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Source. Liu--Sawhney, On further questions regarding unit fractions, arXiv:2404.07113v1, Fact 2.5, p. 8; see the source digest.

Statement, with the quadratic coefficient normalized. Define $e(x)=\exp(2\pi ix)$. Uniformly for $|x|\leq1/2$ and $q\in[0,1]$,

$$\left|(1-q)+qe(x) -e(qx)\bigl(1-2\pi^2q(1-q)x^2\bigr)\right| \ll |x|^3, \tag{1}$$

and

$$|(1-q)+qe(x)|\leq1-8q(1-q)x^2. \tag{2}$$

Source correction. The PDF prints $2\pi$ in (1), but the quadratic Taylor terms displayed in its own proof have coefficient $2\pi^2$. With the stated normalization of $e$, the coefficient in (1) must be $2\pi^2$. Its displayed expansion of $e(-qx)$ also has the wrong sign in the linear term. The proof below makes these local algebraic corrections explicit. The coefficient typo recurs in the proof of Lemma 3.1.

Proof. Multiply the Fourier factor by $e(-qx)$, which has absolute value one. Taylor expansion with a uniform cubic remainder gives

$$\begin{aligned} (1-q)e(-qx) &=(1-q)\bigl(1-2\pi iqx-2\pi^2q^2x^2\bigr)+O(|x|^3),\\ qe((1-q)x) &=q\bigl(1+2\pi i(1-q)x -2\pi^2(1-q)^2x^2\bigr)+O(|x|^3). \end{aligned}$$

The constants add to one, the linear terms cancel, and the quadratic terms add to $-2\pi^2q(1-q)x^2$. This proves (1).

For (2), $\cos(2\pi x)\leq1-8x^2$ on $[-1/2,1/2]$. One way to check this elementary inequality is to use the concavity bound $\sin(\pi|x|)\geq2|x|$ and $1-\cos(2\pi x)=2\sin^2(\pi x)$. Consequently

$$\begin{aligned} |(1-q)+qe(x)|^2 &=(1-q)^2+q^2+2q(1-q)\cos(2\pi x)\\ &\leq1-16q(1-q)x^2. \end{aligned}$$

Here $0\leq16q(1-q)x^2\leq1$. Applying $\sqrt{1-u}\leq1-u/2$ proves (2).

Dependencies. Elementary Taylor expansion and trigonometric inequalities; no external theorem is invoked.

Bears on. #298 and #299, through the Fourier analysis in the quantitative reciprocal-sum criterion.