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I am confused about a passage in the paper by T. Tao A sharp bilinear restriction estimate for paraboloids.

We are in Section 7, near equation (34) (pag.16 of the arxiv).

Notations and conventions $$S=\left\{(\tau, \xi)\in \mathbb{R}\times\mathbb{R}^n\ |\ \tau=-\frac{1}{2}\lvert \xi\rvert^2\right\}\ (\text{paraboloid}),$$ $$B\big( (\tau, \xi), R\big)=\{(\tau, \xi)\in \mathbb{R}\times \mathbb{R}^n\ |\ \tau^2+\lvert\xi\rvert^2< R^2\}\ (\text{space-time ball})$$ $$\widehat{F}(t, x)=\int Fe^{2\pi(\tau t + \xi x)}\ (\text{space-time Fourier transform}) $$ Here $F$ might be a measure. With the symbol $d\sigma$ we will denote the surface measure on $S$. We also have a fixed parameter $R\gg 1$. We adopt the convention that the letter $C$ always denotes an absolute constant whose value may change from one line to another.

From a previous operation (called wave-packet decomposition) we have two functions $$\phi_{T_j}(t, x)=\widehat{f_j d\sigma_j},\qquad j=1, 2, $$ where $f_j$ are smooth functions defined on the following small transversal caps of $S$: $$\left\{ (\tau, \xi)\in S\ :\ \xi=v_j+ O(R^{-1/2})\right\},\qquad v_1=e_1,\ v_2=-e_1.$$ Those functions satisfy a pointwise bound $$\lvert \phi_{T_j}(t, x)\rvert\le C_N R^{-\frac{n}{4}}\left( 1+\frac{\lvert x-( x_j+tv_j)\rvert}{R^{1/2}}\right)^{-N}\ (x_j\in\mathbb{R}^n\ \text{are fixed}),$$ that is, outside (an $R^\varepsilon$-enlargement of) the tubular region $T_j=\{\lvert x-(x_j+tv_j)\rvert<R^{1/2}\}$, the function $\phi_{T_j}$ is very small.

Now, $\phi_{T_j}$ are not $L^2(\mathbb{R}\times\mathbb{R}^n)$ functions, because their Fourier anti-transforms $f_jd\sigma_j$ are singular measures.

Question. The author claims that Fourier anti-transform of the pointwise product $\phi_{T_1}\phi_{T_2}$ is a function supported in a space time ball $$\tag{35}B\left( \left(\frac{-1}{2}(\lvert v_1\rvert^2+\lvert v_2\rvert^2), v_1+v_2\right),\ CR^{-1/2} \right)$$ and satisfying the pointwise bound $\left\lvert(\phi_{T_1}\phi_{T_2})^\vee(\tau, \xi)\right\rvert\le (C_\varepsilon R^\varepsilon) R^{1/2}$. How to prove that?


I understand that, due to the transversality of the caps, the convolution $f_1 d\sigma_1\star f_2 d\sigma_2$ (which is the Fourier anti-transform of $\phi_{T_1}\phi_{T_2}$) is absolutely continuous. This is proved for example in this 1974 paper by D. Ragozin, Theorem 5.1. This also explains the support condition, since the algebraic sum of the two caps is certainly contained in that ball.

But the pointwise bound completely escapes me. With the above approach, I have proved absolute continuity of the convolution $f_1d\sigma_1\star f_2 d\sigma_2$ by using Radon-Nikodym's theorem, therefore losing any control on the pointwise behaviour of the Radon-Nikodym's derivative. That's why I think that I am somewhat off track.

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    $\begingroup$ Your question reminded me to finish up a blog post I had prepared on this sort of thing: terrytao.wordpress.com/2013/07/26/… . Any of the three methods outlined there will give the result. (Incidentally, your pointwise bound on $\phi_{T_i}$ should have a factor of $R^{-n/4}$ rather than $R^{n/4}$.) $\endgroup$
    – Terry Tao
    Commented Jul 26, 2013 at 23:51

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