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Thanks for reading my post. I am trying to prove the following claim:

If we have \begin{equation*} \left\|\hat{f}\right\|_{L^q(N_{1/R}(S))}\lesssim R^{\alpha-1/q}\left\|f\right\|_{L^p(B(0,R))} \end{equation*} then we have \begin{equation*} \left\|\hat{f}|_{S}\right\|_{L^q(S; d\sigma)}\lesssim R^{\alpha}\left\|f\right\|_{L^p(B(0,R))} \end{equation*} in which $S$ is the standard sphere in $\mathbb{R}^n$, $N_{1/R}(S)$ is the $1/R$ neighborhood of the sphere. $R\gg1$. In particular $f$ is supported in the $R$-ball $B(0,R)$.

This is in fact Problem 2.2 in Tao's Recent Progress on the Restriction Conjecture lecures, see arxiv page 26. What I could do so far, is to take a bump funciton $\psi(z/R)$ such that $\psi(z)=1$ when $|z|\leq 1$. Then we use the fact

\begin{equation*} \hat{f}=\hat{f}\ast\hat{\psi}=\int R^n\hat{\psi}(R(\xi-\eta))\hat{f}(\eta)d\eta\hspace{2cm}(1) \end{equation*} to calculate its $L^q(S)$ norm. It is easy to deal with the part when $|\xi-\eta|<1/R$ in $(1)$. Then using the fact $\psi(z)$ is fast decaying when $|z|>1$ we can also easily get rid of part $|\xi-\eta|>1$ ( or a small power of $1/R$ ) in $(1)$. However I am having difficulties in dealing with the part $1/R\leq|\xi-\eta|\leq1$ and especially when $|\xi-\eta|$ is bigger but closed to $1/R$. Should I use some dyadic decomposition or am I working in the wrong direction?


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I could not check the precise wording, because the ArXiv version of lecture notes does not seem to have Problem 2.2. But from looking at the top of page 8 it seems that the assumption $\left\|\hat{f}\right\|_{L^q(N_{1/R}(S))}\lesssim R^{\alpha-1/q}\left\|f\right\|_{L^p(B(0,R))}$ holds for all large $R$, not just for one $R$. In this case yes, it makes sense to decompose $\widehat \psi = \sum_{k=0}^\infty \phi_k$ where $\phi_0$ is supported on $B(0,1)$ and for $k\ge 1$, $\phi_k$ is supported on $2^{k-1}<|x|<2^k$. If $\psi$ is in the Schwarz class, $\|\phi_k\|_{L^1}$ will decay very fast. – user90090 Aug 14 '13 at 22:41
Sorry I used a url for an earlier edition of the lecture notes on arxiv, which does not contain any exercises. I have already updated it. I think I worked it out based on your suggestions (I should have carried out the calculation a little further...) Thank you very much. – Peng Aug 15 '13 at 0:12
I think the decay kicks in immediately. Your $\psi $ is localized at scale $1$, so its Fourier transform has about the same scale. It should be rapidly decreasing outside of the unit ball. More precisely: let $C$ be the $L^2$ norm of the $n$th derivative of $\psi$. This is a number independent of $R$. From $\||\xi|^n \widehat \psi\|_{L^2}=C$ you get $\|\phi_k\|_{L^2}\le 2^{n(1-k)}C$. Choose $n$ big enough, and the contribution of $\phi_k$ decays right away. // OK, I see we posted simultaneously. If you have a solution now, please post it as an answer. – user90090 Aug 15 '13 at 0:12
I made a mistake, to use the estimates you mentioned repeatedly, the function $f$ needs to be supported in $B(0,R/2^k), k=0,1,2,\dots$, which is certainly not possible. So some more work may be needed. Nonetheless thank you for the reminding. It may be the correct way. – Peng Aug 15 '13 at 2:27

As user90090 suggested, consider a decomposition of $\hat\psi=\sum_{k=0}^{\infty}\psi_k$ in which $\psi_k(x), k\geq1$ have support $k<|x|<{k+1}$. Notice that this decomposition is slight different from the original dyadic decomposition. There are mainly two reasons, of different level, to explain this. First, from uncertainty principle, if $f$ lives in a ball $B(0,R)$, then its Fourier transform should have frequency "lives" in a band with width $1/R$, so it is a little more natural to break up the frequency variable into $1/R$ annular bands. Second, from technical level, we need to repeatedly use the following estimate

$$ ||f||_{L^q(N_{1/R}(S))}\lesssim R^{\alpha-1/q}||f||_{L^p(B(0,R))}\hspace{2cm} (2)$$

So simply speaking what we need to do is to break up the $(k/R, (k+1)/R)$ annular area into a group of small $1/R$ balls with finite over-lapping. Notice that there are at most $O(k^n)$ such balls, so the number can be controlled by $k^{-N}$ decay coming from the fact that $\psi$ is fast decaying. Then we shift these small balls back into $N_{1/R}(S)$ with an exponential multiple of $f$, which does not affect the support of it. So now we can safely apply $(2)$ to get what we want.

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