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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!

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2 Answers 2

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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 $R \to \infty$.

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  • $\begingroup$ 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 $\phi$? $\endgroup$ Commented Sep 10, 2022 at 11:39
  • $\begingroup$ With the notation in the answer, the function $h \phi$ belongs to $L^1(\mathbb R^n)$ and then its Fourier transform is bounded. $\endgroup$ Commented Sep 10, 2022 at 11:45
  • $\begingroup$ oh, you are right, I forgot this point. Thank you~ $\endgroup$ Commented Sep 10, 2022 at 11:57
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    $\begingroup$ 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 $\endgroup$ Commented Sep 10, 2022 at 15:11
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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.

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  • $\begingroup$ 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. $\endgroup$ Commented Sep 10, 2022 at 16:31
  • $\begingroup$ "$\phi D_jh−D_jh=(1−\phi)D_jh$ 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? $\endgroup$ Commented Sep 11, 2022 at 0:33

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