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For $h>0,x\in \mathbb{R}$ prove (or disprove) that $|e^{-ihx} - 1|/|h| \leq |x|$.

I tried using Taylor's theorem for complex-valued functions, which says that $e^{-ihx} = \sum_{k\ge 0} (-ihx)^k/k! $, but I can't seem to formally justify why the absolute value of the resulting term would be at most $x$. I know the expression is the magnitude of the complex derivative of $e^{-ihx}$ at $0$ with respect to h, which is $-ix$, but I'm not sure if that's useful.

Fred Jefferson
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1 Answers1

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For all $a \in \mathbb{R}$,

$ |e^{-ia}-1|^2 = |(\cos a - 1) - i\sin a|^2 = (\cos a-1)^2 + \sin^2 a = 2-2\cos a = 4\sin^2(\frac a2)\leq 4(\frac{a}{2})^2 = a^2 \Rightarrow |e^{-ia}-1| \leq |a|.$

Now put $a=h\,x$ and we the desired inequality.

Here I have used the trigonometric equation

$$ 1-\cos a = 2\,\sin^2(\frac a2),$$

and the inequality

$$ \sin^2 a \leq a^2.$$

Dan
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