Estimate (8.18): the $L^\infty$ bound off the strongly major arc

$L^\infty$ estimate off the strongly major arc, eq. (8.18). Let $8.7\times10^{36}\le x\le e^{3100}$ be an integer, $K=1000$, $N_0=4\times10^{14}$, $T_0=3.29\times10^9$. For every $\alpha$ with $\|\alpha\|_{\mathbb R/\mathbb Z}>T_0/(3.6\pi x)$,

This is the pointwise input to the minor-arc bound; combined with Proposition 8.4 by Hölder it bounds the minor-arc contribution to (8.11) by $0.078\times7.09<0.56$ times $\frac{x^2}{K}(N_0/3)^3$, which is what the main term of Proposition 8.3 beats. The paper proves it by symmetry reducing to $T_0/(3\pi x)\le\alpha\le\frac12$ and splitting into regimes:

• weakly major, $\frac{T_0}{3.6\pi x}<\alpha\le\frac{KT_0}{4\pi x}$: Proposition 7.2 plus Lemma 3.3, $|\hat\eta_0(\alpha x/K)|\le 7.2K\log2/T_0\le0.077$;
• $\alpha\ge20/N_0$: the crude bounds $|S|\le1.01\,x/K$ (Lemma 4.3, (5.11)) and $|D_{N_0/3}(\alpha)|\le1/(2\alpha)$;
• (8.19) $\frac{KT_0}{3\pi x}\le\alpha<\frac{20}{N_0}$: reduce to (8.20) $|S_{\eta_0,\sqrt{x/K}\sharp}(x/K,\alpha)|\le0.078x$, write $4\alpha=\frac1q+\mathcal O^*(1/q^2)$ with $\frac{N_0}{80}-1\le q\le\frac{\pi x}{KT_0}$, and apply Theorem 1.3 — (1.12) for $(x/K)^{2/3}\le q$, (1.9) for $(x/K)^{1/3}\le q\le(x/K)^{2/3}$, (1.10) for $q\le(x/K)^{1/3}$ — each giving a bound below $0.078\,x/K$ once $\log x\le3100$ and $T_0=3.29\times10^9$ are inserted.