2
$\begingroup$

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.

$\endgroup$
4
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.