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Assuming that $ \chi\in C_c^{\infty}([-2,2]) $ is a cutoff function such that $\text{supp }\chi\subset[-2,2]$, $\chi\equiv 1 $ in $ [-1,1] $, and $ 0\leq\chi\leq 1 $, suppose that $ f\in C^{\infty}(\mathbb{R}) $ is a function satisfying the generalized $ O(|x|^{-1/2}) $ condition. In other words, for any $ k\in\mathbb{Z}_+ $, we have $$ |(x\partial_x)^kf(x)|\leq C_k|x|^{-1/2}. $$ My question is whether $ \int_{\mathbb{R}}(1-\chi(x))f(x)e^{-2\pi i x\xi}d x $ belongs to $ L_{\xi}^1(\mathbb{R}) $.

Here's my attempt: For $ |\xi|\geq 10 $, we can use integration by parts to obtain \begin{align} &\int_{\mathbb{R}}(1-\chi(x))f(x)e^{-2\pi i x\xi}dx\\ &=-\frac{1}{2\pi i \xi}\int_{\mathbb{R}}\chi'(x)f(x)e^{-2\pi ix\xi}dx+\frac{1}{2\pi i \xi}\int_{\mathbb{R}}(1-\chi(x))f'(x)e^{-2\pi ix\xi}dx\\ &=-\frac{1}{(2\pi i \xi)^2}\int_{\mathbb{R}}\chi''(x)f(x)e^{-2\pi ix\xi}dx\\ &\quad-\frac{1}{(2\pi i \xi)^2}\int_{\mathbb{R}}\chi'(x)f'(x)e^{-2\pi ix\xi}dx\\ &\quad+\frac{1}{(2\pi i \xi)^2}\int_{\mathbb{R}}(1-\chi(x))f''(x)e^{-2\pi ix\xi}dx. \end{align} Using the assumption, we can show that the right-hand side above is of order $ |\xi|^{-2} $. However, I am unsure of how to handle the case when $ |\xi| $ is small. Could you provide me with some hints or references?

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  • $\begingroup$ How about you multiply the integral with $\xi$ and use $(e^{-2\pi ix\xi})'=-2\pi i\xi e^{-2\pi ix\xi}$ so that one factor in the integral gets differentiated. Now you use integration by parts. It follows that your function is $O(|\xi|)$. Repeat this. $\endgroup$
    – Nandor
    Commented May 11 at 5:39
  • $\begingroup$ @Nandor That is what I have down in my attempt and this can only lead the estimates when $ |\xi| $ is large. $\endgroup$
    – Luis Yanka Annalisc
    Commented May 11 at 6:14
  • $\begingroup$ Yes, but what more do you need? $\endgroup$
    – Nandor
    Commented May 11 at 6:18
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    $\begingroup$ @Nandor I want to estimate the order of the function when $ |\xi| $ is small. $\endgroup$
    – Luis Yanka Annalisc
    Commented May 11 at 6:34

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An easy observation is that since $|f'|\lesssim |x|^{-3/2}\in L^1$ and $\int_{-\infty}^{\infty} f'(x)\, dx=0$ (from the bound on $f$), taking the Fourier transform of $f'$ shows that $\widehat{f}(\xi)=g(\xi)/\xi$, with $g\in C(\mathbb R)$ and $g(0)=0$.

While this comes close, it isn't quite good enough to give the desired conclusion $\widehat{f}\in L^1$. However, the $x^{-3/2}$ decay is faster than required to conclude integrability, and we can sacrifice some of this to obtain extra regularity of $g$.

Notice that $x^{1-\delta}f', f'\in L^2$ for $\delta>0$, so $g=\widehat{f'}$ also lies in the Sobolev space $H^{1-\delta}$. This gives Holder continuity of $g$, and hence $\widehat{f}=g/\xi\in L^1(-10,10)$, as desired.

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