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The Question

This question is about Lemma 1.2 on the fifth page of Thomas Wolff's paper, "A sharp bilinear cone restriction estimate", Annals of Mathematics, 153 (2001), 661--698. The Lemma states (the definitions to be given after)

If $x\in Q(1)$ is a smooth $\mu$-fold point for the white tubes with $\mu > \delta^B$ then $x$ is a base-point for an $\eta$-bush of white tubes with cardinality $\gtrsim (\log \frac1\delta)^{-1} \mu (\frac\eta\delta)^M$ for some $\eta \leq \delta^{1-\epsilon}$. Conversely...

(I understand the converse part, and can sketch the proof, so I omit that here.) The Lemma is given without proof in the paper, and left deliberately as an exercise. The problem is I am not even sure if I understand the Lemma correctly! (In particular, I don't see why the restriction $x\in Q(1)$ is necessary at all: the statement seems to be translation invariant.)

Question Can someone supply a sketch of the proof for this Lemma (beyond the one-liner in the paper)? In particular, where does the logarithm loss come from? The converse statement does not require a logarithm. Thanks.

The Definitions

Now, the definitions to make sense of the Lemma. (Do let me know if I missed anything.)

  • $Q(1)\subset \mathbb{R}^d$ is the unit cube centered at the origin.
  • A small constant $\epsilon$ is fixed throughout, and a large constant $B$ depending on dimension $d$ is fixed throughout.
  • For the purpose of this lemma, a white light-ray is a line in $\mathbb{R}^d$ transversal, and making an angle of 45 degrees, with the plane $\{x_d = 0\}$
  • Let $\mathcal{W}$ denote a (finite or countable) collection of white light rays. Fix $\delta > 0$. For an element $W\in \mathcal{W}$, denote by $w$ the set of points $\{ |x - W| \leq \delta\}$. For $w$ the function $\phi_w$ is defined: $ \phi_w(x) = \min (1, \frac{\delta}{|x-w|})^M $ for some fixed large constant $M$ depending on $\epsilon$. $|x-w|$ denotes the Euclidean distance from the point $x$ to the set $w$.
  • $\mathcal{W}$ is assumed to be $\delta$ separated: for each $W$ we can associate a direction in real projective space corresponding to the axis. Then for any $D$ disc in projective space of radius $\delta$, let $\mathcal{W}_D$ denote the subset of light rays whose direction points in $D$, the $\delta$-separation assumption requires that the set $\cap_{W\in \mathcal{W}_D} w$ be a bounded set. In particular this implies two parallel light rays cannot be closer than $2\delta$ apart.
  • We say that $x$ is a smooth $\mu$-fold point for the white tubes if $\sum_{W\in\mathcal{W}} \phi_w(x) \geq \mu$. Roughly speaking this means around $\mu$ light rays get close to $x$. (As far as I can tell, this "roughly speaking" is the content of the Lemma described above.)
  • $x$ is said to be a base-point for an $\eta$-bush $P\subset \mathcal{W}$ if $\sup_P |x-w| \leq \eta$. That is, we have a family of light-rays that all pass within $\delta + \eta$ of $x$.
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2 Answers 2

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It looks like a dyadic pigeonholing argument to me (the presence of the logarithm is a big clue in this regard). One can decompose $\phi_w$ into about $\log \frac{1}{\delta}$ dyadic shells, depending on the magnitude of $|x-w|/\delta$, plus a remainder in which $1+|x-w|/\delta \geq \delta^{-100B}$ (say) which has a negligible contribution. So by the pigeonhole principle, $\gtrsim \frac{1}{\log \frac{1}{\delta}} \mu$ of the multiplicity must be coming from one of the shells. Calling the radius of that shell $\eta$, the result presumably follows.

It is quite likely that one could eliminate the logarithm loss here by exploiting the freedom to degrade the M parameter. But in this sort of work, there are logarithmic factors lost all over the place for other reasons, so one more such loss is pretty much insignificant.

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  • $\begingroup$ I'm accepting this answer. In the end the proof I came up with is slightly different in approach (see below), but the main ideas with the shell decomposition + pigeonhole argument, with the discarding of the faraway pieces came from this answer. $\endgroup$ Dec 20, 2010 at 18:19
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Ah, with Terry's comment it turns out the situation was a lot simpler than I thought.

Step 1: Lack of contribution from distant pieces

Divide the region $\{|y-x| > \delta : y \in \mathbb{R}^d\}$ into concentric shells of thickness $\delta$. Consider the contribution to $\sum \phi_w(x)$ from $W\in\mathcal{W}$ such that $|x-w| \in (k\delta, k+1\delta]$, for $k\in \mathbb{N}$. We can divide the shell up into $\sim k^{d-1}\delta^{1-d}$ balls of size $\delta$. Write $y(W)$ for the point of closest approach on $W$ to $x$. In each ball $V$ of size $\delta$, by the $\delta$-separation assumption, there exists sum large constant $B'$ such that

$$ \left|\{ W\in\mathcal{W} : y(W) \in V \}\right| \leq B \delta^{-B'} $$

So

$$ \sum_{k > \delta^{-\epsilon}} \sum_{k\delta < |y(W)-x| - \delta \leq (k+1)\delta} \phi_w(x) \lesssim \sum_{k > \delta^{-\epsilon}} k^{d-1}\delta^{1-d-B'} k^{-M} $$

For sufficiently large $M$ (and if we fix $\delta < \delta_0 < 1$), the RHS can be bounded $\leq \delta^B$. So we can neglect the contribution from far away pieces up to a $\delta^B$ term.

Step 2: Pigeonhole the inside

In view of Step 1, it suffices to show that there exists a constant $C$ such that if for any $0 < k < \delta^{-\epsilon}$ ($k$ should be though of as $\eta/\delta$) we have

$$ \left| \{ W\in\mathcal{W} : |x-w| < k\delta \} \right| \leq C\frac{1}{\log \frac1\delta} \mu k^M $$

we must have $\sum \phi_w(x) < \mu$. This follows by noting that the above means that there are no rays within distance $\delta$ of $x$ ($k = 0$), and thus we only need to sum over $k > 0$

$$ \sum_{k = 1}^{\delta^{-\epsilon}}\phi_w(x) \lesssim \sum_k \left| \{ W\in\mathcal{W} : |x-w| < k\delta \} \right| \cdot k^{-(M+1)} $$

using that $\phi_w(x) \sim k^{-M}$. Plugging in the hypothesis, we have that the expression is controlled by

$$ C\frac{1}{\log \frac1\delta} \mu \sum_{k = 1}^{\delta^{-\epsilon}} k^{-1} \lesssim C \mu \log_\delta(\delta^{-\epsilon}) = C\epsilon \mu$$

So for sufficiently small $C$, we can bound

$$ \sum_{k = 1}^{\delta^{-\epsilon}}\phi_w(x) \leq \frac12 \mu $$

which gives the desired contradiction.

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