0
$\begingroup$

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.

$\endgroup$
1
  • $\begingroup$ 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

1
$\begingroup$

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 $\sigma$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.