# Product of $C^\infty$ and $\mathcal D$

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 $\psi \phi \in \mathcal D$.

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