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I'm currently reading Bourgain and Demeter's study guide for the $l^2$ decoupling theorem (https://arxiv.org/pdf/1604.06032.pdf). I have some trouble with understanding the proof of Proposition 8.4.

Let me describe the setting. We define the truncated paraboloid $$ \mathbb{P}^{n-1} = \{(\xi_1,\dotsc,\xi_{n-1},\xi_1^2 + \dotsc + \xi_{n-1}^2) : 0 \leq \xi_i \leq 1\} $$ and the $\delta$-neighborhood of $\mathbb{P}^{n-1}$ above some set $U\subset [0,1]^{n-1}$ $$ N_\delta(U)=\{(\xi_1,\dotsc,\xi_{n-1},\xi_1^2+\dotsc+\xi_{n-1}^2+t):(\xi_1,\dotsc,\xi_{n-1})\in U,0\leq t \leq \delta\}. $$

Let us now restrict ourselves to $n=3$ dimensions. Let $L$ be the line in the $(\xi_1,\xi_2)$-plane given by $\eta=0$ (actually the authors choose $\eta = 1$ but this shouldn't change anything due to translation invariance) and let $S_L = \{(\xi_1,\xi_2)\in [0,1]^2:\text{dist}((\xi_1,\xi_2),L)<\frac{C}{K}\}$ be the fattened line for some constant $C>0$. Next, for a function $g:[0,1]^2\rightarrow \mathbb{C}$ we define the extension operator $$ E_{S_L}g(x,y,z)=\int_{S_L}g(\xi_1,\xi_2)e(\xi_1 x + \xi_2y+(\xi_1^2+\xi_2^2)z)d(\xi_1,\xi_2) $$ where $e(z)=e^{2\pi i z}$, which extends $g$ to $\mathbb{P}^2$. Now we fix some $y\in\mathbb{R}$ and define the restricted operator $$ E_{S_L,y}g(x,z):=E_{S_L}g(x,y,z). $$ So now we are in two dimensions.

The issue I have trouble with is the following: In order to obtain a decoupling inequality the authors claim that one can apply Thm. 5.1 to $E_{S_L,y}g$ which requires $E_{S_L,y}g$ being Fourier supported in $N_{C/K}([0,1])$ for some constant $C$. Their argument why this works is, because $E_{S_L,y}g$ is supported in the $O(K^{-1})$ neighborhood of the parabola $\eta=\xi^2$. But $N_{C/K}([0,1])$ is a strict subset of this neighborhood, so this argument is not sufficient to apply Thm. 5.1.

So to be precise, my question is the following:

How can one show, that $E_{S_L,y}g$ has Fourier support in $N_{C/K}([0,1])$?

By the uncertainty principle this is quite clear to me, but I don't know how to make this rigorous. My approach follows the standard ideas: Instead of $E_{S_L,y}g$ I consider $\phi E_{S_L,y}g$ for some Schwartz function $\phi$ which is Fourier supported in the cube $B(0,1/K)$ centered at the origin and with side length $1/K$. I deduce that $\phi E_{S_L,y}g$ is Fourier supported in the $O(K^{-1})$ neighborhood of the parabola $\eta=\xi^2$. And now I would like to apply Thm 5.1 to $\phi E_{S_L,y}g$, which I can't because of the reason discussed above.

Any help is appreciated. Thanks in advance.

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2 Answers 2

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The following should work. Depending on $C$, break $g$ into two pieces, $g_1$ which is supported on $\sim C$ of the $K^{-1} \times K^{-1/2}$ rectangles at the ends of the strip $S_L$ and $g_2$ supported on the rest of the $K^{-1} \times K^{-1/2}$ rectangles covering $S_L$. So for example if $C = 1$ take $g_1$ to be supported on the leftmost and rightmost $K^{-1} \times K^{-1/2}$ rectangles covering $S_L$.

With $\phi$ having Fourier support in $B(0, 1/K)$, the support of $g_2$ and the argument you gave shows that $\phi E_{S_L, y}g_2$ is Fourier supported in $N_{C/K}([0, 1])$ and you can apply Theorem 5.1. For $g_1$ use its support, Cauchy-Schwarz, and that the decoupling constant is $\gtrsim_{E, p} 1$ to obtain that \begin{align*} \|E_{S_L, y}g_1\|_{L^{p}_{x, z}([0, K]^2)} &\lesssim \sum_{U' \text{ at the ends}}\|E_{U'}g\|_{L^{p}_{x, z}([0, K])^{2})}\\ & \lesssim_{E, p} \operatorname{Dec}_{2}(K^{-1}, p, 10E)(\sum_{U}\|E_{U}g\|_{L^{p}_{x, z}([0, K])^{2})}^{2})^{1/2}. \end{align*}

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  • $\begingroup$ Sorry for answering only now. Your argument for $g_1$ seems fine to me. However, I still have a question about your argument concerning $g_2$. If I compute the Fourier support of $\phi E_{S_L,y}g_2$, I find that $\mathcal{F}(\phi E_{S_L,y}g_2)(\xi_1,\xi_2)$ vanishes unless $(\xi_1,\xi_2)\in B_{1/K}(\alpha_1,\alpha_1^2+\alpha_2^2)$ for $(\alpha_1,\alpha_2)$ in the support of $g_2$. But I don't see why this implies that $(\xi_1,\xi_2)\in N_{C/K}([0,1])$. $\endgroup$
    – msaBU
    Feb 18, 2019 at 19:41
  • $\begingroup$ For $\alpha_2=0$ for example, it is still possible that $\xi$ lies under the parabola $\eta=\alpha_1^2$ and therefore not in $N_{C/K}([0,1])$. Or am I missing something? $\endgroup$
    – msaBU
    Feb 18, 2019 at 19:41
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    $\begingroup$ How about this argument: I will take all notation from the study guide (including their definition of $S_L$ and $L$ is the line $\eta = 1$). Let $y$ be a fixed value and let $G_y(x, z) := \int_{1 - 1/K}^{1 + 1/K}g(x, y, z)e(\eta y)e((\eta^2 - 1)z)\, d\eta$. Since $(E_{S_L}g)(x, y, z) = \int_{0}^{1}\int_{1 - 1/K}^{1 + 1/K}g(x, y, z)e(\xi x + \eta y + (\xi^2 + \eta^2)z)\, d\eta\, d\xi$, using our definition of $G_y$ shows that $(E_{S_L}g)(x, y, z) = \int_{0}^{1}G_{y}(x, z)e(\xi x + (\xi^2 + 1)z)\, d\xi$. $\endgroup$
    – Zane Li
    Feb 19, 2019 at 20:54
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    $\begingroup$ Break this up into two integrals $\int_{[0, 1/K^{1/2}] \cup [1 - 1/K^{1/2}, 1]}$ and the remainder. The first piece we can control like $g_1$ as above and the second piece we can multiply by $\phi(x, z)$ as you defined and is then the Fourier transform of $\phi E_{S_L}g$ is supported in a $O(K^{-1})$ neighborhood of the curve $(\xi, \xi^2 + 1)$, that is a $O(K^{-1})$ neighborhood of the curve $\eta = \xi^2+ 1$. $\endgroup$
    – Zane Li
    Feb 19, 2019 at 20:54
  • $\begingroup$ This sounds fine to me. Thank you very much! $\endgroup$
    – msaBU
    Feb 22, 2019 at 10:57
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For any function $f$, since $\hat{f}(x,y,z)=\int f(\xi_1,\xi_2,\xi_3)e(-\langle (x,y,z), \xi\rangle)\,d\xi$, for fixed $y$ we have that the function $(x,z)\mapsto \hat{f}(x,y,z)$ is the Fourier transform of $h_y(\xi_1,\xi_3)=\int f(\xi)e(-y\xi_2)\,d\xi_2$. Apply this to $f(\xi)=(\chi_Lg)(\xi_1,\xi_2)\delta(\xi_3-(\xi_1^2+\xi_2^2))$, to see that $h_y$ is supported in $N_{C/K^2}\subset N_{C/K}$. By the way, for $g$ bounded, $h_y$ blows-up right on the parabola, but it doesn't have any effect at the end.

I have been trying without success to get an alternative to Theorem 5.1. If I get it, I'd like to post it here.

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