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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.

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    $\begingroup$ As you noted, this statement needs to be true for all test functions. So, try looking for test functions which make that sum as simple as possible (without being trivial). $\endgroup$ – Semiclassical Sep 18 at 14:54
  • $\begingroup$ Yes. My problem is finding test functions which make that sum simple. $\endgroup$ – mmath Sep 18 at 15:16
  • $\begingroup$ As Semiclassical suggests, try considering what happens when $\phi = x^i$ for $1\le i \le n-1$. $\endgroup$ – jgon Sep 18 at 18:05
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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 $n$.

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