# Linear combination of Dirac delta distribution and its derivatives

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=&\\ =&\\ =& \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*} $$\forall \;\varphi \in C^{\infty}_{c}(\mathbb{R}).$$

• As Semiclassical suggests, try considering what happens when $\phi = x^i$ for $1\le i \le n-1$. – jgon Sep 18 at 18:05
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 $$n$$.