We denote by $E'(\mathbb{R})$ the set of distribution with compact support , and $\mathcal{D}(\mathbb{R})$ is the set of function $\mathcal{C}^{\infty}$ with a compact support.

1) I want to compute $\langle T * 1 , \varphi\rangle$ where $T \in E'(\mathbb{R}).$

So, let $\varphi \in \mathcal{D}(\mathbb{R}).$ We have $\langle T * 1 , \varphi\rangle = \langle T , \check{1}* \varphi\rangle = \langle T , 1 * \varphi\rangle $ because $1$ is constant

$= \left\langle T , \displaystyle\int\varphi (y) dy\right\rangle$

How I can finish?

And my last question :

2) I also want to compute $T\star e^x$ where $T \in E'(\mathbb{R}).$

So, let $\varphi \in \mathcal{D}(\mathbb{R})$ we have:

$\langle T \star e^x , \varphi\rangle = \langle T , \check{e^x} * \varphi \rangle = \left\langle T , \displaystyle\int e^{y-x} \varphi(y) dy\right\rangle$.

How I can finish?

3) In general, if $T$ and $S$ are two distributions such that $supp T$ or $supp S$ is compact. To compute $S \star T$ we write: $$\langle T * S , \varphi \rangle = \langle T , \check{S} \star\varphi\rangle = \langle T , S \star \check{\varphi}\rangle$$ and how we can finish this formula? Who's $S \star \check{\varphi}$?


  • $\begingroup$ what is $E'(\mathbb{R})$? $\mathcal{D}(\mathbb{R})$? $\endgroup$ – robjohn Apr 27 '13 at 22:10
  • $\begingroup$ i edit my message. Can you help me please? $\endgroup$ – jijiii Apr 27 '13 at 22:44
  • $\begingroup$ Did you mean $\langle T,\int\varphi(y)\,\mathrm{d}y\rangle$? $\endgroup$ – robjohn Apr 27 '13 at 23:22
  • $\begingroup$ yes, sorry. I edit my message. Sow, how we can determine $T * 1$? $\endgroup$ – jijiii Apr 28 '13 at 5:27

I am not sure what you are after. To know a distribution, you just need to know how it acts of appropriate functions.

In 1) T has compact support, hence it can act on all infinitely differentiable functions. You have already calculated that the action of $T*1$ on $\varphi$ is the same as that of $T$ on the constant function $x\mapsto\int\varphi(y) dy$. Pretty much explicit.

In 2) and 3) this is basically the same. Note that in 3) in the case when $T$ has compact support, $S*\check\varphi$ is really a infinitely differentiable function (not explicit in your notation) and, due to the compact support of $T$, it makes sense to let $T$ act on $S*\check\varphi$. In $S$ has compact support, then $S*\check\varphi$ also has compact support, and hence, is a test function.


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