My question is the same as this one, but for the Dirac delta function. I did find several Math SE posts and one Wikipedia article on seemingly the same issue, but upon a closer inspection they all are dealing with $\langle f \delta ', \phi \rangle$ (see, e.g. here), whereas I'm interested in the case when $f$ is within the parentheses with Dirac delta.
I have attempted to expand it the two ways. And in both cases I've got zeros.
The first way.
$$\langle (f \delta)', \phi \rangle = - \langle (f \delta), \phi ' \rangle = - \langle \delta, (f \phi ' ) \rangle = - (f \phi ' ) \rvert_{t=0} = -f(0) \phi(0)' $$
Then I had a look at the first picture on this Wikipedia page. The picture represents a test function. And it is absolutely clear that its derivative at 0 is 0. And I concluded that:
$$-f(0) \phi(0)' = 0$$
The second way.
$$\langle (f \delta)', \phi \rangle = - \langle (f \delta), \phi ' \rangle = - \langle \delta, (f \phi ' ) \rangle = - \langle \delta, ((f \phi)' - f' \phi) \rangle = - \langle \delta, (f \phi)' \rangle + \langle \delta, (f' \phi) \rangle = - \langle \delta, (f' \phi + f \phi ') \rangle + (f' \phi) \rvert_{t=0} = - \langle \delta,(f' \phi) \rangle - \langle \delta,(f \phi ') \rangle + f'(0) \phi(0) = -(f' \phi) \rvert_{t=0} - (f \phi ') \rvert_{t=0} + f'(0) \phi(0) = -f(0)' \phi(0) - f(0) \phi(0)' + f'(0) \phi(0) = - f(0) \phi(0)'$$
And using the same argument as above, I conclude that $$ -f(0) \phi(0)' = 0$$
Here are my questions.
- I'm wondering if my two derivations are correct.
- Derivative of the test function at zero is so obviously zero, yet I haven't encountered examples where it would be equalized to zero. Thus, I'm slightly worried about equating it to zero. Is it ok to do it? Is there an example of a physical problem where it is normally done?