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we have the following proposition: if $u, v$ and $w$ are distributions with convolutifs supports, then $$ u*(v*w)= (u*v)*v $$ where $*$ designate convolution.

As example, it puposed to compare between $1*(\delta' * H)$ and $(1*\delta')*H$, where $H$ is Heaviside, $\delta$ is Dirac.

My question is: why the products 1*(\delta' * H)$ and $(1*\delta')*H$ are well defined and why we can calculate them?

Kin regards

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  • $\begingroup$ Your concern is the discontinuity of the Heaviside function at 0? $\endgroup$ – Dunkel Jun 10 at 16:43
  • $\begingroup$ Or what aspect? $\endgroup$ – Dunkel Jun 10 at 16:44
  • $\begingroup$ Hi, my question is: why the product $1*(\delta' *H)$ exits? $\endgroup$ – mati Jun 10 at 17:12
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The convolution $u*v$ of two distributions $u$ and $v$ exist if (but not only if) at least one of $u$ and $v$ has compact support.

In the convolutions given, $\delta'$ has compact support, while $1$ and $H$ do not. Therefore $\delta'*H$ exists. It equals $(\delta*H)' = H' = \delta$ which also has compact support, so $1*(\delta'*H)$ is defined. This equals $1*\delta = 1$. Thus, $1*(\delta'*H) = 1$. On the other hand, $1*\delta' = (1*\delta)' = 1' = 0,$ so $(1*\delta')*H = 0*H = 0.$ Thus, $1*(\delta'*H)$ and $(1*\delta')*H$ both exist, but they do not coincide.

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  • $\begingroup$ and support f $1*\delta'=0$ is $\emptyset$. Why $\emptyset$ is compact? $\endgroup$ – mati Jun 10 at 19:56
  • $\begingroup$ Let $\{ U_\alpha \mid \alpha \in A \}$ be a family of open sets whose union covers $\emptyset$. Then there is a finite subfamily $\{ U_\alpha \mid \alpha \in B \subset A \}$ whose union covers $\emptyset$; we can take $\beta=\emptyset$ since the empty union covers $\emptyset$. $\endgroup$ – md2perpe Jun 10 at 20:03
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As you may know $\delta(x) \ast H(x) $ is an integral $$ (\delta \ast H)(x) = \int \delta(x-y) H(y) \mathrm{d}y = H(x) $$ More explicitly $H(y)$ is the value of $H$ at $x = y$ and the area of the delta function is assumed to be one, thus $ \int \delta(x) dx = 1 $

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