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I have found the following identity online $f(t)\delta'(t) = f(0)\delta'(t) − f'(0) δ(t)$ however I am quite stuck on to how I would start this to prove this identity, I have an idea that I should use the distribution product rule however I can't seem to find it. Any help on how I can start this will be greatly appreciated hopefully I can continue and prove this identify.

Dirac Delta Yeah
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    Do you know the definition of $\delta'$? – qualcuno Mar 12 '21 at 01:26
  • I think maybe i'm missing that, I know that <δ',Φ> = -Φ(0) ( I hope this is correct) however I think that's for a test function and i'm not sure how I would ammend that , I hope i'm along the right line – Dirac Delta Yeah Mar 12 '21 at 01:35
  • That's right, $\langle \delta',\varphi\rangle = -\varphi(0)$. Now I think you should say what $f$ is... is it another distribution, a test function? How is $f \cdot D$ defined for a distribution $D$? – qualcuno Mar 12 '21 at 01:40
  • I'm not very familiarized on the theory of distribution, but e.g. Wikipedia cites $\varphi D$ with $\varphi$ a test function and $D$ a distribution to be defined as $\langle \varphi D, \psi\rangle := \langle D,\varphi \psi\rangle$. Is this the definition you are considering? – qualcuno Mar 12 '21 at 01:43
  • I'm assuming its a standard function as in the title of the identity it said function times dirac delta , not sure if that makes a difference. I'm unsure how the function f induces the distribution D as the identity was quite simplistic as it just states a function f(t) – Dirac Delta Yeah Mar 12 '21 at 01:43
  • Hm. Maybe you should cite where the identity comes from to add context. Or maybe this is really standard (again, I'm far from knowledgeable on this). But once you know what each thing is, proving both sides to be equal should be a matter of evaluating at a test function and checking that both distributions coincide. – qualcuno Mar 12 '21 at 01:45
  • okay that looks great, instead of D would I use δ and would phi in this case be our standard function with psi as the test function? – Dirac Delta Yeah Mar 12 '21 at 01:45
  • Yes, that's what I meant! But I was actually asking you if that's the definition you are using. Wikipedia cites that as the definition of a test function times a distribution; but I try not to trust Wikipedia too much... – qualcuno Mar 12 '21 at 01:46
  • https://dsp.stackexchange.com/a/68736 , it is the identity (2) in the first answer – Dirac Delta Yeah Mar 12 '21 at 01:47

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By the product rule: $$ (f\delta)' = f'\delta + f\delta' $$

But $f\delta = f(0)\delta$ and $f'\delta = f'(0)\delta$ so $$ (f(0)\delta)' = f'(0)\delta + f\delta' $$ i.e. $$ f\delta' = (f(0)\delta)' - f'(0)\delta = f(0)\delta' - f'(0)\delta. $$

md2perpe
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