2
$\begingroup$

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 $S \star \check{\varphi}$?

thank's

$\endgroup$
  • $\begingroup$ 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
2
$\begingroup$

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.

$\endgroup$

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.