How to prove the inequality $|e^{2 \pi i \alpha x} - e^{2 \pi i \alpha y}| \ll \|\alpha(x-y)\|$.

Question

How to prove the inequality $|e^{2 \pi i \alpha x} - e^{2 \pi i \alpha y}| \ll \|\alpha(x-y)\|$?

Here $\|\alpha x\|$ is the distance from $\alpha x$ to the nearest integer.

My attempt was to rewrite the LHS of the inequality as $|e^{2 \pi i \alpha(x-y)} -1| = |e^{2 \pi i \alpha \frac{x-y}{2}} - e^{2 \pi i \alpha \frac{y-x}{2}}| = 2|\sin(\pi \alpha (x-y))|$.

But the issue is it seems the opposite inequality is true, since I think one can prove that $\sin(\pi \alpha(x-y)) \gg \|\alpha(x-y)\|$.

Answer 1

You have

$$|\exp^{2 \pi i \alpha x} - \exp^{2\pi i \alpha y} | = |e^{2\pi i \alpha(x-y)}-1|$$

Now check that $$|e^{2\pi i t} -1| = |e^{2\pi i (t+ m)} -1|$$ if $m$ is an integer and also $$|e^{2\pi i t} -1|\le 2 \pi |t|$$

Obs: you may be missing a factor of $2\pi$.