This is a followup to my existing question: Question on Heaviside step function (distribution) identities
I was trying to show the equality between two Heaviside step functions using their derivatives.
The old question was:
For $c>0$, and in the sense of distributions:
First:
$$\frac {d}{dt} \theta (ct-z)=c\delta(ct-z)=\frac{1}{c} c\delta(t-\frac{z}{c})=\delta(t-\frac{z}{c})\tag{1}$$
Second:
$$\frac {d}{dt} \theta (t-\frac{z}{c})=\delta(t-\frac{z}{c})\tag{2}$$
So then
$$\theta (ct-z)=\theta (t-\frac{z}{c})\tag{3}$$
to within a constant of integration.
My question was: is that true?
The answer I got was YES.
MY QUESTION NOW IS: Does that mean that $\theta(f(x))$ is identical to $\theta(g(x))$ for any two functions as long as they have identical zeros? (Assuming the signs of their derivatives are the same at the zeros)
The primary necessity for $\theta \circ f$ to equal $\theta \circ g$ is $f(x)$ and $g(x)$ are negative for the same $x$ and positive for the same $x$ (modulo a null set).
If $f$ and $g$ are differentiable then this will occur if they have the same zeros and the same signs of the derivatives at the zeros. I here require the derivatives to be non-zero. The zero case is possible, but then we have to look at non-zero second derivatives. And so on.
The reason I write "non-zero derivatives" is that for example $\sin(x)^2$ and $-\sin(x)^2$ have the same zeros and the same derivatives at the zeros, but they have not the same sign: $\theta(\sin(x)^2)$ is a "square wave" while $\theta(-\sin(x)^2)$ vanishes.