# Problem on multiplication of distributions with test functions

This question comes from Grubb's Distributions and Operators, question 3.8(b):

Consider $u\in \mathcal{D}'(\mathbb{R}^n)$ and $\phi\in C^{\infty}_0(\mathbb{R}^n)$. Find out whether one of the following implications holds for arbitrary $u$ and $\phi$:$$\langle u,\phi\rangle=0\Rightarrow \phi u=0 \:\:\:(1)$$or $$\phi u=0 \Rightarrow \langle u,\phi\rangle=0\:\:\: (2)$$

Not really sure how to approach this. My guess is that $(1)$ is more likely to hold than $(2)$ since there's at least an obvious example of the delta distribution, but I'm not sure how to go about proving/disproving either statement.

• How is $\phi u$ defined?
– frog
Apr 22, 2015 at 6:48
• $\phi u$ is the distribution $\langle \phi u, f\rangle:=\langle u,\phi f\rangle$ for $f$ any test function.
– Moya
Apr 22, 2015 at 6:51

Hint: Start writing down what $\phi u=0$ means. It means that $\phi u$ is the zero distribution, i.e. for all test functions $\psi$ it holds that $(\phi u)(\psi) = 0$ and by definition of the product of smooth functions with distributions that means that for all test functions $\psi$ it holds that $u(\phi\psi) = 0$.
• Right, so I guess then the idea would be to apply this to $\psi$ to be a test function such that $\psi\equiv 1$ on the support of $\phi$? Then would this imply $u(\phi\psi)=u(\phi)=0$?
Another hint for a counterexample to (1): Think of $u$ as point-symmetric function, i.e. $u(x)=-u(-x)$. Let $\phi$ be a non-negative testfunction with $\int\phi=1$ and $\phi(x)=\phi(-x)$. Then $u(\phi)=0$ but it is not difficult to find a test function $\psi$ with $u(\phi\psi)\neq 0$.