For $x \in [0,1]$, let $$\tilde{G}(x) = \frac{1}{2(e - 1)}(e^x + e^{1-x})$$
And let $G(x)$ be the 1-periodic extension of $\tilde{G}(x)$ to all of $\mathbb{R}$.
Show that, in the sense of distributions, we have $$G'' + G = \sum_{z \in \mathbb{Z}} \delta_z.$$
This is a simplification of a question from an exam from a couple years ago on spectral theory. The question appears in the context of showing that the laplacian is self-adjoint with Born-von Karman boundary conditions.
I'm (extra) confused by the fact that $G(x)$ is a function extended to all or $\mathbb{R}$. My feeble attempt is to write $G(x)$ as $$G(x) = \sum_{z \in \mathbb{Z}} \mathbb{1}_{x\in [z,z+1]} \tilde{G}(x-z)$$
And then somehow do something with the Fourier transform: $$\langle \mathfrak{F}(G'' + G), \phi \rangle = \dots = \sum \langle \mathbb{1}, \phi \rangle$$
But honestly I have no idea.
Thanks in advance for any suggestions/solutions.