We know that given a sufficiently regular function $f: \mathbb{R} \to \mathbb{R}$, then its periodisation (say to period $1$) is given by
$$
\begin{align}
F(x) := \sum\_{n\in\mathbb{Z}} f(x + n).\tag{$A$}
\end{align}
$$


Say instead we have a periodic function $F:[0,1] \to \mathbb{R}$ with period 1. Is it possible/is there some procedure by which one can determine the function $f : \mathbb{R} \to \mathbb{R}$ such that we can write $F$ as $(A)$?


**Example**
For example, if $f : \mathbb{R}\backslash\{0\} \to \mathbb{R}$ is defined as $f(x) = \frac{1}{x^2}$, then we know that
$$
\sum\_{n\in\mathbb{Z}}f(x+n) = \sum\_{n\in\mathbb{Z}}\frac{1}{(x+n)^2}=\pi^2\csc^2(\pi x)
$$
and thus $F(x) = \pi^2\csc^2(\pi x)$.


Although, what if we only knew $F$ and didn't a-priori know the summation equation relation, is there a procedure to determine $f$?


**NOTE:** Although a method has been described by Piero D'Ancona, it would be interesting to know whether other such methods/procedures exist.


**NOTE 2 (4/7/22):** Following on from the comments of @Nick S, it may be the case that what I am trying to do is perform a localisation in the sense of [*On the Duality of Regular and Local Functions - Jens Fischer*](https://www.mdpi.com/2227-7390/5/3/41).



How to unperiodise a function
Take a periodic partition of unity, that is to say a test function $\chi(x)$ such that $\sum \chi(x+n)=1$, and define $f(x)=F(x)\chi(x)$.


In case you are wondering, a periodic partition of unity is easy to build: pick a non negative test function $\psi(x)$ which is strictly positive on $[0,1]$ and zero outside $[-1/2,3/2]$ and define $\chi(x)=\psi(x)/\sum\psi(x+n)$.


