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.

$\endgroup$
  • $\begingroup$ 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 '16 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 $B(0,\delta)$. $\endgroup$ – Matt Rosenzweig Jun 19 '16 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 '16 at 4:59

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.