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Rudin in Real and Complex Analysis uses this in a proof near the beginning of chapter 9:

$\displaystyle \left \vert{ \frac {e^{-ixu}-1}{u}}\right\vert \le \vert x \vert$ for all real $u \ne 0$

Why is this true?

Edit: I believe $x$ is real

1 Answers 1

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$$\left\vert\frac{e^{-ixu}-1}{u}\right\vert^2=\frac{2-2\cos(xu)}{u^2}=\frac{4\sin^2(\frac{xu}{2})}{u^2}\leq\frac{4(\frac{xu}{2})^2}{u^2}=x^2$$