I posted on MathStackExchange but there are no answers. So, I arrived at MathOverflow in order to have some comments or any idea.

I am interested in oscillatory integral which has restricted range by supported size $\xi \sim 2^k$. The detail is below.

Let $\psi_k(|\xi|)$ be a bump function which support is $\{ \xi\in\mathbb{R^n}:2^{k} \le |\xi| \le 2^{k+1} \}$.

Then, I guess, for $\sigma \in \mathbb{R}$, the following equation (its inverse Fourier transform) $$\int_{\mathbb{R}^{n}} e^{ix\cdot \xi} \psi_k(|\xi|) \sqrt{1+|\xi|^{2}}^{\sigma} d\xi = C 2^{k\sigma} \int_{\mathbb{R}^{n}} e^{ix\cdot \xi} \psi_k(|\xi|)d\xi$$

is probably true for some $C$ and I want to prove (or disprove) it rigorously.

I tried an inequality of its modulus version, $|(LHS)| \le C|(RHS)|$ by using Holder's inequality or Young's inequality. But, it was failed since the former removes $e^{ix\cdot\xi}$ and the latter removes pointwiseness of $x$ by taking $ \| \cdot \|_{L^{\infty}} $ and it needs to calculate the Fourier inversion of $\sqrt{1+|\xi|^{2}}^{\sigma}$. More precisely, when I use Holder's, $$ |(LHS)| \le \int_{\mathbb{R}^{n}} | e^{ix\cdot \xi} \psi_k(|\xi|)|d\xi\; \cdot \; \| \sqrt{1+|\xi|^{2}}^{\sigma}\|_{L^{\infty}(\text{supp}(\psi_k))} $$ since $\psi_k$ is supported, or use Young's, $$ |(LHS)| = | (\psi_{k})\check{} (x) \ast (\sqrt{1+|\cdot|^{2}}^{\sigma})\check{}(x)| \le \| (\psi_{k})\check{}(x)\|_{L^{\infty}_{x}} \| (\sqrt{1+|\cdot|^{2}}^{\sigma})\check{}(x) \|_{L_x^{1}}. $$ But it needs to calculate $(\sqrt{1+|\cdot|^{2}}^{\sigma})\check{}(x)$ while the bounding by $\| (\psi_{k})\check{}(x)\|_{L^{\infty}_{x}}$ (i.e. uniform in $x$) do not imply pointwiseness.

I think that it looks simple and obvious but I can't prove rigorously.

  • $\begingroup$ What's the relation between $\psi_k$ for different $k$? If you are looking for a proof of an identity, why are you describing attempts to prove inequalities next? $\endgroup$
    – Kostya_I
    Commented Dec 15, 2022 at 15:27

1 Answer 1


I suggest a mild modification of your question. In the first place, let us set $ \psi_k(t)=\psi_1(t2^{-k}). $ Let us also replace $\sqrt{1+\vert \xi\vert^2}$ by $\vert \xi\vert$. We get $$ \int_{\mathbb R^n} e^{ix\cdot\xi}\psi_1(\vert\xi\vert 2^{-k})\vert \xi\vert^\sigma d\xi =\int e^{i2^k x\cdot\xi}\underbrace{\psi_1(\vert\xi\vert)\vert \xi\vert^\sigma}_{=\phi (\xi)} 2^{k\sigma} d\xi 2^{kn} =2^{k(n+\sigma)}\hat \phi(-2^k x). $$ Note that the fonction $\phi$ is "fixed" and depends only on $\sigma$.


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