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i'm lost in the following question: why $(1*\delta')*H$ and why $1*(\delta' * H)$ are well defined? Where $*$ is the product of convolution and $H$ is the function of Heaviside.

Thank you in advance

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In both cases the convolution inside parentheses is well-defined (as a convolution of a compactly supported distribution $\delta' \in \mathcal E'$ with a another distribution).

After that you need to prove that the value of the convolution inside parentheses is also compactly supported.

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  • $\begingroup$ So $\delta'*H$ is well defined because $Supp(\delta')$ is compact. But why $1*(\delta'*H)$ exists? $\endgroup$ – mati Apr 1 at 15:49
  • $\begingroup$ @mati hint: try to find the value of $\delta'*H$ $\endgroup$ – TZakrevskiy Apr 1 at 16:13
  • $\begingroup$ $\delta'*H=\delta*H'=0$ then $Supp(\delta' *H)= \emptyset$. I don't understand how we conclude the existence and we can calculate the product of convolution before prouve it's existence? $\endgroup$ – mati Apr 1 at 16:32
  • $\begingroup$ @mati First, $\delta'*H \ne 0$. Second, every time we calculate the convolution here, we already know that at least one of factors has compact support, hence the product is well-defined $\endgroup$ – TZakrevskiy Apr 1 at 16:43
  • $\begingroup$ Sorry, we have $\delta'*H= \delta*H'= \delta*\delta=\delta$. So, in geneal to say that $T*S$ a sufficient condition is: Supp T or Supp S is compact. But it is an sufficient condition. What's the necessary condition to say that T*S exists? $\endgroup$ – mati Apr 1 at 17:23

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