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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 $\mathcal{D}((0,2\pi))$.

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    $\begingroup$ I think the issue is rather that $T$ is not of compact support and thus cannot be tested directly with functions in $\mathcal{E}$. $\endgroup$ May 2 '19 at 16:20
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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 $n\in\mathbb{Z}$, $$ \int_{\mathbb{R}}\rho(t)e^{2i\pi nt}\ dt =\sum_{m\in\mathbb{Z}}\int_{m}^{m+1} \rho(t)e^{2i\pi nt}\ dt $$ $$ =\sum_{m\in\mathbb{Z}}\int_{0}^{1} \rho(t+m)e^{2i\pi n(t+m)}\ dt $$ $$ =\int_{0}^{1}\left(\sum_{m\in\mathbb{Z}}\rho(t+m)\right)e^{2i\pi nt}\ dt $$ $$ =\int_{0}^{1}e^{2i\pi nt}\ dt\ = \delta_{n,0}\ . $$

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Both the exponentials and any periodic distribution descend to quotients (circles) $\mathbb R/(\tau\cdot \mathbb Z)$, and there $T$ can be legitimately evaluated on the exponentials.

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