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This question is about Daubechies' book "Ten Lectures on Wavelets", section 2.4.
This author says, if $\psi \in L^2(\mathbf{R})$, $\int_{\mathbf{R}}dx\ \psi(x) = 0$, and \begin{align*} \exists \alpha>0 \quad \int_{\mathrm{R}}dx (1+|x|)^\alpha\ |\psi(x)| < \infty \end{align*} holds, then $|\hat{\psi}(\xi)| \leq C|\xi|^\beta$ follows, where $\beta = \min(\alpha, 1)$, $C$ is some constant, and $\hat{\psi}$ is the Fourier transform of $\psi$. And he says that from this results, \begin{align*} 2\pi \int d\xi \ |\xi|^{-1} |\hat{\psi}(\xi)|^2 < \infty \end{align*} follows.

I cant't think of how to derive the former result. And, I think the latter result doesn't follow from the former, because \begin{align*} 2\pi \int d\xi \ |\xi|^{-1} |\hat{\psi}(\xi)|^2 \leq 2\pi C^2 \int d\xi \ |\xi|^{2\beta-1} \end{align*} and the right hand side does't have any upper bound if, say, $\alpha = 2$.

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  • $\begingroup$ The second sentence of the section in my copy of the reference says, "We always suppose that $\psi\in L^{2}(\mathbb{R})$..." $\endgroup$ – Matt Rosenzweig Sep 3 '15 at 12:12
  • $\begingroup$ I'm sorry that I overlooked the condition. I revised my question. $\endgroup$ – dazaga Sep 3 '15 at 13:30
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From the conditions on $\psi$ we have $$ |\psi(\xi)|=\Bigl|\int_\mathbb{R}\psi(x)(e^{-ix\xi}-1)\,dx\Bigr|\le\Bigl(\int_\mathbb{R}|\psi(x)|(1+|x|)^\alpha\,dx\Bigr)\sup_{x}\frac{|e^{-ix\xi}-1|}{(1+|x|)^\alpha}. $$ If $\alpha\ge1$, it is clear that $$\frac{|e^{-ix\xi}-1|}{(1+|x|)^\alpha}\le|\xi|.$$

For the case $0<\alpha<1$, observe that $|e^{-ix\xi}-1|$ is periodic of period $\pi/|\xi|$ as a function o f $x$. Then, since $1/(1+x)^\alpha$ is decreasing, $$\sup_{x}\frac{|e^{-ix\xi}-1|}{(1+|x|)^\alpha}\le\sup_{0\le x\le\pi/|\xi|}\frac{|e^{-ix\xi}-1|}{(1+|x|)^\alpha}\le\sup_{0\le x\le\pi/|\xi|}\frac{x\,|\xi|}{(1+x)^\alpha}.$$ Since $x/(1+x)^\alpha$ is decreasing (because $\alpha<1$), $$\sup_{0\le x\le\pi/|\xi|}\frac{x\,|\xi|}{(1+x)^\alpha}=\frac{\pi\,|\xi|^\alpha}{(|\xi|+\pi)^\alpha}\le\pi^{1-\alpha}\,|\xi|^\alpha.$$

As for the second estimate, are you sure that there is no further condition on $\psi$? If for instance $\psi\in L^2$, then $$ \int_\mathbb{R}\frac{|\hat\psi(|\xi)|^2}{|\xi|}\,d\xi\le C\int_{|\xi|\le1}|\xi|^{-1+2\beta}\,d\xi+\int_{|\xi|>1}|\hat\psi(\xi)|^2\,d\xi<\infty. $$

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  • $\begingroup$ I got $|e^{-ix\xi} - 1| \leq \max(2, |x|\ |\xi|)$, but how do you derive $|\hat{\psi}(\xi)|\leq C|\xi|^\beta$ ? $\endgroup$ – dazaga Sep 1 '15 at 15:24
  • $\begingroup$ If $|x|\le2/|\xi|$, then $$\frac{|e^{-ix\xi}-1|}{(1+|x|)^\alpha}\le\frac{|x|}{(1+|x|)^\alpha}\,|\xi| \le | \xi|.$$ If $|x|>2/|\xi|$, then $$\frac{|e^{-ix\xi}-1|}{(1+|x|)^\alpha} \le \frac{2}{(1+2/|\xi|)^\alpha} =\frac{2\,|\xi|^\alpha}{(|\xi|+2)^\alpha} \le 2^{1-\alpha}\,|\xi|^\alpha.$$ $\endgroup$ – Julián Aguirre Sep 1 '15 at 18:49
  • $\begingroup$ If $\xi = 0.1, x = 3$, and $\alpha = 0.5$, then $|x| \leq 2/|\xi|$, but $\frac{|x|}{(1+|x|)^\alpha}|\xi|\leq |\xi|$ doesn't hold. $\endgroup$ – dazaga Sep 2 '15 at 0:10
  • $\begingroup$ Forget my previous comment and look at the edited answer. $\endgroup$ – Julián Aguirre Sep 3 '15 at 9:37

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