Is this function "$h$" symmetric of the plane $x=y$?

Question

$h=\left\{\begin{matrix} f,x<y\\ g,x\geq y \end{matrix}\right.$

$g=f(y,x)$.

Is $h$ symmetric of $x=y$? Here $g$ is the function that changes all $x$ to $y$ and changes all $y$ to $x$ in $f(x,y)$.

For example, $h=\left\{\begin{matrix} x^2-y^2, x<y\\ y^2-x^2,x\geq y \end{matrix}\right.$.

Is $h$ symmetric of the plane $x=y$?

Answer 1

Yes.

Suppose that $h$ is a function defined as $$ h(x,y) = \begin{cases} f(x, y), & x < y \\ f(y, x), & x \ge y. \end{cases} $$ Notice that $h$ only "sees" the half-plane in $(x, y)$-coordinates where $x < y$. For example, $h(2, 3) = f(2, 3)$ and $h(3, 2) = f(2, 3)$, as well. The value $f(3, 2$) is never called upon, for instance.

This easily generalizes to a proof that $h$ has symmetry about the line $y = x$. If $x > y$, then $h(x, y) = f(y, x)$. In this case, $y < x$, so $h(y, x) = f(y, x)$. In other words, we can conclude that $$ h(y, x) = h(x, y) $$ for any $(x, y)$.

Answer 2

Hint: You are asking whether $h(x,y)=h(y,x)$. You have three cases for the comparison of $x$ and $y$. Take them one at a time, plug into the definition of $h$ and see what happens.