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Originally this was an example stated in class but without proof. When trying to prove it I got a different result for the expression $v \delta'$. Here I presented my calculations which were approved in the comments. In the end it was pointed out by my professor that there was a typo in class. So I changed the question and included my computations as an answer. For the case if some person is stumbles upon this question and is interested in an answer.


As in the title stated I want to show that

$$v \delta'=v(0) \delta' - v'(0) \delta$$

in distributional sense where $v \in C^\infty(\Omega)$ and $\delta \in \mathscr{D}'(\Omega)$ is the Dirac Delta distribution which is defined by $$\langle \delta, \phi \rangle=\phi(0) \quad \forall \phi \in \mathscr{D}(\Omega).$$ Hence we have for the derivative of the Dirac Delta distributional $$\langle \delta',\phi\rangle=-\langle\delta,\phi'\rangle=-\phi'(0).$$

From this we should be able to prove the above relation which involves $\delta'$.

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  • $\begingroup$ Observe that the question is poorly formulated, in the sense that $v(0) \delta$ doesn't have any mathematical sense $\endgroup$ – user5609462 May 22 '16 at 11:21
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    $\begingroup$ It is meant in distributional sense, i.e. we interpret it as $<v(0) \delta, \phi>=<\delta,v(0) \phi>=v(0) \phi(0)$ for all $\phi \in C_0^\infty$. Then it should make sense, I think. $\endgroup$ – Fritz May 22 '16 at 11:33
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    $\begingroup$ Your computations looks fine to me. $\endgroup$ – Sangchul Lee May 22 '16 at 11:40
  • $\begingroup$ @SangchulLee Thank you for your comment. I also didn't notice a mistake, so I will accept that it should be typo in the task. $\endgroup$ – Fritz May 22 '16 at 11:50
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We have$$\begin{aligned} \langle v \delta', \phi \rangle =\langle \delta',v \phi \rangle=-\langle\delta, (v \phi)'\rangle &=-(v\phi)'(0) \\ &=-v(0) \phi'(0)-v'(0) \phi(0) \\ &=v(0) \langle\delta',\phi\rangle-v'(0)\langle\delta,\phi\rangle \\ &=\langle v(0)\delta'-v'(0)\delta, \phi\rangle \end{aligned}$$ for all test function $\phi \in \mathscr{D}(\Omega)$ and hence $$v\delta'=v(0) \delta' - v'(0) \delta.$$

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