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This is an application of: Spectral Measures: Domain Criterion

I'm trying to check the estimate: $$\frac{1}{h}|e^{ixh}-1|\leq C\left(|ix|+1\right)\quad(h\in(-\varepsilon,\varepsilon))$$ for some constant $C\geq0$ and all reals $x\in\mathbb{R}$.

I had some tries but just couldn't get any further. Do you have a hint for me, please?

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3 Answers3

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First simplify: $$|e^{\frac{ia}{n}}-1|^2 = |e^{\frac{ia}{2n}}|\cdot|2\sin(\frac{a}{2n})|^2 = 2\sin^2(\frac{a}{2n}) $$ Now it follows from: $$|\sin(x)| \leq |x| \implies \sin^2(x) \leq x^2 $$ Thus you get: $$ n^2|e^{\frac{ia}{n}}-1|^2 \leq2n^2\cdot(\frac{a}{2n})^2 = \frac{{a}^2}{2} \leq \frac{1}{2}\cdot(a^2 +1) = \frac{1}{2}\cdot(|ia|^2 +1) $$

(I think it's good, unless I've completely misunderstood the question and i is not the complex number..)

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mvggz
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Hint: use the mean value theorem with the function $$ t\to e^{iat} $$

mookid
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  • the inequality $|f(x) - f(a)| \le \sup |f'| \times |x-a|$ – mookid Nov 03 '14 at 18:55
  • Ah ok let me think. – C-star-W-star Nov 03 '14 at 18:55
  • @Freeze_S so, what did you find? – mookid Nov 03 '14 at 19:00
  • So it's simply: $|\frac{1}{t}(e^{ita}-1)|\leq|iae^{ita}|=|ia|\leq1\cdot(|ia|+1)$ – C-star-W-star Nov 03 '14 at 19:02
  • Ah no it's fine now thanks ;) Thanks!! – C-star-W-star Nov 03 '14 at 19:06
  • you are welcome ! – mookid Nov 03 '14 at 19:17
  • I just remembered: Is the mean value theorem still valid for complex functions?? – C-star-W-star Dec 08 '14 at 16:55
  • @Freeze_S the inequality is true (the equality isn't). – mookid Dec 08 '14 at 19:21
  • I'm sorry that I felt I have to unaccept yours. Really! (I didn't want to give an impression of being rude; if I find a way to honor your effort I will do so. Promise!) I didn't accept my answer since it is mine but because it is from my point of view the best answer up to now. There are also cases where I've got an answer meanwhile whereas somebody else a better answer. Of course, I do accept then the other answer. All in all, I try to force myself to be as objective as possible. At least I try! I hope that is fine for you. – C-star-W-star Dec 08 '14 at 19:29
  • Now, the problem is that the MVT fails in general for complex functions although it may still hold by accident or due to special properties. – C-star-W-star Dec 08 '14 at 19:33
  • For some examples see: MVT: Counterexamples – C-star-W-star Dec 08 '14 at 19:52
  • @Freeze_S to me, the mean value theorem is: if $f:[a,b]\to E$ is differentiable, and $|f'(x)|\le M$ then $|f(a) - f(b)| \le M(b-a)$. Note that I already told you that in fi first comment (Nov 3 at 18:55). I'm aware of the counterexamples, thanks for your concern. – mookid Dec 08 '14 at 20:24
  • Formally you're right: That is the mean value theorem. Rigorously it isn't: For Banach-space-valued functions as for example complex functions it needs much much more assumptions. – C-star-W-star Dec 08 '14 at 20:30
  • This is priceless. Your answer is a proof of my statement. But I am sure you have an answer to that as well ;) – mookid Dec 08 '14 at 20:44
  • Shall we leave it for now? I mean actually I don't want to fight right now. :P If you've got something fresh new let me now. ;) – C-star-W-star Dec 08 '14 at 20:50
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I think I got it now circumventing the mean value theorem...

One has by Duhamel's principle: $$\frac{1}{|h|}|e^{ixh}-1|=\frac{1}{|h|}\left|\int_0^h(e^{ixs})'\mathrm{d}s\right|\leq\frac{1}{|h|}\int_0^h\left|(e^{ixs})'\right|\mathrm{d}s=\frac{1}{|h|}|h|\cdot|ix|\leq1(|ix|+1)$$

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