The derivative of the Delta function times another function

Question

Good evening! I can't understand how to prove that $$\alpha(x)\delta'(x)=-\alpha'(0)\delta(x)+\alpha(0)\delta'(x).$$ I tried to use $$(Df,\phi)=-(f, D\phi),$$ also I used that $$(D(hf),\phi)=(h'f+hf',\phi)$$ but I got another result. Thanks for help.

It would be very good, if you advised some books.

IMHO delta functions only make sense inside integrals, so try to stick it into an integral with a good function, $f$, i.e. $\int dx f\left(x\right).\alpha\left(x\right)\delta'\left(x\right)$ and integrate by parts, as already suggested by @AmIAStudent. Book-wise there is M. J. Lighthill "An introduction to Fourier analysis and generalized functions" and M. Rahman "Applications of Fourier Transforms to Generalized Functions" — Cryo, May 14 '19 at 23:09
Possible duplicate of Kac-Moody algebra from WZW model via Poisson brackets — AccidentalFourierTransform, May 14 '19 at 23:27

score 3 · Answer 1 · answered May 14 '19 at 23:06

The derivative of the Dirac delta works exactly through the integration-by-parts scheme you've mentioned, $$(Df,\phi)=-(f, D\phi),$$ i.e. $\delta'(x)$ is the generalized function which returns for arbitrary $f$ $$ \int f(x) \delta'(x) \mathrm dx = -f'(0). $$ Multiplying $\delta'(x)$ by some regular function is essentially equivalent to doing the same with $f(x)$, so that \begin{align} (f, \alpha \, \delta') & = \int f(x) \alpha(x)\delta'(x) \mathrm dx \\ & = -\frac{\mathrm d}{\mathrm dx}\left[ f(x)\alpha(x)\right]_{x=0} \\ & = - \alpha'(0) f(0) - \alpha(0) f'(0) \\ & = - \alpha'(0) \,(f,\delta) + \alpha(0)\, (f,\delta') \\ & = (f,- \alpha'(0)\,\delta + \alpha(0)\,\delta') \end{align} as required.

@GIFT you should probably mark this as the correct answer. – hft May 15 '19 at 00:46 — hft, May 15 '19 at 00:46

The derivative of the Delta function times another function

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