$H(x)$ is defined as the Heaviside step function, so $H(x) = \Bigg\lbrace\begin{array}{ll} 1 & x > 0 \\ 0 & x \leq 0 \end{array}$
The case for $H(0) = 0$ matters for me.
I have the function $H(x+y)$ nested in a product. I need to break up $H(x+y)$ to simplify my overall function further. I know that $H(x+y) \neq H(x) + H(y)$, is there any identity that I can use to break up $H(x+y)$? I have been stumped for hours on this.
The only identity that you can use is $$H(ax)=H(x)\quad,\quad a\in\mathbb R$$ So you can do $$H(x+y)=H\left(\frac{x+y}{\sqrt2}\right)$$ then, you can use linear transform (actually rotating by $45$ degree: $$x=\frac{u-v}{\sqrt2}\quad,\quad y=\frac{u+v}{\sqrt2}$$ which will change $$H(x+y)=H\left(\frac{x+y}{\sqrt2}\right)=H(u)$$ But, be sure to apply the same tranform to the rest of your system.