On my book there is the following statement:

We can define the product of a distribution $u \in \mathcal D'(\Omega)$ and a function $\psi \in C^\infty(\Omega)$ in the following way:

$\langle \psi u, \varphi \rangle = \langle u, \psi \varphi \rangle$ for all $\varphi \in \mathcal D(\Omega)$

Moreover my book defined distributions as linear continuos functional from $\mathcal D(\Omega) \to \mathbb R$

My question is: for this definition to make sense,should't we have $\psi \varphi \in \mathcal D(\Omega)$?

But I don't think that is true; for example $$\varphi(x) = \begin{cases} e^{\frac 1{x^2-1}}, & \text{if $|x|$ < 1} \\ 0, & \text{otherwise} \end{cases}$$

$$\psi(x) = e^{1 - \frac 1{x^2 - 1}}$$

We have $\varphi \in \mathcal D(\mathbb R)$ and $\psi \in C^\infty(\Bbb R)$ but $\varphi \psi = \begin{cases} e, & \text{if |x| < 1} \\ 0 , & \text{otherwise} \end{cases}$

and $\varphi \psi \notin \mathcal D(\mathbb R)$ as it is not continuous. What am I misunderstanding?

P.S. Also, please help with finding better tags! I didn't know what to put.


Your counterexample does not work: $$\lim_{x\to 1,\,x<1}\psi(x) = \lim_{x\to 1,\,x<1} e^{1 - \frac 1{x^2 - 1}}=+\infty,$$ $$\lim_{x\to 1,\,x>1}\psi(x) = \lim_{x\to 1,\,x>1} e^{1 - \frac 1{x^2 - 1}}=0,$$ hence $\psi$ is not a continuous function.

The key point is to understand that

  1. The product of two $C^\infty$ functions is also a $C^\infty$ function.

  2. The support of the product of two continuous functions is the intersection of their supports.

These two facts allow us to say that if $\phi\in \mathcal D$ and $\psi \in C^\infty$, then $\psi \phi \in \mathcal D$.

  • $\begingroup$ I should have thought longer about it. Thank you very much! $\endgroup$
    – Ant
    Jan 8 '15 at 20:54

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.