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
Best Answer
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.