5
$\begingroup$

I've been tackling the following problem for some time,

  • Problem 2.7. (a) Let $S:=\left\{(x, y) \in \mathbf{R}_{+} \times \mathbf{R}_{+}: x^2+y^2=1\right\}$ be a quarter-circle. Let $R \geq 1$, and let $R^{-1 / 2} \leq \theta \lesssim 1$ be an angle. Let $A, B$ be two arcs in $S$ of angle $\sim \theta$ and separation $\sim \theta$. Show that $$ \left\|\chi_{N_{1 / R}(A)} * \chi_{N_{1 / R}(B)}\right\|_{\infty} \lesssim R^{-2} \theta^{-1} . $$ Conclude (by an application of the Cauchy-Schwarz inequality) that $$ \|G * H\|_{L^2\left(\mathbf{R}^n\right)} \lesssim R^{-1} \theta^{-1 / 2}\|G\|_{L^2\left(N_{1 / R}(A)\right)}\|H\|_{L^2\left(N_{1 / R}(B)\right)} $$ for all $L^2$ functions $G, H$ supported on $N_{1 / R}(A)$ and $N_{1 / R}(B)$ respectively. (Hint: estimate $G * H$ pointwise by the geometric mean of $\chi_{N_{1 / R}(A)} *$ $\chi_{N_{1 / R}(B)}$ and $\left.|G|^2 *|H|^2\right)$. In particular, from Hölder's inequality, conclude that \begin{equation} \|G * H\|_{L^2\left(\mathbf{R}^n\right)} \lesssim R^{-3 / 2}\|G\|_{L^4\left(N_{1 / R}(A)\right)}\|H\|_{L^4\left(N_{1 / R}(B)\right)} \,\,(1) \end{equation}
  • (b) For every function $G$ and $H$ defined on $N_{1 / R}(S)$, and any $R^{-1 / 2} \leq$ $\theta \lesssim 1$, define the partial convolution $G *_\theta H$ by $$ G *_\theta H(x)=\int_{y+z=x ; \angle y, z \sim \theta} G(y) H(z) d y . $$ Using (1), show that $$ \left\|G *_\theta G\right\|_{L^2\left(\mathbf{R}^n\right)} \lesssim R^{-3 / 2}\|G\|_{L^4\left(N_{1 / R}(S)\right)}^2 $$ for all $R^{-1 / 2} \leq \theta \lesssim 1$. (Hint: Split $S$ into $\operatorname{arcs} A$ of width $\theta$, and split $G$ accordingly; apply (2.24) to various pieces and then sum. The key point here is that as one varies the arcs, the support of the corresponding portion of $G *_\theta G$ also varies, so that one has plenty of orthogonality).

The final objective is to prove the restriction conjecture for the circle. Up until now I've successively done part a) and b), next Tao ask to show $$\|G * G\|_{L^2\left(\mathbf{R}^n\right)} \lesssim(\log R)^{1 / 2} R^{-3 / 2}\|G\|_{L^4\left(N_{1 / R}(S)\right)}^2 $$

I've tried to decompose the quarter circle into arcs of similar length but I can't see how the partial convolution plays a role. Any guidance would be appreciated.

$\endgroup$
1
  • 1
    $\begingroup$ In the meantime, I've solved this problem. I'm working on typing up a solution guide for most of the proposed problems in this set of lecture notes. It will be publish a draft on my GitHub. Now, to answer my own question, one needs the following observations: By treating $ \theta$ as a dyadic number, one can decompose the convolution into a sum of partial convolutions. Then, one splits the sum into two regimes: one with $R^{-1 / 2} \leq \theta \lesssim 1$ and its complement, and applies the known estimate. For the other term, one needs a similar estimate (which is the trivial one). $\endgroup$ Commented Aug 16, 2024 at 11:14

1 Answer 1

4
$\begingroup$

In the meantime, I've solved this problem. I'm working on typing up a solution guide for most of the proposed problems in this set of lecture notes. It will be publish a draft on my GitHub.

Now, to answer my own question, one needs the following observations: By treating $ \theta$ as a dyadic number, one can decompose the convolution into a sum of partial convolutions.

Then, one splits the sum into two regimes: one with $R^{-1 / 2} \leq \theta \lesssim 1$ and its complement, and applies the known estimate. For the other term, one needs a similar estimate (which is the trivial one).

$\endgroup$
1
  • $\begingroup$ You are doing Gods work for people like me who intend to read the notes sometime! $\endgroup$ Commented Aug 17, 2024 at 1:03

You must log in to answer this question.

Start asking to get answers

Find the answer to your question by asking.

Ask question

Explore related questions

See similar questions with these tags.