2
$\begingroup$

My question comes from Bourgain and Demeter's The proof of the $l^2$ Decoupling Conjecture. A related question was asked about a year ago go on MO, but the author was interested in the reverse implication of my question.

Let $P^{n-1}$ be the truncated paraboloid $$P^{n-1}:=\{(\xi_{1},...,\xi_{n-1},\xi_{1}^2+...+\xi_{n-1}^2)\in \mathbb{R}^{n} : |\xi_{i}| \leq \frac{1}{2}\}$$ Let $\mathcal{N}_{\delta}$ be the $\delta$ neighborhood of $P^{n-1}$, and let $\mathcal{P}_{\delta}$ be a finitely overlapping cover of $\mathcal{N}_{\delta}$ with curved regions $\theta$ of the form $$\theta:=\{(\xi_{1},...,\xi_{n-1},\eta + \xi_{1}^2+...+\xi_{n-1}^2) : (\xi_{1},...\xi_{n-1}) \in C_{\theta}, \; |\eta| \leq 2\delta \},$$ where $C_{\theta}$ runs over all cubes $c+[-\frac{\delta^{1/2}}{2},\frac{\delta^{1/2}}{2}]^{n-1}$ with $c \in \frac{\delta^{1/2}}{2}\mathbb{Z}^{n-1} \cap [-\frac{1}{2}, \frac{1}{2}]^{n-1}$.

For a cap $\tau$ on $P^{n-1}$, we let $g_{\tau}=g1_{\tau}$ be the (spatial) restriction of $g$ to $\tau$. Let $w_{B_{R}}(x)$ be weight function such that it is Fourier supported in $B(0,\frac{1}{R})$ and satisfy $$1_{B_{R}}(x) \leq Cw_{B_{R}}(x) \leq C'(1+\frac{|x-c(B_{R})|}{R})^{-10n}$$

Let $K^{(1)}_{p,n}(\delta)$ be the smallest constant such that $$||\widehat{gd\sigma}||_{L^p(w_{B_{\delta^{-1}}})} \leq K^{(1)}_{p,n}(\delta)(\sum_{\theta : \delta^{1/2}-\mathbb{cap}}||\widehat{g_{\theta}d\sigma}||_{L^p(w_{B_{\delta^{-1}}})}^2)^{1/2}$$ for each $g:P^{n-1}\rightarrow\mathbb{C}$ and each $\delta^{-1}$ ball $B_{\delta^{-1}}$. Let $K^{(2)}_{p,n}(\delta)$ be the smallest constant such that $$||f||_{L^p} \leq K^{(2)}_{p,n}(\delta)(\sum_{\theta \in P_{\delta}}||f_{\theta}||_{L^p}^2)^{1/2}$$

for each $f$ Fourier supported in $\mathcal{N}_{\delta}$ and each $\delta^{-1}$ ball $B_{\delta^{-1}}$. Here, $f_{\theta}$ denote the Fourier restriction of $f$ to $\theta$. On pg. 20, the authors claim that

$$K^{(1)}_{p,n} \leq C_{p,n}K^{(2)}_{p,n},$$ where $C_{p,n}>0$ is some constant which only depends on $p$ and $n$.

Following the argument in this related question, one has \begin{align*} \|\widehat{gd\sigma}\|_{L^{p}(w_{B_{\delta^{-1}}})}&\leq K_{p,n}^{(2)}(\sum_{\theta\in\mathcal{P}_{\delta}}\|\widehat{\chi_{\theta}}\ast(\widehat{gd\sigma}w_{B_{\delta^{-1}}})\|_{p}^{2})^{1/2}\\ &\leq K_{p,n}^{(2)}(\sum_{\theta\in\mathcal{P}_{\delta}}\sum_{\tau\in\mathcal{P}_{\delta}}\|\widehat{\chi_{\theta}}\ast(\widehat{g_{\tau}d\sigma}w_{B_{\delta^{-1}}})\|_{p}^{2})^{1/2}\\ &=K_{p,n}^{(2)}(\sum_{\theta\in\mathcal{P}_{\delta}}\sum_{\tau\in J(\theta)}\|\widehat{\chi_{\theta}}\ast(\widehat{g_{\tau}d\sigma}w_{B_{\delta^{-1}}})\|_{p}^{2})^{1/2} \tag{1} \end{align*} where we use two applications of Minkowski's inequality. One can show that the cardinality of $J(\theta)=\{\tau : \widehat{\chi_{\theta}}\ast (\widehat{g_{\tau}d\sigma}w_{B_{\delta}^{-1}})\}$ is uniformly bounded in $\theta$. Similarly, for each $\tau$ fixed, $|\{\theta: \tau\in J(\theta)\}|$ is uniformly bounded. My problem is how to get rid of the convolution in (1). If I knew that $\|\widehat{\chi_{\theta}}\|_{1}\lesssim_{n}1$, then I could use Young's inequality and the preceding observations to get \begin{align*} (1)&\leq K_{p,n}^{(2)}(\sum_{\theta\in\mathcal{P}_{\delta}}\sum_{\tau\in J(\theta)}(\|\widehat{\chi_{\theta}}\|_{1}\|\widehat{g_{\tau}d\sigma}w_{B_{\delta^{-1}}}\|_{p})^{2})^{1/2}\\ &\lesssim K_{p,n}^{(2)}(\sum_{\tau\in\mathcal{P}_{\delta}}\|\widehat{g_{\tau}d\sigma}w_{B_{\delta^{-1}}}\|_{p}^{2})^{1/2}\tag{2} \end{align*} But I don't see why this should be the case, given that $\chi_{\theta}$ is a rough frequency projection. If it were a bump function adapted to $\theta$, then there would be no problem; however, this is not the case.


Following Prof. Tao's suggestion, introduce a smooth partition of unity $\{\widehat{K_{\theta}}\}_{\theta\in\mathcal{P}_{\delta}}$ of Fourier space such that $\mathrm{supp}(K_{\theta})\subset \frac{9}{10}\theta$, and $\|K_{\theta}\|_{1}\lesssim_{n} 1$ (see pg. 18 of the arXiv version). We introduce the notation $\tilde{f}_{\theta}:=f\ast K_{\theta}$. Note that $\tilde{f}_{\theta}=f_{\theta}\ast K_{\theta}$. Similarly, for $g: P^{n-1}\rightarrow\mathbb{C}$, we write $\tilde{g}_{\theta}:=g\widehat{K_{\theta}}$. For $\delta$ sufficiently small, $\mathrm{supp}((\tilde{g}_{\theta}d\sigma)\ast w_{B_{\delta^{-1}}}^{\vee})\subset\theta$.

Now write $$\mathcal{P}_{\delta}=\bigcup_{j=1}^{N}\mathcal{P}_{\delta}^{j}$$ where the $\mathcal{P}_{\delta}^{j}$ are $\sim\delta^{1/2}$-separated subcollections of caps in $\mathcal{P}_{\delta}$. Note that $N\lesssim_{n}1$. If we choose the separation appropriately, then for any fixed $\tau\in\mathcal{P}_{\delta}$, $$|I_{j}(\tau)|:=|\{\theta\in\mathcal{P}_{\delta}^{j}: \chi_{\tau}((\tilde{g}_{\theta}d\sigma)\ast w_{B_{\delta^{-1}}}^{\vee}\neq 0\}|\leq 1$$

Set $h_{j}:=\sum_{j=1}^{N}\sum_{\theta\in\mathcal{P}_{\delta}^{j}}\tilde{g}_{\theta}$. By linearity and the triangle inequality, $$\|\widehat{gd\sigma} w_{B_{\delta^{-1}}}\|_{p}\leq\sum_{j=1}^{N}\|\widehat{h_{j}d\sigma} w_{B_{\delta^{-1}}}\|_{p}$$ So without loss of generality, we may and will assume that $g=h_{j}$ for $j=1$. Since $\chi_{\theta}=1$ on the support of $(\tilde{g}_{\theta}d\sigma)\ast w_{B_{\delta^{-1}}}^{\vee}$, applying the decoupling inequality gives \begin{align*} \|\widehat{gd\sigma} w_{B_{\delta^{-1}}}\|_{p}&\leq K_{p,n}^{(2)}(\delta)(\sum_{\tau\in\mathcal{P}_{\delta}}\|\widehat{\chi_{\tau}}\ast(\widehat{gd\sigma}w_{B_{\delta^{-1}}})\|_{p}^{2})^{1/2}\\ &=K_{p,n}^{(2)}(\delta)(\sum_{\tau\in\mathcal{P}_{10\delta}}\|\sum_{\theta\in I(\tau)}\widehat{\tilde{g}_{\theta}d\sigma}w_{B_{\delta^{-1}}})\|_{p}^{2})^{1/2}\\ &\leq K_{p,n}^{(2)}(\delta)(\sum_{\theta\in\mathcal{P}_{\delta}^{1}}\|\widehat{\tilde{g}_{\theta}d\sigma}w_{B_{\delta^{-1}}}\|_{p}^{2})^{1/2}\\ &=K_{p,n}^{(2)}(\delta)(\sum_{\theta\in\mathcal{P}_{\delta}^{1}}\|(\widehat{g_{\theta}d\sigma}\ast K_{\theta})w_{B_{\delta^{-1}}}\|_{p}^{2})^{1/2} \tag{3} \end{align*} To get rid of $K_{\theta}$, we argue as follows. By Holder's inequality and Fubini, \begin{align*} \|(\widehat{g_{\theta}d\sigma}\ast K_{\theta})w_{B_{\delta^{-1}}}\|_{p}^{p}&\leq\int_{\mathbb{R}^{n}}(\int_{\mathbb{R}^{n}}|\widehat{g_{\theta}d\sigma}(y)|^{p}|K_{\theta}(x-y)|dy)\|K_{\theta}(x-\cdot)\|_{1}^{p/p'}w_{B_{\delta}^{-1}}(x)dx\\ &\lesssim_{n,p}\int_{\mathbb{R}^{n}}|\widehat{g_{\theta}d\sigma}(y)|^{p}(\int_{\mathbb{R}^{n}}|K_{\theta}(x-y)w_{B_{\delta^{-1}}}(x)dx)dy\\ &=:\|\widehat{g_{\theta}d\sigma}\|_{L^{p}(\tilde{w}_{B_{\delta}^{-1}})}^{p}\ \end{align*} If we construct the $K_{\theta}$ appropriately, then $\tilde{w}_{B_{\delta^{-1}}}$ is another weight, though not necessarily Fourier supported in $B(0,\delta)$. I'm not sure how to fix this issue.

$\endgroup$
3
  • $\begingroup$ You can use the sparsification trick. By partitioning the space of caps into $O(1)$ collections of $\delta^{1/2}$-separated caps and using the triangle inequality (and maybe a smooth partition of unity), one can assume without loss of generality that $g$ is supported on one of these sparsified collections, at which point one can use smooth cutoffs instead of rough ones. $\endgroup$
    – Terry Tao
    Jun 18, 2016 at 17:02
  • $\begingroup$ @TerryTao Thank you; your comment is very helpful. When I follow your suggest, I end up with expressions like $$(\sum_{\theta\in\mathcal{P}_{\delta}}\|(\widehat{g_{\theta}d\sigma}\ast K_{\theta})\|_{L^{p}(w_{B_{\delta^{-1}}})}^{2})^{1/2}$$ on the RHS, where $K_{\theta}$ is the convolution resulting from the partition of unity applied to $g$ (see (3) in my edited answer). In trying to remove the $K_{\theta}$, I end up with the desired RHS with another weight $\tilde{w}_{B_{\delta^{-1}}}$, which is not necessarily Fourier supported in $B(0,\delta)$. $\endgroup$ Jun 19, 2016 at 1:58
  • $\begingroup$ Section 4 of Bourgain and Demeter's study guide arxiv.org/pdf/1604.06032.pdf has some tools to replace one weight with another, more or less for this sort of reason. $\endgroup$
    – Terry Tao
    Jun 19, 2016 at 4:59

0

Your Answer

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

Browse other questions tagged or ask your own question.