 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 $3hl(2x+h+l)$.