# Fourier coefficients of a periodic distribution?

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

• I think the issue is rather that $T$ is not of compact support and thus cannot be tested directly with functions in $\mathcal{E}$. May 2 '19 at 16:20

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 $$n\in\mathbb{Z}$$, $$\int_{\mathbb{R}}\rho(t)e^{2i\pi nt}\ dt =\sum_{m\in\mathbb{Z}}\int_{m}^{m+1} \rho(t)e^{2i\pi nt}\ dt$$ $$=\sum_{m\in\mathbb{Z}}\int_{0}^{1} \rho(t+m)e^{2i\pi n(t+m)}\ dt$$ $$=\int_{0}^{1}\left(\sum_{m\in\mathbb{Z}}\rho(t+m)\right)e^{2i\pi nt}\ dt$$ $$=\int_{0}^{1}e^{2i\pi nt}\ dt\ = \delta_{n,0}\ .$$
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.