Hello. I am trying to understand an estimate on the kernel of the fractional laplace operator in the following article "Well-posedness of the Cauchy problem for the fractional power" by Miao, C., Yuan, B., & Zhang
I have been able to understand how to estimate the integral $I$ but I cannot understand the integral $II$. I attach what I have.
My attempt:
First, $$\partial_{\xi_i}\mathrm{e}^{-|\xi|^{2\alpha}}=-2\alpha |\xi|^{2\alpha-1}\xi_i$$ then $$\nabla_{\xi} \mathrm{e}^{-|\xi|^{2\alpha}}=-2\alpha |\xi|^{2\alpha-1}\mathrm{e}^{-|\xi|^{2\alpha}}\xi$$ therefore $$L^*(\mathrm{e}^{-|\xi|^{2\alpha}})=\frac{1}{i|x|^2}2\alpha|\xi|^{2\alpha-1} \mathrm{e}^{-|\xi|^{2\alpha}} x\cdot \xi$$ implies that
$$|L^*(\mathrm{e}^{-|\xi|^{2\alpha}})|\leq \frac{2\alpha |\xi|^{2\alpha}\mathrm{e}^{-|\xi|^{2\alpha}}}{|x|}$$
Now, with
$$\rho(\xi/\delta)\leq 2\text{ implies }|\xi|\leq 2\delta$$ and $$\frac{|\xi|^{2\alpha}}{\mathrm{e}^{|\xi|^{2\alpha}}}\leq |\xi|^{2\alpha-1}$$ it follows that
$$|I|\leq c_n\int_{|\xi|\leq 2\delta}\frac{1}{|x|}2\alpha |\xi|^{2\alpha}\mathrm{e}^{-|\xi|^{2\alpha}}\,d\xi\\ \leq 2\alpha c_n\int_{|\xi|\leq 2\delta}|\xi|^{2\alpha-1}\,d\xi$$
For $II$, with $(L^*)^{N-1}(\mathrm{e}^{ix\cdot\xi})=\mathrm{e}^{ix\cdot\xi}$ i have this \begin{align} II&=c_n\int_{\mathbb{R}^n}\mathrm{e}^{ix\cdot \xi} (1-\rho(\xi/\delta))L^*(\mathrm{e}^{-|\xi|^{2\alpha}})d\xi\\ &=c_n\int_{\mathbb{R}^n} (L^*)^{N-1}(\mathrm{e}^{ix\cdot\xi}) (1-\rho(\xi/\delta))L^*(\mathrm{e}^{-|\xi|^{2\alpha}})d\xi\\ &=c_n\int_{\mathbb{R}^n} (\mathrm{e}^{ix\cdot\xi})(L^*)^{N-1}\left( (1-\rho(\xi/\delta))L^*(\mathrm{e}^{-|\xi|^{2\alpha}})\right)d\xi \end{align}
How does the following inequality follow? I can't understand it