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Show $\left|\frac{e^{-ixt}-1}{t}\right| \le |x|$ for all real $x \ne 0$ and all real $t \ne 0$.

I tried taylor theorem but as the reminder is complex, I don't see how to get the result.

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    See also: [Prove $|e^{i\theta} -1| \leq |\theta|$](http://math.stackexchange.com/questions/584498/prove-ei-theta-1-leq-theta)2016-03-30

2 Answers 2

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As explained by @Dennis Gulko in a comment, the task is to prove that $|\mathrm e^{-\mathrm is}-1|\leqslant|s|$ for every real number $s$. But the function $u:\mathbb R\to\mathbb C$, $t\mapsto\mathrm e^{-\mathrm it}$, has derivative $u'(t)=-\mathrm i\mathrm e^{-\mathrm it}$, whose modulus is (at most) $1$ everywhere, hence $ |u(s)-u(0)|\leqslant|s-0|\cdot\sup\{|u'(t)|\mid0\leqslant t\leqslant s\}=|s|. $

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HINT

\begin{align} \exp(-ixt) - 1 &= \cos(xt) - i \sin(xt) - 1\\ & = -\left(2 \sin^2(xt/2) + 2i \cos(xt/2) \sin(xt/2)\right)\\ &= -2i \sin(xt/2) \left(\cos(xt/2) - i \sin(xt/2)\right)\\ & = -2i \sin(xt/2) \exp(-ixt/2) \end{align} Hence, $\vert \exp(-ixt) - 1 \vert = \vert -2i \sin(xt/2) \exp(-ixt/2) \vert = 2 \vert \sin(xt/2) \vert$