Vinogradov's Theorem/Minor Arcs/Lemma 1

Lemma
For $\beta \in \R$, define:
 * $\norm \beta := \min \set {\size {n - \beta}: n \in \Z}$

Then:
 * $\ds \forall \alpha \in \R: \size {\sum_{k \mathop = N_1}^{N_2} \map e {\alpha k} } \le \min \set {N_2 - N_1, \frac 1 {2 \norm \alpha} }$

Proof
The bound $N_2 - N_1$ is trivial:

Since $\size {\map e {\alpha k} } = 1$ for all $k$, by the Triangle Inequality:


 * $\ds \size {\sum_{k \mathop = N_1}^{N_2} \map e {\alpha k} } \le \sum_{k \mathop = N_1}^{N_2} 1 = N_2 - N_1$

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

We have:
 * $\map e {\alpha k} = \map e \alpha^k$

So by Sum of Geometric Sequence:

By the polar form of a complex number:


 * $\map e \alpha - 1 = \map e {\alpha / 2} \paren {\map e {\alpha / 2} - \map e {-\alpha / 2} }$

and:


 * $\map e {\alpha / 2} - \map e {-\alpha / 2} = \map \exp {\pi i \alpha} - \map \exp {-\pi i \alpha} = 2 i \map \sin {\pi \alpha}$

Therefore:


 * $\ds \size {\sum_{k \mathop = N_1}^{N_2} \map e {\alpha k} } \le \frac 1 {\size {\map \sin {\pi \alpha} } }$

We know that the sine function is concave on $\closedint 0 {\pi / 2}$.

So:
 * $\map \sin {\pi \alpha} \ge 2 \alpha$

for $\alpha \in \closedint 0 {1/2}$.

By definition there exists $n \in \Z$ such that:
 * $\alpha = n + \norm \alpha$

and:
 * $\norm \alpha \in \closedint 0 {\pi/2}$.

So:

This completes the proof.