Is my derivation of the derivative of a product of the Heaviside function and a function correct? (A follow-up question) That's a follow-up question to the accepted answer to this post.
After some thinking, I ended up deriving the derivative differently. I'm wondering if the dear Math stack exchange community can tell me whether it is wrong or not please.
Here goes my derivation.
Assume we have some continuous function $f(x)$ defined on the interval $[0, \infty)$ and it doesn't go to infinity when $x$ goes to infinity.
I multiply it by the Heaviside function to expand it to the entire $x$ axis: $f(x)H(x)$. Here, $H(x)$ is equal to $1$ at $x=0$.
I want to differentiate that product.
The post that I cited states that I can't do that and that I have to convert the product to a distribution first. However, the post doesn't specify how one does the conversion to preserve the physical meaning of the product $f(x)H(x)$.
I have come up with the following conversion procedure. I'm going to mollify the product: $(f(x)H(x))*\phi(x)$. Here, $\phi(x)$ is the mollifier and $*$ stands for convolution.
Convolution can be considered as a distribution (I read that in Wikipedia).
Now we have the distribution $\langle(f(x)H(x)), \phi \rangle$ that approximates the product $f(x)H(x)$ preserving its physical meaning and we can differentiate it:
$$\langle(f(x)H(x))', \phi \rangle = \int_{-\infty}^\infty (f(x)H(x))' \phi(x)dx$$
Here:

*

*$H(x)=0$ for $x \in (-\infty, 0)$, therefore, I can change the integration limit to $[0,   \infty)$.


*$H(x)=1=const$ for $x \in [0, \infty)$, therefore, $(f(x)H(x))'=H(x)f(x)'$ for $x \in [0, \infty)$.
With that in mind we have the following:
$$\langle(f(x)H(x))', \phi \rangle = \int_{-\infty}^\infty (f(x)H(x))' \phi(x)dx = \int_0^\infty f(x)' \ 1 \ \phi(x)dx = \int_0^\infty f(x)' \phi(x)dx = (f(x=0)' \phi(x=0)) = \langle \delta, (f(x)'\phi(x) \rangle = \int_{-\infty}^\infty \delta(x) f(x)' \phi(x)dx = \int_{-\infty}^\infty (f(x)' \delta(x)) \phi(x)dx = \langle(f(x)' \delta), \phi \rangle$$
I'm wondering if this derivation is correct.
If positive, then I have the follow up questions.

*

*Why my result is not the same as in the post that I cited?

*How can the function $f(x)$ defined on the interval $x \in [0, \infty)$ be under the integral with the integration limits $(-\infty, \infty)$?

*What if the function $f(x)$ were defined on the interval $x \in (0, \infty)$? How would it affect the derivation?

 A: Your error
In your second step, $\int_{-\infty}^\infty (f(x)H(x))' \phi(x)dx = \int_0^\infty f(x)' \ 1 \ \phi(x)dx,$ you are not using the definition of distributional derivative. Remember that the definition is $\langle u', \phi \rangle := -\langle u, \phi' \rangle.$ Do not assume that you can just take the derivative of a function wherever the derivative is defined and use that. If it were that simple you would have $H'=0$ but the correct is $H'=\delta.$
Correct derivation
Let $f \in C^\infty([0, \infty)),$ meaning that $f \in C^\infty((0, \infty))$ and that the limits of all derivatives exist at $0.$ Let $f$ define a distribution by $\phi \mapsto \int_0^\infty f(x) \, \phi(x) \, dx$ for $\phi \in C^\infty_c(\mathbb R).$ Let $f'$ denote the classical derivative of $f$ defined on $(0, \infty).$ Then the distributional derivative of $f,$ denoted by $Df$ is given by
$$
\langle Df, \phi \rangle
= -\langle f, \phi' \rangle
= -\int_0^\infty f(x) \, \phi'(x) \, dx
= -\left[ f(x) \, \phi(x) \right]_0^\infty + \int_0^\infty f'(x) \, \phi(x) \, dx
= f(0) \, \phi(0) + \langle f', \phi \rangle
= \langle f(0)\delta, \phi \rangle + \langle f', \phi \rangle
= \langle f' + f(0)\delta, \phi \rangle,
$$
i.e. $Df = f' + f(0)\delta.$
A: I found an error in my derivation: $$\int_0^{\infty} f(x)' \phi(x) dx \neq (f(x=0)'\phi(x=0))$$
With that in mind, my question doesn't bring any good to Math stack exchange. But md2perpe's answer and comments are very useful to mind which makes me uncertain about deleting my question.
I'll leave it for now.
If the moderators decide to delete it, I'll delete it.
