Convolution and associatuvity

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 • Your concern is the discontinuity of the Heaviside function at 0? – Dunkel Jun 10 at 16:43 • Or what aspect? – Dunkel Jun 10 at 16:44 • Hi, my question is: why the product$1*(\delta' *H)$exits? – mati Jun 10 at 17:12 2 Answers 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. • and support f$1*\delta'=0$is$\emptyset$. Why$\emptyset$is compact? – mati Jun 10 at 19:56 • 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\$. – md2perpe Jun 10 at 20:03

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 $$\int \delta(x) dx = 1$$