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!

  • $\begingroup$ 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

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:


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

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

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


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.