A Classical Functional Equation - Full Version

If $A$ is a proper subset of $\mathbb{R}$ that contains $0$ , it is easy to see that the function $F_A(x,y) \; = \; \left\{ \begin{array}{lll} 1 & \hspace{1cm} & x,y \in A \; \mathrm{or} \; x,y \in A' \\ -1 & & \mathrm{o.w.} \end{array} \right.$ belongs to $\mathcal{F}$ , with $\mathrm{Im}\,F = \{1,-1\}$ , and that $A = \{x \in \mathbb{R} \,|\, F_A(x,0) = 1\}$ .

Suppose now that $F \in \mathcal{F}$ . If there exist $a,b \in \mathbb{R}$ such that $F(a,b) = 0$ , then $F(a,y) = F(a,b)F(b,y) = 0$ for all $y \in \mathbb{R}$ , and so $F(x,y) = F(x,a)F(a,y) = 0$ for all $x,y \in \mathbb{R}$ , so that $F \equiv 0$ , which would imply that $\mathrm{Im}\,F = \{0\}$ only had one element. Thus we deduce that $F(x,y) \neq 0$ for all $x,y \in \mathbb{R}$ . Since $F(x,x)F(x,y) = F(x,y)$ for all $x,y \in \mathbb{R}$ , we deduce that $F(x,x) = 1$ for all $x \in \mathbb{R}$ . Thus the set $\mathrm{Im}\,F$ must be equal to $\{1,u\}$ for some $u \neq 1$ .

Find $a,b \in \mathbb{R}$ such that $F(a,b) = u$ . Then $uF(b,a) = F(a,a) = 1$ , and hence $F(b,a) = u^{-1} \neq 1$ . Thus we deduce that $u = u^{-1}$ , so that $u=-1$ . Thus it follows that $\mathrm{Im}\,F = \{1,-1\}$ .

Define the function $G\,:\, \mathbb{R} \to \mathbb{R}$ by the formula $G(x) \; = \; F(x,0) \hspace{2cm} x \in \mathbb{R}$ Since $F(x,0) = F(x,y)F(y,0)$ for all $x,y \in \mathbb{R}$ , we deduce that $F(x,y) \; = \; \frac{G(x)}{G(y)} \hspace{2cm} x,y \in \mathbb{R}$ and it is clear that $\mathrm{Im}\,G = \{1,-1\}$ . If we define $A = G^{-1}\{1\}$ , then it is clear that $A$ is a proper subset of $\mathbb{R}$ that contains $0$ , and that $F(x,y) \; =\; \left\{ \begin{array}{lll} 1 & \hspace{1cm} & x,y \in A \; \mathrm{or}\; x,y \in A' \\ -1 & & \mathrm{o.w} \end{array} \right.$ so that $F = F_A$ .

It follows that $\mathcal{F}$ is in 1-1 correspondence with the collection of proper subsets of $\mathbb{R}$ that contain $0$ , which is in turn in 1-1 correspondence with the collection of subsets of $\mathbb{R}$ . Thus $\mathcal{F}$ is equicardinal with $2^\mathbb{R}$ .

If $A$ is a proper subset of $\R$ , define the function $f_A:\R\to\{-1,1\}$ such that $f_A(x)=1$ if $x\in A$ and $f_A(x)=-1$ if $x\notin A$ . Then it is easy to see that $F_A:\R^2\to\R$ such that $F_A(x,y)=f_A(x)f_A(y)$ is an element of $\mathcal{F}$ . It's clear that $|\{F_A\}|=|2^{\R}|$ , and since $\{F_A\}\subseteq\mathcal{F}$ , it follows that $|\mathcal{F}|\geq |2^{\R}|$ . On the other hand, $\mathcal{F}$ is a subset of the set of all functions from $\R^2\to\R$ , which is equicardinal with $2^{\R}$ , hence $|\mathcal{F}|\leq |2^{\R}|$ . We conclude, therefore, that $|\mathcal{F}|= |2^{\R}|$ .