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I am unclear as to what precisely is a symmetric function if it is defined piecewise.

For example, (in Maple) is 

         A := (x, y) -> piecewise(x < y, 2*x-y, y < x, x-2*y, x = y, 0)

a symmetric function? And if so, why?.

Also, (in Maple) is

        B := (x, y) -> piecewise(x < y, x-(1-x)*y, y < x, -y+x*(1-y), x = y, 0)

a symmetric function. And if so, why?

My understanding is that a function is symmetric if f(a,b) = f(b,a) for all a, b. Is this correct?

Thanks for your help.

PiE
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1 Answers1

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Yes, defining a symmetric function by $f(a,b)=f(b,a)$ is very reasonable. If a formula for $f$ is given piecewise, $$f(x,y)=\begin{cases} g(x,y) \quad & \text{if } \ x<y; \\ h(x,y) \quad & \text{if } \ x>y; \\ k(x) \quad & \text{if } \ x=y \end{cases} $$ then the symmetry requirement amounts to $g(x,y)=h(y,x)$. What $k$ is does not matter.

The two examples you gave satisfy a different identity, $f(a,b)=-f(b,a)$. Such functions are called antisymmetric. In the above notation, the antisymmetry requirement amounts to $g(x,y)=-h(y,x)$, and $k$ being identically zero.

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