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Could you help me to prove $$ |e^{i\theta} -1| \leq |\theta| $$

I am studying the proof of differentiability of Fourier Series, and my book used this lemma. How does it work?

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    $\begingroup$ Both the good answers can be summarized (in a way) as "Draw a picture!" $\endgroup$ – Jyrki Lahtonen Nov 28 '13 at 12:55
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By the fundamental theorem of calculus $$e^{i\theta}-1=\int_0^\theta ie^{it}\mathrm{d}t$$ Hence...

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  • $\begingroup$ Thank you very much! I got it! $\endgroup$ – Block Jeong Nov 28 '13 at 12:54
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    $\begingroup$ By far the shortest, most elegant and clear proof. +1 $\endgroup$ – DonAntonio Nov 28 '13 at 13:12
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    $\begingroup$ Nice. This approach also works for complex $\theta$. $\endgroup$ – Antonio Vargas Nov 28 '13 at 13:41
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    $\begingroup$ Filling in the "..." (which I suppose is supposed to be obvious): you need to know that the norm of an integral is $<=$ to the integral of the norm of the integrand. Then the integral of the norm reduces to the integral of $|dt|$ from $0$ to $\theta$, which will be $|\theta|$. $\endgroup$ – Mark Lakata Nov 29 '13 at 6:47
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    $\begingroup$ @MarkLakata Yes. It was not meant as "obvious", but rather as "left to the OP". We could also apply FTC like that: $e^{i\theta}-1=i\theta \int_0^1 e^{it\theta}\mathrm{d}t$ which allows to handle the negative case or even, as Antonio Vargas mentioned, the complex case more easily. $\endgroup$ – Julien Nov 29 '13 at 12:04
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Think of it that way : You start at point $1+0i$ and move on the unit circle by an angle of theta. This inequality is just saying that going from $1+0i$ to your point $e^{i\theta}$ by a straight line is shorted than going to it by moving along the circle $r\theta$ (where $r=1$ is the radius of the cicle).

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  • $\begingroup$ Nice, but without a drawing...+1 $\endgroup$ – DonAntonio Nov 28 '13 at 13:29
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And this is somehow a proof without words:

enter image description here $z=e^{i\theta}\\ \theta= \overset{\displaystyle\frown}{AB}\\ AH=\sin(\theta)\\ OH=\cos(\theta)\\ HB=1-\cos(\theta) \\ AB=|e^{i\theta}-1| $

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  • $\begingroup$ I see xavierm02 has already given this answer, so let this be a shape for it. $\endgroup$ – hhsaffar Nov 28 '13 at 13:28
  • $\begingroup$ Very nice!+1 For the ones who tend to be distracted, $\;\theta=\angle AOB\;$ $\endgroup$ – DonAntonio Nov 28 '13 at 13:28
  • $\begingroup$ very nice! Really like this proof! $\endgroup$ – SiXUlm Dec 28 '15 at 21:18
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$$|e^{i\theta}-1|=|\cos\theta-1+i\sin\theta|=\sqrt{\cos^2\theta-2\cos\theta+1+\sin^2\theta}=$$

$$=\sqrt{2(1-\cos\theta)}\le|\theta|\iff 2(1-\cos\theta)\le\theta^2\iff\theta^2+2\cos\theta-2\ge0$$

But if we put

$$g(\theta):=\theta^2+2\cos\theta-2\implies g'(\theta)=2\theta-2\sin\theta\ge0$$

since we know (hopefully) that $\;|\sin\theta|\le|\theta|\;\;\;\forall\;\theta\in\Bbb R$

and then $\;g\;$ is monotone non-decreasing, and since $\;g(0)=0\;$ we get what we want.

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    $\begingroup$ Monotone for $\theta>0$, and even. Apart from that, I like this proof very much! $\endgroup$ – Jean-Claude Arbaut Nov 28 '13 at 12:58
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A little boot-strapping, just to play. We have $|e^{i\theta}-1|\le|\theta|+M\theta^2 $ for some constant $M$ and any $\theta\in\mathbb{R}$ (for some reason depending on your favourite definition of $e^{i\theta}$; e.g., because it is bounded, and the derivative at $\theta=0$ is not larger than $1$ in absolute value).

But then $|e^{i\theta}-1|=|e^{i\theta/2}+1||e^{i\theta/2}-1|\le 2\big(|\theta|/2+M \theta^2/4\big)= |\theta| +M \theta^2/2 $ for all $\theta$, which proves that the infimum of the constants $M$ for which the inequality holds is $M=0$, whence the thesis.

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For real $\theta$ we have $\;|e^{i\theta}-1|^2=\;|\cos \theta +i\sin \theta-1|^2=$ $=(\cos \theta-1)^2+(\sin \theta)^2=1+\cos^2 \theta+\sin^2 \theta-2\cos \theta=$ $=2-2\cos \theta=4((1-\cos \theta)/2)=4\sin^2(\theta /2)\leq 4(\theta$ $ /2)^2=\theta^2.$

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