im reading a book about fourieranalysis and found this inequality ...
$|e^{2\pi imh}-1|\leq \min(2,2\pi|m|h)\leq 2^{1-\gamma}(2\pi)^{\gamma}|m|^{\gamma}h^{\gamma}$ with $m\in\mathbb{Z}$ and $\gamma, h\in[0,1)\subset\mathbb{R}$
Obviously the euler-relation gives us:
$$0\leq|e^{2\pi imh}-1|\leq|e^{2\pi imh}|+|-1|=2$$
The inequality $|e^{i\phi}-1|\leq|\phi|$ seems clear from geometry.
$$\Rightarrow |e^{2\pi imh}-1|\leq \min(2,2\pi|m|h)$$
But what's about the last part? $\min(2,2\pi|m||h|)\leq 2^{1-\gamma}(2\pi)^{\gamma}|m|^{\gamma}h^{\gamma}$ ?
I can not see how to solve this inequality.
Thanks for your help, Tiling
As you already said, the inequality $|e^{i\phi}-1|\leq|\phi|$ is clear from geometry. It can for example be derived as $$ |e^{i\phi}-1|^2 = 2 - 2 \cos \phi = 2 - 2 \cos 2 \frac\phi 2 = 2 - 2 \cos^2 \frac\phi 2 + 2 \sin^2 \frac\phi 2 = 4 \sin^2 \frac\phi 2 \le 4 \big(\frac\phi 2 \bigr)^2 = \phi^2 \, . $$
For $a, b > 0$ and $0 \le \gamma \le 1$ you have $$ a^{1 - \gamma} \ge \min(a, b) ^{1 - \gamma} \\ b^\gamma \ge \min(a, b) ^\gamma $$ because both $t\to t^{1 - \gamma}$ and $t \to t^\gamma$ are increasing functions. Therefore $$ a^{1 - \gamma} b^\gamma \ge \min(a, b) ^{1 - \gamma} \min(a, b) ^\gamma = \min(a, b) \, . $$ With $a = 2$ and $b = 2\pi|m||h|$ you get the desired inequality $$ \min(2,2\pi|m||h|)\leq 2^{1-\gamma}(2\pi)^{\gamma}|m|^{\gamma}h^{\gamma} $$