Weyl differencing

Taking a finite difference lowers the degree of a polynomial phase. For $S(\theta )={\sum }_{x=1}^{n}e(\theta P(x))$, expand $|S{|}^2$ by shifts $h$ and apply the Cauchy-Schwarz inequality to those shift sums. This gives $|S{|}^4\le 2n{\sum }_{h,l}{\sum }_{x\in I_{h,l}}e(\theta {\Delta }_h{\Delta }_{l}P(x))$, where $I_{h,l}$ enforces that $x,x+h,x+l,x+h+l$ lie in $[1,n]$. Although individual terms can be complex, the entire sum is real and nonnegative because it equals a sum of squares. For $P(x)=x^3$, the phase is $3hl(2x+h+l)$.

For a real degree-$k$ polynomial with leading coefficient $\theta$, $k\ge 2$, and coprime integers $a,q$ satisfying $|\theta -a/q|\le q^{-2}$, the displayed bound applies to its exponential sum over $1\le n\le N$. Repeated Weyl differencing reduces the phase to slope $k!\theta h_1\cdots h_{k-1}$. A geometric-sum estimate, the subpower bound for the divisor function and the reciprocal fractional-part sum near a rational complete the estimate. The arbitrary positive exponent $\eta$ absorbs divisor bounds and logarithms.

If $S(\theta )={\sum }_{x\le n}e(\theta x^3)$ and $|\theta -a/q|\le q^{-2}$ with $(a,q)=1$, the displayed estimate holds. Twice applying Weyl differencing bounds $|S{|}^4$ by $O(n^3+n{\sum }_{0<|h|,|l|<n}\min (n,\Vert 6\theta hl{\Vert }^{-1}))$. Group by $6hl$ and use the subpower bound for the divisor function. The reciprocal fractional-part sum near a rational bounds the resulting sum by $O(\log (2q)(n^3/q+n^2+n+q))$. Take fourth roots, absorbing logarithms into $n^{\eta }$ when $q\le n^3$; for larger $q$ the trivial bound suffices. Leiden lecture notes, Section 7.2 develop this cubic differencing estimate.

For $1\le j\le k$ and every $\eta >0$, the displayed estimate bounds a mean value of a polynomial exponential sum. For cubes its eighth-moment bound is $O_{\eta }(n^{5+\eta })$. It complements pointwise Weyl differencing estimates: a ninth moment over a region is bounded by its supremum there times this eighth moment. Leiden lecture notes, Section 8.1 give the cubic formulation.

Write $S(\theta )={\sum }_{x\le n}e(\theta x^3)$ and $I_j={\int }_0^1|S{|}^j$. Character orthogonality gives $I_2=n$. For $I_4$, fix the difference $D=x^3-y^3$. If $D=0$, there are $n^2$ choices for the two equal pairs. If $D\ne 0$, any solution $z^3-w^3=D$ has $z-w\mid D$; fixing this divisor gives a quadratic equation for $w$, hence at most two choices. The subpower bound for the divisor function yields $I_4{\ll }_{\eta }{n}^{2+\eta }$.

The twice-differenced inequality in Weyl differencing reads $|S{|}^4\le 2n{\sum }_v{c}_{v}e(\theta v)$, with $c_v$ counting triples $(h,l,x)$ for which $v=3hl(2x+h+l)$ and all four shifted arguments are in $[1,n]$. Thus $c_0\ll n^2$, and $c_v{\ll }_{\eta }{n}^{\eta }$ for $v\ne 0$: the product $hl$ divides $v$, and the remaining equation determines $x$. The Fourier coefficients $r(v)$ of $|S{|}^4$ are nonnegative counts, with $r(0)=I_4$ and ${\sum }_{v}r(v)=n^4$. Multiply the inequality by $|S{|}^4$ and integrate. It follows that $I_8\ll n(n^2{I}_4+n^{\eta }{n}^4){\ll }_{\eta }{n}^{5+\eta }$, after choosing the divisor-bound exponents small enough.