\begin{equation*} \langle u, \varphi \rangle= \sum_{k=1}^{\infty}\varphi^{(k)}(\frac{1}{k}) \quad \varphi \in C^{\infty}_{c}((0,+\infty)) \end{equation*} define a distribution in $\mathcal{D}'((0,+\infty))$.
I want to show that there isn't $v \in \mathcal{D}'(\mathbb{R})$ such that $$v|_{(0,+\infty)} =u,$$ that is $\langle v, \varphi \rangle=\langle u, \varphi\rangle \quad \forall \, \varphi \in C^{\infty}_{c}((0,+\infty)) \subset C^{\infty}_{c}(\mathbb{R}).$ By contradiction, I thought about finding a sequence in $C^{\infty}_{c}((0,+\infty))$ such that \begin{align*} \varphi_j \rightarrow 0 \quad \mbox{em} \,\, C^{\infty}_{c}(\mathbb{R})\end{align*} but $v(\varphi_j)$ doesn't converges to $0$ in $\mathbb{C}$. However I don't know which sequence I should choose.
I appreciate any answer or suggestion on how to solve this problem. Thank you.