10-weyl-sums.tex

\section{Estimates for Weyl sums}
\cite[Theorem 5.2]{MR1435742}
Let
\[
f_{k}(\vec\alpha; X) := \sum_{x=1}^{X} e(\alpha_{1}x + \dotsb \alpha_{k}x^{k}).
\]
\begin{theorem}
Let $k$ be an integer with $k > 2$, and let $\vec\alpha \in \R^{k}$.
Suppose that there exists a natural number $j$ with $2 \leq j \leq k$ such that, for some $a \in \Z$
and $q \in \N$ with $(a, q) = 1$, one has $\abs{\alpha_{j} - a/q} \leq q^{-2}$ and $q \leq X^{j}$.
Then one has
\[
f_{k} (\vec\alpha; X) \lesssim X^{1+\epsilon} (q^{-1} + X^{-1} + qX^{-j} )^{1/(k(k-1))}
\]
\end{theorem}
% Almost all coefficients: \cite{MR3519125}, \cite{arxiv:1903.07330}

With notation from \cite[Section 5.2]{MR1435742}
\begin{lemma}
For $2 \leq j \leq k$ and integers $1 \leq x,y \leq N$ we have
\[
\norm{(j \dotsm k) \alpha_{j}(x-y)}
\lesssim
\sum_{h=j-1}^{k-1} \norm{ \gamma_{h}(x) - \gamma_{h}(y) } N^{h-j+1}
\]
\end{lemma}
\begin{proof}
By downward induction on $j$.
Suppose that the claim is already known for all larger $j$.
Expand
\begin{align*}
\gamma_{j-1}(x)-\gamma_{j-1}(y)
&=
\sum_{h=j-1}^{k} \alpha_{h} \binom{h}{j-1} ((-x)^{h-j+1}-(-y)^{h-j+1})
\\ &=
\sum_{h=j}^{k} (-1)^{h-j+1} \alpha_{h} \binom{h}{j-1} (x-y)(x^{h-j}+x^{h-j-1}y+\dotsb+y^{h-j})
\end{align*}
Hence
\begin{align*}
\MoveEqLeft
\norm{(j \dotsm k) \alpha_{j} (x-y)}
=
\norm{((j+1) \dotsm k) \alpha_{j} \binom{j}{j-1} (x-y)}
\\ &\leq
((j+1) \dotsm k) \norm{\gamma_{j-1}(x)-\gamma_{j-1}(y)}
\\ &\quad + \sum_{h=j+1}^{k} \norm{((j+1) \dotsm k) \alpha_{h} \binom{h}{j-1} (x-y)(x^{h-j}+\dotsb+y^{h-j})}
\\ &\lesssim
\norm{\gamma_{j-1}(x)-\gamma_{j-1}(y)} + \sum_{h=j+1}^{k} \norm{(h \dotsm k) \alpha_{h} (x-y)} (x^{h-j}+\dotsb+y^{h-j})
\\ &\lesssim
\norm{\gamma_{j-1}(x)-\gamma_{j-1}(y)} + \sum_{h=j+1}^{k} N^{h-j} \sum_{h'=h-1}^{k-1} \norm{\gamma_{h'}(x)-\gamma_{h'}(y)} N^{h'-h+1}
\\ &=
\norm{\gamma_{j-1}(x)-\gamma_{j-1}(y)} + \sum_{h=j+1}^{k} \sum_{h'=h-1}^{k-1} \norm{\gamma_{h'}(x)-\gamma_{h'}(y)} N^{h'-j+1}
\\ &\sim
\sum_{h'=j-1}^{k-1} \norm{\gamma_{h'}(x)-\gamma_{h'}(y)} N^{h'-j+1}.
\qedhere
\end{align*}
\end{proof}