Let $g$ be a compactly supported function on $\mathbb R$, and define $G(x)= \sum_{k\in \mathbb Z} g(x+ 2k\pi )$ such that $G\in L^2(\mathbb T).$
Edit: Put $ H(\xi)=\sum_{k\in \mathbb Z} g(x+ (2k+1)\pi ).$
Can we say that $\hat{H}(-j)= (-1)^j \hat{G}(-j)$? That is, $\int_{\mathbb T} H(\xi) e^{ij \xi} \,d\xi = (-1)^j\int_{\mathbb T} G(\xi) e^{ij \xi} \,d\xi $?
Edit: Note: If the above is true, then the following is also true. I think, this can be shown easily.
Can we expect $$\sum_{n\in \mathbb Z} (-1)^n \hat{G}(n) \overline{\hat{G}(n)} = \int_{\mathbb T}\left( \sum_{k\in \mathbb Z} g(x+ (2k+1)\pi ) \right) \overline{ \left( \sum_{p\in \mathbb Z} g(x+ 2p\pi ) \right)} \,dx \text{?}$$
Side thoughts: I think, I might needs to use the Plancherel theorem, and the motivation for me is that this kind of trick used in the paper I have been reading- roughly speaking this helps us to compute $L^2(\mathbb R)-$norm after some some modifications (e.g., Poisson summation formula)
Edit: Attempt: \begin{eqnarray*} RHS & = & \int_{\mathbb T}\left( \sum_{k\in \mathbb Z} g(x+ (2k+1)\pi ) \right)\overline{ \left( \sum_{p\in \mathbb Z} g(x+ 2p\pi ) \right)} dx\\ & = & \int_{\mathbb T}\left( \sum_{k\in \mathbb Z} g(x+ (2k+1)\pi ) \right)\overline{ \left( \sum_{j \in \mathbb Z} \hat{G}(j) e^{-ij\xi} \right)} \,dx\\ & = & \sum_{j\in \mathbb Z} \overline{\hat{G}(j)} \int_{\mathbb T} H(\xi) e^{ij\xi} \,d\xi. \end{eqnarray*}