Thue–Morse Fourier product (${\sum }_{r<2^j}(-1{)}^{s_2(r)}e(r\theta )={\prod }_{l=0}^{j-1}(1-e(2^l\theta ))$)

Each binary digit independently contributes either $1$ or $-e(2^l\theta )$, proving the product. Two adjacent factors have modulus $8|\cos (\pi t)|(1-{\cos }^2(\pi t))\le 16/(3\sqrt{3})<4$. Consequently the product is $O(2^{\sigma j})$, with $\sigma =2-\frac{3}{4}{\log }_{2}3<1$. This two-factor estimate suffices for a power saving, though its exponent is not optimal.

An integer interval is a disjoint union of aligned binary blocks with at most two of each length. On $[a{2}^j,(a+1)2^j)$ the Thue–Morse sequence sign factors as $f(a)f(r)$, so the Thue–Morse Fourier product bounds its exponential sum by $O(2^{\sigma j})$. Sum the geometric series over block lengths to obtain the uniform bound, for $\sigma =2-\frac{3}{4}{\log }_{2}3$.

Let $f(r)=(-1{)}^{s_2(r)}$ be the Thue–Morse sequence, let $e(t)=e^{2\pi it}$, and let $N\ge 2$. For each positive integer $m\le N^{1/100}$, take an integer interval $I_m\subseteq [1,N/m]$. The range condition ensures that every product $mn$ remains at most $N$.

Choose the integer interval $J_m$ from $m\min I_m$ to $m\max I_m$, omitting an empty $I_m$. Character orthogonality givesThe uniform interval cancellation for binary digit parity bounds each inner exponential sum by $C{N}^{\sigma }$, with $\sigma =2-\frac{3}{4}{\log }_{2}3<1$. The average over $a$ does not increase that bound. ThereforeThe exponent is negative, so this normalized Type I sum bound is smaller than every fixed negative power of $\log N$ for large $N$. No cancellation between the outer terms is needed; each inner progression is controlled separately.