All solutions to $f(a,b)+f(b,a)=0$

Question

I am looking for all functions $f:(0,1)^2\to\mathbb{R}$ that satisfy $f(a,b)+f(b,a)=0$ for every $(a,b)\in(0,1)^2$. I know that $f\equiv 0$, $f(a,b)=c(a-b)$, and $f(a,b)=\pm\log(a/b)$ are such examples. Are there any other solutions? If not, how can I prove it?

Answer 1

Let $$\Omega = \{(x, y)| x,y\in(0,1)\land x>y\}$$

and let $g:\Omega \to\mathbb R$ be any function.

Then, define

$$f(x,y)=\begin{cases} g(x,y);&x>y\\ -g(y,x);&x<y\\ 0;&x=y \end{cases}$$

then $f$ satisfies your condition.

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What is more, you can see that every function that satisfies your condition can be written in that form. In other words, if $F$ is a mapping which transforms $g$ into $f$, then the range of $F$ is the entire set of solutions to your problem.

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What is more more, you can see that $f$ is continuous if and only if $g$ is continuous and the limit of $g$ as you approach the line $x=y$ is $0$.

Answer 2

Take any function $g: \{(a,b): 0<a<1,0<b<1, a\leq b\}$ with the condition $g(a,a)=0$ and define $f(a,b)=g(a,b)$ if $ a\leq b$, $f(a,b)=-g(b,a)$ if $ a> b$. This gives all possible solutions since $f(a,a)=0$ necessarily.

Answer 3

A small variant of the other answers: If $g: (0, 1) \to \Bbb R$ is an arbitrary function then $$ f(a, b) = g(a, b) - g(b, a) $$ satisfies $f(a, b) + f(b, a) = 0$.

All solutions can be written in that way with $g(x, y) = f(x, y)/2$.

Your examples $f\equiv 0$, $f(a,b)=c(a-b)$, and $f(a,b)=\pm\log(a/b)$ are obtained in this way from $g(a, b) = 0$, $g(a, b) = ca$, and $g(a, b) = \pm \log(a)$, respectively.