(b\right),\hfill \end{array}$$

(42)

which actually is Equation (30) with $g_1=f$. Consequently, according to Corollary 2,

$$\begin{array}{c}\hfill f\left(a\right)=h\left(a^2\right)+c_1,a\in [0,\infty ),g_i\left(a\right)=h\left(a^2\right)+c_i,a\in \mathbb{R},i=2,3,\end{array}$$

(43)

where $h:[0,\infty )\to X$ is a solution of Equation (32) and constants $c_1,c_2,c_3\in X$ are such that $c_1=c_2+c_3$. According to (40), this implies the following equalities:

$$\begin{array}{c}\hfill h\left(a^2\right)+c_2=g_2\left(a\right)={\alpha }_{2}f\left({\displaystyle \frac{a}{\sqrt{{\alpha }_1}}}\right)={\alpha }_2\left(h\left({\displaystyle \frac{a^2}{{\alpha }_1}}\right)+c_1\right),a\in \mathbb{R},\end{array}$$

(44)

$$\begin{array}{c}\hfill h\left(a^2\right)+c_3=g_3\left(a\right)={\beta }_{2}f\left({\displaystyle \frac{a}{\sqrt{{\beta }_1}}}\right)={\beta }_2\left(h\left({\displaystyle \frac{a^2}{{\beta }_1}}\right)+c_1\right),a\in \mathbb{R}.\end{array}$$

(45)

Since h is additive,

$$\begin{array}{c}\hfill {\alpha }_2{c}_1=c_2,{\beta }_2{c}_1=c_3,\end{array}$$

(46)

and

$$\begin{array}{c}\hfill h\left({\alpha }_{1}a\right)={\alpha }_{2}h\left(a\right),h\left({\beta }_{1}a\right)={\beta }_{2}h\left(a\right),a\in \mathbb{R}.\end{array}$$

(47)

Note that (46) implies that either ${\alpha }_2+{\beta }_2=1$ or $c_1=0$ (because $c_1=c_2+c_3$). Therefore, we obtain

$$\begin{array}{c}\hfill f\left(a\right)=h\left(a^2\right)+c_1,a\in \mathbb{R},\end{array}$$

(48)

where $h:\mathbb{R}\to X$ is additive and fulfills (47) , and

$$\begin{array}{c}\hfill ({\alpha }_2+{\beta }_2-1)c_1=0.\end{array}$$

(49)

Moreover, it is easy to check that Equation (39) is fulfilled by every $f:\mathbb{R}\to X$ given by (48) with some additive $h:\mathbb{R}\to X$ satisfying (47) and fixed $c_1\in X$ such that (49) holds. Therefore, we have the following corollary.

Corollary 3. 

Function $f:\mathbb{R}\to X$ fulfills functional Equation (39) if and only if there exists an additive $h:\mathbb{R}\to X$ and constant $c_1\in X$ such that (47)–(49) hold.

Remark 3. 

Note that an additive ($h:\mathbb{R}\to X$) satisfies condition (47) if and only if it is a solution to the following equation:

$$\begin{array}{c}\hfill h({\alpha }_{1}a+{\beta }_{1}b)={\alpha }_{2}h\left(a\right)+{\beta }_{2}h\left(b\right).\end{array}$$

(50)

Therefore, very useful information on additive mappings satisfying condition (48) can be derived from [26] (Ch. IV, §12, Theorem 2 and Ch. XIII, §10, Theorem 5).

It was proven in [9] (Theorem 2.3) that for $\alpha +\beta \ne 1$, each solution $g:\mathbb{R}\to X$ to (4) satisfies the quadratic functional equation:

$$\begin{array}{c}\hfill Q(a+b)+Q(a-b)=2Q\left(a\right)+2Q\left(b\right),\end{array}$$

(51)

i.e., g must be a quadratic function (see [24] for more information on such functions). Moreover, it is easy to check that for every additive $h:\mathbb{R}\to X$, the function $f:\mathbb{R}\to X$ with the form of (48), where $c_1=0$, satisfies quadratic Equation (51). Therefore, Corollary 3 is consistent with the result obtained in [9] (Theorem 2.3).

However, not all functions $g:\mathbb{R}\to X$ satisfying (51) must be solutions to (4). For instance, let $\alpha =\beta =\sqrt{2}$, $X=\mathbb{R}$ and $\xi :\mathbb{R}\to \mathbb{Q}$ be additive. Let $g\left(x\right)=\xi {\left(x\right)}^2$ for $x\in \mathbb{R}$. Then, it is easy to check that g is a solution to (51). But $g\left(\sqrt{2}\right),g\left(1\right)\in \mathbb{Q}$ and $g\left(0\right)=0$, which means that $g\left(\sqrt{2}\right)\ne \sqrt{2}g\left(1\right)$, and consequently, (4) does not hold for $a=1$ and $b=0$.

From Corollary 2, we can also derive the following conclusion.

Corollary 4. 

Assume that $n,k,m\in \mathbb{N}$ are odd. Then, functions $f_1,f_2,f_3:\mathbb{R}\to X$ fulfill the functional equation

$$\begin{array}{c}\hfill f_1\left(\sqrt[n]{a^k+b^m}\right)=f_2\left(a\right)+f_3\left(b\right)\end{array}$$

(52)

if and only if there exist $c_1,c_2,c_3\in X$ and a solution $h:\mathbb{R}\to X$ of Equation (32) such that $c_1=c_2+c_3$ and

$$\begin{array}{c}\hfill f_1\left(a\right)=h\left(a^n\right)+c_1,f_2\left(a\right)=h\left(a^k\right)+c_2,f_3\left(a\right)=h\left(a^m\right)+c_3,a\in \mathbb{R}.\end{array}$$

(53)

Moreover, if $X=\mathbb{R}$ and there is $i\in \{1,2,3\}$ such that $f_i$ is continuous at least at one point, then $h\left(a\right)=\alpha a$ for $a\in P_0$, with some real constant α.

Proof. 

Let $f_1,f_2,f_3:\mathbb{R}\to \mathbb{R}$ fulfill Equation (52). Fix $u,v\in \mathbb{R}$ and insert $a=\sqrt[k]{u^n}$ and $b=\sqrt[m]{v^n}$ in (52). Then, we obtain

$$\begin{array}{c}\hfill f_1\left(\sqrt[n]{u^n+v^n}\right)=f_2\left(\sqrt[k]{u^n}\right)+f_3\left(\sqrt[m]{v^n}\right),\end{array}$$

(54)

which is (30) with $g_1=f_1$ and

$$g_2\left(u\right)=f_2\left(\sqrt[k]{u^n}\right),g_3\left(u\right)=f_3\left(\sqrt[m]{u^n}\right),u\in \mathbb{R}.$$

Now, it is enough to use Corollary 2.

The converse is easy to check. □

We also have the following result.

Corollary 5. 

Let $n,k,m\in \mathbb{N}$ be odd and such that it is not true that $n=k=m$. Let $f:\mathbb{R}\to \mathbb{R}$ be continuous at least at one point. Then, f fulfills the functional equation

$$\begin{array}{c}\hfill f\left(\sqrt[n]{a^k+b^m}\right)=f\left(a\right)+f\left(b\right)\end{array}$$

(55)

if and only if $f\left(a\right)=0$ for $a\in \mathbb{R}$.

Proof. 

Let f be a solution to Equation (55). According to Corollary 4 with $f_1=f_2=f_3=f$, condition (53) holds, which (with $a=0$) yields $c_1=c_2=c_3=0$ (because $c_1=c_2+c_3$). Furthermore, $h\left(a\right)=\alpha a$ for $a\in \mathbb{R}$, and $f\left(a\right)=f_1\left(a\right)=\alpha a^n$ for $a\in \mathbb{R}$.

Suppose that there is $a\in \mathbb{R}$ such that $f\left(a\right)\ne 0$. Then, $\alpha \ne 0$, and according to (53), $\alpha 2^n=h\left(2^n\right)=f\left(2\right)=f_2\left(2\right)=h\left(2^k\right)=\alpha 2^k$. This implies that $2^n=2^k$. Analogously we, obtain $2^n=2^m$ and $2^k=2^m$. Thus we have shown that $n=k=m$, which is a contradiction with the assumption that n, k, and m are not equal. Consequently, $f\left(a\right)=0$ for all $a\in \mathbb{R}$.

The converse is easy to check. □

Remark 4. 

Corollaries 4 and 5 show that there is a significant difference between the solutions of Equation (55) and the solutions of its pexiderized form, i.e., Equation (52).

Next, consider the following functional equation:

$$\begin{array}{c}\hfill g_1\left(\sqrt[n]{a^n+b^n}\right)=g_2\left(a\right)g_3\left(b\right),\end{array}$$

(56)

which is a radical version of

$$\begin{array}{c}\hfill f_1\left(a+b\right)=f_2\left(a\right)f_3\left(b\right),\end{array}$$

(57)

i.e., of the pexiderization of the following well known exponential functional equation:

$$\begin{array}{c}\hfill f\left(a+b\right)=f\left(a\right)f\left(b\right).\end{array}$$

(58)

Let $g_1,g_2,g_3:\mathbb{R}\to \mathbb{R}$ fulfill functional Equation (56). First, we study the situation when $g_i\left(\mathbb{R}\right)\ne \left\{0\right\}$ for $i=2,3$. Then, the assumption (b) of Corollary 2 is valid, and consequently, $g_1$, $g_2$, and $g_3$ fulfill functional Equation (56) if and only if there exist solutions $f_1,f_2,f_3:\mathbb{R}\to \mathbb{R}$ of Equation (57) such that

$$\begin{array}{c}\hfill g_1\left(a\right)=f_1\left(a^n\right),a\in P_0,g_i\left(b\right)=f_i\left(b^n\right),b\in \mathbb{R},i=2,3.\end{array}$$

(59)

Clearly,

$$\begin{array}{c}\hfill f_i\left(\mathbb{R}\right)\ne \left\{0\right\},i=2,3,\end{array}$$

(60)

which implies that $f_1\left(P_0\right)\ne \left\{0\right\}$.

Therefore, we focus on functions $f_1,f_2,f_3:\mathbb{R}\to \mathbb{R}$ satisfying Equation (57) such that $f_i\left(\mathbb{R}\right)\ne \left\{0\right\}$ for $i=1,2,3$.

Let $f_1,f_2,f_3:\mathbb{R}\to \mathbb{R}$ be solutions to (57) and $d_i=f_i\left(0\right)$ for $i=1,2,3$. Then, according to (57) (with $a=0$ and/or $b=0$), $d_1=d_2{d}_3$ and

$$\begin{array}{c}\hfill f_1\left(a\right)=f_2\left(a\right)d_3=d_2{f}_3\left(a\right),a\in P_0.\end{array}$$

(61)

Hence, according to (60), $d_1=d_2{d}_3\ne 0$, and consequently, (57) implies that

$$d_1{f}_1(a+b)=f_1\left(a\right)f_1\left(b\right),a,b\in P_0.$$

Consequently, the function ($f:P_0\to \mathbb{R}$) given by $f\left(a\right):=f_1\left(a\right)/d_1$ for $a\in P_0$ satisfies Equation (58). Clearly, according to (61), we have

$$\begin{array}{c}\hfill f_i\left(a\right)=d_{i}f\left(a\right),a\in P_0,i=1,2,3.\end{array}$$

(62)

It is known (see [24], Ch. 3, p. 28–29 and Theorem 5) that every solution $f:P_0\to \mathbb{R}$ of (58) has one of the following forms:

(A)

$f\left(a\right)=exp\left(h\right(a\left)\right)$ for $x\in P_0$, where $h:P_0\to \mathbb{R}$ is additive (i.e., a solution to (32));

(B)

$f\left(x\right)=0$ for $x\in P_0$;

(C)

(only when $P_0=[0,\infty )$, $f\left(0\right)=1$ and $f\left(a\right)=0$ for $a\in P_0$, $a\ne 0$.

Hence, (59) and (62) imply that if n is odd, then

$$\begin{array}{c}\hfill g_i\left(a\right)=d_{i}f\left(a^n\right),a\in \mathbb{R},i=1,2,3,\end{array}$$

(63)

and if n is even,

$$\begin{array}{c}\hfill g_1\left(a\right)=d_{1}f\left(a^n\right),a\in [0,\infty ),g_i\left(a\right)=d_{i}f\left(a^n\right),a\in \mathbb{R},i=2,3,\end{array}$$

(64)

where $f:P_0\to \mathbb{R}$ is depicted by (A)–(C).

Therefore, we have proven the following result.

Corollary 6. 

Functions $g_1,g_2,g_3:\mathbb{R}\to \mathbb{R}$, with $g_i\left(\mathbb{R}\right)\ne \left\{0\right\}$ for $i=2,3$, fulfill functional Equation (56) if and only if

(i)  

in the case where n is odd, condition (63) holds with some real constants ($d_1,d_2,d_3$, $d_1=d_2{d}_3$) and with a function ($f:\mathbb{R}\to \mathbb{R}$) of the form depicted by (A) or (B) with $P_0=\mathbb{R}$;

(ii)  

in the case where n is even, condition (64) holds with some real constants $d_1,d_2,d_3$, $d_1=d_2{d}_3$, and with a function $f:[0,\infty )\to \mathbb{R}$ of the form depicted in one of conditions (A)–(C) with $P_0=[0,\infty )$.

Moreover, if (A) holds and there is $i\in \{1,2,3\}$ such that $g_i$ is continuous at least at one point, then $h\left(a\right)=\alpha a$ for $a\in P_0$, with some real constant α.

Proof. 

The necessary forms of $g_1$, $g_2$, and $g_3$ were determined by the reasoning presented before this corollary. It is also easy to check that if one of conditions (i) and (ii) holds, then $g_1$, $g_2$, and $g_3$ satisfy functional Equation (56).

It remains to notice that if (A) holds and there is $i\in \{1,2,3\}$ such that $g_i$ is continuous at least at one point, then h must be continuous at this point, and consequently, $h\left(a\right)=\alpha a$ for $a\in P_0$, with some real constant $\alpha$ (see, e.g., [24], Ch. 2.1, pp. 15 and 18, Corollaries 5 and 9). □

The following remarks complement Corollary 6.

Let $g_1,g_2,g_3:\mathbb{R}\to \mathbb{R}$ satisfy Equation (56). Assume that $g_2\left(\mathbb{R}\right)=\left\{0\right\}$. Then,

$$g_1\left(\sqrt[n]{a^n+b^n}\right)=g_2\left(a\right)g_3\left(b\right)=0,a,b\in \mathbb{R},$$

which means that $g_1\left(d\right)=0$ for $d\in \{a^n+b^n:a,b\in \mathbb{R}\}=P_0$. Analogously, if $g_3\left(\mathbb{R}\right)=\left\{0\right\}$, then $g_1\left(a\right)=0$ for $a\in P_0$.

On the other hand, if $g_1\left(P_0\right)=\left\{0\right\}$ and $g_3\left(\mathbb{R}\right)=\left\{0\right\}$, then Equation (56) is satisfied with any function $g_2:\mathbb{R}\to \mathbb{R}$. Analogously, if $g_1\left(P_0\right)=\left\{0\right\}$ and $g_2\left(\mathbb{R}\right)=\left\{0\right\}$, then Equation (30) is satisfied with any function $g_3:\mathbb{R}\to \mathbb{R}$.

Therefore, we obtain the following conclusion.

Corollary 7. 

Let $g_1,g_2,g_3:\mathbb{R}\to \mathbb{R}$ and $g_i\left(\mathbb{R}\right)=\left\{0\right\}$ for some $i\in \{2,3\}$. Then, $g_1$, $g_2$, and $g_3$ fulfill functional Equation (56) if and only if $g_1\left(a\right)=0$ for $a\in P_0$.

The following generalized pexiderization of (3):

$$\begin{array}{c}\hfill g_1\left(\sqrt{a^n+b^n}\right)={\displaystyle \frac{g_2\left(a\right)g_3\left(b\right)}{g_2\left(a\right)+g_3\left(b\right)}}\end{array}$$

(65)

is a radical version of the equation:

$$\begin{array}{c}\hfill h_1\left(a+b\right)={\displaystyle \frac{h_2\left(a\right)h_3\left(b\right)}{h_2\left(a\right)+h_3\left(b\right)}}.\end{array}$$

(66)

In this section, we consider (65) only for $g_1,g_2,g_3:[0,\infty )\to (0,\infty )$ (a slightly more involved situation is studied in the next section). Then, the assumptions of Corollary 1 are valid, and consequently $g_1$, $g_2$, and $g_3$ fulfill functional Equation (65) if and only if there exist solutions $f_1,f_2,f_3:[0,\infty )\to (0,\infty )$ of Equation (66) such that

$$\begin{array}{c}\hfill g_i\left(a\right)=f_i\left(a^n\right),a\in [0,\infty ),i=1,2,3.\end{array}$$

(67)

Note that (66) can be rewritten as

$$\begin{array}{c}\hfill {\displaystyle \frac{1}{f_1\left(a+b\right)}}={\displaystyle \frac{1}{f_2\left(a\right)}}+{\displaystyle \frac{1}{f_3\left(b\right)}},\end{array}$$

(68)

which actually is (31) with $h_i\left(a\right)=1/f_i\left(a\right)$ for $a\in [0,\infty )$, $i=1,2,3$. Clearly, $h_1$, $h_2$, and $h_3$ also map $[0,\infty )$ into $(0,\infty )$.

As before, we find that $h_1,h_2,h_3:[0,\infty )\to (0,\infty )$ satisfy (31) if and only if they have the form

$$\begin{array}{c}\hfill h_i\left(a\right)=h\left(a\right)+c_i,a\in [0,\infty ),i=1,2,3,\end{array}$$

(69)

where $h:[0,\infty )\to \mathbb{R}$ is additive (i.e., a solution to (32)) and $c_1,c_2,c_3\in \mathbb{R}$ are constants with $c_1=c_2+c_3$ (see, e.g., [24], Ch. 4.3, p. 43, Theorem 9). It is clear that h must be bounded from below, so (see, e.g., [24], Ch. 2.1, pp. 15 and 18, Corollary 9) there is a real constant $\alpha \ge 0$ such that $h\left(a\right)=\alpha a$ for $a\in [0,\infty )$, which implies that $c_i>0$ for $i=1,2,3$. Hence,

$$\begin{array}{c}\hfill h_i\left(a\right)=\alpha a+c_i,a\in [0,\infty ),i=1,2,3,\end{array}$$

(70)

and consequently,

$$\begin{array}{c}\hfill g_i\left(a\right)={\displaystyle \frac{1}{\alpha a^n+c_i}},a\in [0,\infty ),i=1,2,3.\end{array}$$

(71)

In this way, we obtain the following.

Corollary 8. 

Functions $g_1,g_2,g_3:[0,\infty )\to (0,\infty )$ fulfill functional Equation (65) if and only if there exist $c_1,c_2,c_3\in (0,\infty )$ and $\alpha \in [0,\infty )$ such that $c_1=c_2+c_3$ and (71) holds.

Example 2. 

If we consider Equation (65) for functions $g_1,g_2,g_3:[0,\infty )\to \mathbb{R}$ such that $g_2\left(a\right)+g_3\left(b\right)\ne 0$ for every $a,b\in \mathbb{R}$, they must not necessarily be continuous, as in (71). For instance, for $n=2$, they may have the form of

$$\begin{array}{c}\hfill g_i\left(a\right)={\displaystyle \frac{1}{h\left(a^2\right)+c_i}},a\in [0,\infty ),i=1,2,3,\end{array}$$

(72)

where $h:\mathbb{R}\to \mathbb{R}$ is an injective discontinuous additive function (i.e., an injective discontinuous solution to functional Equation (32)) such that $h\left(\mathbb{R}\right)\ne \mathbb{R}$ and $c_1,c_2,c_3\in \mathbb{R}\setminus h\left(\mathbb{R}\right)$ satisfy the condition $c_1=c_2+c_3$. Then, $h\left(a\right)+c_i\ne 0$ for $a\in [0,\infty )$, $i=1,2,3$, which means that (72) makes sense. It follows from the results in [26] that such additive functions h exist.

Remark 5. 

As we mentioned in the Introduction, the equation

$$\begin{array}{c}\hfill f\left(\sqrt{a^2+b^2}\right)={\displaystyle \frac{f\left(a\right)f\left(b\right)}{f\left(a\right)+f\left(b\right)}}\end{array}$$

(73)

has been studied in [16] for nonzero functions $f:(0,\infty )\to \mathbb{R}$. This has not been clearly stated in [16], but we must exclude in (73) all $a,b\in (0,\infty )$ such that $f\left(a\right)+f\left(b\right)=0$, which means that, actually, we consider the following conditional equation:

$$\begin{array}{c}\hfill f\left(\sqrt{a^2+b^2}\right)={\displaystyle \frac{f\left(a\right)f\left(b\right)}{f\left(a\right)+f\left(b\right)}},f\left(a\right)+f\left(b\right)\ne 0.\end{array}$$

(74)

In [16] (Theorem 3.1), it has been stated that every nonzero solution $f:(0,\infty )\to \mathbb{R}$ of (73), such that the limit

$$\begin{array}{c}\hfill \underset{a\to 0+}{lim}f\left(a\right)a^2\end{array}$$

(75)

is finite, has the form

$$\begin{array}{c}\hfill f\left(a\right)={\displaystyle \frac{c}{a^2}},a\in (0,\infty ),\end{array}$$

(76)

with some real constant c.

It is not clear in this statement what does a nonzero solution means. At first, we could think that the authors assume, in this way, that $f\left(a\right)\ne 0$ for some $a\in (0,\infty )$. But simple calculations show that the function $h:(0,\infty )\to \mathbb{R}$ given by

$$h\left(a\right)=\left\{\begin{array}{cc}{a}^{-1}\hfill & ifa\in \mathbb{N};\hfill \\ 0\hfill & otherwise,\hfill \end{array}\right.$$

satisfies the conditional equation

$$\begin{array}{c}\hfill h(a+b)={\displaystyle \frac{h\left(a\right)h\left(b\right)}{h\left(a\right)+h\left(b\right)}},h\left(a\right)+h\left(b\right)\ne 0,\end{array}$$

(77)

where $f:(0,\infty )\to \mathbb{R}$ is given by $f\left(a\right)=h\left(a^2\right)$ for $a\in (0,\infty )$, is a solution to (74). Therefore, the assumption that f is a nonzero solution in [16] (Theorem 3.1) must mean, in particular, that $f\left(a\right)\ne 0$ for every $a\in (0,\infty )$.

We end this section with results on solutions of (9), i.e., of the following pexiderized radical version of the sine equation:

$$\begin{array}{c}\hfill g_1{\left(\sqrt[n]{{\displaystyle \frac{1}{2}}\left(a^n+b^n\right)}\right)}^2-g_1{\left(\sqrt[n]{{\displaystyle \frac{1}{2}}(a^n-b^n)}\right)}^2=g_2\left(a\right)g_3\left(b\right),\end{array}$$

(78)

of which some particular cases (with $g_1=g_2$, $g_1=g_3$, or $g_1=g_2=g_3$) have been studied, e.g., in [15] for $g_1,g_2,g_3:\mathbb{R}\to \mathbb{C}$ and odd $n\in \mathbb{N}$.

For instance, Lemma 1 in [15] states the following.

Proposition 1. 

Let $n\in \mathbb{N}$ be odd. If $g_1,g_2,g_3:\mathbb{R}\to \mathbb{C}$ satisfy (78), then, as one of the solutions of (78), $g_1$, $g_2$, and $g_3$ have the forms of $g_1\left(x\right)=cos\left(x^n\right)$, $g_2\left(x\right)=sin\left(x^n\right)$, and $g_3\left(x\right)=-sin\left(x^n\right)$ for all $x\in \mathbb{R}$.

The authors of [15] actually has proven the converse, i.e., they have shown that if $g_1,g_2,g_3:\mathbb{R}\to \mathbb{C}$ have the forms specified in Proposition 1, then they fulfill (78).

However, if $\alpha ,\beta ,\gamma$ are complex numbers with ${\alpha }^2=\beta \gamma$ and $g_1\left(x\right)=\alpha cos\left(x^n\right)$, $g_2\left(x\right)=\beta sin\left(x^n\right)$, $g_3\left(x\right)=-\gamma sin\left(x^n\right)$ for $x\in \mathbb{R}$, then clearly, $g_1$, $g_2$, and $g_3$ are also solutions to (78). Moreover, it is easy to check that for $g_1\left(x\right)=\alpha cosh\left(x^n\right)$, $g_2\left(x\right)=\beta sinh\left(x^n\right)$ and $g_3\left(x\right)=\gamma sinh\left(x^n\right)$ for $x\in \mathbb{R}$, Equation (78) is also satisfied, (cf. [15], Lemma 2).

Therefore, Proposition 1 (i.e., [15], Lemma 1) only gives examples of solutions to (78), and it seems that the authors of [15] probably had the following formulation in mind: if $g_1,g_2,g_3:\mathbb{R}\to \mathbb{C}$ have the forms of $g_1\left(x\right)=cos\left(x^n\right)$, $g_2\left(x\right)=sin\left(x^n\right)$, and $g_3\left(x\right)=-sin\left(x^n\right)$ for all $x\in \mathbb{R}$, then they satisfy Equation (78); or eventually: if $g_1,g_2,g_3:\mathbb{R}\to \mathbb{C}$ satisfy (78), then $g_1$, $g_2$, and $g_3$ can have the forms: $g_1\left(x\right)=cos\left(x^n\right)$, $g_2\left(x\right)=sin\left(x^n\right)$, and $g_3\left(x\right)=-sin\left(x^n\right)$ for all $x\in \mathbb{R}$. The same is true for [15] (Lemmas 2 and 3).

Below, we provide a description of solutions to (78) that can be derived from Theorem 2. For the sake of simplicity, we do it only for odd n values, but it can also be easily done for even n.

Corollary 9. 

Let n be odd. Functions $g_1,g_2,g_3:\mathbb{R}\to \mathbb{C}$ fulfill functional Equation (78) if and only if there exist functions $h_1,h_2,h_3:\mathbb{R}\to \mathbb{C}$ satisfying the equation

$$\begin{array}{c}\hfill h_1{({\displaystyle \frac{1}{2}}\left(a+b\right))}^2-h_1({\displaystyle \frac{1}{2}}{(a-b))}^2=h_2\left(a\right)h_3\left(b\right)\end{array}$$

(79)

such that $g_i\left(a\right)=h_i\left(a^n\right)$ for $a\in \mathbb{R}$, $i=1,2,3$.

Clearly, (79) is a pexiderization of the so-called sine equation, i.e.,

$$\begin{array}{c}\hfill h{({\displaystyle \frac{1}{2}}\left(a+b\right))}^2-h({\displaystyle \frac{1}{2}}{(a-b))}^2=h\left(a\right)h\left(b\right),\end{array}$$

(80)

which is sometimes better known (see, e.g., [25], Ch. 15) in the following modified form (with $a=x+y$ and $b=x-y$):

$$\begin{array}{c}\hfill h(x+y)h(x-y)=h{\left(x\right)}^2-h{\left(y\right)}^2.\end{array}$$

(81)

The sine function and the hyperbolic sine function satisfy Equations (80) and (81) (see [25] (Ch. 15) for fuller descriptions of solutions).

It is easy to check that every solution of (32) fulfills (81), i.e., (79) with $h_1=h_2=h_3$. Moreover, it follows from [25] (Corollary 15.5) (cf. [27]) that the only other solutions $h:\mathbb{R}\to \mathbb{C}$ of (81) have the forms of $h\left(a\right)=c\left(f\right(a)-f(-a\left)\right)$ for $a\in \mathbb{R}$, where c is a complex constant and $f:\mathbb{R}\to \mathbb{C}$ is a solution to Equation (58).

Finally, observe that if $h_1:\mathbb{R}\to \mathbb{C}$ is a solution to (58), then (79) holds for all $a,b\in \mathbb{R}$ with $h_2=h_1$ and $h_3\left(b\right)=h_1\left(b\right)-h_1(-b)$ for $b\in \mathbb{R}$. Some further information on solutions to (79) can also be found in [28].

All this shows that the description in Proposition 1 (i.e., in [15], Lemma 1) is not complete.

4. Further General Results on Solutions of Equations of Radical Type

In this section, S is a nonempty set, $(Y,★)$ is a groupoid (i.e., Y is a nonempty set endowed with a binary operation $★:Y^2\to Y$), $(W,\diamond )$ is a cancellative groupoid (i.e., $x=y$ for every $x,y,z$ such that $x\diamond z=y\diamond z$ or $z\diamond x=z\diamond y$), $\mathrm{\Pi }:S\to Y$, $P_0:=\mathrm{\Pi }\left(S\right)$, and $p:P_0\to S$ is a selection of $\mathrm{\Pi }$ (i.e., $\mathrm{\Pi }\left(p\right(x\left)\right)=x$ for $x\in P_0$).

We show how to apply a particular case of Theorem 2.1 in [21] and prove a very simple generalization of it. So, consider Equation (11), which is the conditional functional equation

$$f\left(p\right(\mathrm{\Pi }\left(x\right)★\mathrm{\Pi }\left(y\right)\left)\right)=f\left(x\right)\diamond f\left(y\right),\mathrm{\Pi }\left(x\right)★\mathrm{\Pi }\left(y\right)\in P,$$

(82)

for functions $f:S\to W$, where $P\subset P_0$ is nonempty.

The next theorem is a very simplified version of the main result in [21] (Theorem 2.1).

Theorem 3. 

Let $P\subset P_0$ be nonempty and

$$P\cap (x★P)\ne \varnothing ,P\cap (P★x)\ne \varnothing ,x\in P_0.$$

(83)

Then, $f:S\to W$ satisfies conditional Equation (82) if and only if there exists a solution $A:P_0\to W$ of the conditional equation

$$A(u★v)=A\left(u\right)\diamond A\left(v\right),u★v\in P,$$

(84)

such that $f=A\circ \mathrm{\Pi }$. Moreover, such an A is unique, and $A=f\circ p$.

Now, we show an application of Theorem 3 in finding solutions of a conditional version of a generalization of Equation (3), i.e., of the equation

$$\begin{array}{c}\hfill g\left(\sqrt[n]{a^n+b^n}\right)={\displaystyle \frac{g\left(a\right)g\left(b\right)}{g\left(a\right)+g\left(b\right)}},a+b>1,\end{array}$$

(85)

for functions $g:{\mathbb{R}}_0\to (0,\infty )$, where $n\in \mathbb{N}$, $n>1$ is fixed, and ${\mathbb{R}}_0:=\mathbb{R}\setminus \left\{0\right\}$. As mentioned in the introduction, Equation (3) was considered in [16] for $n=2$ and functions $g:(0,\infty )\to \mathbb{R}$.

Therefore, let $g:{\mathbb{R}}_0\to (0,\infty )$ be a solution to (85). Define $f:{\mathbb{R}}_0\to (0,\infty )$ by $f\left(a\right)=1/g\left(a\right)$ for $a\in {\mathbb{R}}_0$. Then, f satisfies

$$\begin{array}{c}\hfill f\left(\sqrt[n]{a^n+b^n}\right)=f\left(a\right)+f\left(b\right),a+b>1.\end{array}$$

According to Theorem 3 with $S={\mathbb{R}}_0$, $\mathrm{\Pi }\left(a\right)=a^n$ for $a\in {\mathbb{R}}_0$, $P=(1,\infty )$, and $p\left(b\right)=\sqrt[n]{b}$ for $a\in P_0$, there exists a solution $A:P_0\to (0,\infty )$ of the conditional equation

$$A(u+v)=A\left(u\right)+A\left(v\right),u+v>1,$$

(86)

such that

$$f\left(a\right)=A\left(a^n\right),a\in {\mathbb{R}}_0.$$

(87)

Let $A_0$ be a restriction of A to $(1,\infty )$. Clearly, $A_0$ is a solution to the equation

$$A_0(u+v)=A_0\left(u\right)+A_0\left(v\right),$$

and $A_0$ is bounded from below (because A is). Hence, according to [29] (Theorem 2), there is a solution $A_1:\mathbb{R}\to \mathbb{R}$ of the equation

$$A_1(u+v)=A_1\left(u\right)+A_1\left(v\right)$$

such that $A_0\left(a\right)=A_1\left(a\right)$ for $a>1$. Since $A_1$ is bounded from below on $(1,\infty )$, there is a real c such that $A_1\left(a\right)=ca$ for $a\in \mathbb{R}$ (see, e.g., [24], Ch. 2.1, pp. 15 and 18, Corollary 9), and consequently, $A_0\left(a\right)=ca$ for $a\in (1,\infty )$.

Let $a\in P_0$ be fixed. There is $b>1$ with $a+b>1$. Therefore, according to (86), $A\left(a\right)+cb=A\left(a\right)+A_0\left(b\right)=A\left(a\right)+A\left(b\right)=A(a+b)=A_0(a+b)=c(a+b)$, which implies that $A\left(a\right)=ca$. Thus, in view of (87), we have shown that $f\left(a\right)=A\left(a^n\right)=c{a}^n$ for $a\in {\mathbb{R}}_0$. This implies that

$$g\left(a\right)={\displaystyle \frac{1}{c{a}^n}},a\in {\mathbb{R}}_0.$$

(88)

Clearly, if n is odd, then $g\left({\mathbb{R}}_0\right)\not\subset (0,\infty )$. If n is even, then $g\left({\mathbb{R}}_0\right)\subset (0,\infty )$ only for $c>0$. This means that we have obtained the following.

Corollary 10. 

If n is odd, then conditional Equation (85) does not have solutions $g:{\mathbb{R}}_0\to (0,\infty )$.

If n is even, then a function $g:{\mathbb{R}}_0\to (0,\infty )$ satisfies conditional Equation (85) if and only if there exists a real constant $c>0$ such that (88) holds.

Now, we prove the following pexiderized version of Theorem 3.

Theorem 4. 

Let $P\subset P_0$ be nonempty and (83) be valid. Then, $f_1,f_2,f_3:S\to W$ satisfy the conditional equation

$$f_1\left(p(\mathrm{\Pi }\left(a\right)★\mathrm{\Pi }\left(b\right))\right)=f_2\left(a\right)\diamond f_3\left(b\right),\mathrm{\Pi }\left(a\right)★\mathrm{\Pi }\left(b\right)\in P,$$

(89)

if and only if there exist $A_1,A_2,A_3:P_0\to W$ fulfilling the conditional equation

$$A_1(u★v)=A_2\left(u\right)\diamond A_3\left(v\right),u\diamond v\in P,$$

(90)

such that

$$f_1\left(s\right)=A_1\left(\mathrm{\Pi }\left(s\right)\right),s\in p\left(P_0\right),f_i=A_i\circ \mathrm{\Pi },i=2,3.$$

(91)

Moreover, $A_1$, $A_2$, and $A_3$ are unique, and $A_i=f_i\circ p$ for $i=1,2,3$.

Proof. 

Assume that $f_1$, $f_2$, and $f_3$ fulfill (89). Write $A_i=f_i\circ p$ for $i=1,2,3$. We show that (90) holds. To this end, take $u,v\in P_0$ with $u★v\in P$. Then,

$$\begin{array}{c}\hfill A_1(u★v)=f_1\left(p(u★v)\right)=f_1\left(p(\mathrm{\Pi }\left(p\left(u\right)\right)★\mathrm{\Pi }\left(p\left(v\right)\right))\right)\\ \hfill =f_2\left(p\left(u\right)\right)\diamond f_3\left(p\left(v\right)\right)=A_2\left(u\right)\diamond A_3\left(v\right).\end{array}$$

Next, it is easily seen that $f_1\left(p\left(r\right)\right)=A_1\left(r\right)=A_1(\mathrm{\Pi }\left(p\left(r\right)\right)$ for $r\in P_0$, which yields $f_1\left(s\right)=A_1\left(\mathrm{\Pi }\left(s\right)\right)$ for $s\in p\left(P_0\right)$.

Furthermore, let $s,t\in S$ be such that $\mathrm{\Pi }\left(s\right)=\mathrm{\Pi }\left(t\right)$. Clearly, (83) implies that there exists $u,v\in P$ with

$$v★\mathrm{\Pi }\left(s\right),\mathrm{\Pi }\left(s\right)★u\in P.$$

Fix $a,b\in S$ with $u=\mathrm{\Pi }\left(a\right)$ and $v=\mathrm{\Pi }\left(b\right)$. Then, according to (82),

$$\begin{array}{c}\hfill f_2\left(s\right)\diamond f_3\left(a\right)=f_1\left(p(\mathrm{\Pi }\left(s\right)★\mathrm{\Pi }\left(a\right))\right)=f_1\left(p(\mathrm{\Pi }\left(t\right)★\mathrm{\Pi }\left(a\right))\right)=f_2\left(t\right)\diamond f_3\left(a\right),\end{array}$$

$$\begin{array}{c}\hfill f_2\left(b\right)\diamond f_3\left(s\right)=f_1\left(p(\mathrm{\Pi }\left(b\right)★\mathrm{\Pi }\left(s\right))\right)=f_1\left(p(\mathrm{\Pi }\left(b\right)★\mathrm{\Pi }\left(t\right))\right)=f_2\left(b\right)\diamond f_3\left(t\right),\end{array}$$

and consequently, $f_i\left(s\right)=f_i\left(t\right)$ for $i=1,2$, because groupoid $(W,\diamond )$ is cancellative. In this way, we have shown that

$$\begin{array}{c}\hfill f_i\left(s\right)=f_i\left(t\right),s,t\in S,\mathrm{\Pi }\left(s\right)=\mathrm{\Pi }\left(t\right),i=2,3,\end{array}$$

(92)

where

$$\begin{array}{c}\hfill f_i\left(s\right)=f_i\left(p\left(\mathrm{\Pi }\left(s\right)\right)\right)=A_i\left(\mathrm{\Pi }\left(s\right)\right),s\in S,i=2,3.\end{array}$$

(93)

Consequently, (91) holds.

Now, let $A_1,A_2,A_3:P\to W$ fulfill (90) and (91) be valid. Fix $a,b\in S$ with $\mathrm{\Pi }\left(a\right)★\mathrm{\Pi }\left(b\right)\in P$. Then,

$$\begin{array}{cc}\hfill f_1\left(p(\mathrm{\Pi }\left(a\right)★\mathrm{\Pi }\left(b\right))\right)=& A_1\circ \mathrm{\Pi }\left(p(\mathrm{\Pi }\left(a\right)★\mathrm{\Pi }\left(a\right))\right)=A_1(\mathrm{\Pi }\left(a\right)★\mathrm{\Pi }\left(b\right))\hfill \\ \hfill =& A_2\left(\mathrm{\Pi }\left(a\right)\right)\diamond A_3\left(\mathrm{\Pi }\left(b\right)\right)=f_2\left(a\right)\diamond f_3\left(b\right),\hfill \end{array}$$

which means that $f_1$, $f_2$, and $f_3$ satisfy Equation (89).

Finally, observe that (91) implies

$$A_i\left(a\right)=A_i\left(\mathrm{\Pi }\left(p\left(a\right)\right)\right)=f_i\left(p\left(a\right)\right),a\in P_0,i=1,2,3,$$

whence we obtain the form and the uniqueness of $A_1$, $A_2$, and $A_3$. □

We have the following two simple examples of applications of Theorem 4.

Example 3. 

Let $(Y,★)$ be group $(\mathbb{Z},+)$ and $(W,\diamond )$ be group $(\mathbb{R},+)$, $\mathrm{\Pi }\left(a\right)=\left|a\right|$ for $a\in \mathbb{Z}$, $P=P_0={\mathbb{N}}_0$, and $p\left(a\right)=a$ for $a\in {\mathbb{N}}_0$. Then, Equation (89) takes the form of

$$f_1\left(\right|a|+\left|b\right|)=f_2\left(a\right)+f_3\left(b\right).$$

(94)

Let $f_1,f_2,f_3:\mathbb{Z}\to \mathbb{R}$ be solutions to Equation (94). According to Theorem 4, there exist solutions $A_1,A_2,A_3:{\mathbb{N}}_0\to \mathbb{R}$ of the equation

$$A_1(u+v)=A_2\left(u\right)+A_3\left(v\right)$$

(95)

such that

$$f_1\left(s\right)=A_1\left(s\right),s\in {\mathbb{N}}_0,f_i=A_i\left(\right|s\left|\right),s\in \mathbb{Z},i=2,3.$$

(96)

On account of [24] (Ch. 4.3, p. 43, Theorem 9), $A_1$, $A_2$, and $A_3$ must have the forms of

$$\begin{array}{c}\hfill A_i\left(a\right)=A\left(a\right)+c_i,a\in {\mathbb{N}}_0,i=1,2,3,\end{array}$$

(97)

where $A:{\mathbb{N}}_0\to \mathbb{R}$ is additive (i.e., a solution to (32)) and $c_1$, $c_2$, and $c_3\in \mathbb{R}$ are constants with $c_1=c_2+c_3$. Furthermore, it is easily seen that $A\left(a\right)=c_{0}a$ for $a\in {\mathbb{N}}_0$, where $c_0=A\left(1\right)$ can be any real number. Consequently,

$$f_1\left(s\right)=c_{0}s+c_1,s\in {\mathbb{N}}_0,f_i=c_0\left|s\right|+c_i,s\in \mathbb{Z},i=2,3.$$

(98)

Example 4. 

Let $(Y,★)$ be semigroup $\left(\right[0,\infty ),+)$, $(W,\diamond )$ be a group, $P=P_0=[0,\infty )$, and $\mathrm{\Pi }\left(a\right)=\sqrt{a}$ and $p\left(a\right)=a^2$ for $a\in [0,\infty )$. Then, Equation (89) takes the form

$$f_1\left(x+y+2\sqrt{xy}\right)=f_2\left(x\right)\diamond f_3\left(y\right)$$

(99)

for $f_1,f_2,f_3:[0,\infty )\to W$, which is a generalization of Equation (12).

Using [24] (Ch. 4.3, p. 43, Theorem 9) and arguing analogously as in Example 3, we find that $f_1,f_2,f_3:[0,\infty )\to W$ satisfy Equation (99) if and only if there exist constants $c_1,c_2\in W$ and a solution $A:[0,\infty )\to W$ to the equation

$$A(x+y)=A\left(x\right)\diamond A\left(y\right)$$

such that

$$\begin{array}{c}\hfill f_1\left(s\right)=c_1\diamond A\left(\sqrt{s}\right)\diamond c_2,f_2\left(s\right)=c_1\diamond A\left(\sqrt{s}\right),f_3\left(s\right)=A\left(\sqrt{s}\right)\diamond c_2,s\in [0,\infty ).\end{array}$$

Below, we provide two further (less trivial) examples of the application of Theorem 4 to the pexiderized versions of conditional Equations (13) and (14), i.e., to the equations

$$\begin{array}{c}\hfill g_1(\lfloor x\rfloor +\lfloor y\rfloor )=g_2\left(x\right)+g_3\left(y\right),\lfloor x\rfloor +\lfloor y\rfloor \in \mathbb{N},\end{array}$$

(100)

$$\begin{array}{c}\hfill g_1(\left\{x\right\}+\left\{y\right\})=g_2\left(x\right)+g_3\left(y\right),\left\{x\right\}+\left\{y\right\}\in [0,1),\end{array}$$

(101)

for $g_1,g_2,g_3:\mathbb{R}\to \mathbb{R}$, where $\lfloor x\rfloor$ denotes the greatest integer less than or equal to a real number (x) and $\left\{x\right\}:=x-\lfloor x\rfloor$.

Let us start with Equation (100). Then, according to Theorem 4 with $S=W=\mathbb{R}$, $\mathrm{\Pi }\left(x\right):=\lfloor x\rfloor$ for $x\in S$, $\mathrm{\Pi }\left(\mathbb{R}\right)=\mathbb{Z}$, $P=\mathbb{N}$ and $p\left(u\right)=u$ for $u\in \mathbb{Z}$, there exist $A_1,A_2,A_3:\mathbb{Z}\to \mathbb{R}$ fulfilling the conditional equation

$$A_1(u+v)=A_2\left(u\right)+A_3\left(v\right),u+v\in \mathbb{N},$$

(102)

such that

$$g_i=A_i\circ \mathrm{\Pi },i=1,2,3.$$

(103)

Let $k_i=A_i\left(0\right)$ for $i=2,3$ and $k_1=k_2+k_3$. Then, according to (102) (with $u=0$ or $v=0$),

$$\begin{array}{c}\hfill A_1\left(u\right)=A_2\left(u\right)+k_3=k_2+A_3\left(u\right),u\in \mathbb{N}.\end{array}$$

(104)

Hence, (102) yields

$$A_1(u+v)=A_1\left(u\right)+A_1\left(v\right)-k_1,u,v\in \mathbb{N}.$$

Consequently, the function $A_0:\mathbb{N}\to \mathbb{R}$, given by $A_0\left(u\right):=A_1\left(u\right)-k_1$ for $u\in \mathbb{N}$, satisfies the following equation:

$$A_0(u+v)=A_0\left(u\right)+A_0\left(v\right).$$

Clearly, (104) implies that

$$\begin{array}{c}\hfill A_i\left(u\right)=k_i+A_0\left(u\right),u\in \mathbb{N},i=1,2,3.\end{array}$$

(105)

Next, it is easy to show by induction that

$$A_0\left(u\right)=cu,u\in \mathbb{N},$$

(106)

where $c=A_0\left(1\right)$. Hence, according to (105),

$$\begin{array}{c}\hfill A_i\left(u\right)=k_i+cu,u\in \mathbb{N},i=1,2,3.\end{array}$$

(107)

Fix $v\in \mathbb{Z}$ and take $u\in \mathbb{N}$ with $u+v\in \mathbb{N}$. Then, (102) and (107) yield

$$\begin{array}{c}\hfill c(u+v)+k_1=A_1(u+v)=k_2+cu+A_3\left(v\right),\\ \hfill c(v+u)+k_1=A_1(v+u)=A_2\left(v\right)+k_3+cu,\end{array}$$

whence we find that $A_2\left(v\right)=k_2+cv$ and $A_3\left(v\right)=k_3+cv$ for $v\in \mathbb{Z}$. Therefore, according to (103),

$$g_1\left(x\right)=c\lfloor x\rfloor +k_1,g_i\left(y\right)=c\lfloor y\rfloor +k_i,x,y\in \mathbb{R},x\ge 1,i=2,3.$$

(108)

Since it is easy to check that every $g_1,g_2,g_3:\mathbb{R}\to \mathbb{R}$ given by (108), with any real c, satisfies Equation (100), we obtain the following.

Corollary 11. 

Functions $g_1,g_2,g_3:\mathbb{R}\to \mathbb{R}$ satisfy conditional Equation (100) if and only if there exist $c,k_1,k_2,k_3\in \mathbb{R}$ such that $k_1=k_2+k_3$ and (108) holds.

Analogously, one can obtain the following similar result for the equation

$$\begin{array}{c}\hfill g_1(\lceil x\rceil +\lceil y\rceil )=g_2\left(x\right)+g_3\left(y\right),\lceil x\rceil +\lceil y\rceil \in \mathbb{N},\end{array}$$

(109)

with $g_1,g_2,g_3:\mathbb{R}\to \mathbb{R}$, where $\lceil x\rceil$ denotes the least integer greater than or equal to the given real number x.

Corollary 12. 

Functions $g_1,g_2,g_3:\mathbb{R}\to \mathbb{R}$ satisfy conditional Equation (109) if and only if there exist $c,k_1,k_2,k_3\in \mathbb{R}$ such that $k_1=k_2+k_3$ and

$$g_1\left(x\right)=c\lceil x\rceil +k_1,x\in \mathbb{R},x\ge 1,g_i\left(y\right)=c\lceil y\rceil +k_i,y\in \mathbb{R},i=2,3.$$

Now, consider Equation (101). Then, according to Theorem 4 with $f_i=g_i$ for $i=1,2,3$, $\mathrm{\Pi }\left(x\right)=\left\{x\right\}:=x-\lfloor x\rfloor$ for $x\in \mathbb{R}$, $\mathrm{\Pi }\left(\mathbb{R}\right)=[0,1)$, $P=P_0=[0,1)$, and $p\left(x\right)=x$ for $x\in [0,1)$, there exists $A_1,A_2,A_3:[0,1)\to \mathbb{R}$ fulfilling the conditional equation

$$A_1(u+v)=A_2\left(u\right)+A_3\left(v\right),u+v\in [0,1),$$

(110)

such that

$$g_1=A_1\left(\mathrm{\Pi }\left(s\right)\right),s\in [0,1),g_i=A_i\circ \mathrm{\Pi },i=2,3.$$

(111)

Let $k_i=A_i\left(0\right)$ for $i=1,2,3$. Then, according to (110) (with $u=0$ and/or $v=0$), $k_1=k_2+k_3$ and

$$\begin{array}{c}\hfill A_1\left(u\right)=A_2\left(u\right)+k_3=k_2+A_3\left(u\right),u\in [0,1).\end{array}$$

Hence, (110) yields

$$A_1(u+v)=A_1\left(u\right)+A_1\left(v\right)-k_1,u,v\in [0,1),u+v\in [0,1).$$

Consequently, the function ($A:[0,1)\to \mathbb{R}$) given by $A\left(a\right):=A_1\left(a\right)-k_1$ for $a\in [0,1)$ satisfies the following equation:

$$A_0(u+v)=A_0\left(u\right)+A_0\left(v\right),u+v\in [0,1).$$

Clearly,

$$\begin{array}{c}\hfill A_i\left(u\right)=k_i+A\left(a\right),u\in [0,1),i=1,2,3.\end{array}$$

Next, it follows from [30] (Lemma 1.2) and [29] (Theorem 2) that there exists a solution $A:\mathbb{R}\to \mathbb{R}$ of the equation expressed as

$$A(u+v)=A\left(u\right)+A\left(v\right)$$

(112)

such that $A_0\left(u\right)=A\left(u\right)$ for $u\in [0,1)$. Clearly,

$$g_1\left(v\right)=A\left(v\right)+k_1,v\in [0,1),g_i\left(u\right)=A\left(\left\{u\right\}\right)+k_i,u\in \mathbb{R},i=2,3.$$

(113)

Since it is easy to check that every $g_1,g_2,g_3:\mathbb{R}\to \mathbb{R}$ fulfilling (113) are solutions of Equation (101), we obtain the following.

Corollary 13. 

Functions $g_1,g_2,g_3:\mathbb{R}\to \mathbb{R}$ satisfy conditional Equation (101) if and only if there exist an additive $A:\mathbb{R}\to \mathbb{R}$ and $k_1,k_2,k_3\in \mathbb{R}$ such that $k_1=k_2+k_3$ and (113) holds.

5. Conclusions

So-called functional equations of the radical type have been studied by many authors in various situations. An example of such an equations is

$$g\left(\sqrt{a^2+b^2}\right)=g\left(a\right)+g\left(b\right),$$

which can be considered, e.g., for real-valued functions with real variables. Authors sometimes provide some information about the solutions of these equations, although not always in a clear or complete form. In this paper, we provide simple descriptions of solutions to very general functional equations of the radical type and (with appropriate examples) show how to apply them to particular cases of such equations. This research could help in easy correction of such information and avoidance of similar problems in the future. Moreover, this paper shows that there is a significant symmetry between solutions of various functional equations and solutions of equations of the radical type that correspond to them.