Let
$$u=\sum_{k=1}^\infty\delta^{(k)}_{\frac{1}{k}} \quad ,$$
where $\delta_x$ is the standard Dirac delta distribution $\delta_x(\phi)=\phi(x)$.
The distribution above acts on elements of $C^\infty_c((0,\infty))$ via
$$u(\phi)=\sum_{k=1}^\infty(-1)^k\phi^{(k)}\left(\frac{1}{k}\right)\quad\ ,$$
using the standard formula of derivating a distribution via integration by parts.
I'm trying to see why there can't be a distribution $v$ such that $u=v$ in a neighborhood of $0$, that is, such that $u(\phi)=v(\phi)$ for every $\phi$ with compact support contained in say, the compact $[-1,1]$. It seems that if we can find a $\phi$ such that $\phi(x)=\exp(x)$ for $|x|<1$, then
$$u(\phi)=\sum_{k=1}^\infty(-1)^k\exp\left(\frac{1}{k}\right)>\sum_{k=1}^\infty\exp(0)\ ,$$
which shows in particular that such extension cannot satisfy the seminorm condition for a distribution. However, I don't see how we can take such $\phi$ to be smooth and compactly supported on $[-1,1]$ (how do we make $\exp(x)$ fall smoothly outside $[-1,1]$ for example?). Also, I am not sure about the inequality above: the absolute value of $u(\phi)$ is bounded by $\sum_{k=1}^\infty\exp(\frac{1}{k})$, and the sum is actually finite because $\phi$ supposedly has compact support, but the argument still doesn't seem right. What am I missing?