Vinogradov's Theorem/Minor Arcs/Lemma 1

Lemma
For $\beta \in \R$, define:
 * $\left\Vert{\beta}\right\Vert := \min \left\{ {\left\vert{n - \beta}\right\vert : n \in \Z}\right\}$

Then:
 * $\displaystyle \forall \alpha \in \R: \left\vert{\sum_{k \mathop = N_1}^{N_2} e \left({\alpha k}\right)}\right\vert \le \min \left\{ {N_2 - N_1, \frac 1 {2 \left\Vert{\alpha}\right\Vert} }\right\}$

Proof
The bound $N_2 - N_1$ is trivial:

Since $\left\vert{e \left({\alpha k}\right)}\right\vert = 1$ for all $k$, by the Triangle Inequality:


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

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

We have:
 * $e \left({\alpha k}\right) = e \left({\alpha}\right)^k$

So by Sum of Geometric Progression:

By the polar form of a complex number:


 * $e \left({\alpha}\right) - 1 = e \left({\alpha / 2}\right) \left({e \left({\alpha / 2}\right) - e \left({-\alpha / 2}\right)}\right)$

and:


 * $e \left({\alpha / 2}\right) - e \left({-\alpha / 2}\right) = \exp \left({\pi i \alpha}\right) - \exp \left({-\pi i \alpha}\right) = 2 i \sin \left({\pi \alpha}\right)$

Therefore:


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

We know that the sine function is concave on $\left[{0 \,.\,.\, \pi/2}\right]$.

So:
 * $\sin \left({\pi \alpha}\right) \ge 2 \alpha$

for $\alpha \in \left[{0 \,.\,.\, 1/2}\right]$.

By definition there exists $n \in \Z$ such that:
 * $\alpha = n + \left\Vert{\alpha}\right\Vert$

and:
 * $\left\Vert{\alpha}\right\Vert \in \left[{0 \,.\,.\, \pi/2}\right]$.

So:

This completes the proof.