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!!
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.