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I try to understand why the following inequality holds.

$$\left|\int_{|y|<1} e^{iuy}−1−iuy\ \, dy \right| \le \frac{1}{2} \cdot \int_{|y|<1} |uy|^2\ \, dy$$

Due to a hint I'm pretty sure, that the taylor expansion of $$z↦e^{iuz}−1−iuz$$ is part of the solution.

The taylor expansion of the mentioned function is $$e^{iuz}-1-iuz=\sum_{n=2}^\infty\frac{i^nu^nz^n}{n!}$$

Does anyone have an idea how to continue?

Thank you!

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  • $\begingroup$ I just can't understand that $\;v(dy)\;$ notation...If it were just $\;dy\;$ or even $\;(dy)\;$ is fine, but $\;v(dy)\;$ looks a little like Riemann-Stieltjes...what is $\;v\;$ there? $\endgroup$ – DonAntonio May 18 '16 at 6:45
  • $\begingroup$ $v$ is just a borel measure defined on $\mathbb{R}-\{0\}$. Could you explain it in the case of just $dy$? Thanks! $\endgroup$ – FeldO May 18 '16 at 11:03
  • $\begingroup$ Maybe I'm too stupid to see it? I'm literally staring at this inequality for hours... $\endgroup$ – FeldO May 18 '16 at 18:22
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There are (at least) two ways to prove this inequality.

Solution 1: Taylor's formula states that for any nice function $f: \mathbb{R} \to \mathbb{C}$, we have

$$f(y) = f(0)+ f'(0) y + \frac{1}{2} f''(\xi) y^2$$

where $\xi = \xi(y)$ is an intermediate point between $0$ and $y$. Hence,

$$|f(y)-f(0) -f'(0)y| = \frac{1}{2} |f''(\xi)|.$$

Applying this for $f(y) := e^{iuy}$ (with $u$ fixed), we get

$$|e^{iuy}-1-iuy| \leq \frac{1}{2} u^2 y^2 |e^{iu \xi}| \leq \frac{1}{2} u^2 y^2.$$

Using the triangle inequality, this proves the claimed integral inequality.

Solution 2:

Since

$$e^{iuy}-1 =iu \int_0^y e^{iuz} \, dz, \tag{1}$$

we have

$$e^{iuy}-1-iyu \stackrel{(1)}{=} iu \int_0^y (e^{iuz}-1) \, dz \stackrel{(1)}{=} (iu)^2 \int_0^y \int_0^z e^{iuw} \, dw \, dz.$$

As $|e^{iuw}| \leq 1$, this implies

$$|e^{iuy}-1-iyu| \leq |u|^2 \int_0^y \int_0^z 1 \, dw \, dz = \frac{1}{2} y^2 u^2.$$

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