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I am reading a paper and I don't understand one thing in the paper.Consider the convolution operator $Tf=f*\mu$ acting on $f\in L^p(\mathbb{R}^n)$, where $\mu$ is a measure defined by $\int_{\mathbb{R}^n}gd\mu=\int_{-1}^1g(h(t))dt$ and $h(t)=(t,t^2,t^3...t^n)$. The paper says though $\mu$ is a singular measure, its fourier transform satisfies a decay estimate $\hat\mu(\xi)=O(1+|\xi|)^{-1/n}$.

I don't quite know what does he mean by a singular measure, is it mean $\mu\bot\nu$ for some other measure $\nu$? If he means so, why being a singular measure matters? If he means singular to Lebesgue measure, $\mu(E)=\int_{\mathbb{R}^n}X_Ed\mu=\int_{-1}^1X_E(h(t))dt=m(F)$, where $F=(t\in[-1,1]|(t,t^2...t^n)\in E)$, $E$ is any measurable set.

And I don't quite know how to get this estimate of the Fourier transform.

$\begin{equation} \hat\mu(\xi)=\int_{-1}^1e^{-2\pi i\xi\cdot(t,t^2,t^3...t^n)}dt \end{equation}$

I don't know what to do next. Thanks for any help!!

user146507
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    He means singular wrt Lebesgue measure. The point is the Riemann-Lebesgue lemma says that if $\mu$ is absolutely continuous then $\hat\mu$ vanishes at infinity, but this need not be the case for a singular measure. Your calculation of the FT makes no sense to me - i have no idea what $\hat\xi$ is or what $\xi(x)$ is. If the definition of $\mu$ you gave is correct then it seems to me $\hat\mu(\xi)$ is just $\int_{-1}^1e^{-2\pi i x\cdot(t,\dots,t^n}dt$. ??? – David C. Ullrich Aug 12 '15 at 16:05
  • You are right . I made a mistake. It is a measure not a distribution. Thanks! I have edited the question. – user146507 Aug 12 '15 at 16:35
  • Would you please include a reference for the referred to paper? – Matt Rosenzweig Aug 14 '15 at 18:27
  • Here is the paper. http://imrn.oxfordjournals.org/content/1998/19/1033 – user146507 Aug 14 '15 at 20:18

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You have $\hat{\mu} (\xi) = \int_{-1}^1 e^{-i2\pi \xi \cdot (t,\ldots,t^n)} dt = \prod_{k=1}^n \int_{-1}^1 e^{-i2\pi \xi_k t^k} dt$. Estimate the integrals $\int_{-1}^1 e^{-i2\pi \xi_k t^k} dt$ separately. The change of variable $s = \xi t^k$ does the job.

Tomas Persson
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  • Why you can interchange the product and the integral? Thanks! – user146507 Aug 12 '15 at 20:27
  • Sorry, I made a silly mistake. The variables $t^k$ are not independent som we cannot change integral and product. Instead, you might want to use van der Corput's lemma. Check for instance https://en.wikipedia.org/wiki/Van_der_Corput_lemma_(harmonic_analysis) I did not do the details, but I think this gives you what you want. – Tomas Persson Aug 13 '15 at 05:14
  • Thanks! It is a good hint. But I guess it is hard to show $\phi^{(1)}(t)=\xi_1+2\xi_2+...n\xi_n$ is monotone so I can apply the lemma. – user146507 Aug 13 '15 at 22:34
  • $\phi'(t) = \xi_1 + 2\xi_2t + \cdots$ is in general not monotone, but it is not required either. Take $k = n$ in the lemma. – Tomas Persson Aug 14 '15 at 03:51
  • Thanks! I think it should work. What about the $\geq$? I guess the lemma only solve one direction. Thanks! – user146507 Aug 14 '15 at 21:11
  • I have worked it out. Thank you very much for the hint! That lemma is so important. – user146507 Aug 18 '15 at 18:33