Let $D(\mathbb{R})$ be the space of test functions on $\mathbb{R}$, i.e. the set of smooth compactly supported functions on $\mathbb{R}$, and where we say $\phi_k \to \phi$ if $\phi_k$ and all its derivatives converge uniformly to $\phi$.
Question: Is the functional on $D(\mathbb{R})$ defined by $T(\phi)=\Sigma_{n=1}^{\infty}\frac{d^n}{dx^n}\phi(\frac{1}{n})$ continuous?
If I look at $\vert T(\phi_k) - T(\phi)\vert$ I get a series $\vert \phi_k'(1) - \phi'(1)\vert + \vert \phi''(\frac{1}{2}) - \phi''(\frac{1}{2})\vert + \vert \phi_k'''(\frac{1}{3})-\phi'''(\frac{1}{3})\vert$. I thought I could maybe find some creative choice of $\epsilon_n$ for which $\vert \phi_k^n(\frac{1}{n}) - \phi^n(\frac{1}{n})\vert < \epsilon_n$ for $k>N\in \mathbb{N}$ which makes the series converge to less than $\epsilon$, but I'm not sure how to actually do this (or if it's even possible). Any insight on how to attack the above question would be helpful.
The series in the definition of $T$ is not even convergent on $D(\mathbb{R})$. To see that consider $\phi (x)=\varphi(x)\sin x$, where $\varphi\in D(\mathbb{R})$ is such that $\varphi\equiv 1$ on $(0,1)$. Then $T(\phi)=T(\sin x)$ and note that for every $n$ odd we have $$ \Big|\frac{d^n\sin}{dx^n}(1/n)\Big|=|\cos(1/n)|\ge |\cos(1/2)|=1/\sqrt{2}, \qquad n\ge 2. $$