Fourier transform, series, distribution: $F\phi=\sum_{n \in \mathbb{Z}} \phi(n), ~~~~~ \phi \in \mathcal S(\mathbb{R})$

Question

$$F\phi=\sum_{n \in \mathbb{Z}} \phi(n), ~~~~~ \phi \in \mathcal S(\mathbb{R})$$ How can I show that the series converges to a tempered distribution? I tried: $$|F\phi|=|\sum_{n \in \mathbb{Z}} \phi(n)| \le \sum_{n \in \mathbb{Z}} ||\phi||$$ But I don't think that's enough...

Another thing: If I wanted the Fourier transform could I write: $$(\mathcal F F)\phi= F(\mathcal F \phi)=\sum_{n \in \mathbb{Z}} \mathcal F \phi(n)$$ ?

Answer 1

Since $\phi \in \mathcal S(\mathbb R)$ the function $(1+x^2)\phi$ is bounded. Therefore, there exists a constant $C$ such that $$ \left| \phi(x) \right| \leq \frac{C}{1+X^2}$$

This gives $$\sum_{n \in \mathbb{Z}} |\phi(n)| \leq \sum_{n \in \mathbb{Z}}\frac{1}{1+n^2}$$

For the second question, look for Poisson Summation Formula.