How do I compute the Fourier Coefficients of a Riesz Product?

Question

Let $(\alpha_n)_{n\in \mathbb{N}}$ be a sequence of non-zero real numbers such that $\sum_{n=1}^\infty \lvert \alpha_n \rvert^2 < \infty$. We consider the Riesz Product \begin{align*} \prod_{n=1}^\infty \left (1+2i \alpha_n \sin(2 \pi 3^nx) \right), \quad x\in \mathbb{R}. \end{align*} I want to write this product as a Fourier series \begin{align*} \sum_{k=1}^\infty a_k e^{2\pi ik x}. \end{align*} I already know that the coefficients $a_k$ are $0$ unless there exists an integer $N$ and $\eta_1, \ldots, \eta_N \in \{-1,0,1\}$ with $\eta_N \neq 0$ such that $k=\sum_{j=1}^N \eta_j3^j$ (and that $N$ and the $\eta_j$ are uniquely determined). But i do not know how to prove this. I am also unsure whether the product and the sum above are equal or if it is easier to use means of Fourier Analysis in order to compute the coefficients $a_k$. I feel like I need to use the fact that $\frac{3^{j+1}}{3^j} \geq 3 >1$ for all $j\in \mathbb{N}$, but i do not know how to proceed from here. I have also tried using the expression of $\sin$ as a sum of exponentials, but I wasn't able to do much with that. Any help is appreciated!

Answer 1

Every nonzero integer $k$ has the unique representation of the form $$k=\sum_{n=0}^N \eta_n 3^n,\quad \eta_n\in \{-1,0,1\},\ \eta_N\neq 0\quad (1)$$ This follows from the representation of $k$ in the ternary system, with digits $-1,0,1$ instead of $0,1,2.$ The sequence $\eta_n$ can be determined in the following way. We have $k=\eta_0+3k_1, $ for the unique $\eta_0\in \{-1,0,1\}.$ Then $|k_1|<|k|.$ Next $k_1=\eta_1+3k_2$ for the unique $\eta_1\in \{-1,0,1\}.$ We continue this procedure until we get $k_N=\pm 1 $ In this way we obtain $(1).$ For example $$-11=1+3\,(- 4), \ k_1=-4\\ -4=-1+3\,(-1) ,\ k_2=-1\\ -11=1-1\cdot 3-1\cdot 3^2$$

For simplicity we will put a stronger assumption $\sum |\alpha_n|<\infty.$ Then $$|2i\alpha_n\sin(2\pi 3^nx)|\le 2|\alpha|_n$$ By the stronger assumption we get $$\prod_{n=1}^\infty (1+2|\alpha_n|)<\infty$$ Therefore the product $$\prod_{n=1}^\infty (1+2i\alpha_ne\sin(2\pi i3^nx)= \prod_{n=1}^\infty (1+\alpha_ne^{2\pi i3^nx}-\alpha_ne^{-2\pi i3^nx})$$ is uniformly convergent. Hence $$f(x):=\prod_{n=1}^\infty (1+\alpha_ne^{2\pi i3^nx}-\alpha_ne^{-2\pi i3^nx})=1+\sum_{k\neq 0}^\infty a_ke^{ikx}$$ is a continuous function. Fix $k$ with representation $(1).$ We want to determine the coefficient $a_k$ in the expansion of $f.$ Let $$f_N(x):=\prod_{n=1}^N (1+\alpha_ne^{2\pi i3^nx}-\alpha_ne^{-2\pi i3^nx})\qquad \ \ \ (2)\\ R_{N,M}(x):=\prod_{n=N+1}^M (1+\alpha_ne^{2\pi i3^nx}-\alpha_ne^{-2\pi i3^nx}),\quad M>N$$ We have $$a_k=\langle f,e^{2\pi i kx}\rangle =\lim_M\langle f_M,e^{2\pi i kx}\rangle =\lim_M\langle R_{N,M}f_N,e^{2\pi i kx}\rangle\\ =\lim_M\langle f_N,\overline{R_{N,M}}e^{2\pi i kx}\rangle =\langle f_N,e^{2\pi i kx}\rangle \qquad (3)$$ as $e^{\pm 2\pi i 3^nx}$ for $0\le n\le N$ is orthogonal to $e^{\pm 2\pi i3^mx}e^{2\pi ikx}$ for $m\ge N+1.$ The latter follows from $\pm 3^n\neq \pm 3^m+k$ for $0\le n\le N$ and $m\ge N+1.$ Indeed the representation of $\pm 3^m+k$ is of the form $$\sum_{j=0}^N\eta_j3^j\pm 3^m$$ hence it cannot be equal $\pm 3^n$ for $0\le n\le N.$ The formulas $(1)$, $(2)$ and $(3)$ imply $$a_k=\prod_{n=0}^N\eta_n\alpha_n$$