# How do I show $\|(e^{-2\pi ihx_j} - 1)/h \| \leq 2 \pi \|x\|$?

How do I show $\|(e^{-2\pi ihx} - 1)/h \| \leq 2 \pi \|x\|$? for each h and x? I thought about using taylor expansion of Euler's formula, but it did not work out. Thank you in advance.

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The inequality $|e^{i\theta}-1| \le |\theta|$ is clear from geometry: The LHS is the euclidean length between $e^{i\theta}$ and $1$ as complex numbers on the unit circle, while the RHS is the radian measure. Now set $\theta=-2\pi h x$ and you are done.
I'm assuming $x,h \in \mathbf{R}$. Multiply both sides by $|h|$ and set $y = 2\pi hx$ to get the equivalent inequality $$|e^{-i y} - 1| \le |y|, \quad y \in \mathbf{R}.$$ Now this can be equivalently written as $$|\cos(y) - 1 - i \sin(y)| \le |y|.$$ Square both sides and expand to get the equivalent inequality $$\cos^2(y) - 2 \cos(y) + 1 + \sin^2(y) \le y^2.$$ When you simplify this, you get $$f(y)=\frac{1}{2}y^2 + \cos(y) - 1 \ge 0.$$ Now at $y=0$ the inequality holds. We'll show that $y=0$ is a global minimum for the function $f$. Indeed notice that as $y \to \pm \infty$, $f(y) \to \infty$. Differentiating $f$ we get $$f'(y) = y - \sin(y).$$ We wish to solve $f'(y) = 0$, i.e. the equation $$\sin(y) = y.$$ It is well known that $\sin(y) = y$ if and only if $y=0$, so we are done.
(The last fact is also easy to prove: $x \mapsto \sin(x) - x$ is strictly decreasing by a simple inspection of the derivative.)