Let $\tau>0$, and let $T\in \mathcal{D}'(\mathbb{R})$ be a $\tau$-periodic distribution (that is, 
 $
\langle T, \varphi(\cdot+\tau)\rangle= \langle T,\varphi\rangle
$ 
for all $\varphi \in \mathcal{D}(\mathbb{R})$). Then 
$$
T=\sum\_{n\in \mathbb{Z}} c\_n e^{i 2\pi t/\tau},
$$
for some $c\_n\in \mathbb{C}$, and where the equality means that the symmetric partial sums of the series on the right hand side converge in $\mathcal{D}'(\mathbb{R})$ to $T$. What are the $c\_n$s in terms of $T$? One would think that they are given by $c\_n=\langle T|\_{(0,2\pi)}, e^{-in2\pi /\tau}\rangle/\tau$, but $e^{-in2\pi/\tau}$ is not a test function in $\mathcal{D}((0,2\pi))$. 



Fourier coefficients of a periodic distribution?
Just a quick complement to what Paul said, in order to explain more concretely what "descends" means. Take $\tau=1$ for simplicity. Let $\rho$ be a function in $\mathscr{D}(\mathbb{R})$ ($\mathscr{S}(\mathbb{R}$) would work too) which gives a partition of unity
of the form
$$
\sum\_{n\in\mathbb{Z}}\rho(t+n)=1
$$
for all $t\in\mathbb{R}$. then $c\_n$ can be extracted as
$$
c\_n=\langle T,\rho(t)e^{-2i\pi n t}\rangle\ .
$$
Indeed if $n\in\mathbb{Z}$,
$$
\int\_{\mathbb{R}}\rho(t)e^{2i\pi nt}\ dt
=\sum\_{m\in\mathbb{Z}}\int\_{m}^{m+1} \rho(t)e^{2i\pi nt}\ dt
$$
$$
=\sum\_{m\in\mathbb{Z}}\int\_{0}^{1} \rho(t+m)e^{2i\pi n(t+m)}\ dt
$$
$$
=\int\_{0}^{1}\left(\sum\_{m\in\mathbb{Z}}\rho(t+m)\right)e^{2i\pi nt}\ dt
$$
$$
=\int\_{0}^{1}e^{2i\pi nt}\ dt\ = \delta\_{n,0}\ .
$$


