Is this a continuous linear form on $\mathcal{D}(\mathbb{R})$ ($C^\infty$ with compact support) such that for any sequence $\varphi_n\to0 \Rightarrow \langle T, \varphi_n \rangle \to 0$.
$\int_\mathbb{R}\sum_{m \in \mathbb{N}} e^{ -(x-m)^2 } \varphi_n(x) dx \to 0$ as $n \to \infty$
I am guessing that it is not, so I could try to find a function for which this does not work, this is how far I have got, am I on the right track, should I try to use the translation of $\varphi_n$ for the counter example?
$\int_\mathbb{R} e^{ -x^2 } \sum_{m \in \mathbb{N}} \varphi_n(x+m) dx $
Your $f$ is bounded, hence locally integrable (so it is a distribution ). For any x,
$$\sum_{n\in \Bbb N} \exp( -(x-n)^2 ) \leq \sum_{n\in \Bbb Z} \exp( -(x-n)^2 ) $$
But $x-n = \{x\} + \lfloor x\rfloor - n = \{x\} + k$ (where $ \{x\}$ is the fractional part of $x$) $$\leq \sum_{k\in \Bbb Z} \exp( -(k+ \{x\})^2 ) $$
$$\leq \sum_{k\in \Bbb Z} \exp( -k^2 ) $$
$$\leq 2 \sum_{n \in \Bbb N} e^{-n^2} = C < +\infty $$
And every bounded function is a distribution :
Take a sequence $\varphi_n \to 0$ in $\mathcal{D}$, there exists a compact $K$ such that $\forall n, \text{Supp}(\phi_n) \subset K$, and we have
$$\left| \langle T_f, \varphi \rangle \right| = \left| \int_{\Bbb R} f(x) \varphi_n(x) dx \right| = \left| \int_{K} f(x) \varphi_n(x) dx \right| \leq \| \varphi_n \|_\infty \| f \|_\infty \mu( K ) \to 0 $$