Vinogradov's Theorem/Minor Arcs/Lemma 1

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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:

\(\ds \size {\sum_{k \mathop = N_1}^{N_2} \map e {\alpha k} }\) \(=\) \(\ds \size {\frac {\map e {\alpha \paren {N_2 + 1} } - \map e {\alpha N_1} } {\map e \alpha - 1} }\)
\(\ds \) \(\le\) \(\ds \frac 2 {\size {\map e \alpha - 1} }\)


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:

\(\ds \size {\map \sin {\pi \alpha} }\) \(=\) \(\ds \size {\sin {\pi \norm \alpha} }\) Sine and Cosine are Periodic on Reals
\(\ds \) \(=\) \(\ds \map \sin {\pi \norm \alpha}\) as $\sin x > 0$ for $x \in \closedint 0 {\pi/2}$
\(\ds \) \(\ge\) \(\ds 2 \norm \alpha\)

This completes the proof.

$\blacksquare$