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CONTEXT: I was trying to find a method of solving the functional equation $$f(x,y)+f(y,x)=g(x,y) \tag{1}$$ for some given function $g$ satisfying $g(x,y)=g(y,x)$, with $f,g:\mathbb R^2\mapsto\mathbb R^2$. To solve it, I defined $$h(x,y):= f(x,y)-f(y,x)$$ so that I could solve this equation and the functional equation like a system, yielding $$f(x,y)=\frac{g(x,y)+h(x,y)}{2}$$ I noticed then that, as shown, $f$ must be in this form, and that $f$ satisfies the functional equation (1) as long as $h(x,y)=-h(y,x)$.

QUESTION: Is there a name for a function of two variables that satisfies the following? $$h(x,y)=-h(y,x)\tag{2}$$ What kind of properties are known about this type of function?

Most importantly, can this type of function be expressed in terms of other more commonly-known types of functions? For example, any involution from reals to reals can be written in the form $$a^{-1}(-a(x))$$ for some function $a$. Can I decompose $h$ into a combination of simpler types of functions? I thought initially that maybe any function $h$ satisfying (2) could be expressed in the form $$h(x,y)=O(x-y)$$ where $O$ is an odd function. This doesn't work though (consider the counterexample $h=x^2-y^2$). I then settled on $$h(x,y)=O(b(x)-b(y))$$ where $O$ is any odd function from reals to reals and $b$ is any function from reals to reals. Am I correct that any function $h$ satisfying (2) can be written in this form? If not, what is a counterexample, and how might I successfully break $h$ down into simpler functions?

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  • $\begingroup$ "Is there a name for a function of two variables that satisfies the following $h(x,y)=-h(y,x)$" Anti-symmetric? At least this is used for tensors having this property ( en.wikipedia.org/wiki/Antisymmetric_tensor ). Not sure how common it is used for functions like here $\endgroup$ – Winther Apr 3 '18 at 23:22
  • $\begingroup$ The keywords antisymmetric, skew-symmetric and anticommutative may be helpful. $\endgroup$ – mucciolo Apr 4 '18 at 2:26

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