# convolution-distribution

i want to compute the product of convolution $1 * (\delta' * H)$ where $\delta$ is distribution of Dirac and $H$ is function of Heaviside.

first, we compute $\delta' * H.$ We have by definition that $\delta' * H (x) = \displaystyle\int_{-\infty}^x \delta' (y) dy$ how we can finish?

Thank's

• Take a test function $\varphi$ and try to compute $\langle \delta' * H,\varphi\rangle$ instead... – Philippe Malot Apr 27 '13 at 13:04
• ok, so let $\varphi \in \mathcal{D}$ then, $$(\delta ' * H , \varphi) = (\delta', H * \varphi) = (\delta' , \displaystyle\int H(y-x) \varphi(y) dy) = (\delta' , \displaystyle\int_{-\infty}^x \varphi(y) dy)$$ how we can finish ? – lili Apr 27 '13 at 13:12
• @lili: Apply $(S*T)' = S'*T$ for any convolutable distributions $S,T$. – Vobo Apr 27 '13 at 13:39

In fact, this can easily be done if you already know some properties of distributions.

The distribution $\delta$ is the identity element of the convolution and it's easy to see that:

$\hspace{3cm}\delta^{(k)}*T=(\delta*T)^{(k)}=T^{(k)}$ for all $k\in\Bbb N$ and $T\in\cal{D}'(\Bbb R)$.

Now using the fact that $H'=\delta$, you get : $$1*(\delta'*H)=1*H'=1*\delta=1$$

If you don't know all this, you take $\varphi\in\cal{D}(\Bbb R)$, and start like you did, except you made an error :

$$\langle \delta'*H,\varphi\rangle=\langle\delta',H*\tilde{\varphi}\rangle$$

where $\tilde{\varphi}:x\mapsto\varphi(-x)$.

For the part where you were stuck, don't forget that $\langle T',\varphi\rangle=-\langle T,\varphi'\rangle$.

• okay, so $$< \delta' * H , \varphi> = - < \delta , H * \tilde{\varphi}'>$$ and for the rest? and i have another question, in general if $T, S \in \mathcal{D}'$ then $T * g \in \mathcal{D}'?$ and if $T \in \mathcal{D}'$ and $f \in \mathcal{C}^{\infty}$ then $T * f \in$? – lili Apr 27 '13 at 16:19
• Dear lili, maybe you should read a course before asking questions! – Philippe Malot Apr 27 '13 at 16:34