Consider the Heaviside Theta function
$$H(x) = \begin{cases} 1,& x>0,\\ 1/2,& x=0,\\ 0,& x>0. \end{cases}$$
I know that for $x\neq 0$, the condition $H(x)H(-x)=0$ holds true. Moreover, from the definition proposed for the Heaviside function, for $x=0$ the expression $H(x)H(-x)$ should be $1/4$, which would result in a new discontinuous function.
I have an expression of the form
$$H(x-q_1(t))H(-x+q_2(t)),$$
for differentiable functions $q_1$ and $q_2$ (the domains can be taken as needed), which should vanish identically through the curve $x=q_1(t)=q_2(t)=:q(t)$.
During some conversations, a friend has used approximation arguments to state that if $x=q$, then we must have
$$H(0)H(0) = H(0^+)H(0^-) =: \left(\lim\limits_{x\to 0^+}H(x)\right)\left(\lim\limits_{x\to 0^-}H(x)\right) = 0,$$
but I do not exactly agree because a) the Heaviside is not continuous and b) it is defined at $0$, so $H(0)H(0) = 1/4$ (???).
The whole issue here is that for the expression we are working with to make sense, the multiplication must vanish everywhere. But I do not understand how.
Does anyone have an idea that might shed some light on where I (or he/we) am going wrong?
OBS: I used Mathematica to try to understand. Although the plot seems to have $H(x)H(-x) = 0$ for all $x\in\mathbb{R}$, it fails to calculate $H(0)H(0)$: https://i.stack.imgur.com/GeOq4.png
