Given the linear form $$ \langle u, \phi\rangle = \sum_{k=1}^\infty \partial^k \phi(1/k) $$ which is a distribution on $(0, \infty)$, show that there is no $v \in \mathscr{D}(\mathbb{R})$ whose restriction to $(0, \infty)$ equals $u$.
This question is similar to On the extension of distribution but I didn't understand that solution and I'm wondering if there is an argument that involves making a partition of unity.
My attempt so far:
For any test function $\phi \in C_0^\infty((0, \infty))$, for some $N$, $\operatorname{supp} (\phi) \subset (\frac{1}{N}, \infty)$. So,
$$\left|\langle u, \phi\rangle\right| \leq \sum_{k=1}^{N+1} \left|\partial^k \phi(1/k)\right|.$$
Using that estimate it is not hard to show that $u \in \mathscr{D}((0, \infty))$. Suppose $v$ is a distribution on $\mathbb{R}$ that restricts to $u$ on $(0, \infty)$. Let $\rho$ be a test function identically equal to 1 on $[-1, 1]$ and support in a small neighborhood of that set. For all $n$, let $\mathbb{O_n} = \{(-2, \frac{1}{n-1}), (\frac{1}{n}, 2)\}$ and let $\psi_1$ and $\psi_2$ be the partition of unity of $supp(\rho e^x)$ with respect to $\mathbb{O_n}$. Then $$\langle v, \rho e^x\rangle = \langle v, \psi_1 \rho e^x\rangle +\langle v, \psi_2 \rho e^x\rangle = \langle v, \psi_1 \rho e^x\rangle + \sum_{k=1}^{n-1} e^{1/k}.$$
The last sum goes off to infinity as $n \to \infty$, but that doesn't finish the problem and I'm not sure what to do.