# A proof of Bernstein's inequality

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.

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

The result is true, as stated. Changing slightly the notation we have to prove that the Fourier transform of $$x^\beta |x|^{-s} \phi$$ is in $$L^1(\mathbb R^n)$$ for $$\phi \in C_0^\infty (\mathbb R^n)$$, $$\phi=1$$ in a neighborhhod of 0.

The function $$h(x)=x^\beta |x|^{-s}$$ is homogenuous of degree $$0<|\beta|-s<1$$. Integrating by parts $${\cal F}(h\phi)(\xi)=\frac{c}{\xi_j}\cal F(D_j (h\phi))=\frac{c}{\xi_j}\cal F(\phi D_j h)+...$$ where the dots indicate terms which decay faster then polynomials as $$|\xi| \to \infty$$ (the function $$hD_j \phi$$ is smooth).

Let us now consider the Fourier transform of $$\phi D_j h$$. Since $$D_j h$$ is of the form $$a_1x^{\gamma_1} |x|^{-s}+x^{\gamma_2}|x|^{-s-2}$$ with $$|\gamma_1|=|\beta|-1$$, $$|\gamma_2|=|\beta|+1$$ we can use Stein-Weiss "Introduction to Fourier Analysis in Euclidean spaces" Theorem 4.1, Chapter 4 pag. 160 to get for large $$|\xi|$$ $$| {\cal F}(D_j h(\xi)| \leq \frac{C}{|\xi|^{n+|\beta|-s-1}}.$$ Note that Theorem 4.1 requires negative exponents $$\gamma_1-s$$, etc. and this is the reason for integrating by parts. Also it is stated for spherical harmonics but applies to homogenuous polynomials, by Theorem 2.1 p. 139. The Fourier transform of $$\phi D_j h$$ differs from that of $$D_j h$$ by a fast decaying function (see (*) below) and has the same decay. Finally $$| {\cal F}(h\phi )| \leq \frac{C}{|\xi|^{n+|\beta|-s}}$$ which gives integrability at infinity.

(*) EDIT $$\phi D_j h-D_j h=(1-\phi)D_j h$$ is smooth but not integrable at infinity. Its (distributional) Fourier transform $$g$$ is a function, being the difference of the Fourier transforms of $$\phi D_j h$$ and $$D_j h$$. Fix $$N \in \mathbb N$$; then $$|\xi|^{2N} g$$ is the Fourier transform of $$\Delta^N [(1-\phi)D_j h]$$ and, distributing derivative and using that $$D^\alpha \phi$$ has compat support for $$|\alpha| \geq 1$$, it follows that $$\Delta^N [(1-\phi)D_j h] \in L^1(\mathbb R^n)$$ if $$N$$ is sufficiently large. Then $$|\xi|^{2N}g$$ is bounded. This argument is from the book of Grafakos, classical Fourier analysis, Example 2.4.9 pag 142.

Another possibility (which simplifies some parts) is to differentiate n-times instead of 1 at the beginning of the proof then getting ah homogenuous function $$D^\alpha h$$, $$|\alpha|=n$$, of degree $$-n<|\beta|-s-n<-n+1$$. If $$n \geq 3$$ then $$D^\alpha h \in L^1+L^2$$ and its Fourier transform is a function which is homogenuous of degree $$s-|\beta|$$ (but local boundedness needs an argument). The last part concerning the decay of the Fourier transform of $$(1-\phi)D^\alpha h$$ is easier since, using the definition, one can integrate by parts on large balls $$B_R$$ and the boundary terms tend to 0 as $$R \to \infty$$.

• Thank you very much for your answer, but how about the integrablity near the origin? Can we work out this obstacle without any addtional condition on $\phi$? Commented Sep 10, 2022 at 11:39
• With the notation in the answer, the function $h \phi$ belongs to $L^1(\mathbb R^n)$ and then its Fourier transform is bounded. Commented Sep 10, 2022 at 11:45
• oh, you are right, I forgot this point. Thank you~ Commented Sep 10, 2022 at 11:57
• I deleted my original comment which was wrong. I misread the formula as $\frac{\xi^\beta}{|\xi|^\beta}$ with the same exponent at numerator and denominator :D Commented Sep 10, 2022 at 15:11

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 $$$$K(x)=\int \frac{\xi^\beta}{|\xi|^s}\hat\phi(\xi) e^{ix\cdot\xi}\,d\xi=\int\frac{\xi^\beta}{|\xi|^s}e^{ix\cdot\xi}\,d\xi-\int \frac{\xi^\beta}{|\xi|^s}[1-\hat\phi(\xi)] e^{ix\cdot\xi}\,d\xi$$$$ For the second term, let $$|\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.

• See the Edit in my answer to fill an argument. It is important that the FT is a function,, it could have Dirac $\delta$ at 0. Commented Sep 10, 2022 at 16:31
• "$\phi D_jh−D_jh=(1−\phi)D_jh$ is smooth but not integrable at infinity. Its (distributional) Fourier transform $g$ is a function". Is it correct to understand it as a distribution $W$ but coinicides with a function $g$ away from the origin and, then, multiplying $|\xi|^{2N}$ on $g$ will eliminate the dirac mass at the 0? Commented Sep 11, 2022 at 0:33