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I already know that $\delta^{2}(t)$ is not defined.

However consider $A(t) = \delta(t) \cdot \delta(t-t_0)\ $ with $ t_0 \neq 0 $. Can I claim that $A(t) = 0$ ? If that is the case, is there a way to prove it besides just using the non rigorous argument that $\delta(t) = 0$, $\forall t \neq 0$ and $\delta(0) = \infty$. Even then we would have $@t = 0$ for example $0 \cdot \infty$ which is indeterminate.

Anonymous
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    Yes. Since they have disjoint distributional support they can be multiplied. I would suggest this nice pedagogical note https://arxiv.org/abs/1404.1778 – Alan Garbarz Dec 14 '21 at 12:26
  • How do you define $A(t)$? At the end, the fact that this distribution is $0$ is almost the definition of the product in this case ... – LL 3.14 Dec 14 '21 at 13:39
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    If $u$ and $v$ have disjoint singular support then the product of them can be defined. – md2perpe Dec 14 '21 at 14:56
  • Thank you, I think with those links I am satisfied @LL3.14 A(t) is just a notation I used to represent this product. I have not studied distributions and generalised functions in depth. In the previous year I came across with dirac delta and now in some proof I needed to know if this computed to zero. Any course recommendation from Mathematics Dept. that deals with distributions in more detail is appreciated. Thank you again for the help – Anonymous Dec 15 '21 at 19:25

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