Which of these two ways to take the derivative of a delta function times another function is correct?

Question

A well known identity of the Dirac delta function is that for any function $f(x)$: $$ \delta(x) f(x) = \delta(x) f(0). $$ If we take the derivative of the right hand side we get: $$ \delta'(x) f(0). $$ But if we take the derivative of the left hand side we get $$ \delta'(x) f(x) + \delta(x) f'(x) = \delta'(x) f(0) + \delta(x) f'(0) $$ Which one is correct?

P.S. I know that this problem has something to do with the fact that the delta function is not really a function, but rather a generalized function. However, the delta function and its derivatives are useful in calculations (especially in physics), and I want to know the correct rules for using them.

Answer 1

Any distribution $T$ can be multiplied with a $C^{\infty}$-function $f$ by the formula $$ (f T)(\varphi) = T(f\varphi) $$ for a test function $\varphi$. And if $T$ has order $0$ like the $\delta$-distribution, $f$ may be a $C^0$-function. And in this sense, the identity $$ f\delta = f(0) \delta$$ is perfectly true.

Applying this multiplication to your left hand side, you have
$$ (f'\delta)(\varphi) + (f\delta')(\varphi) = \delta(f'\varphi) + \delta (-(f\varphi)') = \delta (-f\varphi') = f(0) \delta' (\varphi),$$ your right hand side. So you were perfectly right, you just need to interpret the expression $f\delta'$ correctly as $$ f\delta' = f(0) \delta' - f'(0) \delta.$$

Answer 2

You should define the delta on a proper space of test functions. So,

$$\int_{-\infty}^\infty \delta'(x)f(x) \,dx =\left.\delta(x)f(x)\right|_{-\infty}^\infty-\int_{-\infty}^\infty \delta(x)f'(x)\,dx =-f'(0)$$

after integration by parts and where use has been made of the fact that test function $f(x)$ goes enough rapidly to 0 at $\pm\infty$. Manipulations on distribution are only meaningful in this sense.

Answer 3

The way you wrote the first equation is very misleading and it seems that it caused your confusion. I would prefer to write $\delta(f)=f(0)$. The correct formular for the derivative of $\delta$ is the following $$ \delta'(f)=-f'(0). $$ For an explanation, see here.

Answer 4

OK I think I know the answer to my own question, and that I can put it in simple terms without getting to much into the mathematical background of delta functions. My mistake was that I thought the similarly to this identity: $$ \delta(x) f(x) = \delta(x) f(0), $$ which is correct, the following identity is also true: $$ \delta'(x) f(x) = \delta'(x) f(0), $$ but this is wrong.
The correct identity is: $$ \delta'(x) f(x) = \delta'(x) f(0) - \delta(x) f'(0). $$ If I use this identity when comparing the two results in the question, they turn out to be the same - meaning both ways of taking the derivative are correct.
In fact, the derivation in the question is a proof of this identity.
BTW it is a generalization of the well known identity: $\delta'(x) x = -\delta(x)$