Functional Equation for Two-Variable Functions

For any function $g\,:\mathbb{R} \to \mathbb{R}$ , the definition $F_g(x,y) \; = \; \left\{ \begin{array}{lll} 1 & \hspace{1cm} & x \neq y \\ g(x) & & x = y \end{array}\right.$ defines a function such that $(y-x)F_g(x,y) = y-x$ for all $x,y$ , and hence $F_g$ is a function with the desired property. It is clear that $F_g \neq F_h$ whenever $g \neq h$ , and hence that the desired cardinality is at least that of $\mathbb{R}^{\mathbb{R}}$ , so is certainly greater than that of $\mathbb{R}$ .

Let $G(x,y)=(y-x)F(x,y),$ then the given functional equation becomes $G(x,y)+G(y,z)=G(x,z).$ Let us prove that $G(x,y)=-G(y,x).$ Indeed, $G(x,z)=G(x,y)+G(y,z)=G(x,y)+G(y, x)+G(x,z).$ Therefore, $G(x,y)=-G(y,x).$

Now, we can get that $G(x, y)=G(x,0)+G(0,y)=G(0,y)-G(0,x).$ If we consider that $G(0, \cdot)$ is an arbitrary funcion $f$ from $\mathbf R$ to $\mathbf R,$ then we have obtained that $G(x, y)=f(y)-f(x).$ Therefore, the solutions of the given functional equations can be written in the form $F(x,y)=\frac{f(y)-f(x)}{y-x},$ where $f(x)$ represents an arbitrary function from $\mathbf R$ to $\mathbf R,$ and for any real number $x$ the value of $F(x,x)$ is whatever real number depending on $x$ .

Now it is easy to see that the cardinality of the sets of solutions of the given equation is greater than the cardinality of $2^\mathbf R$ that is greater than the cardinality of $\mathbf R.$ Actually, with any subset $A$ of $\mathbf R,$ we can associate a function $\chi_A$ by the formula $\chi_A(x) = \begin{cases} 1 & \text{if} \; x\in A \\ 0 & \text{if}\; x\notin A \\ \end{cases}$ Then we define the function $F_A(x,y)$ in the following way, $F_A(x,y) = \begin{cases} \frac{ \chi_A(y)- \chi_A(x) }{y-x} & \text{if} \; x\neq y \\ 0 & \text{if}\; x=y \\ \end{cases}$ Then, it is easy to see that if two subsets of the real numbers $A$ and $B$ are distinct, then $F_A \neq F_B.$ This implies that the cardinality of the set of solutions of the given equations is larger than or equal to the cardinality of the set of all subsets of $\mathbf R,$ that is represented by $2^\mathbf R,$ and in turn the cardinality $2^\mathbf R$ is larger than the cardinality of $\mathbf R.$ Using transitivity, it follows that the cardinality of the set of solutions is greater than the cardinality of $\mathbf R.$ Then the cardinality of the set of solutions of the given functional equation is larger than the cardinality of $\mathbf R.$