Step function properties!

Question

Is there an expression for $\theta (x^2-y^2)$ in term of $\theta(x-y)$? Where $\theta(x)$ is step function.

And is $\theta(ax-y)=\theta(x-y/a)$?

Answer 1

$$\theta (xy)~=~\theta (x)\theta (y)+\theta (-x)\theta (-y)$$ almost everywhere. Hence $$\theta (x^2-y^2)~=~\theta (x-y)\theta (x+y)+\theta (y-x)\theta (-x-y),$$ and $$\theta (ax-y)~=~\theta (a)\theta (x-y/a)+\theta (-a)\theta (y/a-x), \qquad a~\neq~0.$$