# Condition to the convolution product well defined

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.

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).
• So $\delta'*H$ is well defined because $Supp(\delta')$ is compact. But why $1*(\delta'*H)$ exists? – mati Apr 1 at 15:49
• @mati hint: try to find the value of $\delta'*H$ – TZakrevskiy Apr 1 at 16:13
• $\delta'*H=\delta*H'=0$ 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? – mati Apr 1 at 16:32
• @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 – TZakrevskiy Apr 1 at 16:43
• 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? – mati Apr 1 at 17:23