I'm studying the Meyer's book, "Wavelets and operators", and I'm confused about a proof of Bernstein's inequality at page 47, which is stated below:


"The function $\frac{\xi^\beta}{|\xi|^s}\hat\phi(\xi)$ is the Fourier transform of an integrable function." Here $\hat\phi(\xi)\in C^\infty\_0(\mathbb{R}^n)$ is a bump function, which is equal to 1 on $|\xi|\leq \frac{1}{2}$, 0 on $|\xi|\geq 1$, $|\beta|\geq 1\in\mathbb{N}$ and $|\beta|-1<s<|\beta|$.


Is it correct? How to prove it? Thanks in advance!



A proof of Bernstein's inequality
Well, I find that there could be another answer. We still only need prove the integrablity at inifity. Considering the homogeneous function $\frac{\xi^\beta}{|\xi|^s}$, by the proposition 2.4.8 in "Classical Fourier Analysis", its original function, denoted by $g(x)$, is smooth on $\mathbb{R}^n\backslash\{0\}$ and homogeneous of degree $-n-|\beta|+s$, that is, $g(x)\sim\frac{g(x')}{|x|^{n+|\beta|-s}}$. Then, we have
\begin{equation}
K(x)=\int \frac{\xi^\beta}{|\xi|^s}\hat\phi(\xi) e^{ix\cdot\xi}\,d\xi=\int\frac{\xi^\beta}{|\xi|^s}e^{ix\cdot\xi}\,d\xi-\int \frac{\xi^\beta}{|\xi|^s}[1-\hat\phi(\xi)] e^{ix\cdot\xi}\,d\xi
\end{equation}
For the second term, let $|\alpha|=m$ large enough, then
$$|x^\alpha K(x)|\lesssim \int \left|\sum\_{\alpha\_1+\alpha\_2=\alpha}D\_\xi^{\alpha\_1}\frac{\xi^\beta}{|\xi|^s}D\_\xi^{\alpha\_2}[1-\hat\phi(\xi)]\right|\,d\xi\lesssim 1,$$
since, for each $|\alpha\_2|\geq 1$, $D\_\xi^{\alpha\_2}[1-\hat\phi(\xi)]$ has compact support away from the origin and $[1-\hat\phi(\xi)]D\_\xi^{\alpha}\frac{\xi^\beta}{|\xi|^s}$ is integrable on $\mathbb{R}^n$. Therefore, choosing $m\geq n+2(|\beta|-s)$, we obtain
$$|K(x)|\lesssim \frac{|g(x')|}{|x|^{n+|\beta|-s}}+\frac{1}{|x|^{m-|\beta|+s}}\lesssim \frac{1}{|x|^{n+|\beta|-s}},\,\mathrm{when}\,|x|\geq 1$$
which conculdes the proof.


