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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.

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up vote 6 down vote accepted

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 $$ f\delta' = f(0) \delta' - f'(0) \delta.$$

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It took me a while to understand your answer, but I see now that you are right - thank you. I hope you don't mind I wrote my own answer that explains the same thing in terms that are easier for physicists to understand, anyway I have accepted your answer. – Joe Jan 19 '12 at 18:01
@Vovo: Dear Vobo, I think it would be helpful if in your answer you also explained that the same method verifies the OP's identity $\delta' f = f(0)\delta' - f'(0) \delta.$ (Of course, it follows from what you wrote and the OP's own manipulations, but it would be good to make it clear.) Regards, – Matt E Jan 19 '12 at 18:08
@Joe: We typed our answers at the same time, so I didn't see yours, thanks for accepting it anyway, and Matt: I added your hint. – Vobo Jan 20 '12 at 7:47

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: $\delta'(x) x = -\delta(x)$

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I consider this the best answer :)) – BCLC May 20 '14 at 17:26

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.

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I understand that when you write $\delta'(f)=-f'(0)$ you mean what I would write as $\delta'(x)f(x)=-\delta(x)f'(0)$, but in fact there is another factor of $\delta'(x)f(0)$ which you left out (see my answer). – Joe Jan 19 '12 at 17:54
Dear Rasmus, I have downvoted his answer because it does not deal with the OP's question. When you write $\delta(f)$, this is what the OP would denote by $\int \delta(x) f(x) dx$. Hence your formula for $\delta'(f)$ is what the OP would denote by $\int \delta'(x) f(x) dx$. However, the OP is not asking about the values of these expressions after integrating them; he wants to know the actual values of the expressions themselves. Regards, – Matt E Jan 19 '12 at 18:01
@Joe: Dear Joe, Hopefully my comment above will clarify what Rasmus is saying in his answer, and how it relates to what you are asking. Regards, – Matt E Jan 19 '12 at 18:02
Yes it does, thank you @MattE. – Joe Jan 19 '12 at 18:05
Dear @Matt, thanks for clarifying. I thought I guessed correctly on what the OP was asking, but obviously I didn't. I'm not even sure I do now--what is the meaning of the symbol $\delta(x)$ for $x\in\mathbb R$? This seems to be similar to the usual confusion concerning the difference between a function and its values. Only that in this context, the value isn't even defined. – Rasmus Jan 19 '12 at 19:47

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.

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Dear Jon, You are not answering the OP's question. See my comment on Rasmus's answer. Regards, – Matt E Jan 19 '12 at 18:02
@MattE: Sorry, my hands were faster than my mind. – Jon Jan 19 '12 at 18:09
Dear Jon, No worries! Best wishes, – Matt E Jan 19 '12 at 18:10

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