I have given a functional $l$ on $C_c^\infty(\mathbb{R}^n)$. Now let's assume that for any $p \in \mathbb{R}^n$ we have a neighborhood $V_p$ and a $2\pi$-periodic $C^\infty$-function $u_p$ on $\mathbb{R}^n$, such that
$ \forall \varphi \in C_c^\infty(V_p) $ (compact support in $V_p$)$ \colon \, l(\varphi) = \langle u_p , \varphi \rangle := 1/(2\pi)^n \int u_p \varphi$
So locally the functional is given by $u_p$. If I have overlapping neighborhoods $V_p$ and $V_q$ one can easily conclude that $l = \langle u_p , \cdot \rangle = \langle u_q , \cdot \rangle$ on $C_c^\infty(V_p \cap V_q)$. But since $u_p,u_q$ are not compactly supported on $V_p \cap V_q$ I can not conclude directly $u_p = u_q$ on $V_p \cap V_q$.
Am I right so far? How can I show that $u_p = u_q$ on the overlapping area?
In fact there is an easier way to look at the problem: Let $U$ be such an overlapping area.
Consider the pairing $\, C^\infty(U) \times C_c^\infty(U) \to \mathbb{C}\\ \, (u,\phi) \mapsto \langle u,\phi \rangle$
All we need to show is that this is non degenerate: (by contraposition) If $u \neq 0$, then $u$ does not vanish on an open set $V$. Take any bump function $\phi$ with compact support in $V$, then $\langle u,\phi \rangle \neq 0 $.