I'm lost, can you help me please.

How compute the product convolution between two distributions $T$ and $S$? (we suppose that $T * S$ exist)?

How we compute the product convolution between an distribution $T$ with an function $\varphi \in C^{\infty}$?

Example: how we compute $\delta_a * \delta_b$ and $\delta'*H$

where $\delta$ is Dirac distribution in 0, and $H$ is Heaveaside.


Assume for a moment, $S$ and $T$ are smooth functions with compact support. Then you want to have for a test function $\varphi$ $$ (S\ast T)(\varphi) = \int (S\ast T)(x)\varphi(x) dx = \int \int S(y) T(x-y) dy \varphi(x) dx \\ = \int S(y) \int T(x-y) \varphi(x) dx \, dy = S(\check{T}\ast\varphi) $$ using Fubini, where $\check{T}(z):=T(-z)$. Now if $S$ and $T$ are convolvable distributions (e.g. have compact support), this is precisely what you want to have. So defining first the convolution of a distribution $S$ with a test function $\varphi$ at a point $x$ as $$ (S\ast\varphi)(x) = S(\varphi(x-\cdot)), $$ meaning $S$ evaluated for the function $y\mapsto\varphi(x-y)$, you can easily verify that $S\ast\varphi$ is a smooth function with $D^\varkappa(S\ast\varphi) = S\ast (D^\varkappa\varphi) = (D^\varkappa S)\ast\varphi$. Now define for a distribution $T$ the distribution $\check{T}(\varphi):=T(\check{\varphi}) = T(\varphi(-\cdot))$ (meaning $T$ evaluated for the function $y\mapsto\varphi(-y)$).

So you are ready to define for distributions $S,T$ one having compact support their convolution as $$ (S\ast T)(\varphi) = S(\check{T}\ast\varphi) $$ in a meaningful way as motivated above. Applying all this to $\delta_a\ast\delta_b$, you have for a test function $\varphi$ $$ (\check{\delta_b}\ast\varphi)(x)=\check{\delta_b}(\varphi(x-\cdot)=\delta_b(\varphi(x+\cdot) = \varphi(x+b), $$ hence $(\delta_a\ast\delta_b)(\varphi) = \delta_a(\check{\delta_b}\ast\varphi) = \varphi(a+b)$. Finally $\delta'\ast H = \delta$ is easy to see using the formula $$ D^\varkappa(S\ast T) = S\ast (D^\varkappa T) = (D^\varkappa S)\ast T $$ in the sense of distributions, which is not difficult to verify.


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