Another Heaviside step function question

Question

I previously asked the question:

Question on Heaviside step functions

Is $θ(f(x))$ is identical to $θ(g(x))$ for any two functions as long as they have identical zeros? (Assuming the signs of their derivatives are the same at the zeros)

If the answer is yes then their derivatives should be the same, ie.

$$\frac{d}{dx}\theta(f(x))= \frac{d}{dx}\theta(g(x))$$

which means that

$$\theta_x(f(x))f_x(x)= \theta_x(g(x))g_x(x)$$

or

$$\delta(f(x))f_x(x)= \delta(g(x))g_x(x)$$

MY QUESTION IS: is this correct?

Answer 1

It is not obvious that the chain rule is valid for distributions. But in this case it actually is.

Assume that $f \in C^1(\mathbb{R})$ changes sign at $x_i$ for $i \in I$ (finite or countably infinite). Then $$ (\theta \circ f)'(x) = \sum_{i \in I} \operatorname{sign}(f'(x_i)) \, \delta(x-x_i) . $$

On the other hand, formally by the chain rule, in a neighborhood of $x_0$ we expect to have $$ (\theta \circ f)'(x) = (\theta' \circ f)(x) \, f'(x) = (\delta \circ f)(x) \, f'(x) . $$

Are these right hand sides equal? Actually they are. By a formula that can be found in Dirac Delta Function of a Function we have $$ (\delta \circ f)(x) \, f'(x) = \left( \sum_{i \in I} \frac{\delta(x-x_i)}{|f'(x_i)|} \right) \, f'(x) = \sum_{i \in I} \frac{\delta(x-x_i)}{|f'(x_i)|} \, f'(x_i) \\ = \sum_{i \in I} \frac{f'(x_i)}{|f'(x_i)|} \,\delta(x-x_i) = \sum_{i \in I} \operatorname{sign}(f'(x_i)) \,\delta(x-x_i) . $$

Thus, to answer your question: Yes, it is correct that $\delta(f(x)) \, f'(x) = \delta(g(x)) \, g'(x)$ if $f,g \in C^1(\mathbb{R})$ and $\theta \circ f = \theta \circ g$ (e.g. if $f$ and $g$ have the same zeros, and $f'$ and $g'$ have the same (nonzero) signs at the zeros).