Riesz-product witnesses for a Hadamard-lacunary set

Choose $r$ with $q^r\ge 3$ and split the original indices by residues [given, algebra] modulo $r$. Each resulting frequency class has successive ratio at least $q^r\ge 3$. Each factor of $P_{\nu }$ equals $\mathrm{\mid }1+{\epsilon }_{\lambda }{e}_{\lambda }{\mathrm{\mid }}^2\mathrm{/}2$ and is nonnegative.

On one such class, a nonempty signed sum with coefficients in [step 1.1, algebra] $\{-1,0,1\}$ cannot be zero: its largest frequency exceeds the sum of all smaller possible frequencies, by the ratio-three geometric bound. Therefore the product expansion has constant term only when every factor contributes its $1$. Its integral is consequently $1$.

The same largest-frequency argument says that frequency $-\lambda$ in [step 1.1, step 2.1, algebra] the expansion occurs only by taking $\overline{{\epsilon }_{\lambda }}{e}_{-\lambda }\mathrm{/}2$ from the $\lambda$ factor and $1$ elsewhere. Hence ${\widehat{P}}_{\nu }(-\lambda )=\overline{{\epsilon }_{\lambda }}\mathrm{/}2$, as claimed.