Tube Doubling Problem and the Kakeya Problem

Recall ball doubling problem: Suppose that $B_i$ is a finite list of balls in ${\mathbb{R}}^n$. Let $2B_i$ be the ball with the same center as $B_i$ and twice the radius. Is there a universal constant $C_n$ so that for any finite set of balls in ${\mathbb{R}}^n$, $|{\cup }_{i}2B_i|\le C_n|{\cup }_i{B}_i|$ ? From Vitali’s Covering Lemma, we get following estimate for the ball doubling problem.

Finite ball doubling

If $\{B_i{\}}_{i\in I}$ is a finite collection of balls, then $|{\cup }_{i}2B_i|\le 6^n|{\cup }_i{B}_i|$

Vitali Covering Lemma for infinite collections of balls

Suppose $\{B_i{\}}_{i\in I}$ is a collection of balls in ${\mathbb{R}}^n$. Suppose that there is some finite constant $M$ so that any disjoint subset of the balls $\{B_i{\}}_{i\in I}$ has total volume at most $M$. Then there exists a subcollection $J\subset I$ such that $\{B_j{\}}_{j\in J}$ are disjoint but${\cup }_{i\in I}{B}_i\subset {\cup }_{j\in J}4B_j$.

Then we move on to a similar question about possible intersection patterns of cylindrical tubes in Euclidean space. Suppose that $T_i\subset {\mathbb{R}}^n$ are cylindrical tubes of radius 1 and length $N$. Let $2T_i$ be the concentric tube of radius 2 and length $2N$ formed by dilating $T_i$ around its center by a factor of 2. For a single tube T, we note that $|2T|=2^n|T|$. Is there a constant $C_n$ so that for any $N$, for any set of $1\times N$ tubes $T_i$ in ${\mathbb{R}}^n$, $|{\cup }_{i}2T_i|\le C_n|{\cup }_i{T}_i|$? The counterexample was given by Besicovitch, showing that $|{\cup }_{i}2T_{i|\sim }\frac{\log N}{\log \log N}|{\cup }_i{T}_i|.$ refined by Schoenberg, showing that $|{\cup }_{i}2T_i|\sim \log N|{\cup }_i{T}_i|.$

Now we give the example of Besicovitch. We will construct a set of $N$ rectangles in the plane, $R_j$, with width $1/N$ and length $1$, with slopes changing evenly between $0$ and $1$, and with a lot of overlap. For integers $1\le j\le N$, let $l_j:[0,1]\to \mathbb{R}$ be a list of affine linear functions of the form

$$l_j(x)=\frac{j}{N}x+H(j).$$

Let $R_j$ be the $1/N$ neighborhood of the graph of $l_j$, which contains a rectangle of width $1/N$ and length $1$. This is a multiscale argument. In order to talk about the different scales, we expand $j/N$ in base $A$: $\frac{j}{N}={\sum }_{a=1}^{A}j(a)A^{-a}.$In this equation $j(a)$ are the digits in the base $A$ decimal expansion of $j/N$. The first term $j(1)$, is the first digit after the decimal, and it contributes the largest amount to $j/N$. The different values of $a$ represent different scales in the problem.

The property below is basically considering restricted lines to be “close enough” when the slop is “close enough”. We can think of the choice of $H(a)$ as, classifying all the lines with slope from $[0,1]$ with

Property

Suppose $1\le b\le A$. If $j(a)=j^{\prime }(a)$ for $1\le a\le b-1$, then for all $x\in \left[\frac{A-b}{A},\frac{A-b+1}{A}\right]$,
$|l_j(x)-l_{j^{\prime }}(x)|\le 4A^{-b}.$

We will write $H(j)$ as a sum of $A$ different terms with different orders of magnitude. $H(j)={\sum }_{a=1}^{A}h(a)j(a)A^{-a},$ where $h(a)\in [-1,1]$ is a constant that we can choose later. The constant $h(a)$ will be designed to make the property hold for $b=a$. Plugging this expression into the formula for $l_j(x)$, we see that $l_j\left(\frac{A-b}{A}\right)=\frac{A-b}{A}\cdot \frac{j}{N}+H(j)={\sum }_{a=1}^A\left(\frac{A-b}{A}+h(a)\right)j(a)A^{-a}.$ Therefore, $\left|l_j\left(\frac{A-b}{A}\right)-l_{j^{\prime }}\left(\frac{A-b}{A}\right)\right|\le {\sum }_{a=1}^A\left|\frac{A-b}{A}+h(a)\right||j(a)-j^{\prime }(a)|A^{-a}.$ For the rest of the proof, let us suppose that $j(a)=j^{\prime }(a)$ for $1\le a\le b-1$. These equalities imply that, in the last inequality, the first $b-1$ terms on the right-hand side vanish. We now choose $h(a):=-\frac{A-a}{A}$. With this choice, the $b^{th}$ term on the right-hand side vanishes also. We can easily bound the other terms by noting that $|h(a)|\le 1$ and $|j(a)-j^{\prime }(a)|\le A$ , yielding $\left|l_j\left(\frac{A-b}{A}\right)-l_{j^{\prime }}\left(\frac{A-b}{A}\right)\right|\le {\sum }_{a=b+1}^A{A}^{-a+1}\le 2A^{-b}.$ Now for any $x\in \left[\frac{A-b}{A},\frac{A-b+1}{A}\right]$, we see that \begin{align*} |l_j(x) - l_{j'}(x)| &\le \left| l_j \left( \frac{A-b}{A} \right) - l_{j'} \left( \frac{A-b}{A} \right) \right| + \left| \frac{j}{N} - \frac{j'}{N} \right| \left| x - \frac{A-b}{A} \right| \\ &\le 2A^{-b} + A^{-b+1} A^{-1} = 3A^{-b}. \end{align*}

Theorem

Suppose that $N$ is an integer of the form $A^A$ for some large integer $A$. Let $R_j$ be the rectangles described above. If we choose the constants $H(j)$ correctly, then $|{\cup }_j{R}_j|\lesssim A^{-1}\lesssim \frac{\log \log N}{\log N}.$

Note that ${\sum }_j|R_j|\sim 1$, and so this inequality represents a compression by a factor $\frac{\log \log N}{\log N}$.

Fix a value of $b$. For a given choice of $j(1),\dots ,j(b-1)$, consider all the rectangles $R_{j^{\prime }}$ where $j^{\prime }(a)=j(a)$ for $1\le a\le b-1$. By above property, each $R_{j^{\prime }}\cap \left[\frac{A-b}{A},\frac{A-b+1}{A}\right]\times \mathbb{R}$ lies in the parallelogram defined by the inequalities $\frac{A-b}{A}\le x\le \frac{A-b+1}{A}\text{ and }|y-l_j(x)|\le 5A^{-b}.$ This parallelogram has area $10A^{-b-1}$. Since there are $A^{b-1}$ possible values of $j(1),\dots ,j(b-1)$, we see that $\left|({\cup }_j{R}_j)\cap \left(\left[\frac{A-b}{A},\frac{A-b+1}{A}\right]\times \mathbb{R}\right)\right|\le 10A^{-2}.$ Summing over all choices of $b$ in the range $1\le b\le A$, we see that $|({\cup }_j{R}_j)\cap ([0,1]\times \mathbb{R})|\le 10A^{-1}.$ The rectangles $R_j$ may stick out a little from $[0,1]\times \mathbb{R}$, but it’s straightforward to see that $|{\cup }_j{R}_j|\le 11A^{-1}$.

For a rectangle $R_j$ as above let $v(R_j)$ be the direction of $R_j$. Recall that $v(R_j)$ is only defined up to sign, and we make the choice so that the $x$-component of $v(R_j)$ is negative. Now we define $R_j^{+}$ to be the translation of $R_j$ by $10v(R_j)$. We note that since $R_j$ has length $\sim 1$, $R_j^{+}$ is contained in $100R_j$. We claim that for the values of $H(j)$ constructed in the proof above, the rectangles $R_j^{+}$ are disjoint. We state this refined result as a corollary.

Corollary

Suppose that $N$ is an integer of the form $A^A$ for some large integer $A$. Let $R_j$ be the rectangles described in previous theorem. If we choose the constants $H(j)$ correctly, then $|{\cup }_j{R}_j|\lesssim \frac{\log \log N}{\log N}{\sum }_j|R_j|,$ and yet $R_j^{+}$ are disjoint.

Let $H(j)$ be as in the proof of theorem. Recall that we expanded $j/N$ in base $A$ as $\frac{j}{N}={\sum }_{a=1}^{A}j(a)A^{-a}.$ In terms of the digits $j(a)$, we define

$$H(j):=-\sum _{a=1}^A\frac{A-a}{A}j(a)A^{-a}.$$

For any $1\le j<j^{\prime }\le N$, $H(j^{\prime })<H(j)$ since $H(j^{\prime })-H(j)={\sum }_{a=b}^A\frac{A-a}{A}(j(a)-j^{\prime }(a))A^{-a}\le -A^{-b}+{\sum }_{a=b+1}^A\frac{A-a}{A}(j(a)-j^{\prime }(a))A^{-a}$ $\le -A^{-b}+{\sum }_{a=b+1}^A(A-1)A^{-a}<0$

Therefore, for $x\le -5$, and $1\le j<j^{\prime }\le N$, we have $l_{j^{\prime }}(x)-l_j(x)=\left(\frac{j^{\prime }}{N}-\frac{j}{N}\right)x+(H(j^{\prime })-H(j))\le -5\left|\frac{j^{\prime }}{N}-\frac{j}{N}\right|\le -5/N.$ Therefore, $R_j^{+}$ and $R_{j^{\prime }}^{+}$ are disjoint. From this corollary we see that $|{\cup }_{j}100R_j|\ge |{\cup }_j{R}_j^{+}|={\sum }_j|R_j|\gtrsim \frac{\log N}{\log \log N}|{\cup }_j{R}_j|.$With a little bit more care, it is also possible to choose $1\times N$ rectangles $R_j$ so that $|{\cup }_{j}2R_j|\gtrsim \frac{\log N}{\log \log N}|{\cup }_j{R}_j|$.

Tube doubling conjecture

For any dimension $n$, for any $ϵ>0$, there is a constant $C_n(ϵ)$, so that the following estimate holds for any $N$. If $T_i$ are tubes of radius 1 and length $N$, then $|{\cup }_{i}2T_i|\le C_n(ϵ)N^{ϵ}|{\cup }_i{T}_i|.$

Kakeya set of tubes

For a tube $T\subset {\mathbb{R}}^n$, we write $v(T)\in S^{n-1}$ for a unit vector parallel to the axis of symmetry of $T$. Suppose that$T_i\in {\mathbb{R}}^n$ are tubes of length $N$ and radius 1. $\{T_i\}$ is a Kakeya set of tubes if $\{v(T_i)\}$ is $\frac{1}{N}$-separated and $\frac{2}{N}$-dense in $S^{n-1}$. (Where$\frac{1}{N}$-separated means $\forall i\ne j,|v(T_i)-v(T_j)|\ge \frac{1}{N}$ and $\frac{2}{N}$-dense means $\forall v\in S^{n-1}$, $\exists i,s.t.|v-v(T_i)|\le \frac{2}{N}$.)

Kakeya Conjecture, tube version

In any dimension $n\ge 2$, for any $\epsilon >0$, there is a constant $C_{n,\epsilon }$ so that for any $N$ the following holds. For any Kakeya set of tubes $T_i\subset {\mathbb{R}}^n$ of dimensions $1\times N$, $|\cup T_i|\ge C_{n,ϵ}\cdot N^{n-ϵ}.$