Sum of Fourier transforms is a trigonometric polynomial I'm interested in the following problem: Suppose that the Fourier transform of an $L^2$ function $g(x)$ is resembled by $\hat g(\gamma)$, and that $\langle g,T_ng\rangle = \int_{\mathbb R} g(x) \overline{g(x-n)} dx $. Then if $g\in L^2$ then
a) The sum $\sum_n \vert \hat g(\gamma + n)\vert^2$ converges in $L^1$ on $[0,1].$
b) $$\sum_n \vert \hat g(\gamma + n)\vert^2 = \sum_n \langle g,T_ng\rangle e^{-2\pi i n\gamma}. $$
c) Conclude that if $g\in L^2$ and has compact support, then $\sum_n \vert \hat g(\gamma + n)\vert^2$ is a trigonometric polynomial.
To prove b, since I see no other way, I take the direct approach of invoking the definition of FT and placing it in the sum
$$  \sum_n \vert \hat g(\gamma + n)\vert^2 = \sum_n \vert \int_{\mathbb{R}}g(x)e^{-2\pi i (\gamma + n) x}dx\vert^2   $$ which leads to nothing. Alternatively, starting from the right side of b,
$$ \sum_n \langle g,T_ng\rangle e^{-2\pi i n\gamma} = \sum_n \int_{\mathbb{R}} g(x) \overline{g(x-n)}e^{-2\pi i n \gamma} dx = \sum_n \int_{\mathbb{R}} g(x) \overline{g(x-n)e^{2\pi i n \gamma}} dx $$
which, again, leads nowhere.
Any help will be appreciated!
 A: I came across the same exercise 7.10 & 7.11 in David F. Walnut's Introduction to Wavelet Analysis. The solution is basically going the reverse direction of Lemma 7.4 in the book.
Notice the RHS of b) is a Fourier series. The idea is to use uniqueness of Fourier series and show that LHS's Fourier coefficient is ${\langle g, T_kg\rangle}$
\begin{align}
\int_0^1\sum_n\left|\hat g (\gamma+n)\right|^2e^{2\pi ik\gamma} d\gamma &= \sum_n\int_0^1\left|\hat g (\gamma+n)\right|^2e^{2\pi ik\gamma} d\gamma \\ &=
\sum_n\int_n^{n+1}\left|\hat g(\gamma)\right|^2e^{2\pi ik\gamma}d\gamma \\ &=
\int_{\mathbf{R}}\hat g(\gamma)\overline{\hat g(\gamma)e^{-2\pi ik\gamma}}d\gamma \\ &=
\int_{\mathbf{R}} g(t)\overline{g(t-k)}dt \\ &= \langle g, T_k g\rangle
\end{align}
Interchange of sum and integral can be justified by monotonic convergence theorem. Alternatively, a) gives a shortcut to it.
Since $\sum_n\left|\hat g (\gamma+n)\right|$ converges in $L^1_{[0,1)}$, $lim_{N\to\infty}\int_0^1\left|\sum_{|n|> N}\left|\hat g (\gamma+n)\right|^2\right| d\gamma = 0$
\begin{align}
& lim_{N\to\infty}\left|\int_0^1\sum_n\left|\hat g (\gamma+n)\right|^2e^{2\pi ik\gamma} d\gamma - \sum_{|n|\leq N}\int_0^1\left|\hat g (\gamma+n)\right|^2e^{2\pi ik\gamma} d\gamma \right| \\ 
=& lim_{N\to\infty}\left|\int_0^1\sum_n\left|\hat g (\gamma+n)\right|^2e^{2\pi ik\gamma} d\gamma - \int_0^1\sum_{|n|\leq N}\left|\hat g (\gamma+n)\right|^2e^{2\pi ik\gamma} d\gamma \right| \\
\leq& lim_{N\to\infty}\int_0^1\left|\sum_{|n|> N}\left|\hat g (\gamma+n)\right|^2e^{2\pi ik\gamma}\right| d\gamma \\
=& lim_{N\to\infty}\int_0^1\left|\sum_{|n|> N}\left|\hat g (\gamma+n)\right|^2\right| d\gamma \\
=& 0
\end{align}
