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