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How does one prove that the Dirac Delta function satisfy this property ?

$$f(x)\delta'(x-y) =f(y)\delta'(x-y) - f(y)'\delta(x-y)$$

This is stated after this property

$$ f(x)\delta(x-y) =f(y)\delta(x-y)$$

which has been explained in this forum before in for example Why does the Dirac delta function satisfy $f(x)\delta(x-a) = f(a)\delta(x-a)$?

I have tried to use integration to prove it but all I got is $-f'(x)$ as final result, reference was https://www.reed.edu/physics/faculty/wheeler/documents/Miscellaneous%20Math/Delta%20Functions/Simplified%20Dirac%20Delta.pdf.

It seems closer to what one obtains with the for the Dirac delta derivative identity as in here Dirac delta derivative identity

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    $\begingroup$ You should read this post that I wrote very carefully. Dirac delta is NOT a function, and you definitely must understand what is written in that post otherwise it's simply impossible to truly understand the dirac delta is. $\endgroup$
    – user21820
    Jan 22 at 4:48
  • $\begingroup$ Thank you @user21820, do you have any recommended readings on this topic ? I have both a physics and applied maths background, I have tried to look around online but the place where I have learned the most has been on this forum. $\endgroup$
    – karthikT3
    Jan 23 at 16:02
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    $\begingroup$ Unfortunately, I don't have a recommended reference, because I wrote that entire post from my own understanding so as to give a very clear concrete sense of what we are doing when dealing with dirac delta. But if you have any question on that post, feel free to ask over there! $\endgroup$
    – user21820
    Jan 23 at 16:24

2 Answers 2

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Write $\delta_y(x) = \delta(x-y)$. Let $\phi\in \mathcal{D}(\mathbb{R})$ be a test function. Then we have $$ \langle f\delta_y',\phi\rangle = \langle \delta_y',f\phi\rangle = - \langle \delta_y,(f\phi)'\rangle = -\langle \delta_y,f'\phi + f\phi'\rangle = -f'(y)\phi(y)-f(y)\phi'(y). $$ Now, clearly $\phi(y)=\langle \delta_y,\phi\rangle$ and $\phi'(y)=-\langle \delta_y',\phi\rangle$, so we obtain $$ \langle f\delta_y',\phi\rangle = \langle -f'(y)\delta_y+f(y)\delta_y',\phi\rangle, $$ which means $$ f\delta_y' = -f'(y)\delta_y + f(y)\delta_y', $$ or, equivalently, $$ f(x)\delta'(x-y) = f(y)\delta'(x-y) - f'(y)\delta(x-y). $$

Note: If you think in terms of integrals, just write $$ \langle \delta_y,g\rangle = \int_{\mathbb{R}} \delta(x-y)g(x) \; dx $$ in the above.

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  • $\begingroup$ Thanks you sir this is very neat. I am just starting with this topic but it seems powerful, do you have any recommended readings on this topic? I have both a physics and applied maths background. $\endgroup$
    – karthikT3
    Jan 23 at 15:59
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$$ f(x) \delta'(x-y) = - f(x) \frac{\partial}{\partial y}\delta(x-y) = -\frac{\partial}{\partial y} \left( f(x)\delta(x-y) \right) \\ = -\frac{\partial}{\partial y} \left( f(y)\delta(x-y) \right) \\ = -f'(y)\delta(x-y) - f(y)\frac{\partial}{\partial y}\delta(x-y) \\ = -f'(y)\delta(x-y) + f(y)\delta'(x-y) $$

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  • $\begingroup$ Thanks this was less intuitive to me. I am not sure I follow please correct me if I am wrong, going into the second to third line on the last term you change the derivative back as $-\frac{\partial}{\partial y} \delta(x-y)= \frac{\partial}{\partial x} \delta(x-y) $ ? $\endgroup$
    – karthikT3
    Jan 23 at 16:10
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    $\begingroup$ @karthikT3. That's correct. $\endgroup$
    – md2perpe
    Jan 23 at 16:20
  • $\begingroup$ Upon further reflection on both answers here, there is only one thing that confuses me which is the step where you move the partial derivative (first line, last term). Why are we allowed to write it this way ? With two normal functions order of differentiation would matter and this would not hold ? As per other posts dirac delta is not a function but the connection here is still not clear to me. $\endgroup$
    – karthikT3
    Jan 23 at 16:44
  • $\begingroup$ @karthikT3. Are you asking about $- f(x) \frac{\partial}{\partial y}\delta(x-y) = -\frac{\partial}{\partial y} \left( f(x)\delta(x-y) \right)$? $\endgroup$
    – md2perpe
    Jan 23 at 17:03
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    $\begingroup$ @karthikT3. Since the first factor $f(x)$ doesn't depend on $y$ it commutes with $\frac{\partial}{\partial y}.$ $\endgroup$
    – md2perpe
    Jan 23 at 17:16

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