Definition. Let
$u :C^{\infty}_{c}(\Omega) \rightarrow \mathbb{C}$ a linear continuous functional, that is, $u$ is a distribution. We define the support of $u$ as $$\Omega- \{x \in \Omega; \exists \; \mbox{an open}\; V_x \; \mbox{such that}\; \; u|_{V_x}=0\}.$$ Here $u|_{V_x}$ means $$u|_{V_x}(\varphi)=u(\varphi)\quad \forall \varphi \in C_{c}^{\infty}(V_x),$$ where $\Omega \subset \mathbb{R}^{n}$ is open.
How can I find the support of the distribution \begin{equation} \label{1}u(\varphi)=\sum_{n=1}^{\infty}[\varphi (\frac{1}{n})-\varphi(0)-\frac{1}{n}\varphi'(0)] \quad \forall \varphi \in C^{\infty}_{c}(\mathbb{R})? \end{equation}
I would like an answer or a tip on how to solve this question.
I observe this serie is convergent, because using the Taylor Formula $$\varphi(\frac{1}{n})=\varphi(0)+\frac{1}{n}\varphi'(0)+\frac{1}{n^2}\alpha(\frac{1}{n})$$ we conclude $u(\varphi)$ is convergent.
Best Answer
For which $x \in \mathbb{R}$ are the value of $\varphi(x)$ used?