I am reading a proof of a theorem. In that proof there is the following inequality without proof
$$|\exp(i \xi \cdot x) - \exp(i \xi \cdot y) | \leq C |x-y|^\alpha |\xi|^\alpha,$$
where $0<\alpha<1$ is given, $x,y,\xi \in \mathbb{R}^n$ and $C$ is a positive constant which not depend of $x,y,\xi$.
How can I prove that inequality?. Can you help me, please?.
Clearly, $$|e^{iz}-1| \leq \min\{2,|z|\} \tag{1}$$ for any $z \in \mathbb{R}$. If $\alpha \in (0,1]$, then $$|w| \leq |w|^{\alpha} \qquad \text{for all $|w| \leq 1$}$$ and $$1 \leq |w|^{\alpha} \quad \text{for all $|w|>1$}.$$ Using these estimates for $w:=\frac{z}{2}$ we get $$|e^{iz}-1| \leq \min \left\{2 \left| \frac{z}{2} \right|^{\alpha}, 2 \left| \frac{z}{2} \right|^{\alpha} \right\} = 2^{1-\alpha} |z|^{\alpha}. \tag{2}$$ Thus, $$|e^{i \xi \cdot x}-e^{i \xi \cdot y}| = |e^{i \xi \cdot (x-y)}-1| \leq 2^{1-\alpha} |\xi \cdot (x-y)|^{\alpha} \leq 2^{1-\alpha} |\xi|^{\alpha} |x-y|^{\alpha}.$$