Vinogradov's Theorem/Minor Arcs/Lemma 1

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Lemma

For $\beta \in \R$, define $\|\beta\| = \min\{|n - \beta| : n \in \Z\}$.

Then for any $\alpha \in \R$,

$\displaystyle \left| \sum_{k = N_1}^{N_2} e(\alpha k) \right| \leq \min\left\{ N_2 - N_1, \frac1{2\|\alpha\|} \right\}$

The bound $N_2 - N_1$ is trivial; since $|e(\alpha k)| = 1$ for all $k$, by the Triangle Inequality we have:

$\displaystyle \left| \sum_{k = N_1}^{N_2} e(\alpha k) \right| \leq \sum_{k = N_1}^{N_2}1 = N_2 - N_1$

To show the second bound we evaluate the sum as a geometric series.

We have $e(\alpha k) = e(\alpha)^k$, so by Sum of Geometric Progression,

	$\displaystyle $	$\displaystyle $	$\displaystyle $	$\displaystyle \left\vert \sum_{k = N_1}^{N_2} e(\alpha k) \right\vert$	$=$	$\displaystyle \left\vert \frac{e(\alpha(N_2 + 1)) - e(\alpha N_1)}{e(\alpha) - 1} \right\vert$	$\displaystyle $	$\displaystyle $	$\displaystyle $
	$\displaystyle $	$\displaystyle $	$\displaystyle $	$\displaystyle $	$\leq$	$\displaystyle \frac{2}{\vert e(\alpha) - 1 \vert}$	$\displaystyle $	$\displaystyle $	$\displaystyle $

$e(\alpha) - 1 = e(\alpha/2) [e(\alpha/2) - e(-\alpha / 2)]$

and

$e(\alpha/2) - e(-\alpha / 2) = \exp(\pi i \alpha) - \exp(-\pi i \alpha) = 2i\sin(\pi \alpha)$

Therefore we have

$\displaystyle \left\vert \sum_{k = N_1}^{N_2} e(\alpha k) \right\vert \leq \frac{1}{|\sin(\pi \alpha)|}$

We know that the sine function is concave on $[0,\pi/2]$, so $\sin(\pi \alpha) \geq 2\alpha$ for $\alpha \in [0,1/2]$.

By definition there is $n \in \Z$ such that $\alpha = n + \|\alpha\|$, and $\|\alpha\| \in [0,1/2]$, so

$\displaystyle $	$\displaystyle $	$\displaystyle $	$\displaystyle \vert \sin(\pi \alpha) \vert$	$=$	$\displaystyle \vert \sin(\pi \Vert\alpha\Vert) \vert$	$\displaystyle $	$\displaystyle $	$\displaystyle $	By Sine and Cosine are Periodic on Reals
$\displaystyle $	$\displaystyle $	$\displaystyle $	$\displaystyle $	$=$	$\displaystyle \sin(\pi \Vert\alpha\Vert)$	$\displaystyle $	$\displaystyle $	$\displaystyle $	Since $\sin(x) > 0$ for $x \in [0,\pi/2]$
$\displaystyle $	$\displaystyle $	$\displaystyle $	$\displaystyle $	$\geq$	$\displaystyle 2 \Vert\alpha\Vert$	$\displaystyle $	$\displaystyle $	$\displaystyle $

This completes the proof.

$\blacksquare$

	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \left\vert \sum_{k = N_1}^{N_2} e(\alpha k) \right\vert\)	\(=\)	\(\displaystyle \left\vert \frac{e(\alpha(N_2 + 1)) - e(\alpha N_1)}{e(\alpha) - 1} \right\vert\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)
	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\leq\)	\(\displaystyle \frac{2}{\vert e(\alpha) - 1 \vert}\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)

\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \vert \sin(\pi \alpha) \vert\)	\(=\)	\(\displaystyle \vert \sin(\pi \Vert\alpha\Vert) \vert\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	By Sine and Cosine are Periodic on Reals
\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(=\)	\(\displaystyle \sin(\pi \Vert\alpha\Vert)\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	Since $\sin(x) > 0$ for $x \in [0,\pi/2]$
\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\geq\)	\(\displaystyle 2 \Vert\alpha\Vert\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)