Let $f(x)=x$, $u=\sum_{j=1}^{n}a_j\delta^{(j)} \in \mathcal{D}'(\mathbb{R})$, where $a_j \in \mathbb{C}$ and $\delta$ is the Dirac delta distribution. Show that if $fu=0$ then $a_1=a_2=\ldots=a_n=0$.
In fact, $fu=0$ implies
\begin{align*}
0=&<fu, \varphi>\\
=&<u, f\varphi>\\
=& \left<\sum_{j=1}^{n}a_j \delta^{(j)}, f\varphi\right>\\
=&\sum_{j=1}^{n}a_j < \delta^{(j)}, f\varphi>\\
=&\sum_{j=1}a_j (-1)^{j}<\delta, (f\varphi)^{(j)}>\\
=&\sum_{j=1}^{n}a_j(-1)^{j}(f\varphi)^{(j)}(0)\\
=&\sum_{j=1}^{n}(-1)^{j}j a_j \varphi^{(j-1)}(0)
\end{align*} $\forall \;\varphi \in C^{\infty}_{c}(\mathbb{R}).$
How can I continue this? Thanks for your answer.
Best Answer
Try $\phi(x)=x^n\psi(x)$ where $\psi$ is a standard bump function that is constant equal to 1 near the origin, and vary the $n$.